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https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Antidepressants

Antidepressant drugs are used to restore mentally depressed patients to an improved mental status. Depression results from a deficiency of norepinephrine at receptors in the brain. Mechanisms that increase their effective concentration at the receptor sites should alleviate depression. Antidepressant drugs act by one or more of the following stimulation type mechanisms: The tricyclic antidepressants are the most effective drugs presently available for the treatment of depression. These act by increasing the release of norepinephrine. Amphetamine and cocaine can also act in this manner. Imipramine, amitriptylin, and other closely related drugs are among the drugs currently most widely used for the treatment of major depression. The activity of the tricyclic drugs depends on the central ring of seven or eight atoms which confers an angled or twisted conformation. The side chain must have at least 2 carbons although 3 appear to be better. The amine group may be either tertiary or secondary. All tricyclic antidepressants block the re-uptake of norepinephrine at nerve terminals. However, the potency and selectivity for the inhibition of the uptake of norepinephrine, serotonin, and dopamine vary greatly among the agents. The tertiary amine tricyclics seem to inhibit the serotonin uptake pump, whereas the secondary amine ones seem better in switching off the NE pump. For instance, imipramine is a potent and selective blocker of serotonin transport, while desipramine inhibits the uptake of norepinephrine. Serotonin (5-hydroxytryptamine or 5-HT) is a monoamine neurotransmitter found in cardiovascular tissue, in endothelial cells, in blood cells, and in the central nervous system. The role of serotonin in neurological function is diverse, and there is little doubt that serotonin is an important CNS neurotransmitter. Although some of the serotonin is metabolized by monoamine oxidase, most of the serotonin released into the post-synaptic space is removed by the neuron through a re-uptake mechanism inhibited by the tricyclic antidepressants and the newer, more selective antidepressant re uptake inhibitors such as fluoxetine and sertraline. In recent years, selective serotonin reuptake inhibitors have been introduced for the treatment of depression. Prozac is the most famous drug in this class. Clomiprimine, fluoxetine (Prozac), sertraline and paroxetine selectively block the re uptake of serotonin, thereby increasing the levels of serotonin in the central nervous system. Note the similarities and differences between the tricyclic antidepressants and the selective serotonin re uptake inhibitors. Clomipramine has been useful in the treatment of obsessive-compulsive disorders. Monoamine oxidase (MAO) causes the oxidative deamination of norephinephrine, serotonin, and other amines. This oxidation is the method of reducing the concentration of the neurotransmitter after it has sent the signal at the receptor site. A drug which inhibits this enzyme has the effect of Most MAO inhibitors are hydrazine derivatives. Hydrazine is highly reactive and may form a strong covalent bond with MAO with consequent inhibition for up to 5 days. These drugs are less effective and produce more side effects than the tricyclic antidepressants. For example, they lower blood pressure and were at one time used to treat hypertension. Their use in psychiatry has also become very limited as the tricyclic antidepressants have come to dominate the treatment of depression and allied conditions. Thus, MAOIs are used most often when tricyclic antidepressants give unsatisfactory results. Phenelzine is the hydrazine analog of phenylethylamine, a substrate of MAO. This and several other MAOIs, such as isocarboxazide, are structurally related to amphetamine and were synthesized in an attempt to enhance central stimulant properties.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.1%3A_Intermolecular_Forces

As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase “intermolecular attraction” to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions. Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter: The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its (IMFs) and the kinetic energies (KE) of its molecules. s are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles’ KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure \(\Page {1}\) illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE, of a given substance. As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure \(\Page {2}\). We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, C H , is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter’s fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure \(\Page {3}\). Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter. Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not molecular forces. molecular forces are those the molecule that keep the molecule together, for example, the bonds between the atoms. molecular forces are the attractions molecules, which determine many of the physical properties of a substance. Figure \(\Page {4}\) illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy—430 kilojoules. All of the attractive forces between neutral atoms and molecules are known as , although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module.   One of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the in honor of German-born American physicist Fritz who, in 1928, first explained it. This force is often referred to as simply the . Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron’s location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an . These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species—a so-called dispersion force like that illustrated in Figure \(\Page {5}\). Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F and Cl are gases at room temperature (reflecting weaker attractive forces); Br is a liquid, and I is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table \(\Page {1}\). The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as . A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces. Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH , SiH , GeH , and SnH . Explain your reasoning. Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH , SiH , GeH , and SnH are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH is expected to have the lowest boiling point and SnH the highest boiling point. The ordering from lowest to highest boiling point is expected to be CH < SiH < GeH < SnH A graph of the actual boiling points of these compounds versus the period of the group 14 elements shows this prediction to be correct:   Order the following hydrocarbons from lowest to highest boiling point: C H , C H , and C H . All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore C H < C H < C H . The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers -pentane, isopentane, and neopentane (shown in Figure \(\Page {6}\)) are 36 °C, 27 °C, and 9.5 °C, respectively. Even though these compounds are composed of molecules with the same chemical formula, C H , the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for -pentane and least for neopentane. The elongated shape of -pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip’s contact, the stronger the connection. Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way. Geckos’ toes are covered with hundreds of thousands of tiny hairs known as , with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called . The huge numbers of spatulae on its setae provide a gecko, shown in Figure \(\Page {7}\), with a large total surface area for sticking to a surface. In 2000, Kellar , who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight. In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex and Congcong at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications. Recall from the chapter on chemical bonding and molecular geometry that molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a . Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a —the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure \(\Page {8}\). The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F molecules. Both HCl and F consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances. Predict which will have the higher boiling point: N or CO. Explain your reasoning. CO and N are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N molecules, so CO is expected to have the higher boiling point. A common method for preparing oxygen is the decomposition Predict which will have the higher boiling point: \(\ce{ICl}\) or \(\ce{Br2}\). Explain your reasoning. ICl. ICl and Br have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.   Nitrosyl fluoride (ONF, molecular mass 49 amu) is a gas at room temperature. Water (H O, molecular mass 18 amu) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to with these atoms. Molecules with F-H, O-H, or N-H moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called . Examples of hydrogen bonds include ⋯HF, H O⋯HOH, and H N⋯HNH , in which the hydrogen bonds are denoted by dots. Figure \(\Page {9}\) illustrates hydrogen bonding between water molecules. Despite use of the word “bond,” keep in mind that hydrogen bonds are attractive forces, not attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to 10% as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces. Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group 15 (NH , PH , AsH , and SbH ), group 16 hydrides (H O, H S, H Se, and H Te), and group 17 hydrides (HF, HCl, HBr, and HI). The boiling points of the heaviest three hydrides for each group are plotted in Figure \(\Page {10}\). As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily. For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3, 4, and 5. If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect NH to boil at about −120 °C, H O to boil at about −80 °C, and HF to boil at about −110 °C. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure \(\Page {10}\). The stark contrast between our naïve predictions and reality provides compelling evidence for the strength of hydrogen bonding. Consider the compounds dimethylether (CH OCH ), ethanol (CH CH ), and propane (CH CH CH ). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning. The VSEPR-predicted shapes of CH OCH , CH CH OH, and CH CH CH are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH CH CH is nonpolar, it may exhibit dispersion forces. Because CH OCH is polar, it will also experience dipole-dipole attractions. Finally, CH CH OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH CH CH < CH OCH < CH CH OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C. Ethane (CH CH ) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH NH ). Explain your reasoning. The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH CH and CH NH are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C. Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure \(\Page {10}\). Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure \(\Page {12}\) The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication. The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient kinetic energy to move past each other. Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one dipolar molecule for the partial positive end of another. The temporary dipole that results from the motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/The_Polymerization_of_Ethene

This module guides you through the mechanism for the polymerisation of ethene by a free radical addition reaction. We are going to talk through this mechanism in a very detailed way so that you get a feel for what is going on. You will remember that during the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator. The oxygen reacts with some of the ethene to give an organic peroxide. Organic peroxides are very reactive molecules containing oxygen-oxygen single bonds which are quite weak and which break easily to give free radicals. You can short-cut the process by adding other organic peroxides directly to the ethene instead of using oxygen if you want to. The type of the free radicals that start the reaction off vary depending on their source. For simplicity we give them a general formula: \(Ra ^{\bullet}\) In an ethene molecule, CH =CH , the two pairs of electrons which make up the double bond aren't the same. One pair is held securely on the line between the two carbon nuclei in a bond called a sigma bond. The other pair is more loosely held in an orbital above and below the plane of the molecule known as a \(\pi\) bond. It would be helpful - but not essential - if you read about the before you went on. If the diagram above is unfamiliar to you, then you certainly ought to read this background material. Imagine what happens if a free radical approaches the \(\pi\) bond in ethene. Don't worry that we've gone back to a simpler diagram. As long as you realise that the pair of electrons shown between the two carbon atoms is in a \(\pi\) bond - and therefore vulnerable - that's all that really matters for this mechanism. The sigma bond between the carbon atoms isn't affected by any of this. The free radical, Ra , uses one of the electrons in the \(\pi\) bond to help to form a new bond between itself and the left hand carbon atom. The other electron returns to the right hand carbon. You can show this using "curly arrow" notation if you want to: If you aren't sure about about you can follow this link. This is energetically worth doing because the new bond between the radical and the carbon is stronger than the \(\pi\) bond which is broken. You would get more energy out when the new bond is made than was used to break the old one. The more energy that is given out, the more stable the system becomes. What we've now got is a bigger free radical - lengthened by CH CH . That can react with another ethene molecule in the same way: So now the radical is even bigger. That can react with another ethene - and so on and so on. The polymer chain gets longer and longer. The chain does not, however, grow indefinitely. Sooner or later two free radicals will collide together. That immediately stops the growth of two chains and produces one of the final molecules in the poly(ethene). It is important to realise that the poly(ethene) is going to be a mixture of molecules of different sizes, made in this sort of random way. The over-all process is known as Because chain termination is a random process, poly(ethene) will be made up of chains of different lengths. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Chemical_Equations

are a way to show what happens in a chemical reaction. A chemical equation looks something like this: \[A \rightarrow B\] In this case, A represents a reactant or reagent, and B represents a product. Usually one element doesn't turn into another element, so A and B might represent different molecules that are too complicated to just write the formulas. There might also be multiple reactants and products, like this: \[A + B \rightarrow C + D\] In this class, we will mostly study reactions of simple molecules, so we will use their formulas in equations, like this: \[Cl_{2} + Mg \rightarrow MgCl_{2}\] Often, we might want to show the state of the reactants and products, so we can use (g, l, s, or aq) to show if it is a gas, liquid, solid, or in a water solution (solutions in water are called solutions). For the example above, this becomes, assuming we do the reaction dry \[Cl_{2}(g) + Mg(s) \rightarrow MgCl_{2}(s)\] Notice that the numbers showing how many atoms of each element are in the formula go after the element, as a subscript . When we write chemical equations, usually we want them to be , which means that they have the same number of each kind of particle on each side of the equation. Here's an example of an unbalanced equation: \[Br_{2}(l) + Na(s) \rightarrow NaBr(s)\] In this example, there are 2 bromine atoms on the left, and only 1 on the right. We always have to balance the number of each type of nucleus on each side, like this: \[Br_{2}(l) + 2Na(s) \rightarrow 2NaBr(s)$$OR like this:$$\frac{1}{2}Br_{2}(l) + Na(s) \rightarrow NaBr(s)\] Notice that we have put the number of sodium atoms needed for the reaction in front of the symbol for sodium, and not in a after the symbol. This is called a , and it is different because it tells us how many of a molecule we need, not how many atoms are in the molecule, like this: \[2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)\] We also have to balance the number of electrons on each side. The easiest way to do this is usually to make sure the charges on both sides add up to the same number. For example, here's an equation that isn't balanced for electrons, even though it is balanced for nuclei: \[Mg(s) + Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + Ag(s)\] To balance this equation we have to make both the charges and the nuclei balanced, like this: \[Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)\] This means that the number of each type of particle must be the same on both sides of the equation, because particles (nuclei or electrons) can't appear or disappear (except under special circumstances, which we call nuclear chemistry, so don't worry about that right now). This is why equations need to be balanced. The second important thing is that the formulas in the equation need to match the actual molecules that are used or produced in the reaction. So if you are given the formulas, and you change them instead of changing the coefficients when you balance the equations, the equation and the reaction it represents has changed! You can't do this. On the other hand, if you are trying to write a chemical equation but you aren't sure what the formulas are, you can definitely use balancing to help you decide, and in this case you could change the formulas and the coefficients, as long as the formulas you use match all the information you have. Here's an example of how balancing equations can help you avoid mistakes, based on some wrong answers my students gave on an exam last year. The question was, what reaction happens between calcium ions and carbonic acid in the ocean (a water solution)? To answer this, we have to translate it into formulas. Calcium ions: this is an alkaline earth metal, so Ca . Carbonic acid is H CO . Carbonic acid will dissociate a little bit, making some hydrogen ion (H ), some bicarbonate ion (HCO ), and some carbonate ion (CO ). Calcium carbonate is insoluble, so it will form a solid, ionic material. You could write the reaction like this: \[Ca^{2+}(aq) + H_{2}CO_{3}(aq) \rightarrow CaCO_{3}(s) + 2H^{+}(aq)\] This equation is balanced. What if you wrote it like this? \[Ca^{2+}(aq) +H_{2}CO_{3}(aq) \rightarrow CaCO_{3}(s)+ H_{2}(aq)\] Now it isn't balanced, because the charge on the left side is 2+ and the charge on the right side is 0. This means that there are 2 more electrons on the right than on the left, which is a problem! They can't come from nowhere, so this isn't a complete equation: we are missing the source of the electrons. The mistake may have been confusing 2H with H (the difference between 2 before H and 2 after H is important!), or it might have been confusing Ca with Ca metal. For instance, this reaction is fine: \[Ca(s) + H_{2}CO_{3}(aq) \rightarrow CaCO_{3}(s)+ H_{2}(aq)\] However, in the context of the question, it would be very surprising to have Ca(s) in the ocean, because alkaline earth metals, like alkali metals, react with water. So we have lots of Ca , Mg , and Na ions in the ocean, but no atomic Ca, Mg, or Na in the ocean.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Ionic_and_Covalent_Bonds

There are many types of chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons. Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal. Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero. This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron. This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion. The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion. In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0. In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0. If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds. In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule. Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle. Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life. 1. Are these compounds ionic or covalent? 2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded. a) b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium? c)

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Triglycerides are esters of fatty acids and a trifunctional alcohol - glycerol (IUPAC name is 1,2,3-propantriol). The properties of fats and oils follow the same general principles as already described for the fatty acids. The important properties to be considered are: melting points and degree of unsaturation from component fatty acids. Since glycerol has three alcohol functional groups, three fatty acids must react to make three ester functional groups. The three fatty acids may or may not be identical. In fact, three different fatty acids may be present. The synthesis of a triglyceride is another application of the ester synthesis reaction. To write the structure of the triglyceride you must know the structure of glycerol and be given or look up the structure of the in the table. Since glycerol, (IUPAC name is 1,2,3-propantriol), has three alcohol functional groups, three fatty acids must react to make three ester functional groups. The three fatty acids may or may not be identical. In fact, three different fatty acids may be present. nThe synthesis of a triglyceride is another application of the ester synthesis reaction. To write the structure of the triglyceride you must know the structure of glycerol and be given or look up the structure of the in Table \(\Page {1}\) - find lauric acid. The simplified reaction reveals the process of breaking some bonds and forming the ester and the by product, water. Refer to the graphic on the left for the synthesis of . First, the -OH (red) bond on the acid is broken and the -H (red) bond on the alcohol is also broken. Both join to make HOH, a water molecule. Secondly, the oxygen of the alcohol forms a bond (green) to the acid at the carbon with the double bond oxygen. This forms the ester functional group. This process is carried out three times to make three ester groups and three water molecules. As you can see from the graphic on the left, the actual molecular model of the triglyceride does not look at all like the line drawing. The reason for this difference lies in the concepts of molecular geometry. Trilauroylglycerol. All of the above factors contribute to the apparent "T" shape of the molecule. 1 Practice writing out a triglyceride of stearic acid. Again look up the formula of stearic acid and use the structure of glycerol. 2. Write down your answers. Then check the answers from the drop down menu.

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The glass transition is probably the most commonly-cited characteristic of a polymeric material. At the glass transition temperature, the physical nature of the material changes subtly. It goes from being a rubbery, flexible material at higher temperature, above the glass transition, to a glassy, harder material at a lower temperature. The material is still a solid either way, but there is a definite change in how it responds to stimuli. A material could certainly be useful in either state. We might want a plastic to be more rigid, like a water bottle or a rod for a shower curtain. Alternatively, we might want it to be softer, like a seat cushion. Either way, it might be helpful to know the temperature at which the material will change from one type to the other. The glass transition is a little like what happens to gummy bears when you put them in ice cream. Straight from the bag, gummy bears are chewy, but they don't feel like they will break your teeth. Put them on ice cream and that changes. They become much harder to chew. The classic explanation for the glass transition is based on the idea of . Polymers are long-chain molecules and, given a little energy, the chains move around. They wiggle. They undergo bond rotations, switching from one conformation to another. A block of material contains piles of chains, like a nest of snakes. Chain ends and loops squirm past each other continuously. Chain flow allows a material to adapt when forces are exerted on it. We can bend an eraser because the chains in the rubber slide over each other and adopt a new shape. At least, that's what happens at room temperature. It might not work if the eraser were accidentally dropped in some liquid nitrogen, which is very, very cold. As a block of material cools down, it contracts. The molecules become packed more closely together. At some point, the free volume -- that's the amount of volume in the material that is not actually taken up by the molecules -- becomes too small to allow chains to move past each other. All of those chains need a little room around them if they are going to undergo conformational changes, and conformational changes are how polymer chains move. Without that extra room, the material suddenly becomes less flexible. If we start in the glassy state and increase the temperature, the volume of the material is expanding, and the free volume is increasing, too. At some point, the free volume becomes great enough that chains can slip past each other. The material becomes more flexible. It becomes rubbery. Now, this transition is not the same thing as melting. The chains are not completely overcoming their interactions with each other and gaining freedom of movement in any direction. The chains are still highly entangled. Portions of the chain are sliding past each other but overall things have not changed that drastically. The block of polymer does not turn into a gooey puddle of liquid. Well, if a material can change from glassy to rubbery at a certain temperature, there will be consequences in how the material behaves. For example, the tires on your car are meant to be rubbery and flexible; that factor helps them grip the road. If the weather gets too cold and your tires become glassy, the tires no longer have the same amount of flexibility, and you don't have as much traction. "All-weather tires" are composed of a rubber that has a very low glass transition temperature, helping to avoid this problem. ("Snow tires" are a different thing; they have patterns in the treads that help channel away the snow in order to improve traction.) Knowing when this change will occur would be very useful. So, how do we determine the glass transition? Phase changes, such as melting points, are measured using , so let's start there. When a solid is heated, its temperature increases. That seems simple enough. Temperature is basically a measure of heat content, so as heat flows in, the measured temperature rises. This simple relationship breaks down at the melting point. At that point, heat flowing into the material is consumed by the breaking of intermolecular forces. Overcoming these attractions costs extra energy. Consequently, there is a point at which the temperature rise in the material stalls out temporarily while it melts. This extra heat needed to melt the material is called the heat of melting or, more commonly, the heat of fusion. Heat of fusion actually refers to the opposite process as the material is cooled and frozen. As heat is carried away from a cooling material, it cools down gradually, but there comes a point at which those strong intermolecular interactions form, giving off some extra heat. The fusion temperature is the same as the melting temperature, and the heat of fusion is the same as the heat of melting, but in one case the heat is added and in another case the heat is given off. In principle, if we just heat something up and look for that stalling point in the temperature, we can find the temperature of the phase transition. (DSC) is a technique that is commonly used to measure phase transitions, including the glass transition temperature. It is based on the same ideas but the experiment is run in a slightly different way. In DSC, we have two tiny sample chambers side by side. One contains the material we are interested in and the other (empty) one is used as a reference. The instrument heats both samples at a constant rate, all the while maintaining both of them at the same temperature. Consequently, it may actually add more heat to one chamber than the other, so that they both reach 30.5 °C at the same time, then 30.6 °C, and so on. Once we reach the melting point of the sample of interest, heat flow into that sample must be increased so that it can keep up with the reference. Past that point, the heat flow drops back to a more normal level. If we look at a DSC scan, we see a graph of heat flow on the y-axis and temperature on the x-axis. The heat flow usually stays pretty constant as the temperature increases. At the melting point, heat flow increases, but then it drops back down once the melting point has been overcome. It may not drop back to the same level as before, however, because the heat flow that is recorded is related to the heat capacity of the material. The solid and the liquid usually respond differently to heat. In general, because the molecules in a liquid can freely rotate, they have something else to do with added heat. Liquids thus have a slightly higher heat capacity than their corresponding solids. All of that explanation hopefully prepares us for a DSC study of the glass transition temperature. It works the same way, but only up to a point. If we look at a DSC scan, we still see heat flow on the y-axis and temperature on the x-axis. At the glass transition, heat flow bumps up slightly -- and remains there. It looks very different from a melting point. A DSC scan of a melting point shows a "peak" at the transition temperature. A DSC scan of a glass transition point shows only a little step, like a riverbank Why is it different? Well, the glass transition is not really a phase change like the melting point or boiling point. It does not involve a physical change of state. It was a solid before and it is a solid afterward. No intermolecular interactions must suddenly be overcome to free the molecules from each other. Instead, it's just a volume change. The free volume became great enough that the chains can slip past each other, but the chains are still clinging together in a solid-state. Now, that additional movement does have consequences. The material becomes more flexible. If heat flows in, there is more freedom of motion into which the heat can be distributed. In other words, there is a slight increase in heat capacity, and that's what we observe in DSC. One practical note: DSC scans can actually be displayed in two different ways. The y-axis can either display heat flowing in or heat flowing out. That means melting points can look like peaks or like valleys, depending on how the data is displayed. Often the data are labeled with an arrow that says "endo" to tell you which direction along the y-axis means more heat is flowing in (or possibly "exo", meaning which direction means heat is flowing out). You need to look carefully if things seem backward. For each DSC trace, state what sort of transition is occurring at what temperature. Why does the glass transition temperature vary from one kind of polymer to another? What structural factors influence the glass transition temperature? This structure-property relationship isn't a straightforward one, as there seem to be a number of different variables involved. However, the simplest of these factors is just molecular weight. The higher the molecular weight of a polymer, the higher its glass transition temperature. This relationship is true only up to a certain point, however. The non-linear dependence of the glass transition temperature of molecular weight is described by the : \[T_g = T_{g(∞)} - \dfrac{K}{ M_n}\] Here, refers to the glass transition temperature of an infinitely long chain of the polymer. is a constant for a particular polymer, such as polystyrene or polyethylene. A plot of resembles a saturation curve; the line rises sharply, gradually stalling out and continuing parallel to the x-axis. In other words, although this relationship of increasing glass transition temperature with increasing molecular weight holds true at relatively low molecular weights, the glass transition temperature remains constant once a threshold molecular weight has been reached.

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When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure \(\Page {1}\), have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly). The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table \(\Page {1}\) shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases.   The various IMFs between identical molecules of a substance are examples of . The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface—that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure \(\Page {2}\), because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical. is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table \(\Page {2}\). Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively “tough skin” that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure \(\Page {3}\), even though they are denser than water, move on its surface because they are supported by the surface tension. The IMFs of attraction between two molecules are called . Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not “wet” the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure \(\Page {4}\)). If you place one end of a paper towel in spilled wine, as shown in Figure \(\Page {5}\), the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of —when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity. Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many −OH groups. Water molecules are attracted to these −OH groups and form hydrogen bonds with them, which draws the H O molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers. Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure \(\Page {6}\). If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away. The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the following equation: \[h=\dfrac{2T\cosθ}{rρg} \label{10.2.1} \] where When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of 0°. This is the situation for water rising in a glass tube. At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm? For water, = 71.99 mN/m and = 1.0 g/cm . The liquid will rise to a height given by Equation \(\ref{10.2.1}\) : \[h=\dfrac{2T\cosθ}{rρg} \nonumber \] The Newton is defined as a kg m/s , and so the provided surface tension is equivalent to 0.07199 kg/s . The provided density must be converted into units that will cancel appropriately: = 1000 kg/m . The diameter of the tube in meters is 0.00025 m, so the radius is 0.000125 m. For a glass tube immersed in water, the contact angle is = 0°, so cos = 1. Finally, acceleration due to gravity on the earth is = 9.8 m/s . Substituting these values into the equation, and cancelling units, we have: \[h=\mathrm{\dfrac{2(0.07199\:kg/s^2)}{(0.000125\:m)(1000\:kg/m^3)(9.8\:m/s^2)}=0.12\:m=12\: cm} \nonumber \] Water rises in a glass capillary tube to a height of 8.4 cm. What is the diameter of the capillary tube? diameter = 0.36 mm Many medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure \(\Page {7}\). When your finger is pricked, a drop of blood forms and holds together due to surface tension—the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrow-diameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising. The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension (elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise.

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Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. Acids and bases are of interest not only to the chemically inclined; they play a major role in our modern industrial society — so anyone who participates in it, or who is interested in its history and development, needs to know something about them. Five of the major acids and bases fall into the "Top 20" industrial chemicals manufactured in the world. The following table shows year-2000 figures for the U.S: Sulfuric acid - Lime (CaO) Phosphoric acid - Ammonia Sodium hydroxide - Nitric acid - This term refers to any inorganic acid, but its common use is usually limited to the major strong acids plus phosphoric acid. The major mineral acids— sulfuric, nitric, and hydrochloric— have been known since medieval times. Their discovery is usually credited to the Persian alchemist Abu Musa Jabir ibn Hayyan, known in the West by his Latinized name Geber. Jabir also invented , the mixture of nitric and hydrochloric acids that has the unique ability to dissolve gold. More sulfuric acid is manufactured than any other industrial chemical, and it is the cheapest industrial acid worldwide. It has been continuously manufactured in the U.S. since 1793 and in Europe for much longer. 2 H SO → H SO + HSO H SO + 2 NaCl → 2 Na + SO + HCl C H O → 12 C + 11 H O H SO + H SO → HS O + H O Sulfur trioxide, the anhydride of sulfuric acid, is the immediate precursor. Gaseous SO reacts vigorously with water, liberating much heat in the process: \[SO_{3(g)} + H_2O_{(l)} → H_2SO_{4(l)}\] Industrial manufacture of the acid starts with sulfur dioxide, prepared from burning elemental sulfur or obtained as a byproduct from roasting sulfide ores. The oxidation of SO to SO looks simple \[SO_{2(g)} + ½ O_{2(g)} → SO_{3(g)}\] but there are several complications: \[H_2SO_{4(l)} + SO_{3(g)} → H_2S_2O_{7(l)}\] \[H_2S_2O_{7(l)} + H_2O_{(l)} → 2 H_2SO_{4(l)}\] Sulfuric acid has a broad spectrum of industrial uses, and the annual tonnage follows the economic cycle quite closely. - Combustion of fossil fuels which contain organic sulfur compounds releases SO into the atmosphere. Photochemical oxidation of this compound to SO , which rapidly takes up moisture, leads to the formation of H SO , a major component of acid rain. results when sediments of the very common , FeS2, are exposed to air and are oxidized: FeS + 7/2 O + H O → Fe + 2 SO + 2 H further oxidation of the iron to Fe results in additional reactions. The resulting drainage liquid is often orange-brown in color and can a have a pH of below zero. Anhydrous HNO is a colorless liquid boiling at 82.6°C, but "pure" HNO only exists as the solid which melts at –41.6°C. In its liquid and gaseous states, the acid is always partially decomposed into nitrogen dioxide: 2 HNO → 2 NO + ½ O + H O This reaction, which is catalyzed by light, accounts for the brownish color of HNO solutions. [ ] The simplest method, which was used industrially before 1900, was by treatment of sodium nitrate ("Chile saltpeter", NaNO ) with sulfuric acid. The direct synthesis of the acid from atmospheric nitrogen and oxygen is thermodynamically favorable ½ N + 5/4 O + ½ H O → HNO but is kinetically hindered by an extremely high activation energy, a fact for which we can be most thankful (see sidebar.) The first industrial nitrogen fixation process, developed in 1903, used this reaction to produce nitric acid, but it required the use of an electric arc to supply the activation energy and was therefore too energy-intensive to be economical. If it were not for the high activation energy required to sustain this reaction, all of the oxygen in the atmosphere would be consumed and the oceans would be a dilute solution of nitric acid. The modern Ostwald process involves the catalytic oxidation of ammonia to nitric oxide NO, which is oxidized in a further step to NO ; reaction of the latter with water yields HNO . This route, first developed in 1901, did not become practical until the large-scale production of ammonia by the in 1910. The major industrial uses of nitric acid are for the production of ammonium nitrate , and in the manufacture of . On a much small scale, the acid is used in metal pickling, etching semiconductors and electronic circuit boards, and in the manufacture of acrylic fibers. In the laboratory, the acid finds use in a wide variety of roles. High-temperature combustion processes (in internal combustion engines, power plants, and incinerators) can oxidize atmospheric nitrogen to nitric oxide (NO) and other oxides ("NO "); the NO is then photooxidized to NO , which reacts with water to form HNO which is a major component of . NO is the major precursor of . Unlike the other major acids, there is no such substance as "pure" hydrochloric acid; what we call "hydrochloric acid" is just an aqueous solution of hydrogen chloride gas (bp –84°C). But in a sense it is more "pure" than the acids discussed above, since there is no autoprotolysis; hydronium and chloride ions are the only significant species in the solution. Hydrochloric acid is usually sold as a 32-38% (12M) solution of HCl in water; concentrations greater than this are known as . Hydrochloric acid is still sometimes sold under its older name muriatic acid for cleaning bricks and other household purposes. The name comes from the same root as marine, reflecting its preparation from salt. The acid has been known to chemists (and alchemists), and used for industrial purposes since the middle ages. Its composition HCl was demonstrated by Humphrey Davy in 1816. The uses of hydrochloric acid are far too many to enumerate individually, but the following stand out: The ancient method of treating salt with sulfuric acid to release HCl has long since been supplanted by more efficient processes, including direct synthesis by "burning" hydrogen gas in chlorine: \[H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}\] Most hydrochloric acid production now comes from reclaiming byproduct hydrogen chloride gas from other processes, especially those associated with the production of industrial organic compounds. The word comes from the Arabic , which refers to the ashes from which sodium and potassium hydroxides ( , "ashes remaining in the pot", and the origin of the element name ) were extracted as a step in the making of soap. Pure sodium hydroxide is a white solid consisting of Na and OH ions in a crystal lattice. Although it is widely thought of as an ionic solid, make a substantial contribution to its stability. Sodium hydroxide is now manufactured by the electrolysis of brine solutions, and along with chlorine, is one of the two major products of the . Electrolysis of aqueous NaCl produces Cl at the anode, but because H O can be reduced more readily than Na , the water is decomposed to H and OH at the cathode, leaving a solution of NaOH. An older reduces the Na to Na within a mercury amalgam (alloy), and the metallic sodium is then combined with water to produce NaOH and hydrogen. The net reaction for the reduction step is the same for both methods: \[2 Na^+ + 2 H_2O + 2e^– \rightarrow H_{2(g)] + 2 NaOH\] The resulting solution is usually evaporated to such a high concentration that it solidifies at ordinary temperatures. It is commonly shipped in rail cars or barges which can be heated with steam to liquefy the mixture for removal. (It is obviously uneconomical to ship large quantities of water across the country!) In contrast to the extremely diverse applications of sodium hydroxide which makes the demand for this commodity relatively immune to the ups and downs of the economic cycle, the consumption of chlorine is directly dependent on the economy as reflected in the demand for polyvinyl chloride products that are now widely used in the the construction and home furnishings industries. Because chlorine, being a gas, is expensive to store, the output of the chloralkali industry is governed largely by the demand for this commodity. When times are good this presents no problem; caustic is then largely a by-product and can easily be stockpiled if supply exceeds demand. But during an economic downturn, the demand for chlorine declines, limiting its production along with that of caustic. But because the demand for caustic tends to decline much less, it becomes scarce and its price rises, thus tending to drive the industrial economy into even deeper trouble. This compound is known industrially as , and domestically as . The common form is the , Na CO ·7 H O. The white crystals of this substance spontaneously lose water ( ) when exposed to the air, forming the monohydrate. Most of the world's sodium carbonate is made by the ammonia-soda "Solvay" process developed in 1861 by the Belgian chemist Ernest Solvay (1838-1922) whose patents made him into a major industrialist and a rich philanthropist. This process involves a set of simple reactions that essentially converts limestone (CaCO ), ammonia NH and brine (NaCl) into sodium bicarbonate NaHCO and eventually Na CO , recycling several of the intermediate products in an ingenious way. A minor source of soda ash (but quite significant in some countries, such as the U.S.) is the mining of natural (the remains of ancient lakes), such as the found in Southern California. Ammonia NH is of course not a true alkali, but it is conveniently included in this section for discussion purposes. Most people are familiar with the pungent odor of this gas, which can be detected at concentrations as low as 20-50 ppm. Tradition dies slowly: a non-existent chemical available in bottles! Ammonia is made by direct synthesis from the elements: \[N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}\] ... a simple-looking reaction, but one that required some very creative work to implement; the is considered to be the most important chemical synthesis of the 20th Century. Most acids are organic— there are millions of them. The acidic function is usually a hydroxyl group connected to a carbon that is bonded to an electron-withdrawing oxygen atom; the combination is the well-known , –COOH. Here are a few that are part of everyone's life. This is next to formic acid in being the simplest of the organic acids, and in the form of vinegar (a 5-8% solution in water) its characteristic odor is known to everyone. The pure acid is a colorless liquid above 16.7°C; below this temperature it forms a crystalline solid, hence the term "glacial acetic acid" that is commonly applied to the pure substance. The name of the acid comes from , the Latin word for vinegar. Slightly less than half of the world production of acetic acid goes into the production of . The end product visible to most people would be the flexible plastic bottles in which drinking water is sold. Other uses are related mostly to the production of other chemicals, mainly , but also including . Bacterial fermentation of sugars has been the source of vinegar since ancient times, and is still accounts for most food-grade acetic acid and vinegar, but it now amounts to only about 10% of total acetic acid production: C H O → 3 CH COOH There are several important synthetic routes to acetic acid production, but the major one is by treating methanol with carbon monoxide: CH OH + CO → CH COOH (Methanoic acid) HCOOH mp 8.4°C. HOOC–COOH Known by Arabic alchemists in the 8th Century; first isolated by Scheele in 1784 No carboxyl groups here, but the –OH group one carbon away from the double bond is still fairly acidic. One of its geometric isomers, L-ascorbic acid, is more widely known as ; discovery of the essential role of this substance in preventing the disease (from which its name is derived) yielded two Nobel prizes in 1937. Most animals are capable of synthesizing their own Vitamin C, but primates (along with guinea pigs) seem to have lost the required gene somewhere along the way, so we must depend on fruits and veggies for our supply. (Because ascorbic acid is soluble in water, it tends to be leached out of vegetables when they are boiled, so it is much healthier to steam them.) The major industrial use of ascorbic acid is as an , so it is often added to foods and other materials as a preservative. 2-hydroxybenzoic acid is found in willow trees Does chemistry give you a headache? If so, pull a small shoot off of a willow tree and chew on it for a while. This has been a folk-remedy for pains and fever since ancient times. Willow bark ( in Latin) contains, and lent its name to the active principle, salicylic acid. The acid itself turned out to be a bit too much of an irritant to the stomach's lining, so the German firm Bayer began marketing a tamed version, (ASA) under the name Aspirin in 1899, and it has been going strong ever since. Interestingly, the detailed chemistry of its pain-and-fever reliving action was not discovered until 1970. This generic term refers to carboxylic acids built from a chain of 4 to 22 carbon atoms. Fatty acids, as the name implies, are derived from fats, in which they are bound to glycerol in the form of triglycerides. Fats, which occur in all animal tissues, are the most efficient means of storing metabolic energy. Vegetable oils are another source. There are two general categories of fatty acids: The human body is able to synthesize most of the fatty acids it needs, but two classes of unsaturated acids, known as the , can be obtained only from foods. The compounds are known as ω-3 and ω-6 fatty acids; the ω (omega)- notation means that the double bond is located carbons away from the terminal CH group of the molecule. Most of the unsaturated fatty acids found in nature have cis configurations around their double bonds. The trans fatty acids in our foods that we hear much about are made from vegetable oils that have had some of their double bonds changed to single bonds by a chemical process called hydrogenation. The object of this is to make the original liquid fatty acids into solid forms that are more suitable for use in foods (as in margarine) and particularly as shortenings in baking. Unfortunately, the hydrogenation process also changes the remaining cis double bonds into their trans forms which are believed to be implicated in cardiac disease. Amino acids are the building blocks of proteins, which is what you are largely made of. There are twenty of them, of which about half can be manufactured by the body; we must depend on our food to supply the remainder, which are known as the . Each one comes in two mirror-image forms; only the "left" forms are found in most proteins. All amino acids have the basic structure shown above; they differ in the nature of the group of atoms designated by "R" in the diagram. Although we call them "acids", the amino acids are really chemical hermaphrodites; you will recall that amines are weak bases. The balance between their acidic and basic properties can be shifted simply by changing the pH. The carboxylic acid part of one amino acid can react with the amine part of another to form a peptide bond (an linkage) shown here. The polymeric chains that result are known as peptides if they are fairly short (2 to about 20-50 amino acid units); longer chains, or aggregate made up of multiple peptide units, are proteins. Owing to their very large size (500 amino acid residues is quite common, but some have as many as 1500), proteins are able to fold in various ways, so the amino acid sequence alone is not sufficient to determine their properties.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.07%3A_Calorimetry

At one time, calories in foods were measured with a bomb calorimeter. A weighed amount of the food would be placed in the calorimeter and the system was then sealed and filled with oxygen. An electric spark ignited the food-oxygen mixture. The amount of heat released when the food burned gave an idea of the calories present within the food. Today, calories are calculated from the protein, carbohydrate, and fat content of food (all determined by chemical analysis). is the measurement of the transfer of heat into or out of a system during a chemical reaction or physical process. A is an insulated container that is used to measure heat changes. The majority of reactions that can be analyzed in a calorimetry experiment are either liquids or aqueous solutions. A frequently used and inexpensive calorimeter is a set of nested foam cups fitted with a lid to limit the heat exchange between the liquid in the cup and the air in the surroundings (see figure below). In a typical calorimetry experiment, specific volumes of the reactants are dispensed into separate containers and the temperature of each is measured. They are then mixed into the calorimeter, which starts the reaction. The reactant mixture is stirred until the reaction is complete, while the temperature of the reaction is continuously monitored. The key to all calorimetry experiments is the assumption that there is no heat exchange between the insulated calorimeter and the room. Consider the case of a reaction taking place between aqueous reactants: the water in which the solids have been dissolved is the surroundings, while the dissolved substances are the system. The temperature change that is measured is the temperature change that is occurring in the surroundings. If the temperature of the water increases as the reaction occurs, the reaction is exothermic. Heat was released by the system into the surrounding water. An endothermic reaction absorbs heat from the surroundings, so the temperature of the water decreases as heat leaves the surroundings to enter the system. The temperature change of the water is measured in the experiment and the specific heat of water can be used to calculate the heat absorbed by the surroundings \(\left( q_\text{surr} \right)\). \[q_\text{surr} = m \times c_p \times \Delta T\nonumber \] In the equation, \(m\) is the mass of the water, \(c_p\) is the specific heat of the water, and \(\Delta T\) is \(T_f - T_i\). The heat absorbed by the surroundings is equal, but opposite in sign, to the heat released by the system. Because the heat change is determined at constant pressure, the heat released by the system \(\left( q_\text{sys} \right)\) is equal to the enthalpy change \(\left( \Delta H \right)\). \[q_\text{sys} = \Delta H = -q_\text{surr} = - \left( m \times c_p \times \Delta T \right)\nonumber \] The sign of \(\Delta H\) is positive for an endothermic reaction and negative for an exothermic reaction. In an experiment, \(25.0 \: \text{mL}\) of \(1.00 \: \text{M} \: \ce{HCl}\) at \(25.0^\text{o} \text{C}\) is added to \(25.0 \: \text{mL}\) of \(1.00 \: \text{M} \: \ce{NaOH}\) at \(25.0^\text{o} \text{C}\) in a foam cup calorimeter. A reaction occurs and the temperature rises to \(32.0^\text{o} \text{C}\). Calculate the enthalpy change \(\left( \Delta H \right)\) in \(\text{kJ}\) for this reaction. Assume the densities of the solutions are \(1.00 \: \text{g/mL}\) and that their specific heat is the same as that of water. The volume and density can be used to find the mass of the solution after mixing. Then calculate the change in enthalpy by using \(\Delta H = q_\text{sys} = -q_\text{surr} = - \left( m \times c_p \times \Delta T \right)\). \[\begin{align*} m &= 50.0 \: \text{mL} \times \frac{1.00 \: \text{g}}{\text{mL}} = 50.0 \: \text{g} \\ \Delta H &= - \left( m \times c_p \times \Delta T \right) = - \left( 50.0 \: \text{g} \times 4.18 \: \text{J/g}^\text{o} \text{C} \times 7.0^\text{o} \text{C} \right) = -1463 \: \text{J} = -1.5 \: \text{kJ} \end{align*}\nonumber \] The enthalpy change is negative because the reaction releases heat to the surroundings, resulting in an increase in the temperature of the water.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.01%3A_Molecular_Weight

Molecular weight is one of the most central aspects of polymer properties. Of course, all molecules have molecular weights of their own. It might seem obvious that molecular weight is an essential property of any molecular compound. In polymers, molecular weight takes on added significance. That's because a polymer is a large molecule made up of repeating units, but how many repeating units? Thirty? A thousand? A million? Any of those possibilities might still be considered a representative of the same material, but their molecular weights would be very different, and so would their properties. That variation introduces some unique aspects of polymer molecular weight. Because polymers are assembled from smaller molecules, the length (and consequently the molecular weight) of a polymer chain depends on the number of monomers that have been enchained into the polymer. The number of enchained monomers in an average polymer chain in a material is called the (DP). Notice that key point: it is just an average. In any given material, there will be some chains that have added more monomers and some chains that have added fewer. Why the difference? First of all, polymer growth is a dynamic process. It requires monomers to come together and react. What if one monomer starts reacting, forming a growing chain, before any of the others get started? With its head start, this chain will become longer than the rest. What if something goes wrong with one of the growing chains, and it can no longer add new monomers? That chain experienced an early death, and it will never grow as long as the others. As a result, when we are speaking about the molecular weight of a polymer, we are always talking about an average value. Some chains in the material will be longer (and heavier) and some chains in the material will be shorter (and lighter). As with any group of measurements, it's helpful to know how widely distributed the individual values really are. In polymer chemistry, the width of the distribution of molecular weights is described by the (Ð, also called, in older texts, the polydispersity or the polydispersity index, PDI). The dispersity of a polymer sample if often between 1 and 2 (although it can be even higher than 2). The closer it is to 1, the narrower the distribution. That is, a dispersity of 1.0 would mean that all of the chains in a sample are exactly the same length, with the same molecular weight. The original idea of dispersity was based on alternative methods of measuring the molecular weight (or the chain length) of a polymer sample. One set of methods gave something called the (symbol ). These methods essentially took the weight of a sample, counted the molecules in a sample, and therefore found the average weight of each molecule in that sample. A classic example of this approach is a colligative properties experiment, such as a freezing point depression. You know that impurities in a liquid tend to disrupt intermolecular interactions and lower the freezing point of the liquid. You may also know that the amount by which the freezing gets lowered depends on the number of molecules or ions that get dissolved. Hence, if you weigh a sample of polymer, dissolve it in a solvent, and measure the freezing point, you could figure out the number of molecules dissolved and consequently arrive at . That's not so easy in practice; freezing point depressions are very small. They're not used very often anymore. A very common example of the kind of measurement widely used to determine today is end group analysis. In end group analysis, we use H NMR measurements to determine the ratio of a specific proton in the repeat units to a specific proton in the end group. Remember, the end group might be something like the initiator, which only added onto the first monomer to get the polymerization going. By the end of the polymerization, it is still found at the end of the polymer chain, so it is an end group. There is only one of them per chain, whereas there are lots of monomers enchained in the polymer, so the ratio of those enchained monomers to the end group tells us how long the chain is. The other set of methods upon which dispersity was based gave something called the (symbol ). The classic example was a light-scattering experiment. In this experiment, a solution of polymer was exposed to a beam of light and the resulting scattered light -- coming from the sample in different directions -- was analyzed to determine the size of the polymer chains in the solution. The results were more heavily influenced by the larger molecules in solution. As a result, this measurement of molecular weight was always higher than measurements based on counting every single molecule. The resulting ratio, Ð = / , became known as the polydispersity index or, more recently, the dispersity. Because was always more strongly influenced by longer chains, it was a little bigger than and therefore the dispersity was always bigger than 1.0. Nowadays, both molecular weight and dispersity are most commonly measured using gel permeation chromatography (GPC), synonymous with size-exclusion chromatography (SEC). This method is a high-performance liquid chromatography (HPLC) technique. Solvent containing a sample of polymer is pumped through a specialized chromatography column capable of separating molecules based on their size differences. As sample emerges from the column, it is detected and recorded. Most commonly, the presence of sample in the solvent emerging from the column causes a slight change in the refractive index. A graph of refractive index versus time presents a record of the amount of sample emerging from the column at a given time. Because the column separated molecules based on size, the time axis corresponds indirectly with chain length of molecular weight. How can the column separate molecules based on size? The column is packed with a porous material, usually insoluble polymer beads. The pore sizes vary. These pores a crucial to separation because molecules flowing through the column may tarry in the pores. Smaller molecules could become delayed in any of the pores in the material, whereas larger molecules will only be delayed in the very largest pores. Consequently, a longer elution time corresponds to a lower molecular weight. If you injected a series of different polymers into a GPC, each having a different molecular weight distribution, you would observe each one eluting at a different time. What's more, each peak may be broader or narrower, depending on the dispersity of that particular sample. The wider the peak in GPC, the broader the distribution of molecular weights; the narrower the peaks, the more uniform are the chains. Normally a software package analyzes the curve to determine the dispersity. Note that the x axis on a GPC trace is most commonly labeled as "elution time" and it normally runs left to right. However, often the x axis is labeled "molecular wright" because that is really the quantity we are interested in. In fact, sometimes the axis is reversed, so that peaks with higher molecular weights appear to the right, because it can feel more natural to look at it that way. You need to look carefully at the data to see how it is displayed. There are some problems with relying on GPC for molecular weight measurements. The main difficulty is that polymers in solution tend to coil into balls, and those coils will contain greater or lesser amounts of solvent, depending on how strongly the polymer and solvent interact with each other. If it interacts more strongly with the solvent, it will pull lots more solvent molecules inside its coils. The coil has to get bigger to make room for those internal solvent molecules. If it doesn't interact strongly with the solvent, it will mostly just stick to itself, blocking the solvent molecules out. There is a broad range of behaviors in between. As a result, different polymers may swell to different extents in different solvents. That matters because GPC is really using size of the polymer coil as an index of its molecular weight, so comparing GPC traces of two different kinds of polymers has to be done with caution. In each of the following cases, state which polymer has the higher molecular weight, and which one has a narrower dispersity Calculate the molecular weight of the following samples. Use NMR end group analysis to determine the degrees of polymerization in the following samples.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.02_Acid_Strength

All acids and bases do not ionize or dissociate to the same extent. This leads to the statement that acids and bases are not all of equal strength in producing H and OH ions in solution. The terms "strong" and "weak" give an indication of the strength of an acid or base. The terms strong and weak describe the ability of acid and base solutions to conduct electricity. If the acid or base conducts electricity strongly, it is a strong acid or base. If the acid or base conducts electricity weakly, it is a weak acid or base. The instructor will test the conductivity of various solutions with a light bulb apparatus. The light bulb circuit is incomplete. If the circuit is completed by a solution containing a large number of ions, the light bulb will glow brightly indicating a strong ability to conduct electricity as shown for HCl. If the circuit is completed by a solution containing large numbers of molecules and either no ions or few ions, the solution does not conduct or conducts very weakly as shown for acetic acid. An acid or base which strongly conducts electricity contains a large number of ions and is called a and an acid or base which conducts electricity only weakly contains only a few ions and is called a . The bond strengths of acids and bases are implied by the relative amounts of molecules and ions present in solution. The bonds are represented as: where A is a negative ion, and M is a positive ion Acids or bases with strong bonds exist predominately as molecules in solutions and are called "weak" acids or bases. Acids or bases with weak bonds easily dissociate into ions and are called "strong" acids or bases. Acids and bases behave differently in solution based on their strength. Acid or base "strength" is a measure of how readily the molecule ionizes in water. Some acids and bases ionize rapidly and almost completely in solution; these are called strong acids and strong bases. For example, hydrochloric acid (HCl) is a strong acid. When placed in water, virtually every HCl molecule splits into a H ion and a Cl ion in the reaction. \[\ce{HCl(aq) + H2O(l) <=> H3O^{+}(aq) + Cl^{-}(aq)} \nonumber\] For a strong acid like HCl, if you place 1 mole of HCl in a liter of water, you will get roughly 1 mole of H 0 ions and 1 mole of Cl ions. In a weak acid like hydrofluoric acid (HF), not all of the HF molecules split up, and although there will be some H and F ions released, there will still be HF molecules in solution . A similar concept applies to bases, except the reaction is different. A strong base like sodium hydroxide (NaOH) will also dissociate completely into water; if you put in 1 mole of NaOH into water, you will get 1 mole of hydroxide ions. \[\ce{NaOH(aq) + H2O(l) <=> Na^{+}(aq) + OH^{-}(aq) + H2O(l)} \nonumber\] The terms "strong" and "weak" in this context do not relate to how corrosive or caustic the substance is, but only its capability to ionize in water. The ability of a substance to eat through other materials or damage skin is more of a function of the properties of that acid, as well as its concentration. Although, strong acids are more directly dangerous at lower concentrations a strong acid is not necessarily more dangerous than a weak one. For example, hydrofluoric acid is a weak acid , but it is extremely dangerous and should be handled with great care. Hydrofluoric acid is particularly dangerous because it is capable of eating through glass, as seen in the video in the links section . The percent dissociation of an acid or base is mathematically indicated by the acid ionization constant (K ) or the base ionization constant (K ) . These terms refer to the ratio of reactants to products in equilibrium when the acid or base reacts with water. For acids the expression will be K = [H O ,A ]/[HA] where HA is the concentration of the acid at equilibrium, and A is the concentration of its conjugate base at equilibrium and for bases the expression will be \[K_b = \dfrac{[\ce{OH^{-}},\ce{HB^{+}}]}{\ce{B}}\] where B is the concentration of the base at equilibrium and HB is the concentration of its conjugate acid at equilibrium The stronger an acid is, the lower the pH it will produce in solution. pH is calculated by taking the negative logarithm of the concentration of hydronium ions. For strong acids, you can calculate the pH by simply taking the negative logarithm of its molarity as it completely dissociates into its conjugate base and hydronium. The same goes for strong bases, except the negative logarithm gives you the pOH as opposed to the pH. For weak acids and bases, the higher the K or K , the more acidic or basic the solution. To find the pH for a weak acid or base, you must use the K equation and a RICE table to determine the pH. All acids have a conjugate base that forms when they react with water, and similarly, all bases have a conjugate acid that reacts when they form with water. You can judge the relative strength of a conjugate by the \(K_a\) or \(K_b\) value of the substance because \(K_a \times K_b\) is equal to the ionization constant of water, K which is equal to \(1 \times 10^{-14}\) at room temperature. The higher the Ka, the stronger the acid is, and the weaker its conjugate base is. Similarly, the higher the K , the stronger the substance is as a base, and the more weakly acidic its conjugate acid is. For an acid that reacts with water in the reaction \[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}\] \[K_a = \dfrac{[H_3O^+,A^-]}{[HA]}\] where each bracketed term represents the concentration of that substance in solution. Relation of K , K , K \[K_w = K_a \times K_b \nonumber\] Partial List of Strong Acids: Hydrochlroic acid (HCl), Nitric Acid (HNO ), Perchloric Acid (HClO ), Sulfuric Acid (H SO ) Partial List of Strong Bases: Sodium Hydroxide (NaOH), Barium Hydroxide (Ba(OH) ), Calcium Hydroxide (Ca(OH) ), Lithium Hydroxide (LiOH) (Hydroxides of Group I and II elements are generally strong bases) Partial List of Weak Acids: Acetic Acid (CH COOH), Carbonic Acid (H CO ), Phosphoric Acid (H PO ) Partial List of Weak Bases: Ammonia (NH ), Calcium Carbonate (CaCO ), Sodium Acetate (NaCH COO) Find the pH of 0.5 grams of HCl disolved into 100 ml of water: First find moles of acid: grams / molar mass = moles 0.5 grams / (36.5 g/mole) = 0.014 moles HCl Then find molarity: moles / volume = molarity 0.014 moles / 0.100 L = 0.14 M HCl is a strong acid and completely dissociates in water, therefore the pH will be equal to the negative logarithm of the concentration of HCl pH = -log(H O ) pH = -log(0.14) = 0.85 The K value for acetic acid is 1.76*10 , and the K value for benzoic acid is 6.46*10 , if two solutions are made, one from each acid, with equal concentrations, which one will have the lower pH? The K value is a measure of the ratio between reactants and products at equilibrium. For an acid, the reaction will be HA + H O --> A + H O . PH is based on the concentration of the hydronium ion (H O ) which is a product of the reaction of acid and water. A higher K value means a higher ratio of reactants to products, and so the acid with the higher K value will be producing more hydronium, and therefore have a lower pH. Therefore the solution of benzoic acid will have a lower pH. The K value of ammonium (NH ) is 5.6*10 , the K value of ammonia (NH 1.8*10 , is ammonium more strongly acidic than ammonia is basic? The relative strength of an acid or base depends on how high its K or K value is, in this case, the K value is far lower than the K value so the ammonia is more strongly basic than ammonium is acidic.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/History_of_Thermodynamics

For a long time, physicists and chemists debated whether heat was a fluid (like a mysterious liquid) or came from the motion of particles. Many early scientists, like Newton, had thought that heat might be caused by small movement of particles, and greater heat meant greater velocities or kinetic energies. Lavoisier, however, thought that heat was a massless fluid that he called "caloric." Count Rumford observed that the process of boring cannon (drilling the hole in the middle of the brass cannon) produced a lot of heat, especially when the drill was dull or blunt. He showed that the heat produced was related to the amount of mechanical work done by the drill. Davy (mentioned earlier ) showed that even at 0°C, two ice cubes would melt when rubbed together. This frictional heating is also a way that people sometimes start fires in the wilderness. Other scientists liked the fluid theory. thought heat was a fluid that caused the atoms it surrounded to separate (which is why, he said, density usually decreases as you heat a substance). One important contribution he made was to show that the heat generated by human or animal metabolism (oxidizing food with oxygen from breathing) produces the same amount of energy as combusting the food (which is often if not always true). Carnot, who will be very important later (in the development of the second law of thermodynamics), also thought that heat was a liquid, because like liquids it "flows downhill" from hot objects to cold objects. He thought that like power generation from a waterfall, the amount of heat that moves and the distance it falls (change in temperature) determine the available power. However, later he realized that some of the heat is lost when it is converted to mechanical energy (work), which means it can't be a fluid like water (water isn't lost when falling water is used to drive a motor). Mayer used data other people collected on heat capacities of air at constant pressure or constant volume to calculate the relationship between the energy defined as force x distance (like the modern unit joule) and energy defined by change in temperature of a substance (like the modern unit calorie, the energy needed to raise the temperature of 1g of water 1°C). Imagine heating a sample of air in a fixed volume container or in a chamber with a piston, so that it is always at atmospheric pressure. One sample does work when heated (by expanding against atmospheric pressure) and the other does not. The difference in heat required to get the same temperature change in the 2 containers must be equivalent to the work done by the system with the piston. Mayer argued that heat, work, and chemical energy are all interconvertible, meaning they are all energy in different forms. Joule (for whom the unit joule is named) was an English beer-brewer who did the studies mentioned that lead to the first law. You might have learned Joule's law in a physics class, that heat produced by electricity, Q, is \[Q\; \alpha\; I^{2}R\] where I is current and R is resistance. He compared heat produced by electricity and heat produced by mechanical work (heating water using a paddlewheel powered by a falling weight) and thus showed the equivalence of mechanical work and heat. In other words, Joule and Carnot had showed that heat can be used to generate work (like in a power plant) and work can be used to generate heat (like with the cannon drills or paddlewheel). Kelvin combined these ideas and used them to propose the Kelvin temperature scale (but the details of that can wait until we study Carnot's discoveries in more detail).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.12%3A_Isotopes

The presence of neutrons in atomic nuclei accounts for the occurrence of samples of an element whose atoms contain different numbers of neutrons and hence exhibit different "nuclidic masses". The is the mass of a "nuclide", where a nuclide is the term used for any atom whose nuclear composition (Number of protons and neutrons) is defined. For example, naturally occurring hydrogen has two stable nuclides, \(\ce{^{1}_{1}H}\) a d \(\ce{^{2}_{1}H}\), which also are isotopes of one another. More than 99.98 percent is “light” hydrogen, . This consists of atoms each of which has one proton, one electron, and zero neutrons. The rest is “heavy” hydrogen or deuterium, \(\ce{^{2}_{1}H}\), which consists of atoms which contain one electron, one proton, and one neutron. Hence the nuclidic mass of deuterium is almost exactly twice as great as for light hydrogen. By of lithium, it is also possible to obtain a third isotope, tritium, \(\ce{^{3}_{1}H}\). It consists of atoms whose nuclei contain two neutrons and one proton. Its nuclidic mass is about 3 times that of light hydrogen. The discovery of isotopes and its explanation on the basis of an atomic structure built up from electrons, protons, and neutrons required a change in the ideas about atoms which John Dalton had . For a given element all atoms are identical in all respects―especially with regard to mass. It is the number and distribution of electrons which occupy most of the volume of an atom which determines the chemical behavior of atoms. The number of protons in the nucleus of each element is important in determining its chemical properties, because the total positive charge of the nucleus determines how the electrons are distributed. , but different isotopes have different nuclidic masses.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Lipids/Non-glyceride_Lipids/Wax

A wax is a simple lipid which is an ester of a long-chain alcohol and a fatty acid. The alcohol may contain from 12-32 carbon atoms. Waxes are found in nature as coatings on leaves and stems. The wax prevents the plant from losing excessive amounts of water. Carnuba wax is found on the leaves of Brazilian palm trees and is used in floor and automobile waxes. Lanolin coats lambs, wool. Beeswax is secreted by bees to make cells for honey and eggs. Spermaceti wax is found in the head cavities and blubber of the sperm whale. Many of the waxes mentioned are used in ointments, hand creams, and cosmetics (read the ingredients lists). Paraffin wax, used in some candles, is not based upon the ester functional group, but is a mixture of high molecular weight alkanes. Ear wax is a mixture of phospholipids and esters of cholesterol. The waxes with their component alcohols and fatty acids are listed below. Simple esters are made from an organic acid and an alcohol. The ester functional group is of primary importance in the biochemical group of compounds called waxes, triglycerides, and phospholipids. The simplified reaction reveals the process of breaking some bonds and forming the ester and the by product, water. Refer to the graphic on the left for the synthesis of carnuba wax. First, the -OH (red) bond on the acid is broken and the -H (red) bond on the alcohol is also broken. Both join to make HOH, a water molecule. Secondly, the oxygen of the alcohol forms a bond (green) to the acid at the carbon with the double bond oxygen. This forms the ester functional group. The long carbon chains do not participate in the reaction, but are just part of the final molecule. Lipstick consists of a suspension of coloring agents in high molecular weight hydrocarbons, waxes, and/or fats. The color usually comes from a dye precipitated by a metal ion such as Fe (III), Ni(II), or Co(II) ions. An ingredients list may be: dye (4-8%); castor oil, paraffin, or fats to dissolve dye (50%); lanolin (25%); carnauba and/or beeswax as a stiffening agent (36%); perfume (1.5%). The lipstick is made by first dispersing the dye in the castor oil. Then the other waxes and lanolin are added as the mixture is heated and stirred. The molten waxes are then cast in suitable forms to harden. Eyebrow pencils are very much like lipstick but contain lamp black (carbon soot) as a black coloring agent. A different mixture of waxes may be used to give the desired melting point. Brown pencils are made by adding iron oxide (rust) as a pigment. A water-resistant mascara has a mixture of waxes, fats, oils, and soap. Other coloring agents in addition to blacks and browns may be chromic oxide (dark green) and ultramarine (blue pigment of sodium and aluminum silicate).

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photoreceptors/Vision_and_Light

Vision is such an everyday occurrence that we seldom stop to think and wonder how we are able to see the objects that surround us. Yet the vision process is a fascinating example of how light can produce molecular changes. The retina contain the molecules that undergo a chemical change upon absorbing light, but it is the brain that actually makes sense of the visual information to create an image. Light is one of the most important resources for civilization, it provides energy as it pass along by the sun. Light influence our everyday live. Living organisms sense light from the environment by photoreceptors. Light, as waves carry energy, contains energy by different wavelength. In vision, light is the stimulus input. Light energy goes into eyes stimulate photoreceptor in eyes. However, as an energy wave, energy is passed on through light at different wavelength. Light energy can convert chemical to other forms. , also known as retinol, anti-dry eye vitamins, is a required nutrition for human health. The predecessor of vitamin A is present in the variety of plant carotene. Vitamin A is critical for vision because it is needed by the retina of eye. Retinol can be convert to retinal, and retinal is a chemical necessary for rhodopsin. As light enters the eye, the 11- -retinal is isomerized to the all-"trans" form. The molecule cis-retinal can absorb light at a specific wavelength. When visible light hits the cis-retinal, the cis-retinal undergoes an , or change in molecular arrangement, to all-trans-retinal. The new form of trans-retinal does not fit as well into the protein, and so a series of geometry changes in the protein begins. The resulting complex is referred to a bathrhodopsin (there are other intermediates in this process, but we'll ignore them for now). The reaction above shows Lysine side-chain from the opsin react with 11-cis-retinal when stimulated. By removing the oxygen atom form the retinal and two hydrogen atom form the free amino group of the lysine, the linkage show on the picture above is formed, and it is called Schiff base. As the protein changes its geometry, it initiates a cascade of biochemical reactions that results in changes in charge so that a large potential difference builds up across the plasma membrane. This potential difference is passed along to an adjoining nerve cell as an electrical impulse. The nerve cell carries this impulse to the brain, where the visual information is interpreted. The light image is mapped on the surface of the retina by activating a series of light-sensitive cells known as rods and cones or photoreceptors. The rods and cones convert the light into electrical impulses which are transmitted to the brain via nerve fibers. The brain then determines, which nerve fibers carried the electrical impulse activate by light at certain photoreceptors, and then creates an image. The retina is lined with many millions of photoreceptor cells that consist of two types: 7 million cones provide color information and sharpness of images, and 120 million rods are extremely sensitive detectors of white light to provide night vision. The tops of the rods and cones contain a region filled with membrane-bound discs, which contain the molecule cis-retinal bound to a protein called opsin. The resulting complex is called rhodopsin or "visual purple". In human eyes, rod and cones react to light stimulation, and a series of chemical reactions happen in cells. These cells receive light, and pass on signals to other receiver cells. This chain of process is class signal transduction pathway. Signal transduction pathway is a mechanism that describe the ways cells react and response to stimulation.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/22%3A_Carbonyl_Alpha-Substitution_Reactions

When you have completed Chapter 22, you should be able to Alpha‑substitution reactions are the third major type of reaction that you will study in your investigation of the chemistry of carbonyl compounds. As you will see, these reactions proceed through the formation of the enol form of the carbonyl compound. After a brief review of keto‑enol tautomerism, we begin our discussion of alpha‑substitution reactions by looking at the methods in which the enol form of the carbonyl compound is directly involved. After discussing the factors that influence the formation and stability of enolate anions, we will examine some halogenation reactions in which an enolate ion is formed as an intermediate. The chapter concludes with a study of the alkylation of enolate anions. These reactions are of tremendous use in organic syntheses, as they provide a method of forming new carbon‑carbon bonds, and hence facilitate the laboratory preparation of increasingly complex compounds.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.07_Effusion_and_Diffusion

We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. As you have learned, the molecules of a gas are stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses At a given temperature, heavier molecules move more slowly than lighter molecules. During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of \(\ce{^{235}U}\). Naturally occurring uranium is only 0.720% \(\ce{^{235}U}\), whereas most of the rest (99.275%) is \(\ce{^{238}U}\), which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound \(UF_6\) (boiling point = 56°C). isotopic content of naturally occurring uranium and atomic masses of U and U ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF   The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation 6.8.1: \[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043\] Thus passing UF containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% UF . In this case, 0.990 = (0.00720)(1.0043) , which can be rearranged to give \[1.0043^n=\dfrac{0.990}{0.00720}=137.50\] Thus at least a thousand effusion steps are necessary to obtain highly enriched U. Figure \(\Page {2}\) shows a small part of a system that is used to prepare enriched uranium on a large scale. Helium consists of two isotopes: He (natural abundance = 0.000134%) and He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant He by a process of gaseous effusion. a. ratio of effusion rates = 1.15200; one step gives 0.000154% He; b. 96 steps Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}\] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}\] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}\] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation 6.8.4 shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure \(\Page {3}\) for some common gases. Because all gases have the same average kinetic energy, according to the , molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure \(\Page {4}\). The average distance traveled by a molecule between collisions is the . The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 m (about 6 million miles). The the gas, the the mean free path. Calculate the rms speed of a sample of -2-butene (C H ) at 20°C. compound and temperature rms speed Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 6.8.5 to calculate the rms speed of the gas. To use Equation 6.8.4, we need to calculate the molar mass of -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C H , so its molar mass is 56.11 g/mol. Thus \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}=361\;m/s\] Calculate the rms speed of a sample of radon gas at 23°C. 1.82 × 10 m/s (about 410 mi/h) The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. In Section 10.8, we explore how the behavior of real gases differs from that of ideal gases.  Graham’s law of Diffusion and Effusion: is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The of a molecule is the average distance it travels between collisions.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Virtual_Textbook_of_OChem_(Reusch)_UNDER_CONSTRUCTION/32%3A_Exercises

The practice problems provided as part of this text are chiefly interactive, and should provide a useful assessment of the reader's understanding at various stages in the development of the subject. Some of these problems make use of a drawing application created by Peter.Ertl. To practice using this editor Here. Since problem solving is essential to achieving an effective mastery of the subject, it is recommended that many more problems be worked. Most organic chemistry textbooks contain a broad assortment of suitable problems, and paperback collections of practice problems are also available. In addition, a large collection of multiple choice problems may be viewed . The following button will activate a collection of problems concerning the reactivity of common functional groups. Towson University-reaction quizzes and summaries For a useful collection of study materials, including links to other sites, visit the . Prepared by Bob Hanson, St. Olaf College

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/10%3A_Strategies_in_Prostaglandins_Synthesis

Chemical Synthesis of Prostaglandins witnessed phenomenal activity during the 1960’s and 70’s. During this period, organic chemistry saw intensive development in ‘disconnection’ and ‘Logic’ as primary tools for synthesis. This period also saw development of several new reagents for stereoselective synthesis. The complexity of the structure of PG skeleton posed a great challenge for synthesis. The fact that molecules belonging to this family held great potential as drug candidates but were available only in minute quantities from natural sources was the main reason for the intense activity in the chemical synthesis and skeletal modifications for SAR studies. The numbering system on the skeleton and the main structural features of this family of molecules could be seen in Figure 10.1 Placing a keto- group at C9 is a delicate operation because such β- hydroxy ketones would readily undergo dehydration to give a PGA skeleton. On a five membered ring, PGA system could again undergo ready isomerisation to PGB, a stable ring system, probably via a PGC skeleton. On reduction with sodium borohydride, the keto- group at C9 is reduced to a mixture 9α- and the unnatural 9β- epimers. The natural PGF skeleton has an 9α- configuration for the –OH group. In the natural PGF skeleton we have four asymmetric centers on a five membered ring. As you know well, a five membered ring is conformationally very flexible. Hence setting up precise stereochemistry on this ring posed a great challenge during the 60’s. There could be three main strategies for the construction of five membered ring systems . Let us look at a few outstanding synthetic efforts that successfully met these challenges in Prostaglandin chemistry. Setting up a series of asymmetric centers on an open chain is as great a challenge as setting them up on a five membered ring. Corey’s early attempt in 1968 was to design a suitable six membered ring, open up the ring to a chain and then cyclize the chain regiospecifically to the required five membered ring ( J. Am. Chem. Soc., 90, 3245 (1968)). Their retroanalysis and synthesis are shown in Figure 10.3. The first DA reaction set two crucial stereocentres that guide the remaining stereopoints on the five membered ring formed by aldol reaction. This synthesis is a good example for . The starting materials come by two different routes shown in Figure 10.4. The Kojima’s disconnection of the cyclopentane ring at C8 – C12 bond provided an open chain. Careful planning of the functional groups on the proposed chain enabled his group to plan the stereocentres in a more direct way. Their retroanalysis and synthetic route are shown in Figure 10.5. The difficult problem of setting up stereocentres on a five membered ring was solved elegantly by Corey et.al.,(Tet. Lett., 311 (1970)) . Observe the artistic precision with which the substituents are woven into a five membered ring with the aid of iodolactonisation and epoxide ring formation reactions. An example for mastery in the Art of Synthesis. F.S. Alvarez et.al., took advantage of the steric constraints due to eclipsing strains in five membered rings and wove three asymmetric centers in a row on a five membered ring (J. Am. Chem. Soc., 94, 7823 (1972)) . Starting from all cis-cyclohexan1,3,5-triol, Woodward’s school demonstrated their excellence in the Art of Organic Synthesis. The first step is the differential protection with glyoxalic acid . Note an interesting architectural design aspect. This chain of two carbon protecting group is eventually incorporated into the main structure. Note that this also accomplishes another very difficult task viz., placing a reactive 2 carbon chain into the crowded concave phase. The solvolysis of the mesylate is assisted by the neighbouring olefin. Note the elegant planning of the ring contraction step. In 1973, Corey revealed a synthesis (Tet. Lett., 309 (1973)) based on six membered ring involving a ring expansion and a ring contraction to achieve the goal . There are several syntheses based on such strategy to derive the stereochemical advantage of a bicycloalkane ring system. Here we shall discuss the Bicycloheptane strategy of Corey (J. Am. Chem. Soc., 91, 5675 (1969): Ann. N.Y. Acad. Sci., 180, 24 (1971)). This route provides entry to all natural and unnatural PGs. It provides facility for separation of enantiomers at a very early stage using Amphetamine salt procedure on the first chiral acid-alcohol. This scheme has been scaled up to multigram-scales. Corey’s retroanalysis is shown in Figure 10.10. This retroanalysis led to DA reaction. The first problem was the synthesis of a cyclopentadiene with the alkyl group at the methylene carbon. Anion routes for alkylation of cyclopentadiene mostly led to a mixture of cyclopentadienes, due to easy isomerisation of the olefin. The problem was solved by alkylation with thallium anion . The desired cyclopentadiene was obtained in about 97% yield. The next challenge was in the DA reaction. Ketenes do not undergo Diels-Alder cycloaddition. They give cyclobutane products even with dienes. Using a ‘masked ketene precursor’ this problem was solved as shown in Figure 10.12. The bicyclo[2.2.1]heptane thus formed underwent Baeyer-Villiger oxidation as expected. Saponification of the lactone gave a five membered ring with three asymmetric centers in place. The fourth center was generated through an iodolactonisation reaction. Note the stereochemistry of this iodolactonisation step. After removal of the halogen, the –OH at C11 was protected as p-phenylbenzoyl ester. This ester not only gave a crystalline derivative for purification (a useful feature during large scale reactions) but also served as a shielding agent to induce C15-(S)– OH at a later stage. Hydrogenolysis of the C13 – OR to – OH followed by Collins oxidation gave the C13 aldehyde suitable for a Wittig reaction. The α,β-unsaturated ketone was reduced selectively to C15-(S)– OH using Zinc borohydride. This selectivity has been attributed the shielding effect of the bulky aroyl group from the front side, allowing the hydride to enter stereoselectively. Reduction of the ester group with trialkylaluminium hydride gave the hemiacetal for the final Wittig. Note the differential protections used in this synthesis at different stages.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Faraday's_Law

In every electrochemical process, whether spontaneous or not, a certain amount of electric charge is transferred during the oxidation and reduction. The half-reactions we have written for electrode processes include the electrons which carry that charge. It is possible to measure the rate at which the charge is transferred with a device called an ammeter. An ammeter measures the current flowing through a circuit. The units of current are amperes (A) (amps, for short). Unlike a voltmeter, ammeters allow electrons to pass and essentially "clock" them as they go by. The amount of electric charge which has passed through the circuit can then be calculated by a simple relationship: Charge = current x time OR Coulombs = amps x seconds This enables us to connect reaction stoichiometry to electrical measurements. The principles underlying these relationships were worked out in the first half of the 19th century by the English scientist, Michael Faraday. The diagram shows how voltage and current might be measured for a typical galvanic cell but the arrangment is the same for any electrochemical cell. Notice that the voltmeter is placed across the electron conduit (i.e., the wire) while the ammeter is part of that conduit. A good quality voltmeter can be used in this way even though it might appear to be "shorting out" the circuit. Since electrons cannot pass through the voltmeter, they simply continue along the wire. Both the voltmeter and ammeter are polarized. They have negative and positive terminals marked on them. Electrons are "expected" only in one direction. This is important in measurements of direct current (DC) such as comes out of (or goes into) electrochemical cells.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalysts

A is a substance that speeds up a reaction without being consumed by it. More specifically, a catalyst provides an alternative, lower activation energy pathway between reactants and products. As such, catalysts are vitally important to chemical technology; approximately 95% of industrial chemical processes involve catalysts of various kinds. In addition, most biochemical processes that occur in living organisms are mediated by , which are catalysts made of proteins. It is important to understand that a catalyst affects only the of a reaction; it does alter the thermodynamic tendency of the reaction to occur. Therefore, is the same )

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/RNA/Types_of_RNA

Three general types of RNA exist: messenger, ribosomal, and transfer. Messenger RNA (mRNA) is synthesized from a gene segment of DNA which ultimately contains the information on the primary sequence of amino acids in a protein to be synthesized. The genetic code as translated is for m-RNA not DNA. The into the cytoplasm where protein synthesis occurs. Each gene (or distinct segment) on DNA contains instructions for making one specific protein with order of amino acids coded by the precise sequence of heterocyclic amines on the nucleotides. Since proteins have a variety of functions including those of enzymes mistakes in the primary sequence of amino acids in proteins may have lethal effects. How can a polymeric nucleotide with only four different heterocyclic amines specify the sequence of 20 or more different amino acids? If each nucleotide coded for a single amino acid, then obviously only 4 of the 20 amino acids could be accommodated. If the nucleotides were used in groups of two, there are 16 different combinations possible which is still inadequate. It has been determined that the genetic code is actually based upon of nucleotides which provide 64 different codes using the 4 nucleotides. During the 1960's, a tremendous effort was devoted to proving that the code was read as triplets, and also to solving the genetic code. The genetic code was originally translated for the bacteria E. Coli, but its universality has since been established. The genetic code is "read" from a type of RNA called . , can be "read" and translated into an amino acid to be incorporated into a protein being synthesized. The genetic code is shown in Figure 7. Several distinctive features of the genetic code are clearly evident. First, all of the 64 codons or triplets have a known function, with 61 coding for amino acids and the other 3 serving as a stop or termination signal for protein synthesis. Secondly, the code is degenerate, meaning that there are usually several codons for each amino acid. Only methionine and tryptophan have a single codon. More specifics on the importance of the degeneracy of the genetic code will be discussed in a later section. In the cytoplasm, and protein combine to form a nucleoprotein called a ribosome. In the graphic on the left, the ribosome is shown as made from two sub units, 50S and 30 S. There are about equal parts rRNA and protein. The far left graphic shows the complete ribosome with three tRNA attached. The ribosome attaches itself to m-RNA and provides the stabilizing structure to hold all substances in position as the protein is synthesized. Several ribosomes may be attached to a single RNA at any time. In upper right corner is the 30S sub unit with mRNA and tRNA attached. Note: The coordinates used in this display have only the alpha carbons of the proteins (CA) and the DNA backbone atoms. Transfer RNA (tRNA) contains about 75 nucleotides, three of which are called anticodons, and one amino acid. There are at least 20 different tRNA's - one for each amino acid. The basic structure of a tRNA is shown in the left graphic. Part of the tRNA doubles back upon itself to form several double helical sections. On one end, the amino acid, phenylalanine, is attached. On the opposite end, a specific base triplet, called the , is used to actually "read" the codons on the mRNA. The tRNA "reads" the mRNA codon by using its own anticodon. The actual "reading" is done by matching the base pairs through hydrogen bonding following the base pairing principle. Each codon is "read" by various tRNA's until the appropriate match of the anticodon with the codon occurs. In this example, the tRNA anticodon (AAG) reads the codon (UUC) on the mRNA. The UUC codon codes for phenylalanine which is attached to the tRNA. Remember that the codons read from the mRNA make up the genetic code as read by humans.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/ATP_ADP-Gutow_Draft

Adenosine-5'-triphosphate (ATP) is comprised of an adenine ring, a sugar, and three phosphate groups. ATP is often used for energy transfer in the cell. The enzyme ATP synthase produces ATP from ADP or AMP + P using energy produced from in the mitochondria. ATP has many uses. It is used as a , in , for example. ATP is also found in in the processes of replication and transcription. In a neutral solution, ATP has negatively charged groups that allow it to chelate metals. Usually, Mg stabilizes it. ATP is a molecule which can hydrolyze to ADP and inorganic phosphate when it is in water. The formation of solvated ADP and hydrogen phosphate from solvated ATP and water has a of -30.5 kJ/mol. The negative ∆G means that the reaction is spontaneous (given an infinite amount of time it will proceed) and produces a net release of energy. However, because it energy to rupture the P-O bond connecting the phosphate that leaves ATP, ATP molecules do not instantly fall apart and can be used to transport useful energy around the cell. The energy required to rupture the bond contributes to the that prevents the reaction from happening instantly. At pH 7 the balanced reaction for hydrolysis is: \[ATP ^{4-} + H_2O \rightleftharpoons ADP^{3-} + HPO_4^{2-} + H^+\] ∆G = -30.5 kJ ATP is the primary energy transporter for most energy-requiring reactions that occur in the cell. The continual synthesis of ATP and the immediate usage of it results in ATP having a very fast turnover rate. This means that ADP is synthesized into ATP very quickly and vice versa. For example, it takes only a few seconds for half of the ATP molecules in a cell to be converted into ADP to be used in driving endergonic (non-spontaneous) reactions and then converted back into ATP using exergonic (spontaneous) reactions. ATP is useful in many cell processes such as , , , , active transport across cell membranes (as in the ), and synthesis of macromolecules such as .

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.04%3A_Conjugate_Pairs_and_Buffers

Make sure you thoroughly understand the following essential concepts: We often tend to regard the pH as a quantity that is dependent on other variables such as the concentration and strength of an acid, base or salt. But in much of chemistry (and especially in biochemistry), we find it more useful to treat pH as the "master" variable that controls the relative concentrations of the acid- and base-forms of one or more sets of conjugate acid-base systems. In this lesson, we will explore this approach in some detail, showing its application to the very practical topics of buffer solutions, as well as the use of a simple graphical approach that will enable you to estimate the pH of a weak monoprotic or polyprotic acid or base without doing any arithmetic at all! If we add 0.2 mol of sodium hydroxide to a solution containing 1.0 mol of a weak acid HA, then an equivalent number of moles of that acid will be converted into its base form A . The resulting solution will contain 0.2 mol of A and 0.8 mol of HA. Note that because we are discussing stoichiometry here, we are interested in quantities (moles) of reactants, not concentrations of reactants. The important point to understand here is that we will end up with a “partly neutralized” solution in which both the acid and its conjugate base are present in significant amounts. Solutions of this kind are far more common than those of a pure acid or a pure base, and it is very important that you have a thorough understanding of them. To a solution containing 0.010 mole of acetic acid (HAc), we add 0.002 mole of sodium hydroxide. If the volume of the final solution is 100 ml, find the values of , , and the total system concentration . The added hydroxide ion, being a strong base, reacts completely with the acetic acid, leaving 0.010 – 0.002 = 0.008 mole of HAc and 0.002 mole of acetate ion Ac . The final concentrations are \[C_a = \dfrac{0.008 \;mol}{0.10\; L} = 0.08 \; M \nonumber\] \[C_b = \dfrac{0.002 \; mol}{ 0.10 \; L} = 0.02 M \nonumber\] \[C_t = \dfrac{0.010 \;mol}{0.10\; L} = 0.10\; M. \nonumber\] Note that this solution would be indistinguishable from one prepared by combining = 0.080 mole of acetic acid with = 0.020 mole of sodium acetate and adjusting the volume to 100 ml. Thus starting with a solution of a pure weak acid or weak base in water, we can add sufficient strong base or strong acid, respectively, to adjust the ratio of the conjugate species — that is, the ratios [HA]/[A ] in the case of an acid, or [B]/[BH ] for a base, to any value we want. To express the relative concentrations of the protonated and deprotonated forms of an acid-base system present in a solution, we could use the simple ratio [HA]/[A] (or its inverse), but this suffers from the drawback of yielding an indeterminate result when the concentration in the denominator is zero. For many purposes it is more convenient to use the \[\alpha_0 = \dfrac{[HA]}{[HA]+A^-]} = \dfrac{[HA]}{C_a} \label{1-7a}\] \[\alpha_1 = \dfrac{[A^-]}{[HA]+A^-]} = \dfrac{[A^-]}{C_a} \label{1-7b}\] The fraction \(\alpha_1\) is also known as the of the acid. By making appropriate substitutions using the relation \[[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{1-8}\] we can express the ionization fractions as functions of the pH: \[\alpha_0 = \dfrac{[H^+]}{K_a+[H^+]} \label{1-9a}\] \[\alpha_1 = \dfrac{K_a}{K_a+[H^+]} \label{1-9b}\] Notice that the the values for both of these functions are close to zero or unity except within the pH range ± 1 (Figure \(\Page {1}\)). Plots of the \(\alpha\) functions vs. pH for several systems are shown below. Notice the crossing points where [HA] = [A] when [H ] = that corresponds to unit value of the quotient in the . The ionization fractions of a series of acids over a wide pH range can conveniently be summarized as shown below. If you know the of an acid, you can easily sketch out its ionization fraction plot. The extreme tops and bottoms can only be estimated, but the rest of the plots are essentially straight lines. As valuable as these plots are for showing how the distribution of conjugate species varies with the pH, they suffer from two drawbacks: Both of these limitations are readily overcome by the use of easily-constructed logarithmic plots which we describe in the following section. The word has several meanings in English, most of them referring (in its verb form) to cushion, shield, protect, or counteract an adverse effect. In chemistry, it refers specifically to a solution that resists a change in pH when acid or base is added. A (or ) solution is one that resists a change in its pH when H or OH ions are added or removed owing to some other reaction taking place in the same solution. Buffer solutions are essential components of all living organisms. The essential component of a buffer system is a conjugate acid-base pair whose concentration is fairly high in relation to the concentrations of added H or OH it is expected to buffer against. A simple buffer system might be a 0.2 solution of sodium acetate; the conjugate pair here is acetic acid HAc and its conjugate base, the acetate ion Ac . The idea is that this conjugate pair "pool" will be available to gobble up any small (≤ 10 ) addition of H+ or OH that may result from other processes going on in the solution. Sketch out a similar diagram showing what happens when you H or OH . When this happens, the ratio [HAc]/[Ac ], will remain substantially unchanged, as will the pH, as you will see below. You can also think of the process depicted above in terms of the Le Chatelier principle: addition of H to the solution suppresses the dissociation of HAc, partially counteracting the effect of the added acid, as illustrated by the equation at the bottom left of the above diagram. To develop this more quantitatively, we will consider the general case of a weak acid HA to which a quantity of strong base has been added; think, for example, of acetic acid which has been partially neutralized by sodium hydroxide, yielding the same conjugate pair described above, although not necessarily identical concentrations. The mass balance for such a system would be \[[HA] + [A^–] = C_a + C_b = C_T \label {2-1}\] in which denotes the total concentration of all species in the solution. Because we added a (completely dissociated) base NaOH to the acid, we also note that \[C_b = [Na^+] \label{2-2}\] Recalling the equilibrium expression for a weak acid \[K_a = \dfrac{[H^+,A^-]}{[HA]} \label{2-3}\] We can solve this for [H ]: \[[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{2-4}\] Re-writing this in terms of negative logarithms, this becomes \[-\log [H_3O^+] = -\log K_a - log [HA] + \log[A^-] \label{2-5}\] or, since \(pK_a = –\log K_a\), we invert the ratio to preserve the positive sign: \[pH = pK_a + \log \dfrac{[A^-]}{[HA]} \label{2-6}\] This equation is known as the . It tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system. It is interesting to note that the H-H equation was not developed by chemists! You may wonder why these two equations, whose derivation we now consider almost trivial, should have immortalized the names of these two scientists. The answer is that the theory of chemical equilibrium was still developing in the early 1900's, and had not yet made its way into chemistry textbooks. Even the concept of pH was unknown until Sørenson's work appeared in 1909. It was the mystery (and medical necessity) of understanding why shortness of breath made the blood more alkaline, and too-rapid breathing made it more acidic, that forced the work of H&H into modern Chemistry. Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its . According to the above equation, when pH = , the log term becomes zero, so that the ratio [AB ] / [HA] = 10 = 1, meaning that [HA] = [AB ]. In other words, when the pH of a solution is set to the value of the of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically on a proton-free energy diagram: This says, in effect, that at when the pH of a solution containing both the acid and it conjugate base is made identical to the acid , the forms HA and A possess identical free energies, and will therefore be present in equal concentrations. When the pH of a solution is set to the value of the of an acid-base pair, the concentrations of the acid- and base forms will be identical. Buffers are generally most effective when the buffer system is not too far from the target pH. Under these conditions, both [HA] and [A ] are large enough to compensate for the withdrawal or addition, respectively, of hydrogen ions. Equation \(\ref{2-4}\) above and its logarithmic equivalent in Equation \(\ref{2-6}\) are of limited use in calculations because the exact values [A ] and [HA] are known only for the special case when the pH of the solution is identical to the . Most buffer solutions will be adjusted to other pH's, and of course, once the buffered solution begins doing it work by counteracting the effect of additions of H or OH , both [A ] and [HA] will have changed. These two equations are so widely used in practical chemistry (and especially in biochemistry) that they are worth committing to memory. \[[H^+] \approx K_a \dfrac{C_a}{C_b} \label{2-7}\] \[pH \approx pK_a + \log \dfrac{C_b}{C_a} \label{2-8}\] Although these are approximations, they are usually justified because useful buffer systems are always significantly more concentrated than those responsible for adding or removing hydrogen ions. Also, at these higher ionic concentrations, the of the buffer system will seldom be precisely known anyway. Don't expect actual buffer pH's to match calculations to better than 5%. Compare the effects of adding 1.0 mL of 2.5 HCl to (From Table \(\Page {1}\), the of H PO is 7.21) Note that in both cases the pH is reduced (as it must be if we are adding acid!), but the change is far less in the buffered solution. How would you prepare 200 mL of a buffer solution whose pH is 9.0? The first step would be to select a conjugate acid-base pair whose is close to the desired pH. Looking at the of values, either boric acid or ammonium ion would be suitable. For this example, we will select the NH /NH system, = 5.5E–10, = 9.25. We will use Equation \(\ref{2-7}\) to determine the ratio / required to set [H ] to the desired value of 1.00 × 10 . To do this, write out the complete equilibrium constant expression for the acid NH : NH → NH + H , and then solve it to find the required concentration ratio: \[K_a = \dfrac{[NH_3,H^+]}{[NH_4^+]} = 5.5 \times 10^{-10} \nonumber\] \[ \dfrac{C_a}{C_b} = \dfrac{[NH_3]}{[NH_4^+]} = \dfrac{5.5 \times 10^{-10}}{[H^+]} = \dfrac{5.5 \times 10^{-10}}{1.00 \times 10^{-9}} = \dfrac{5.5 \times 10^{-10}}{10.00 \times 10^{-10}} = 0.55 \nonumber \] The easiest (if not particularly elegant) way to work this out is to initially assume that we will dissolve 1.00 mole of solid NH Cl in water, add sufficient OH to partially neutralize some of this acid according to the reaction equation NH + OH → NH + H O, and then add sufficient water to make the volume 1.00 L. mass of ammonium chloride: (1 mol) × (53.5 g ol) = 53.8 g let = moles of NH that must be converted to NH . Then / = (1- )/ = .55; solving this gives x = .64, 1–x = .36. So to make 1 L of the buffer, dissolve .36 × 53.8 g = 19.3 g of solid NH Cl in a small quantity of water. Add .64 mole of NaOH (most easily done from a stock solution), and then sufficient water to make 1.00 L. To make 200 mL of the buffer, just multiply each of the above figures by 0.200. The Henderson-Hasselbalch Approximation is widely used in practical calculations. What most books do not tell you is that Eqution \(\ref{2-4}\) is no more than an “approximation of an approximation” which can give incorrect and misleading results when applied to situations in which the simplifying assumptions are not valid. An exact treatment of conjugate acid-base pairs, including a correct derivation of the Henderson-Hasselbalch equation, is given in the chapter on . The Henderson-Hasselbalch Approximation is only valid for fairly high concentrations The approximations that lead to the H-H equation limit its reliable use to values of and that are within an order of magnitude of each other, and are fairly high. Also, the of the acid should be moderate. The shaded portion of this set of plots indicates the values of and that yield useful results. Clearly, the smaller the buffer concentration, the narrower the range of useable acid s. Most buffer solutions tend to be fairly concentrated, with and typically around 0.01 - 0.1 . Thus a buffer based on a .01 solution of an acid such as chloric (HClO ) with of 1.9 will fall just outside the "safe" boundary near the upper left part of the diagram. \[\underbrace{[H^+] = K_a \dfrac{[HA]}{[A^-]}}_{\text{exact}} \label{2-9a}\] Equation \(\ref{2-9a}\) is simply a re-writing of the equilibrium constant expression, and is therefore always true. Of course, without knowing the actual equilibrium values of [HA] and [A ], this relation is of little direct use in pH calculations. \[\underbrace{[H^+] \approx K_a \dfrac{C_a}{C_b}}_{\text{approximate}} \label{2-9b}\] Equation \(\ref{2-9b}\) is never true, but will yield good results if the acid is sufficiently weak in relation to its concentration to keep the [H ] from being too high. Otherwise, the high [H ] will convert a significant fraction of the A into the acid form HA, so that the ratio [HA]/[A ] will differ from  / in the above two equations. Consumption of H by the base will also raise the pH above the predicted value as we saw in the preceding problem example. The terms and are commonly employed as synonyms for buffer index, but in some contexts, buffer capacity denotes the quantity of strong acid or strong base which alters the buffer's pH by 1 unit. (see below). How effective is a given buffer system in resisting changes in the pH? The most direct expression of this is the rate of change of the pH as small quantities of strong acid or base are added to the system: Δ /Δ(pH). Expressed in calculus notation, this is the , defined as \[\beta = \left| \dfrac{dC}{d(pH)} \right| \label{2-19}\] here refers to the concentration of strong acid or base added to the solution. Because added acid or base affects the pH in opposite ways, we take the absolute value of this function in order to ensure that β is always positive. The value of β can be calculated analytically from , , , and [H ]. By taking the second derivative of β, it can be shown that the buffer index has a maximum value when the pH = . This buffer index plot for a 0.10 M solution of sodium acetate is typical, and confirms that buffering is most efficient within about ±1 pH unit of . But what about the even greater buffering that apparently occurs at the two extremes of pH (orange shading)? This is due to the buffering associated with the water itself, and will be seen in all aqueous buffer solutions. is buffered by the H O /H O conjugate pair at very low pH, and by H O/OH- at high pH. This is easily understood if you think about adding some strong acid acid to pure water; even one drop of HCl will send the pH shooting down toward 0. If you continue adding acid, the pH will not drop significantly below 0, because there won't be enough free water molecules remaining to hydrate the HCl to produce H O ions — thus the solution is strongly buffered. Note that if we were to subtract the effect of the NaAc buffering from the above plot, the remaining plot for water itself would exhibit a minimum at pH 7, where both [H O ] and [OH ] share a common minimum value. This alternative view shows how the distribution fractions of HAc and Ac relate to the effective buffering range of this conjugate pair, which is conventionally defined as ±1 pH unit of the . The term is an alternative means of expressing the ability of a buffer system to absorb the addition of strong acid or base without causing the pH to deviate by more than one unit from that of the pure buffer. In this example, the buffer capacities for addition of acid and base will differ because the buffer pH has been adjusted to a value that differs from its . Buffer systems must be appreciably more concentrated than the concentrations of strong acid or base they are required to absorb while still remaining within the desired pH range. Once the added acid or base has consumed most of one or other of the conjugate species comprising the buffer, the pH will no longer be stabilized. And of course, very small buffer concentrations will approach the pH of pure water. These plots are often referred to as . (1916-1970) was a Swedish chemist who studied the distribution of ionic species in aqueous solutions and especially in the oceans. Because the concentrations of conjugate species can vary over many orders of magnitude, it is far more useful to express them on a logarithmic scale. Since pH is already logarithmic, one can obtain a "bird's-eye view" of an acid-base system in a compact log-log plot. Even better, these plots are easily constructed without any calculations or arithmetic — or even any graph paper — any scrap paper, even the back of an envelope, will be sufficient. A ruler or other straightedge will, however, give more accurate results. In addition, you can use these plots to estimate the pH of a solution of a monoprotic or polyprotic acid or base without contending with quadratic or higher-order equations — or doing any arithmetic at all! The results will not be as precise as you would get from a proper numerical solution, but given the uncertainties of how equilibrium constants are affected by the presence of other ions in the solution, this is rarely a problem. Aside from these advantages, the use of log-C vs pH plots will afford you an insight into the chemistry of acid-base systems that cannot be obtained simply by doing numerical calculations. The basic form of the plot, and the starting point for any use of such plots, looks like this: If you examine this plot carefully, you will see that it is nothing more than a definition of pH and pOH, as well as a definition of a neutral solution at pH 7. Notice, for example, that when the pH is 4.0, [H ] = 10 and [OH ] = 10 . Here is a log C plot for a 10 solution of acetic acid ("HAc") in water. Although it may look a bit complicated at first, it is really very simple. The heavy maroon line on the left plots the concentration of the acid HAc as a function of pH. The blue line on the right shows how the concentration of the base Ac depends on pH. The horizontal parts of these lines are aligned with "3" on the –log- axis, corresponding to the 10 nominal concentration ( ). How do we know the shapes and placement of the plots for the concentrations of acetic acid [HAc] and the acetate ion [Ac ] ? Although both of these concentrations will of course vary with the pH, their deviations from 10 are too small to reveal themselves on a logarithmic plot until the pH approaches the . Of special interest in acid-base chemistry are the pH values of a solution of an acid and of its conjugate base in pure water; as you know, these correspond to the beginning and equivalence points in the titration of an acid with a strong base. Continuing with the example of the acetic acid system, we show below another plot that is just like the one above, but with a few more lines and numbers added. Suppose you would like to find the pH of a 0.001 M solution of acetic acid in pure water. From the equation \[HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{Acetate 2-1} \] we know that equal numbers of moles of hydronium and acetate ions will be formed, so the concentrations of these species should be about the same: \[[H^+] \approx [Ac^–] \label{Acetate 2-2}\] will hold. The equivalence of these two concentrations corresponds to the point labeled on the log -pH plot; this occurs at a pH of about 4, and this is the pH of a 0.001 M solution of acetic acid in pure water. A 0.001 M solution of NaAc in water corresponds to the composition of a solution of acetic acid that has been titrated to its equivalence point with sodium hydroxide. The acetate ion, being the conjugate base of a weak acid, will undergo hydrolysis according to \[Ac^– + H_2O \rightleftharpoons HAc + OH^– \label{Acetate 2-3}\] which establishes the approximate relation \[[HAc] \approx [OH^–] \label{Acetate 2-4}\] This condition occurs where the right sloping part of the line representing [HAc] intersects the [OH ] line at in the plot above. As you would expect for a solution of the conjugate base of a weak acid, this corresponds to an alkaline solution, in this case, at about pH = 7.8. It is interesting to show the Sillén plot for a weak acid system with its titration curve, both on the same pH scale. In this example, the titration plot has been turned on its side and reversed in order to illustrate the correspondence of its ƒ=0 (initial), ƒ=0.5 and ƒ=1 (equivalence) points with the corresponding points on the upper plot. Ammonia, unlike most gases, is extremely soluble in water. Solutions of NH in water are properly known as , NH ; the name "ammonium hydroxide" is still often used, even though there is no evidence to suggest that a species NH OH exists. Ammonia is the conjugate base of the NH . The fact that ammonia is a base has no special significance insofar as the construction and interpretation of the log-C vs pH diagram is concerned; such diagrams always refer to conjugate acid-base , rather than to individual acids and bases. Thus the system point defines the NH /NH pair ( 9.3) at 25°C and a total concentration \[C_T = C_a + C_b = 0.0010\; M. \label{NH3 2-5}\] To estimate the pH of a solution of ammonia in pure water, we make use of the charge balance requirement (known as the ) \[[NH_4^+] + [H^+] = [OH^-] \label{NH3 2-5.5}\] which we simplify by assuming that, because NH is a weak acid, [NH ] >> [H ] and we can drop the [H ] term; thus the approximation \[[NH_4^+] \approx [OH^-] \label{NH3 2-6}\] which corresponds to the crossing point and a pH of 10.0. Similarly, at the crossing point , [NH ] . This comes from the proton condition \[[NH_3] + [H^+] – [OH^–] = 0 \label{NH3 2-7}\] by dropping the [OH ] term on the assumption that its value is negligible compared to [H ]. Equation \(\ref{NH3 2-7}\) is referred to as the "proton condition". Sorry to sneak it in without warning, but we hope that having piqued your curiosity, you might take the time to look at the following discussion of this important tool for estimating the pH of a solution of a pure acid, base, or a salt. In any aqueous solution of an acid or base, certain conservation conditions are strictly observed. Together with concentration and the or  , these put certain constraints on the system that determine the state of the system. The most fundamental of these are conservation of mass and of electric charge, which of course apply to chemical changes of all kinds. We commonly express these as and , respectively. (The latter of these is sometimes referred to as the .) Using the above example of the ammonium system as an illustration of this, we are interested in two particular instances of practical importance: what conditions apply to solutions made by dissolving pure ammonia, or ammonium chloride, in water? Sillén diagram OK so far? That's really all we need, but it's usually more convenient to combine these with mass balances on the protons alone, so that we have a single equation for each of the two solutions. The resulting equations are known as the for the two solutions. To avoid a lot of algebra, there is a simple short-cut for writing a proton condition equation: Let's try it for our ammonia system. The proton reference level (PRL) is defined by NH and H O. Proton condition: [H O ] + [NH ] = [OH ] (Notice that the substances that the PRL (in case, H O and NH ) never appear in the proton condition equation.) (or of any other strong-anion ammonium salt) in water: The PRL is defined by NH and H O. Proton condition: [H O ] = [NH ] + [OH ] (The chloride ion has nothing to do with protons, so it does not appear here.) Good question: the answer is that what seems complicated at first sometimes turns out to be simpler and easier to understand than any alternative. Think of it this way: acid-base reactions involve the transfer of protons: some protons jump up to higher proton-free energy levels (e.g., NH + H O → NH + OH ), others drop down to proton-vacant levels (H O → NH → H O + NH ). But no matter what happens, the total number of "available" protons (that is, all "dissociable" protons, including those in H O ) must be conserved — thus the "mass balance on protons". So let's look again at the log C - pH plot for the ammonium system, which we reproduce here for your convenience: Consider first the solution of ammonium chloride, whose proton condition is given by [H O ] = [NH ] + [OH ]. We know that NH Cl, being the salt of a weak base and a strong acid, will give a solution that is slightly acidic. In such a solution, [OH ] will be quite small so that we can neglect it without too much error. We can then write the proton condition as [H ] [NH ]. On the plot above, this corresponds to point where the lines representing these quantities cross. At this pH of around 6.2, [OH ] is about two orders of magnitude smaller than [NH ], so we are justified in dropping it. Similarly, for a solution of ammonia in water, the proton condition [H O ] + [NH ] = [OH ] that we worked out above can be simplified to [NH ] [OH ] (point ) with hardly any error at all, since [H ] is here about six orders of magnitude smaller than [NH ]. In our discussion of the plot for the acetic acid system, we arrived at the proton conditions [H ] [Ac ] (for HAc in water) and [H ] [OH ] (for NaAc) by simply using the stoichiometries of the reactions. This can work in the simplest systems, but not in the more complicated ones involving polyprotic systems. In general, it is far safer to write out the complete proton condition in order to judge what concentrations, if any, can be dropped. Having looked at the Sillén diagrams for acetic acid and ammonia, let's examine the log C-pH plot for this salt of acetic acid and ammonia. You will recall that we dealt with solutions of the salts sodium acetate and ammonium chloride in the two examples described above. What is different here is that components of the salt CH COONH ("NH Ac") are "weak" in the sense that their conjugate species NH and HAc are also present in significant quantities. This means that we are really dealing with acid-base systems — the ones shown previously — on the same plot. Each system retains its own system point, for the acetate system and for the ammonium system. Note that the two plots previously shown were for 10 solutions, so the non-system points for this more concentrated solution occur at different pH values. The pH's of 10 solutions of HAc , NaAc , and NH are located by using the same proton conditions as before. The only new thing here is the point , which corresponds to the ammonium acetate solution in which we are interested. The proton condition for this solution is [H ] + [HAc] = [NH ] + [OH ]. Inspection of the plot reveals that [H ] << [HAc] and [OH ] << [NH ], so the proton condition can be simplified to [HAc] [NH ], corresponding to the intersection of these two lines at . Now that you are able to find your way around log C-pH plots that encompass two acid-base systems, the polyprotic systems that we describe below should be no trouble at all. A thorough understanding of the carbonic acid-carbonate system is essential for understanding the chemistry of natural waters and the biochemistry of respiration in humans and other animals. There are two special things about this system that you need to know: This Sillén plot for the H CO * closed system is typical of other diprotic acid systems in that there are three conjugate species and two system points. The 10 concentration plotted here is typical of what is found in groundwaters that are in contact with carbonate-containing sediments. The pH of a 10 solution of CO , NaHCO and Na CO are indicted by the blue annotations. You should take a few moments to verify the approximations in the rightmost column. Notice the factors of 2 that multiply some of the concentrations. For example, the "2[CO ]" term in the first line of the table means that the CO species is two steps away from the proton reference level H CO *, so it would take two hydrogen ions to balance a single carbonate ion. Another new feature introduced here is the doubling of the slopes of the lines representing [H CO *] and [CO ] where they cross the pH values corresponding to the s that are one step removed from their own system points and near –log C = 7. This triprotic system is widely used to prepare buffer solutions in biochemical applications. In its fundamental form, there is nothing new here, other than an additional system point corresponding to the of H PO . The pH of a 10 solutions of H PO , NaH PO , Na HPO and Na PO can be estimated by setting up the proton conditions as follows. Unfortunately, these estimates become a bit more complicated owing to the proximity of the three system points. But it's far easier than doing a numerical solution which requires solving a firth-degree polynomial! [H ] + [H PO ] ≈ [HPO ] disodium hydrogen phosphate (Na HPO ) Because the phosphate plot is rather crowded, we show here a modified one in which the details of the pH estimates for the solutions of H PO and its salts are emphasized. Estimation of the pH of 10 solutions of phosphoric acid and trisodium phosphate , based on the approximations in the above table are straightforward. Those for NaH PO and Na HPO , however, are complicated by the presence of two concentration terms on the right side of the respective proton condition approximations. Unless you are in an advanced course, you may wish to skip the following details. The important thing to know is that graphical estimates of these more problematic systems are fundamentally possible. Thus for sodium dihydrogen phosphate, if we were to follow the pattern of the other systems, the approximation at the crossing point would be [H ] = [HPO ], corresponding to point . But inspection of the plot shows that at that pH (4.0), the concentration of H PO is ten times greater than that of H (point ). A rough estimate of the pH can be made by constructing a line (shown in red) that is parallel to those for H PO and H , but is raised up by a factor of 10. This results in the new crossing point . Similarly, the pH of a disodium hydrogen phosphate solution does not correspond to the crossing point because of the significant value of [HPO ] at this pH. Again, we construct a line above this crossing point that predicts a pH corresponding to point . Although these stratagems may seem rather crude, it should be noted that the uncertainties associated with them tend to be minimized on a logarithmic plot. In addition, it should be borne in mind that in a solution as concentrated as 0.1M, the values found in tables are not applicable. The alternative of solving for the pH algebraically, taking into consideration activity coefficients for the high ionic concentrations, is rarely worth the effort. The relation between the pH of a solution and the concentrations of weak acids and their conjugate species is algebraically rather complicated. But over most of the pH range, simplifying assumptions can be made that, when expressed in logarithmic form, plot as straight lines having integral slopes of 0, ±1, ±2, etc. It is only within the narrow pH range near the that these simplifying assumptions break down, but on a logarithmic plot, one can draw a smooth curve that covers this range without introducing significant error. Here is your basic stationery — a plot showing the two ions always present in any aqueous solution. In practice, you can usually cut off the bottom part where concentrations smaller than \(10^{–10}\, M\). For any given acid-base system, you need to know the and nominal concentration CT. Find the point at which these two lines intersect. Locate and mark the on the line. It will be 0.3 of a log-C unit below the concentration line. This is where [HA] = [A ] = ½  . (Recall that log.5 = .3) For pH < , [HA] plots as a line through the system point with slope = 0. For pH > , the line assumes a slope of –1. Use the [H ] and [OH ] lines as guides. Suppose you have a solution of NaA whose pH is 8. Estimate the concentration of the acid HA in the solution. All you need do is look at the plot. The [HA] line crosses the line marking this pH at – log C = 6, or [HA] = 10 . This quantity too small to significantly reduce [A ], which remains at approximately CT = 10   .

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When considering compounds having two or more double bonds in a molecule, it is useful to identify three distinct ways in which these functions may be oriented with respect to each other. First, the double bonds may be separated by one or more sp -hybridized carbon atoms, as in 1,5-hexadiene. In this circumstance each double bond behaves independently of the other, and we refer to them as . A second relationship has the double bonds connected to each other by a single bond, as in 1,3-hexadiene, and we refer to this arrangement as . Finally, two double bonds might share a carbon atom, as in 1,2-hexadiene. The central carbon atom in such a system is sp-hybridized, and we call such double bonds . These three isomers are shown in the following diagram. Another stereoisomeric factor associated with conjugated dienes exists. Rotation about the single bond joining the two double bonds (colored blue) converts a trans-like conformation to its form. The energy barrier to this conformational isomerization is normally low, and the s-trans conformer is often more stable than the s-cis conformer, as shown in the diagram. These categories are based on more than obvious structural variations. We find significant differences in the chemical properties of dienes depending on their structural type. For example, catalytic hydrogenation converts all the dienes shown here to the alkane hexane, but the heats of reaction (heat of hydrogenation) reflect characteristic differences in their thermodynamic stability. This is illustrated in the diagram below. Taking the heat of hydrogenation of 1-hexene (30.1 kcal/mole) as a reference, we find that the isolated diene, 1,5-hexadiene, as expected, generates double this heat of reaction on conversion to hexane. The cumulated diene, 1,2-hexadiene, has a 6 kcal/mole higher heat of reaction, indicating it is less stable than the isolated diene by this magnitude. On the other hand, conjugation of double bonds seems to stabilize a diene by about 5 kcal/mole. The increase in stability of 2,4-hexadiene over 1,3-hexadiene (both are conjugated) is due to the increased double bond substitution of the former, a factor noted earlier for simple alkenes. The stabilization of dienes by conjugation is less dramatic than the aromatic stabilization of benzene. Nevertheless, similar resonance and molecular orbital descriptions of conjugation may be written. A resonance description, such as the one shown here, involves charge separation, implying a relatively small degree of stabilization. CH =CH-CH=CH H -CH=CH- H A molecular orbital model for 1,3-butadiene is shown below. Note that the lobes of the four p-orbital components in each pi-orbital are colored differently and carry a plus or minus sign. This distinction refers to different phases, defined by the mathematical wave equations for such orbitals. Regions in which adjacent orbital lobes undergo a phase change are called . Orbital electron density is zero in such regions. Thus a single p-orbital has a node at the nucleus, and all the pi-orbitals shown here have a nodal plane that is defined by the atoms of the diene. This is the only nodal surface in the lowest energy pi-orbital, π . Higher energy pi-orbitals have an increasing number of nodes. ),

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The mechanism of a reaction is a sequence of elementary steps proceeding from the starting materials through a series of intermediates and eventually to the products. Each step has an activation energy barrier. Each intermediate has some measure of stability. The energy changes along a reaction pathway can be tracked with a reaction progress diagram. The speed of a reaction can be altered by flooding the system with reactant (via the ) or by adding more (thermal) energy to help overcome the barrier (via the ), but the reaction follows this same energy pathway. However, this is not the case with a catalyst. A catalyst introduces a completely different pathway/mechanism that was inaccessible before. Some reactions simply cannot proceed without a catalyst; the energy barrier is too high. This may be true even if the overall reaction is (\(\Delta{G}\)) and thus thermodynamically favored to proceed. Consider the reaction progress diagram above. If reactants approach this barrier from the left, they encounter a very large energy barrier. They cannot react, despite the fact that it would be energetically favorable for them to convert to products. The molecules are stuck. The same is true if molecules approach from the right. They cannot overcome the high barrier, though the energy difference between the left and the right is small. The molecules are again stuck. If the barrier were removed, the molecules could react freely in both directions, eventually coming to a , as in the diagram below: A catalyst does not fully remove the barrier, but it offers a new pathway with a lower barrier. The reaction is able to proceed back and forth. The reaction progress diagram plots the changes in energy along one particular coordinate of interest to the molecule. It may track a chemical bond as it lengthens and breaks through the course of a reaction. However, there are always other energetic possibilities unseen in this diagram. This is a single cross-section of a potential energy surface. A potential energy surface is like a landscape, a mountain range in which elevation corresponds to energy. The reaction progress diagram depicts a single pathway from one energetic valley to another. In one of these valleys is the reactant; the product is in the other. The path from one valley to another leads uphill, over a mountain pass, and down into the other valley again. Suppose a traveler living in a valley must visit a friend in another valley. Each day the traveler takes the same path to the friend's house. He takes the easiest route, over the lowest mountain pass. One day, instead of visiting the friend on foot, the traveler takes the train. The train takes a completely different pathway than the usual. It does not traverse the same mountain pass; it may take another route that was inaccessible by foot, or it may simply tunnel through the mountain. When the train reaches its destination, it picks up more passengers, repeating the process over and over. In molecular terms, it is the reactant that takes the train, a low-energy pathway, on its way to the product. The train is a catalyst, and it has several important features: That recycling of the catalyst is sometimes referred to as "turnover". The turnover number of a catalyst is the number of times the catalyst is able to return and carry out the reaction again (eventually something may occur to make the catalyst stop working). The reactant in a catalytic reaction is called the substrate, particularly in cases in which the catalyst is an enzyme of a transition metal catalyst. The speed at which the catalyst is able to carry out the reaction on new substrates is called the turnover frequency. These are important parameters in describing the efficiency of a catalyst. Polymerization catalysts take small molecules called monomers and connect them together using the same reaction over and over again to make a long polymer chain. What was the minimum turnover number of the catalyst used to make each of the following polymers? (Mn = number average molecular weight, a statistical estimate of the average size of polymer chain.) Catalysis often begins with a binding step, followed by one or more subsequent steps needed to carry out the reaction. For the following catalyzed reactions, draw the binding step using curved arrows. ,

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Because of its nature (i.e., acts as both an acid or a base), water does not always remain as \(H_2O\) molecules. In fact, two water molecules react to form and hydroxide ions: \[ \ce{ 2 H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^{−} (aq)} \label{1}\] This is also called the self-ionization of water. The concentration of \(H_3O^+\) and \(OH^-\) are equal in pure water because of the 1:1 stoichiometric ratio of Equation \(\ref{1}\). The molarity of H O and OH in water are also both \(1.0 \times 10^{-7} \,M\) at 25° C. Therefore, a constant of water (\(K_w\)) is created to show the equilibrium condition for the self-ionization of water. The product of the molarity of hydronium and hydroxide ion is always \(1.0 \times 10^{-14}\) (at room temperature). \[K_w= [H_3O^+,OH^-] = 1.0 \times 10^{-14} \label{2}\] Equation \(\ref{2}\) also applies to all aqueous solutions. However, \(K_w\) does change at different temperatures, which affects the pH range discussed below. \(H^+\) and \(H_3O^+\) is often used interchangeably to represent the hydrated proton, commonly call the . Equation \ref{1} can also be written as \[ H_2O \rightleftharpoons H^+ + OH^- \label{3}\] As expected for any equilibrium, the reaction can be shifted to the reactants or products: Because the constant of water, K is \(1.0 \times 10^{-14}\) (at 25° C), the \(pK_w\) is 14, the constant of water determines the range of the pH scale. To understand what the pK is, it is important to understand first what the "p" means in pOH and pH. The addition of the "p" reflects the negative of the logarithm, \(-\log\). Therefore, the pH is the negative logarithm of the molarity of H, the pOH is the negative logarithm of the molarity of \(\ce{OH^-}\), and the \(pK_w\) is the negative logarithm of the constant of water: \[ \begin{align} pH &= -\log [H^+] \label{4a} \\[4pt] pOH &= -\log [OH^-] \label{4b} \\[4pt] pK_w &= -\log [K_w] \label{4c} \end{align}\] At room temperature, \[K_w =1.0 \times 10^{-14} \label{4d}\] So \[\begin{align} pK_w &=-\log [1.0 \times 10^{-14}] \label{4e} \\[4pt] &=14 \end{align}\] Using the properties of logarithms, Equation \(\ref{4e}\) can be rewritten as \[10^{-pK_w}=10^{-14}. \label{4f}\] The equation also shows that each increasing unit on the scale decreases by the factor of ten on the concentration of \(\ce{H^{+}}\). Combining Equations \ref{4a} - \ref{4c} and \ref{4e} results in this important relationship: \[pK_w= pH + pOH = 14 \label{5b}\] Equation \ref{5b} is correct only at room temperature since changing the temperature will change \(K_w\). The pH scale is logarithmic, meaning that an increase or decrease of an integer value changes the concentration by a tenfold. For example, a pH of 3 is ten times more acidic than a pH of 4. Likewise, a pH of 3 is one hundred times more acidic than a pH of 5. Similarly a pH of 11 is ten times more basic than a pH of 10. From the simple definition of pH in Equation \ref{4a}, the following properties can be identified: It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper nor lower bound, since as defined above, the pH is an indication of concentration of H . For example, at a pH of zero the hydronium ion concentration is one molar, while at pH 14 the hydroxide ion concentration is one molar. Typically the concentrations of H in water in most solutions fall between a range of 1 M (pH=0) and 10 M (pH=14). Hence a range of 0 to 14 provides sensible (but not absolute) "bookends" for the scale. One can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar. Figure \(\Page {1}\) depicts the pH scale with common solutions and where they are on the scale. If the concentration of \(NaOH\) in a solution is \(2.5 \times 10^{-4}\; M\), what is the concentration of \(H_3O^+\)? We can assume room temperature, so \[1.0 \times 10^{-14} = [H_3O^+,OH^-] \nonumber\] to find the concentration of H O , solve for the [H O ]. \[\dfrac{1.0 \times 10^{-14}}{[OH^-]} = [H_3O^+]\] \[\dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} = [H_3O^+] = 4.0 \times 10^{-11}\; M\] \[[H^+]= 2.0 \times 10^{-3}\; M \nonumber\] \[pH = -\log [2.0 \times 10^{-3}] = 2.70 \nonumber\] \[ [OH^-]= 5.0 \times 10^{-5}\; M \nonumber\] \[pOH = -\log [5.0 \times 10^{-5}] = 4.30 \nonumber\] \[pK_w = pH + pOH \nonumber\] and \[pH = pK_w - pOH \nonumber\] then \[pH = 14 - 4.30 = 9.70 \nonumber\] If moist soil has a pH of 7.84, what is the H concentration of the soil solution? \[pH = -\log [H^+] \nonumber\] \[7.84 = -\log [H^+] \nonumber\] \[[H^+] = 1.45 \times 10^{-8} M \nonumber\] Place -7.84 in your calculator and take the antilog (often inverse log or 10 ) 1.45 x 10 M The pH scale was originally introduced by the Danish biochemist S.P.L. Sørenson in 1909 using the symbol p . The letter is derived from the German word meaning power or exponent of, in this case, 10. In 1909, S.P.L. Sørenson published a paper in Biochem Z in which he discussed the effect of H ions on the activity of enzymes. In the paper, he invented the term pH ( to describe this effect and defined it as the \(-\log[H^+]\). In 1924, Sørenson realized that the pH of a solution is a function of the "activity" of the H ion and not the concentration. Thus, he published a second paper on the subject. A better definition would be \[pH = -\log\,a\{\ce{H^{+}}\}\] where \(a\{H^+\}\) denotes the (an effective concentration) of the H ions. The activity of an ion is a function of many variables of which concentration is one. Because of the difficulty in accurately measuring the activity of the \(\ce{H^{+}}\) ion for most solutions the International Union of Pure and Applied Chemistry (IUPAC) and the National Bureau of Standards (NBS) has defined pH as the reading on a pH meter that has been standardized against standard buffers. The following equation is used to calculate the pH of all solutions: \[\begin{align} pH &= \dfrac{F(E-E_{standard})}{RT\;\ln 10} + pH_{standard} \label{6a} \\[4pt] &= \dfrac{5039.879 (E-E_{standard})}{T} + pH_{standard} \label{6b} \end{align}\] with The activity of the H ion is determined as accurately as possible for the standard solutions used. The identity of these solutions vary from one authority to another, but all give the same values of pH to ± 0.005 pH unit. The historical definition of pH is correct for those solutions that are so dilute and so pure the H ions are not influenced by anything but the solvent molecules (usually water). When measuring pH, [H ] is in units of moles of H per liter of solution. This is a reasonably accurate definition at low concentrations (the dilute limit) of H . At very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H Cl , thus reducing the concentration of “available” ions to a smaller value which we will call the . It is the of H and OH that determines the pH and pOH. The pH scale as shown above is called sometimes "concentration pH scale" as opposed to the "thermodynamic pH scale". The main difference between both scales is that in thermodynamic pH scale one is interested not in H concentration, but in H activity. What a person measures in the solution is just activity, not the concentration. Thus it is thermodynamic pH scale that describes real solutions, not the concentration one. For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H ] and pOH = –log[OH ] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00! However, this definition is only an approximation (albeit very good under most situations) of the proper definition of pH, which depends on the of the hydrogen ion: \[pH= -\log a\{H^+\} \approx -\log [H^+] \label{7}\] The activity is a measure of the "effective concentration" of a substance, is often related to the true concentration via an activity coefficient, \(\gamma\): \[a{H^+}=\gamma [H^+] \label{8}\] Calculating the activity coefficient requires detailed theories of how charged species interact in solution at high concentrations (e.g., the ). In most solutions the pH differs from the -log[H ] in the first decimal point. The following table gives experimentally determined pH values for a series of HCl solutions of increasing concentration at 25 °C. While the pH scale formally measures the of hydrogen ions in a substance or solution, it is typically approximated as the concentration of hydrogen ions; this approximation is applicable only under low concentrations. Molecules that make up or are produced by living organisms usually function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). Many biological solutions, such as blood, have a pH near neutral. pH influences the structure and the function of many enzymes (protein catalysts) in living systems. Many of these enzymes have narrow ranges of pH activity. Cellular pH is so important that death may occur within hours if a person becomes acidotic (having increased acidity in the blood). As one can see pH is critical to life, biochemistry, and important chemical reactions. Common examples of how pH plays a very important role in our daily lives are given below:

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is an essential tool for chemists. I have provided a simple version . It shows the symbols, atomic numbers, and average atomic masses for each element. If you point at a symbol, it will show the element name. You can search for an element by name or symbol and it will be highlighted so it's easy to find. How do you use the periodic table? It can help you predict many important properties of elements. To make these predictions, you will need to know a little about the different families or , which are the columns of the table. The term for the rows is . Here is some info about the important groups. To learn your way around the table, try going to this much fancier . Notice how metals are on the left and bottom of the periodic table, while non-metals are on the right and top. are shiny, and they conduct heat and electricity. Non-metals don't conduct, and are often softer or easier to break than metals. Some elements are called because they are in between metals and non-metals, and you can see that the metalloids are also in between the metals and non-metals in the periodic table. The most reactive elements are on the edges of the table (groups 1 and 17), and the most reactive non-metals are O and F, in the top right corner.

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List common applications of radioactive isotopes Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a (or ). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more. Radioisotopes have revolutionized medical practice, where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 \(\ce{(^{99}_{43}Tc)}\), thallium-201 \(\ce{(^{201}_{81}Tl)}\), iodine-131 \(\ce{(^{131}_{53}I)}\), and sodium-24 \(\ce{(^{24}_{11}Na)}\). Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure \(\Page {1}\)) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood. Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure \(\Page {2}\)). The parent nuclide Mo-99 is part of a molybdate ion, \(\ce{MoO4^2-}\); when it decays, it forms the pertechnetate ion, \(\ce{TcO4-}\). These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests. Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. is the use of high-energy radiation to damage the of cancer cells, which kills them or keeps them from dividing (Figure \(\Page {3}\)). A cancer patient may receive delivered by a machine outside the body, or from a radioactive substance that has been introduced into the body. Note that is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells. Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is: \[\ce{^{59}_{27}Co + ^1_0n⟶ ^{60}_{27}Co⟶ ^{60}_{28}Ni + ^0_{−1}β + 2^0_0γ} \nonumber \] The overall decay scheme for this is shown graphically in Figure \(\Page {4}\). Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants. For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is: \[\ce{6CO2}(g)+\ce{6H2O}(l)⟶\ce{C6H12O6}(s)+\ce{6O2}(g), \nonumber \] but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO containing a high concentration of \(\ce{^{14}_6C}\). At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction. Commercial applications of radioactive materials are equally diverse (Figure \(\Page {5}\)). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil. Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure \(\Page {6}\)). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm. Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.

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A characteristic feature of the transition metals is their ability to form a group of compounds called , or sometimes simply . A typical coordination compound is the intensely blue solid substance Cu(NH ) SO which can be crystallized from solutions of CuSO to which a very large excess of concentrated NH has been added. These crystals contain two polyatomic ions, one of which is the sulfate ion, SO , and the other of which is the Cu(NH ) which is responsible for the blue color. We can regard a complex ion such as Cu(NH ) as being the result of the interaction of NH acting as a Lewis base with the Cu ion acting as a Lewis acid. Each NH molecule can be considered as donating a pair of electrons to a central Cu , thus forming four to it: Most coordination compounds contain a complex ion similar to Cu(NH ) . This ion can be either positively charged like Cr(H O) , or it can be negatively charged like CoCl . Neutral complexes like Pt(NH ) Cl are also known. All these species contain a ion attached by coordinate covalent bonds to several . These ligands are invariably Lewis bases. Some typical examples of ligands are H O, NH , Cl , OH , CN , Br , and SCN . The number of ligands attached to the central metal ion is said to be its and is usually 2, 4, or 6. The group of ligands bonded to the metal taken collectively is said to constitute the metal’s . When writing the formula of a coordination compound containing complex ions, square brackets are usually used to enclose the coordination sphere. Examples are When such compounds are dissolved in H O, each of the ions present in the solid becomes an independent species with its own chemical and physical properties. Thus, when 1 mol [Cr(H O) ]Cl crystal is dissolved in H O the solution contains 1 mol Cr(H O) ion which can be recognized by its characteristic grayish-violet color and 3 mol Cl which can be detected by the precipitate of AgCl which forms when AgNO is added to the solution. An even better example of how the various ions in a coordination compound can behave independently when dissolved in water is provided by the set of Pt(II) complexes shown in the table. The first of these compounds contains the complex ion [Pt(NH ) ] and in each subsequent compound one of the NH ligands in the coordination sphere of the Pt is replaced by a Cl ligand. As a result each compound contains a Pt complex of different composition and also of different charge, and when dissolved in H O, it shows just the conductivity and other properties we would expect from the given formula. When 1 mol [Pt(NH ) Cl]Cl is dissolved in H O, it furnishes 1 mol Pt(NH ) Cl ions and 1 mol Cl ions. The strongest evidence for this is the molar conductivity of the salt (1.2 A V dm mol ), which is very similar to that of other electrolytes like NaCl (1.3 A V dm mol ) which also yield a +1 ion and a –1 ion in solution, but very different from that of electrolytes like MgCl (2.5 A V dm mol ) which yield one + 2 ion and two –1 ions in solution. The conductivity behavior also suggests that the Pt atom is part of a cation, since the Pt moves toward the cathode during electrolysis. The addition of AgNO to the solution serves to confirm this picture. One mol AgCl is precipitated immediately, showing 1 mol free Cl ions. After a few hours a further mole of AgCl is precipitated, the Cl this time originating from the coordination sphere of the Pt atom due to the slow reaction \[ \text{[Pt(NH}_{3} \text{)}_{3} \text{Cl]}^{+} (aq) + \text{Ag}^{+} (aq) + \text{H}_{2} \text{O} \rightarrow \text{[Pt(NH}_{3} \text{)}_{3} \text{H}_{2} \text{O]}^{2+} (aq) + \text{AgCl} (s) \nonumber \] It is worth noting that in all these compounds, Pt has an oxidation number of + 2. Thus the combination of Pt with one NH ligand and three Cl ligands yields an overall charge of 2(for Pt) – 3(for Cl) + 0(for NH ) = –1. The ion is thus the anion [PtNH Cl ] found in compound 4. What is the oxidation state of Pt in the compound Ca[Pt(NH )Cl ] ? Since there are two complex ions for each Ca ion, the charge on each must be –1. Adding the charge on each ligand, we obtain –5(for Cl ) + 0(for NH ) = –5. If the oxidation number of Pt is , then – 5 must equal the total charge on the complex ion: \[ x \text{ } – \text{ } 5 = \text{ } –1 \\ ~~ \\ \nonumber \] or \[ x =+4 \nonumber \] The compound in question is thus a Pt(IV) complex.

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Perhaps the most important function of the periodic table is that it helps us to predict the chemical formulas of commonly occurring compounds. At the top of each group, Mendeleev provided a general formula for oxides of the elements in the group.   The heading R O above group I, for example, means that we can expect to find compounds such as H O, Li O, Na O etc. Similarly, the general formula RH above group V suggests that the compounds NH , PH ,VH , and AsH (among others) should exist. To provide a basis for checking this prediction, formulas are shown in the following table for compounds in which H, O, or Cl is combined with each of the first two dozen elements (in order of atomic weights). Even among groups of elements whose descriptive chemistry we have not discussed, you can easily confirm that most of the predicted formulas correspond to compounds which actually exist. Conversely, more than 40 percent of the formulas for known O compounds agree with Mendeleev’s general formulas, and are shaded in color in the table. * For each element compounds are listed in order of decreasing stability. In some cases additional compounds are known, but these are relatively unstable. † A great many stable compounds of carbon and hydrogen are known, but space limitations prevent listing all of them. The periodic repetition of similar formulas is even more pronounced in the case of Cl compounds. This is evident when a list is made of subscripts for Cl in combination with each of the first 24 elements. Consulting the above table, we find HCl (subscript 1), no compound with He (subscript 0), LiCl (subscript 1),and so on.   With only the two exceptions indicated in italics, at least one formula for a compound of each element fits a sequence of subscripts which fluctuate regularly from 0 up to 4 and back to 0 again. (The unusual behavior of K and Ar will be discussed a bit later.) The number of Cl atoms which combines with one atom of each other element varies quite regularly as the atomic weight of the other element increases. The experimentally determined formulas in the first table and the general formulas in Mendeleev’s periodic table both imply that each element has a characteristic chemical combining capacity. This capacity is called , and it varies periodically with increasing atomic weight. The noble gases all have valences of 0 because they almost never combine with any other element. H and Cl both have the same valence. They combine with each other in a 1:1 ratio to form HCl, each combines with Li in the same 1:1 ratio (LiH and LiCl), each combines with Be in the same ratio (BeH , BeCl ), and so on. Because H and Cl have the same valence, we can predict that a large number of H compounds will have formulas identical to those of Cl compounds, except, of course, that the symbol H would replace the symbol Cl. The correctness of this prediction can be verified by studying the formulas surrounded by gray shading in the first table. The combining capacity, or valence, of O is apparently twice that of H or Cl. H atoms combine with one O atom in H O So do two Cl atoms or two Li atoms (Cl O and Li O). The number of atoms combining with a single O atom is usually as the number which combined with a single H or Cl atom. (Again, consulting the gray shaded formulas in the first table will confirm this statement.) After careful study of the formulas in the table, it is also possible to conclude that none of the elements (except the unreactive noble gases) have smaller valences than H or Cl. Hence we assign a valence of 1 to H and to Cl. The valence of O is twice as great, and so we assign a value of 2. Use the data in the first table to predict what formula would be expected for a compound containing (a) sodium and fluorine; (b) calcium and fluorine. From the table we can obtain the following formulas for the most common sodium compounds: All of these would imply that sodium has a valence of 1. For fluorine compounds we have which imply that fluorine also has a valence of 1. Therefore the formula is probably We already know that the valence of fluorine is 1. For calcium the formulas argue in favor of a valence of 2. Therefore the formula is most likely In some cases one element can combine in more than one way with another. For example, you have already encountered the compounds HgBr and Hg Br . There are many other examples of such variable valence in the first table. Nevertheless in its most common compounds, each element usually exhibits one characteristic valence, no matter what its partner is. Therefore it is possible to use that valence to predict formulas. Variable valence of an element may be looked upon as an exception to the rule of a specific combining capacity for each element. The experimental observation that a given element usually has a specific valence can be explained if we assume that each of its atoms has a fixed number of valence sites. One of these sites would be required to connect with one site on another atom. In other words, a noble-gas atom such as Ar or Ne would not have any combining sites, H and Cl atoms would have one valence site each, an O atom would have two, and so on. Variable valence must involve atoms in which some valence sites are more readily used than others. In the case of the F compounds of Cl (ClF, ClF , ClF ), for example, the formulas imply that at least five valence sites are available on Cl. Only one of these is used in ClF and in most of the chlorine compounds in the table. The others are apparently less readily available. Mendeleev’s inclusion of general formulas above the columns of his periodic table indicates that the table may be used to predict valences of the elements and formulas for their compounds. Two general rules may be followed: For groups V to VII, the group number gives the valence only when the element in question is combined with oxygen, fluorine, or perhaps one of the other halogens. Otherwise 8 minus the group number is the rule. Use the modern periodic to predict the formulas of compounds formed from (a) aluminum and chlorine; (b) phosphorus and chlorine. Use the table on this page to verify your prediction. Aluminum is in group III and so rule 1 predicts a valence of 3. Chlorine is in group VII and is not combined with oxygen or fluorine, and so its valence is 8 – 7 = 1 by rule 2. Each aluminum has three valence sites, while each chlorine has only one, and so it requires three chlorine atoms to satisfy one aluminum, and the formula is AlCl . Again chlorine has a valence of 1. Phosphorus is in group V and might have a valence of 5 or of 8 – 5 = 3. Therefore we predict formulas PCl or PCl . All three predicted formulas appear in the table on this page.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.07%3A_Timeline_of_Battery_Development

Although the development practical batteries largely paralelled the expansion of electrical technology from about the mid-19th century on, it is now thought that a very primitive kind of battery was apparently in use more than 2000 years ago. The brief popularity of electrically powered automobiles in the 1920's encouraged storage battery development. The widespread use of portable "personal" electrical devices has kept the search for better batteries very much alive. Earthenware jars containing an iron rod surrounded by a copper cylinder were discovered near Baghdad in 1938. They are believed to have been used by the Parthian civilization that occupied the region about 2000 years ago as a source of electricity to plate gold onto silver. (Welsh) Grove was best known in the 19th century for his "nitric acid battery" which came into wide use in early telegraphy. Now, however, he is most famous for his "gas voltaic battery" in which discovered "reverse electrolysis": the recombination of H and O following electrolysis of water at platinum electrodes. This was the first demonstration of what we now know as the hydrogen-oxygen (see below.) (French) (French) Edison, who was as much a chemist as an all-around inventor, thought that the lead in Planté-type cells made them too heavy, and that having acid in contact with any metal was an inherently bad idea. After much experimentation, he developed a successful alkaline battery. The uses an iron anode, nickel oxide cathode, and KOH electrolyte. This cell is extremely rugged and is still used in certain industrial applications, but it was never able to displace the lead-acid cell as Edison had hoped. This was one of the first "button"-type cells which were widely used in cameras and hearing aids. The constancy of the 1.34 v output made them popular for use in sensitive instruments and cardiac pacemakers. The net cell reaction is Zn + HgO → ZnO + Hg Most countries have outlawed sales of these cells in order to reduce mercury contamination of the environment. The hydride ion H would be an ideal cathode material except for the fact that its oxidation product H is a gas. The discovery that certain compounds such as LiNi and ZrNi can act as "hydrogen sponges" made it practical to employ metal hydrides as a cathode material. One peculiarity of Ni-MH cells is that recharging them is an exothermic process, so that proper dissipation of heat must be allowed for. These batteries are widely used in cell phones, computers, and portable power tools. The electrode reactions take place in a concentrated KOH electrolyte: Cathode (+): NiOOH + H O + → Ni(OH) + OH Anode (-): (1/x) MH + OH → (1/x) M + H O + Lithium is an ideal anode material owing to its low density and high reduction potential, making Li-based cells the most compact ways of storing electrical energy. Lithium cells are used in wristwatches, cardiac pacemakers and digital cameras. Both primary (non-rechargeble) and rechargeable types have been available for some time. More recent applications are in portable power tools and— perhaps most importantly, in electric-powered or hybrid automobiles. Modern lithium cells operate by transporting Li ions between electrodes into which the ions can be inserted or intercalated. Cathodes are lithium transition-metal oxides such as LiCoO , while anodes are lithium-containing carbon, LiC . The species that undergoes oxidation-reduction is not lithium, but the transition metal, e.g. Co(III)-Co(IV). There have been numerous reports of fires and explosions associated with lithium batteries. In 2006, the Dell Corporation had to recall 4.1 million Sony batteries that had been shipped with Dell's laptop computers and were judged to be at risk owing to a manufacturing defect. This illustrates the difficulty of concentrating a large amount of chemical energy into a small package, which is of course the goal of all battery developers eager to meet commercial demands ranging from consumer personal electronics to electrically-powered cars. The fully-charged Li -deficient lithium cobalt oxide cathodes are inherently unstable, held in check only by a thin insulating membrane which, if accidentally breached, can lead to thermal runaway involving gaseous oxygen, carbon, organic solvents, and (in some cases) lithium chlorate— all the components necessary for a fierce fire. Much research has gone into the development of fail-safe membranes. In one type, made by ExxonMobil and targeted at the automotive market, the pores are designed to close up and thus inhibit the passage of lithium ions when the temperature rises above a safe level. Finally, we should mention the biological batteries that are found in a number of . The "electric organs" of these fish are modified muscle cells known as which are arranged in long stacks. A neural signal from the brain causes all the electrocytes in a stack to become polarized simultaneously, in effect creating a battery made of series-connected cells. Most electric fish produce only a small voltage which they use for navigation, much in the way that bats use sound for echo-location of prey. The famous electric eel, however, is able to produce a 600-volt jolt that it employs to stun nearby prey.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Liquids

In general, liquids are harder to describe than gases (in which interactions between particles are simple collisions) and solids, in which particles stay mostly still in an organized arrangement. So we will only describe some properties here. Certain properties of liquids also depend on the intermolecular forces, like the and . These roughly describe the shapes liquids take when poured, or as droplets, etc. means how thick or sticky a liquid is. For instance, water pours easily and quickly, so it is pretty low viscosity. Honey is a thick, sticky liquid that pours slowly, so it has higher viscosity. Viscosity depends on how easily the molecules can flow past each other. The smaller they are, and the weaker the forces between them, the easier they flow. If the molecules are big and flexible, they might be able to get a bit tangled together, and that could make them flow more slowly. means how much the liquid wants to minimize its surface area. If the intermolecular forces are big, then molecules would rather be inside the liquid where they have favorable intermolecular interactions instead of being on the surface. This could make the liquid pull itself into rounded shapes to make the surface area smaller. You've probably seen water do this, like on a non-stick pan, because water has strong hydrogen-bonds. In this case, they are called , which means forces that pull the material together. (A cohesive team is very close and works well together.) On the other hand, water can have good interactions with glass surfaces, so it doesn't mind so much spreading out on glass. In fact, the forces that make water stick to glass, called are bigger than the cohesive forces in water, which is why the water in a glass tube is higher around the edges. In contrast, mercury doesn't spread out over glass, and in a glass tube, it is higher in the middle. This means that mercury has bigger cohesive forces than adhesive forces toward glass.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/3._Proteins_as_Enzymes

This page is an introduction to how proteins can work as enzymes - biological catalysts. Enzymes are mainly globular proteins - protein molecules where the tertiary structure has given the molecule a generally rounded, ball shape (although perhaps a very squashed ball in some cases). The other type of proteins (fibrous proteins) have long thin structures and are found in tissues like muscle and hair. We aren't interested in those in this topic. These globular proteins can be amazingly active catalysts. You are probably familiar with the use of catalysts like manganese(IV) oxide in decomposing hydrogen peroxide to give oxygen and water. The enzyme catalase will also do this - but at a spectacular rate compared with inorganic catalysts. One molecule of catalase can decompose almost a hundred thousand molecules of hydrogen peroxide every second. That's very impressive! This is a model of catalase, showing the globular structure - a bit like a tangled mass of string: An important point about enzymes is that they are very specific about what they can catalyse. Even small changes in the reactant molecule can stop the enzyme from catalysing its reaction. The reason for this lies in the active site present in the enzyme . . . Active sites are cracks or hollows on the surface of the enzyme caused by the way the protein folds itself up into its tertiary structure. Molecules of just the right shape, and with just the right arrangement of attractive groups (see later) can fit into these active sites. Other molecules won't fit or won't have the right groups to bind to the surface of the active site. The usual analogy for this is a key fitting into a lock. For the key to work properly it has to fit exactly into the lock. In chemistry, we would describe the molecule which is actually going to react (the purple one in the diagram) as the reactant. In biology and biochemistry, the reactant in an enzyme reaction is known instead as the substrate. You mustn't take this picture of the way a substrate fits into its enzyme too literally. What is just as important as the physical shape of the substrate are the bonds which it can form with the enzyme. Enzymes are protein molecules - long chains of amino acid residues. Remember that sticking out all along those chains are the side groups of the amino acids - the "R" groups that we talked about on the page about protein structure. Active sites, of course, have these "R" groups lining them as well - typically from about 3 to 12 in an active site. The next diagram shows an imaginary active site: Remember that these "R" groups contain the sort of features which are responsible for the tertiary structure in proteins. For example, they may contain ionic groups like -NH or -COO , or -OH groups which can hydrogen bond, or hydrocarbon chains or rings which can contribute to van der Waals forces. Groups like these help a substrate to attach to the active site - but only if the substrate molecule has an arrangement of groups in the right places to interact with those on the enzyme. The diagram shows a possible set of interactions involving two ionic bonds and a hydrogen bond. The groups shown with + or - signs are obvious. The ones with the "H"s in them are groups capable of hydrogen bonding. It is possible that one or more of the unused "R" groups in the active site could also be helping with van der Waals attractions between them and the substrate. If the arrangement of the groups on the active site or the substrate was even slightly different, the bonding almost certainly wouldn't be as good - and in that sense, a different substrate wouldn't fit the active site on the enzyme. This process of the catalyst reacting with the substrate and eventually forming products is often summarised as: . . . where E is the enzyme, S the substrate and P the products. The formation of the complex is reversible - the substrate could obviously just break away again before it converted into products. The second stage is shown as one-way, but might be reversible in some cases. It would depend on the energetics of the reaction. So why does attaching itself to an enzyme increase the rate at which the substrate converts into products? It isn't at all obvious why that should be - and most sources providing information at this introductory level just gloss over it or talk about it in vague general terms (which is what I am going to be forced to do, because I can't find a simple example to talk about!). Catalysts in general (and enzymes are no exception) work by providing the reaction with a route with a lower activation energy. Attaching the substrate to the active site must allow electron movements which end up in bonds breaking much more easily than if the enzyme wasn't there. Strangely, it is much easier to see what might be happening in other cases where the situation is a bit more complicated . . . What we have said so far is a major over-simplification for most enzymes. Most enzymes aren't in fact just pure protein molecules. Other non-protein bits and pieces are needed to make them work. These are known as cofactors. In the absence of the right cofactor, the enzyme doesn't work. For those of you who like collecting obscure words, the inactive protein molecule is known as an apoenzyme. When the cofactor is in place so that it becomes an active enzyme, it is called a holoenzyme. There are two basically different sorts of cofactors. Some are bound tightly to the protein molecule so that they become a part of the enzyme - these are called prosthetic groups. Some are entirely free of the enzyme and attach themselves to the active site alongside the substrate - these are called coenzymes. Prosthetic groups can be as simple as a single metal ion bound into the enzyme's structure, or may be a more complicated organic molecule (which might also contain a metal ion). The enzymes carbonic anhydrase and catalase are simple examples of the two types. The ideal gas law is easy to remember and apply in solving problems, as long as you get the Zinc ions in carbonic anhydrase Carbonic anhydrase is an enzyme which catalyses the conversion of carbon dioxide into hydrogencarbonate ions (or the reverse) in the cell. (If you look this up elsewhere, you will find that biochemists tend to persist in calling hydrogencarbonate by its old name, bicarbonate!) In fact, there are a whole family of carbonic anhydrases all based around different proteins, but all of them have a zinc ion bound up in the active site. In this case, the mechanism is well understood and simple. We'll look at this in some detail, because it is a good illustration of how enzymes work. The zinc ion is bound to the protein chain via three links to separate histidine residues in the chain - shown in pink in the picture of one version of carbonic anhydrase. The zinc is also attached to an -OH group - shown in the picture using red for the oxygen and white for the hydrogen. The structure of the amino acid histidine is . . . . . . and when it is a part of a protein chain, it is joined up like this: If you look at the model of the arrangement around the zinc ion in the picture above, you should at least be able to pick out the ring part of the three molecules. The zinc ion is bound to these histidine rings via dative covalent (co-ordinate covalent) bonds from lone pairs on the nitrogen atoms. Simplifying the structure around the zinc: The arrangement of the four groups around the zinc is approximately tetrahedral. Notice that I have distorted the usual roughly tetrahedral arrangement of electron pairs around the oxygen - that's just to keep the diagram as clear as possible. So that's the structure around the zinc. How does this catalyse the reaction between carbon dioxide and water? A carbon dioxide molecule is held by a nearby part of the active site so that one of the lone pairs on the oxygen is pointing straight at the carbon atom in the middle of the carbon dioxide molecule. Attaching it to the enzyme also increases the existing polarity of the carbon-oxygen bonds. If you have done any work on organic reaction mechanisms at all, then it is pretty obvious what is going to happen. The lone pair forms a bond with the carbon atom and part of one of the carbon-oxygen bonds breaks and leaves the oxygen atom with a negative charge on it. What you now have is a hydrogencarbonate ion attached to the zinc. The next diagram shows this broken away and replaced with a water molecule from the cell solution. All that now needs to happen to get the catalyst back to where it started is for the water to lose a hydrogen ion. This is transferred by another water molecule to a nearby amino acid residue with a nitrogen in the "R" group - and eventually, by a series of similar transfers, out of the active site completely. . . . and the carbonic anhydrase enzyme can do this sequence of reactions about a million times a second. This is a wonderful piece of molecular machinery! Remember the model of catalase from further up the page . . . At the time, I mentioned the non-protein groups which this contains, shown in pink in the picture. These are heme (US: heme) groups bound to the protein molecule, and an essential part of the working of the catalase. The heme group is a good example of a prosthetic group. If it wasn't there, the protein molecule wouldn't work as a catalyst. The heme groups contain an iron(III) ion bound into a ring molecule - one of a number of related molecules called porphyrins. The iron is locked into the centre of the porphyrin molecule via dative covalent bonds from four nitrogen atoms in the ring structure. There are various types of porphyrin, so there are various different heme groups. The one we are interested in is called heme B, and a model of the heme B group (with the iron(III) ion in grey at the centre) looks like this: The reaction that catalase carries out is the decomposition of hydrogen peroxide into water and oxygen. A lot of work has been done on the mechanism for this reaction, but I am only going to give you a simplified version rather than describe it in full. Although it looks fairly simple on the surface, there are a lot of hidden things going on to complicate it. Essentially the reaction happens in two stages and involves the iron changing its oxidation state. An easy change of oxidation state is one of the main characteristics of transition metals. In the lab, iron commonly has two oxidation states (as well as zero in the metal itself), +2 and +3, and changes readily from one to the other. In catalase, the change is from +3 to the far less common +4 and back again. In the first stage there is a reaction between a hydrogen peroxide molecule and the active site to give: The "Enzyme" in the equation refers to everything (heme group and protein) apart from the iron ion. The "(III)" and "(IV)" are the oxidation states of the iron in both cases. This equation (and the next one) are NOT proper chemical equations. They are just summaries of the most obvious things which have happened. The new arrangement around the iron then reacts with a second hydrogen peroxide to regenerate the original structure and produce oxygen and a second molecule of water. What is hidden away in this simplification are the other things that are happening at the same time - for example, the rest of the heme group and some of the amino acid residues around the active site are also changed during each stage of the reaction. And if you think about what has to happen to the hydrogen peroxide molecule in both reactions, it has to be more complicated than this suggests. Hydrogen peroxide is joined up as H-O-O-H, and yet both hydrogens end up attached to the same oxygen. That is quite a complicated thing to arrange in small steps in a mechanism, and involves hydrogen ions being transferred via amino acids residues in the active site. So do you need to remember all this for chemistry purposes at this level? No - not unless your syllabus specifically asks you for it. It is basically just an illustration of the term "prosthetic group". It also shows that even in a biochemical situation, transition metals behave in the same sort of way as they do in inorganic chemistry - they form complexes, and they change their oxidation state. And if you want to follow this up to look in detail at what is happening, you will find the same sort of interactions around the active site that we looked at in the simpler case of carbonic anydrase. (But please don't waste time on this unless you have to - it is seriously complicated!) Coenzymes are another form of cofactor. They are different from prosthetic groups in that they aren't permanently attached to the protein molecule. Instead, coenzymes attach themselves to the active site alongside the substrate, and the reaction involves both of them. Once they have reacted, they both leave the active site - both changed in some way. A simple diagram showing a substrate and coenzyme together in the active site might look like this: It is much easier to understand this with a (relatively) simple example. NAD+ as coenzyme with alcohol dehydrogenase Alcohol dehydrogenase is an enzyme which starts the process by which alcohol (ethanol) in the blood is oxidised to harmless products. The name "dehydrogenase" suggests that it is oxidising the ethanol by removing hydrogens from it. The reaction is actually between ethanol and the coenzyme NAD+ attached side-by-side to the active site of the protein molecule. NAD+ is a commonly used coenzyme in all sorts of redox reactions in the cell. NAD+ stands for led vitamin B3, niacin or nicotinic acid. Several important coenzymes are derived from vitamins. Ethanol is oxidised by a reaction with NAD+ helped by the active site of the enzyme. At the end of the reaction, ethanal (acetaldehyde) is formed, and the NAD+ has been converted into another compound known as NADH. As far as the NAD+ is concerned, it has picked up a hydrogen atom together with an extra electron which has neutralised the charge. Both major products - ethanal and NADH - leave the active site and are processed further in other cell reactions. The very poisonous ethanal is oxidised at once to ethanoic acid using a different enzyme, but again using NAD+ as the coenzyme. And the ethanoic acid from that reacts on through a whole set of further enzyme-controlled reactions to eventually end up as carbon dioxide and water. What about the NADH? This is a coenzyme in its own right, and takes part in reactions where something needs reducing. The hydrogen atom and the extra electron that it picked up from the ethanol are given to something else. In the process, of course, the NADH gets oxidised back to NAD+ again. In general terms, for a substrate S which needs reducing:

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.08%3A_Fermented_Vegetables

Vegetables may be preserved by fermentation or acidification. The most common commercial fermented vegetables include cucumbers, cabbage, and olives, but there are many other vegetables that have been used. Definitions: Typical Process for Vegetable Fermentation: Fresh cabbage contains about 4-8% fermentable sugars: glucose, fructose, and sucrose. Cucumbers have much lower amounts of these fermentable sugars. There are many complex polysaccharides in vegetables that are not fermentable or easily metabolized. This is often called fiber. Cellulose is a linear chain of thousands of linked D-glucose units. What type of linkages are used in this polysaccharide? Circle the correct designations. Pectin is a polysaccharide made from a mixture of monosaccharides. While many distinct polysaccharides have been identified and characterized within these ‘pectic polysaccharide family’, most contain stretches of linear chains of \(\alpha\)-(1–4)-linked D-galacturonic acid. Yeast and many microorganisms are usually present on surface of raw vegetables. Salt, either as a solid or as a brine solution, is added to the vegetable. Shredded cabbage or other suitable vegetables are placed in a jar. Salt, either as a solid or as a brine solution, is added to the vegetable so that is fully submerged. Mechanical pressure is applied to the cabbage to expel the juice, which contains fermentable sugars and other nutrients suitable for microbial activity. Salt, primarily NaCl, serves several major roles in the preservation of fermented vegetables: In addition, the salt can prevent the pectinolytic or cellulolytic enzymes from working. The fermentation of vegetables usually involves naturally occurring lactic acid bacteria (LAB). This is considered to be a wild fermentation as the LAB bacteria are found naturally on the vegetables. At the start, there are many bacteria that colonize the fresh vegetable; these organisms will compete. As the LAB begin to excrete lactic acid, the pH will decrease, and most other organisms will die. As the pH drops, the environment becomes too acidic for these bacteria to survive and they die out. In the second stage, Lactobacillus species that are better adapted to acidic environments will begin to flourish. Lactobacillus will continue to anaerobically ferment the remaining sugars into lactic acid until the pH reaches about 3. In sauerkraut, converts the vegetable sugars, typically glucose, to lactic and acetic acids and carbon dioxide. also uses fructose as an electron acceptor, reducing it to mannitol. Fructose can be used as an electron acceptor being reduced to mannitol; this reaction contributes to the replenishment of the cells’ NAD pool. Given enough time, will continue to ferment mannitol to lactic acid. Sauerkraut consumption has decreased in the US. In taste comparisons of partially fermented European vs American sauerkraut vs fully fermented sauerkraut, most consumers preferred the flavors of the partially fermented European sauerkraut. The primary chemical differences were higher levels of remaining sugars, mannitol and ethanol (probably from post-processing addition of wine). Mannitol is sweet and has a desirable cooling effect often used to mask bitter tastes. However, ‘partially fermented’ sauerkraut can cause problems in bulk storage; remaining sugars allow spoilage organisms to thrive (and gas evolution). Fully fermented sauerkraut has no remaining sugars, so it does not need further processing. Many strains of and can ferment malic acid (naturally found in vegetables) to lactic acid. The (MLF) involves the conversion of malic acid into lactic acid and carbon dioxide. Some LAB bacteria convert the malic acid in one step; while others utilize these steps that include intermediates from the TCA cycle. Sulfur aromas and flavors are strongly associated with cruciferous vegetables such as cabbage, radishes, kale, and broccoli. S-Methyl cysteine sulfoxide (SMCSO) naturally occurs in large quantities in fresh cabbage. Sauerkraut flavors are characterized mostly by salty, sour, and sulfur notes. The sulfur character of sauerkraut can lend both desirable flavors, as well as unfavorable aromas and flavors. This is often dependent upon concentration levels. Many of the compounds (shown below) found in sauerkraut are derived from the enzymatic degradation of SMCSO. DMTS and MMTSO appear to be the most critical compounds for the sauerkraut sulfur flavor. Caraway spiced commercial sauerkraut is known to be less sulfurous and milder in flavor than traditional sauerkraut, was found to contain no DMTS and the level of the DMDS was also lower. Caraway seeds appear to remove the precursor to these molecules, methanethiol. Spices, wines, and other ingredients may be added to the pickles to augment its flavor. After fermentation and removal from brine storage, cucumbers may be desalted or rinsed to reduce acid content. Many pickle and sauerkraut products undergo pasteurization in their glass containers before they are sold. Fleming HP, McFeeters RF. Residual sugars and fermentation products in raw and finished commercial sauerkraut In Sauerkraut Seminar, 1985, N. Y. State Agric. Expt. Sta. Special Report No. 56:25-29. Johanningsmeier, et. al. Chemical and Sensory Properties of Sauerkraut, 5), 343-349. Pérez-Díaz IM, Breidt F, Buescher RW, Arroyo-Lopez FN, Jimenez-Diaz R, Bautista-Gallego J, Garrido-Fernandez A, Yoon S, Johanningsmeier SD. 2014. Chapter 51: Fermented and Acidified Vegetables. In: Pouch Downes F, Ito KA, editors. Compendium of Methods for the Microbiological Examination of Foods, 5th Ed. American Public Health Association.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/06%3A_Properties_of_Gases/6.01%3A_Observable_Properties_of_Gas

Make sure you thoroughly understand the following essential ideas: The study of gases allows us to understand the behavior of matter at its simplest: individual particles, acting independently, almost completely uncomplicated by interactions and interferences between each other. This knowledge of gases will serve as the pathway to our understanding of the far more complicated (liquids and solids) in which the theory of gases will no longer give us correct answers, but it will still provide us with a useful that will at least help us to rationalize the behavior of these more complicated states of matter. First, we know that a gas has ; a gas will fill whatever volume is available to it. Contrast this to the behavior of a liquid, which always has a distinct upper surface when its volume is less than that of the space it occupies. The other outstanding characteristic of gases is their , compared with those of liquids and solids. One mole of liquid water at 298 K and 1 atm pressure occupies a volume of 18.8 cm , whereas the same quantity of water vapor at the same temperature and pressure has a volume of 30200 cm , more than 1000 times greater. The most remarkable property of gases, however, is that to a very good approximation, in response to changes in temperature and pressure, expanding or contracting by predictable amounts. This is very different from the behavior of liquids or solids, in which the properties of each particular substance must be determined individually. We will see later that each of these three characteristics of gases follows directly from the view— that is, from the nature of matter. The molecules of a gas, being in continuous motion, frequently strike the inner walls of their container. As they do so, they immediately bounce off without loss of kinetic energy, but the reversal of direction ( ) imparts a to the container walls. This force, divided by the total surface area on which it acts, is the of the gas. The pressure of a gas is observed by measuring the pressure that must be applied externally in order to keep the gas from expanding or contracting. To visualize this, imagine some gas trapped in a cylinder having one end enclosed by a freely moving piston. In order to keep the gas in the container, a certain amount of weight (more precisely, a ) must be placed on the piston so as to exactly balance the force exerted by the gas on the bottom of the piston, and tending to push it up. The pressure of the gas is simply the quotient , where is the cross-section area of the piston. The unit of pressure in the SI system is the (Pa), defined as a force of one newton per square meter (1 Nm = 1 kg m s .) At the Earth's surface, the force of gravity acting on a 1 kg mass is 9.81 N. Thus if the weight is 1 kg and the surface area of the piston is 1 M , the pressure of the gas would be 9.81 Pa. A 1-gram weight acting on a piston of 1 cm cross-section would exert a pressure of 98.1 pA. (If you wonder why the pressure is higher in the second example, consider the number of cm contained in 1 m .) In chemistry, it is often common to express pressures in units of or : 1 atm = 101325 Pa = 760 torr. The older unit (mm Hg) is almost the same as the torr and is defined as one mm of level difference in a mercury barometer at 0°C. In meteorology, the pressure unit most commonly used is the : 1 bar = 10 N m = 750.06 torr = 0.987 atm. The pressures of gases encountered in nature span an exceptionally wide range, only part of which is ordinarily encountered in chemistry. Note that in the chart below, the pressure scales are logarithmic; thus 0 on the atm scale means 10 = 1 atm. The column of air above us exerts a force on each 1-cm of surface equivalent to a weight of about 1034 g. The higher into the air you go, the lower the mass of air above you, hence the lower the pressure (right). This figure is obtained by solving Newton's law \(\textbf{F} = m\textbf{a}\) for \(m\), using the acceleration of gravity for \(\textbf{a}\): \[ m = \dfrac{101375\; kg\, m^{-1} \, s^{-2}}{9.8 \, m \, s^{-2}} = 10340 \, kg\, m^{-1} =1034\; g \; cm^{-2}\] If several kilos of air are constantly pressing down on your body, why do you not feel it? Because every other part of your body (including within your lungs and insides) also experiences the same pressure, so there is no net force (other than gravity) acting on you. This was the crucial first step that led eventually to the concept of gases and their essential role in the early development of Chemistry. In the early 17th century the Italian Evangelista Torricelli invented a device — the — to measure the pressure of the atmosphere. A few years later, the German scientist and some-time mayor of Magdeburg Otto von Guericke devised a method of pumping the air out of a container, thus creating which might be considered the opposite of air: the . As with so many advances in science, idea of a vacuum — a region of nothingness — was not immediately accepted. Torricelli's invention overturned the then-common belief that air (and by extension, all gases) are weightless. The fact that we live at the bottom of a sea of air was most spectacularly demonstrated in 1654, when two teams of eight horses were unable to pull apart two 14-inch copper hemispheres (the "Magdeburg hemispheres") which had been joined together and then evacuated with Guericke's newly-invented vacuum pump. The classical barometer, still used for the most accurate work, measures the height of a column of liquid that can be supported by the atmosphere. As indicated below, this pressure is exerted directly on the liquid in the reservoir, and is transmitted hydrostatically to the liquid in the column. Metallic , being a liquid of exceptionally high density and low vapor pressure, is the ideal barometric fluid. Its widespread use gave rise to the "millimeter of mercury" (now usually referred to as the "torr") as a measure of pressure. How is the air pressure of 1034 g cm related to the 760-mm height of the mercury column in the barometer? What if water were used in place of mercury? The density of Hg is 13.6 g cm , so in a column of 1-cm cross-section, the height needed to counter the atmospheric pressure would be (1034 g × 1 cm ) / (13.6 g cm ) = 76 cm. The density of water is only 1/13.6 that of mercury, so standard atmospheric pressure would support a water column whose height is 13.6 x 76 cm = 1034 cm, or 10.3 m. You would have to read a water barometer from a fourth-story window! Water barometers were once employed to measure the height of the ground and the heights of buildings before more modern methods were adopted. A modification of the barometer, the , provides a simple device for measuring the pressure of any gas in a container. The U-tube is partially filled with mercury, one end is connected to container, while the other end can either be opened to the atmosphere. The pressure inside the container is found from the difference in height between the mercury in the two sides of the U-tube. The illustration below shows how the two kinds of manometer work. The manometers ordinarily seen in the laboratory come in two flavors: closed-tube and open-tube. In the closed-tube unit shown at the left, the longer limb of the J-tube is evacuated by filling it with mercury and then inverting it. If the sample container is also evacuated, the mercury level will be the same in both limbs. When gas is let into the container, its pressure pushes the mercury down on one side and up on the other; the difference in levels is the pressure in torr. For practical applications in engineering and industry, especially where higher pressures must be monitored, many types of mechanical and electrical are available. If two bodies are at different temperatures, heat will flow from the warmer to the cooler one until their temperatures are the same. This is the principle on which is based; the temperature of an object is measured indirectly by placing a calibrated device known as a in contact with it. When thermal equilibrium is obtained, the temperature of the thermometer is the same as the temperature of the object. A makes use of some temperature-dependent quantity, such as the density of a liquid, to allow the temperature to be found indirectly through some easily measured quantity such as the length of a mercury column. The resulting scale of temperature is entirely arbitrary; it is defined by locating its zero point, and the size of the unit. At one point in the 18 century, 35 different temperature scales were in use! The Celsius temperature scale (formally called centigrade) locates the zero point at the freezing temperature of water; the Celsius degree is defined as 1/100 of the difference between the freezing and boiling temperatures of water at 1 atm pressure. The older scale placed the zero point at the coldest temperature it was possible to obtain at the time (by mixing salt and ice.) The 100° point was set with body temperature (later found to be 98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The Fahrenheit scale is a finer one than the Celsius scale; there are 180 Fahrenheit degrees in the same temperature interval that contains 100 Celsius degrees, so 1 F° = 5/9 C°. Since the zero points are also different by 32F°, conversion between temperatures expressed on the two scales requires the addition or subtraction of this offset, as well as multiplication by the ratio of the degree size. In 1787 the French mathematician and physicist Jacques Charles discovered that for each Celsius degree that the temperature of a gas is lowered, the volume of the gas will diminish by 1/273 of its volume at 0°C. The obvious implication of this is that if the temperature could be reduced to –273°C, the volume of the gas would contract to zero. Of course, all real gases condense to liquids before this happens, but at sufficiently low pressures their volumes are linear functions of the temperature ( ), and extrapolation of a plot of volume as a function of temperature predicts zero volume at -273°C. This temperature, known as , corresponds to the total absence of thermal energy. The temperature scale on which the zero point is –273.15°C was suggested by Lord Kelvin, and is usually known as the scale. Since the size of the Kelvin and Celsius degrees are the same, conversion between the two scales is a simple matter of adding or subtracting 273.15; thus room temperature, 20°, is about 293 K. Because the Kelvin scale is based on an absolute, rather than on an arbitrary zero of temperature, it plays a special significance in scientific calculations; most fundamental physical relations involving temperature are expressed mathematically in terms of absolute temperature. In engineering work, an absolute scale based on the Fahrenheit degree is sometimes used; this is known as the . The of a gas is simply the space in which the molecules of the gas are free to move. If we have a mixture of gases, such as air, the various gases will coexist within the same volume. In these respects, gases are very different from liquids and solids, the two condensed states of matter. The volume of a gas can be measured by trapping it above mercury in a calibrated tube known as a . The SI unit of volume is the cubic meter, but in chemistry we more commonly use the and the (mL). The (cc) is also frequently used; it is very close to 1 milliliter (mL). It's important to bear in mind, however, that the volume of a gas varies with both the temperature and the pressure, so reporting the volume alone is not very useful. A common practice is to measure the volume of the gas under the ambient temperature and atmospheric pressure, and then to correct the observed volume to what it would be at standard atmospheric pressure and some fixed temperature, usually 0° C or 25°C. )

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Drug_Activity

A very broad definition of a drug would include "all chemicals other than food that affect living processes." If the affect helps the body, the drug is a medicine. However, if a drug causes a harmful effect on the body, the drug is a poison. The same chemical can be a medicine and a poison depending on conditions of use and the person using it. Another definition would be "medicinal agents used for diagnosis, prevention, treatment of symptoms, and cure of diseases." Contraceptives would be outside of this definition unless pregnancy were considered a disease. A disease is a condition of impaired health resulting from a disturbance in the structure or function of the body. Diseases may be classified into the following major categories: In most cases, a drug bearing a generic name is equivalent to the same drug with a brand name. However, this equivalency is not always true. Although drugs are chemically equivalent, different manufacturing processes may cause differences in pharmacological action. Several differences may be crystal size or form, isomers, crystal hydration, purity-(type and number of impurities), vehicles, binders, coatings, dissolution rate, and storage stability. One major problem of pharmacology is that no drug produces a single effect. The primary effect is the desired therapeutic effect. Secondary effects are all other effects beside the desired effect which may be either beneficial or harmful. Since the differences may not be very great, drugs may be nonspecific in action and alter normal functions as well as the undesirable ones. This leads to undesirable side effects. The biological effects observed after a drug has been administered are the result of an interaction between that chemical and some part of the organism. Mechanisms of drug action can be viewed from different perspectives, namely, the site of action and the general nature of the drug-cell interaction. Chemotherapeutic agents act by killing or weakening foreign organisms such as bacteria, worms, viruses. The main principle of action is selective toxicity, i.e. the drug must be more toxic to the parasite than to the host. Drugs act by stimulating or depressing normal physiological functions. Stimulation increases the rate of activity while depression reduces the rate of activity. Drugs act within the cell by modifying normal biochemical reactions. Enzyme inhibition may be reversible or non reversible; competitive or non-competitive. Antimetabolites may be used which mimic natural metabolites. Gene functions may be suppressed. Drugs act on the cell membrane by physical and/or chemical interactions. This is usually through specific drug receptor sites known to be located on the membrane. A receptor is the specific chemical constituents of the cell with which a drug interacts to produce its pharmacological effects. Some receptor sites have been identified with specific parts of proteins and nucleic acids. In most cases, the chemical nature of the receptor site remains obscure. Drugs act exclusively by physical means outside of cells. These sites include external surfaces of skin and gastrointestinal tract. Drugs also act outside of cell membranes by chemical interactions. Neutralization ofstomach acid by antacids is a good example.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Fundamentals_of_Chirality

Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existance of these molecules are determined by concept known as . The word "chiral" was derived from the Greek word for hand, because our hands display a good example of chirality since they are non-superimposable mirror images of each other. The opposite of chiral is . Achiral objects are superimposable with their mirror images. For example, two pieces of paper are achiral. In contrast, chiral molecules, like our hands, are non superimposable mirror images of each other. Try to line up your left hand perfectly with your right hand, so that the palms are both facing in the same directions. Spend about a minute doing this. Do you see that they cannot line up exactly? The same thing applies to some molecules A Chiral molecule has a mirror image that cannot line up with it perfectly- the mirror images are non superimposable. The mirror images are called . But why are chiral molecules so interesting? A chiral molecule and its enantiomer have the same chemical and physical properties(boiling point, melting point,polarity, density etc...). It turns out that many of our biological molecules such as our DNA, amino acids and sugars, are chiral molecules. It is pretty interesting that our hands seem to serve the same purpose but most people are only able to use one of their hands to write. Similarily this is true with chiral biological molecules and interactions. Just like your left hand will not fit properly in your right glove, one of the enantiomers of a molecule may not work the same way in your body. This must mean that enantiomers have properties that make them unique to their mirror images. One of these properties is that they cannot have a or an internal mirror plane. So, a chiral molecule cannot be divided in two mirror image halves. Another property of chiral molecules is optical activity. As mentioned before, chiral molecules are very similar to each other since they have the same components to them. The only thing that obviously differs is their arrangement in space. As a result of this similarity, it is very hard to distinguish chiral molecules from each other when we try to compare their properties such as boiling points, melting points and densities. However, we can differentiate them by their . When a plane-polarized light is passed through one of the 2 enantiomers of a chiral molecule that molecule rotates light in a certain direction. When the same plane polarized light is passed through the other enantiomer, that enantionmer rotates light by the same amount but in the opposite direction. If one enantiomer rotates the light counterclockwise, the other would rotate it clockwise. Because chiral molecules are able to rotate the plane of polarization differently by interacting with the electric field differently, they are said to be optically active. In general molecules that rotate light in differen directions are called Another property of chiral molecules is called (CD). This pertains to their differential absorption of left and right circularly polarized light. When left and right circularly polarized light passes through chiral molecules, the absorption coefficients differ so that the change in absorption coefficients does not equal zero. where ΔA is the difference between absorbance of left circularly polarized (LCP) and right circularly polarized (RCP) light (this is what is usually measured). = molar concentration of the sample and is the path length. The CD signals of chiral molecules can give important information and this information can be used for visible and ultraviolet spectroscopy. Every chiral molecule shows a particular CD spectrum. By looking at the distinctive spectra of molecules such as proteins and DNA, we can obtain useful information about their secondary structures and see how they differ. An example of a CD spectrum comparing the differences between .

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.4%3A_Work

One definition of energy is the capacity to do work. There are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work and focus on the work done during changes in the pressure or the volume of a gas. The easiest form of work to visualize is mechanical work (Figure \(\Page {1}\)), which is the energy required to move an object a distance d when opposed by a force F, such as gravity: \[w=F\,d \label{7.4.1}\] Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (\(a\)), Equation \ref{7.4.1} can be rewritten to: Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Hence for works against gravity (on Earth), \(a\) can be set to \(g=9.8\; m/s^2)\). Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the opposing force of gravity. The amount of work done ( ) and thus the energy required depends on three things: To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions. Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure P and initial volume V (Figure \(\Page {2}\)). If \(P_{ext} = P_{int}\), the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston (P ) is less than P , however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume (V ) will be greater than \(V_i\). If \(P_{ext} > P_{int}\), then the gas will be compressed, and the surroundings will perform work on the system. If the piston has cross-sectional area \(A\), the external pressure exerted by the piston is, by definition, the force per unit area: \[P_{ext} = \dfrac{F}{A}\] The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height (V = Ah). Rearranging to give F = P A and defining the distance the piston moves (d) as Δh, we can calculate the magnitude of the work performed by the piston by substituting into Equation 7.4.1: \[w = F d = P_{ext}AΔh \label{7.4.3}\] The change in the volume of the cylinder (ΔV) as the piston moves a distance d is ΔV = AΔh, as shown in Figure \(\Page {3}\). The work performed is thus \[ w = P_{ext}ΔV \label{7.4.4}\] The units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m ) and volume has units of cubic meters, so \[w=\left(\dfrac{F}{A}\right)_{\textrm{ext}}(\Delta V)=\dfrac{\textrm{newton}}{\textrm m^2}\times \textrm m^3=\mathrm{newton\cdot m}=\textrm{joule}\] If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R: \[R=\dfrac{0.08206\;\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}=\dfrac{8.314\textrm{ J}}{\mathrm{mol\cdot K}}\] Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J. Whether work is defined as having a positive sign or a negative sign is a matter of convention. Heat flow is defined from a system to its surroundings as negative; using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings. This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because ΔV > 0 for an expansion, Equation 7.4.4 must be written with a negative sign to describe PV work done by the system as negative: \[ w = −P_{ext}ΔV \label{7.4.5}\] The work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, ΔV < 0 and the work is positive because work is being done on a system by its surroundings. Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation. In contrast to internal energy, work is not a state function. We can see this by examining Figure \(\Page {4}\), in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from (\(V_1\), \(P_1\)) to (\(V_2\), \(P_2\)), which means that work is a state function. Internal energy is a state function, whereas work is not. A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat. : final volume, compression ratio, and external pressure : work done : : A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: V = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is V = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative. w = −P ΔV = −(40.0 atm)(0.400 L − 0.0400 L) = −14.4 L·atm Converting from liter-atmospheres to joules, \[w=-(14.4\;\mathrm{L\cdot atm})[101.3\;\mathrm{J/(L\cdot atm)}]=-1.46\times10^3\textrm{ J}\] In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well. Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising? : −0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal). We have stated that the change in energy (Δ ) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called , (or just . Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows: \[ Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_{3})_{2(aq)} + 2H_{2}O_{(l)} + 2NO_{2(g)} \] If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure \(\Page {5}\)). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., ). We find the amount of work done by multiplying the external pressure by the change in volume caused by movement of the piston (Δ ). At a constant external pressure (here, atmospheric pressure) \[w = −PΔV \label{7.4.6}\] The negative sign associated with work done indicates that the system loses energy. If the volume increases at constant pressure (Δ > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (Δ < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy. The symbol \(U\) in resents the internal energy of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy ( ) (from the Greek , meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy \(U\) plus the product of its pressure and volume : \[H =U + PV \label{7.4.7}\] Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. If a chemical change occurs at constant pressure (i.e., for a given , Δ = 0), the change in enthalpy (Δ ) is \[ ΔH = Δ(U + PV) = ΔU + ΔPV = ΔU + PΔV \label{7.4.8}\] Substituting + for ΔU (Equation 5.2.2) and − for Δ (Equation 7.4.6), we obtain \[ΔH = ΔU + PΔV = q_p + w − w = q_p \label{7.4.9}\] The subscript \(p\) is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation 7.4.9 we see that at constant pressure the change in enthalpy, Δ of the system, defined as − , is equal to the heat gained or lost. \[ΔH = H_{final} − H_{initial} = q_p \label{7.4.10}\] Just as with ΔU, because enthalpy is a state function, the magnitude of Δ depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does occur at constant pressure. To find \(ΔH\) , measure \(q_p\). 1. W = − pΔV ΔV = Vfinal - VInitial = 5 L - 2 L = 3 L Convert 750 mmHg to atm: 750 mmHg * 1/760 (atm/mmHg) = 0.9868 atm. W = − pΔV = -(.9868 atm)(3 Liters) = -2.96 L atm. 2. First we must find the final volume using the idela gas law: pv = nRT or v = (nRT)/P = [(.54 moles)(.082057(L atm)/ (mol K))(303K)] / (1.3 atm) = 10.33 L ΔV = Vfinal - Vinitial = 10.3 Liters - 8 Liters = 2.3 Liters W = − pΔV = - (1.3 atm)(2.3 Liters) = -3 L atm. 3. \(W = - p * ΔV\) = - 1.8 atm * ΔV. Given \(p_1\),\(V_1\), and \(p_2\), find \(V_2\): \(p_1V_1=p_2V_2\) (at constant \(T\) and \(n\)) \(V_2= (V_1* P_1) / P_2\) = (1.56 L * 1.7 atm) / 1.8 atm = 1.47 L Now, \(ΔV = V_2 - V_1=1.47 L - 1.56 L = -0.09\) W = - (1.8 atm) * (-0.09 L) = 0.162 L atm.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.2%3A_Heat

is kinetic energy associated with the random motion of atoms and molecules. is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE, and we say that the object is “cold” (Figure \(\Page {1}\)). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease. Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure \(\Page {2}\). The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes. is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance (L). The atoms and molecules in H have a higher average KE than those in L. If we place substance H in contact with substance L, the thermal energy will flow spontaneously from substance H to substance L. The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance L will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure \(\Page {3}\)). Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an . For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process—this process also releases energy in the form of light as evidenced by the torch’s flame (Figure \(\Page {4}\)a). A reaction or change that absorbs heat is an . A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold. Historically, energy was measured in units of . A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m /s , which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules. We now introduce two concepts useful in describing heat flow and temperature change. The of a body of matter is the quantity of heat ( ) it absorbs or releases when it experiences a temperature change (Δ ) of 1 degree Celsius (or equivalently, 1 kelvin) \[C=\dfrac{q}{ΔT} \label{7.2.1}\] Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an property—its value is proportional to the amount of the substance. For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,150 J of energy to raise the temperature of the pan by 50.0 °C \[C_{\text{small pan}}=\dfrac{18,140 J}{50.0\; °C} =363\; J/°C \label{7.2.2}\] The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change: \[C_{\text{large pan}}=\dfrac{90,700\; J}{50.0\;°C}=1814\; J/°C \label{7.2.3}\] The of a substance, commonly called its “specific heat,” is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin): \[c = \dfrac{q}{m\Delta T} \label{7.2.4}\] Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore: \[c_{iron}=\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C \label{7.2.5}\] The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron: \[c_{iron}=\dfrac{90,700 J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C \label{7.2.6}\] Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure \(\Page {5}\)). The heat capacity of an object depends on both its and its . For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity ( ) is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of are thus J/(mol•°C). The specific heat ( ) is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change either via moles (Equation 7.3.2) or mass (Equation 7.3.3): \[q = nC_pΔT \label{7.3.2}\] e \[q = mC_sΔT \label{7.3.3}\] The specific heats of some common substances are given in Table \(\Page {1}\). Note that the specific heat values of most solids are less than 1 J/(g•°C), whereas those of most liquids are about 2 J/(g•°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g•°C), is one of the highest known. Liquid water has a relatively high specific heat (about 4.2 J/g °C); most metals have much lower specific heats (usually less than 1 J/g °C). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table \(\Page {1}\). The value of is intrinsically a positive number, but Δ and can be either positive or negative, and they both must have the sign. If Δ and are positive, then . If Δ and are negative, then . If we know the mass of a substance and its specific heat, we can determine the amount of heat, , entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost: \[\begin{align}q&=\ce{(specific\: heat)×(mass\: of\: substance)×(temperature\: change)}\label{7.2.7}\\q&=c×m×ΔT=c×m×(T_\ce{final}−T_\ce{initial})\end{align}\] In this equation, \(c\) is the specific heat of the substance, is its mass, and Δ (which is read “delta T”) is the temperature change, − . If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, − has a positive value, and the value of is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, − has a negative value, and the value of is negative. The Movement of Heat in a Substance: A flask containing \(8.0 \times 10^2\; g\) of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb? To answer this question, consider these factors: The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C). This can be summarized using the equation: \[ \begin{align} q &=c×m×ΔT=c\times m \times (T_\ce{final}−T_\ce{initial}) \\ &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−20)°C}\\ &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(65)°\cancel{C}}\\ &=\mathrm{210,000\: J(=210\: kJ)} \end{align}\] Because the temperature increased, the water absorbed heat and \(q\) is positive. How much heat, in joules, must be added to a \(5.00 \times 10^2 \;g\) iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C. \(5.05 \times 10^4\; J\) Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced. A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity). Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship: \[q=c \times m \times \Delta T=c \times m \times (T_{final}−T_{initial})\] Substituting the known values: \[6,640\; J=c \times (348\; g) \times (43.6 − 22.4)\; °C\] Solving: \[c=\dfrac{6,640\; J}{(348\; g) \times (21.2°C)} =0.900\; J/g\; °C\] Comparing this value with the values in Table \(\Page {1}\), this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum. A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity. \(c = 0.45 \;J/g \;°C\); the metal is likely to be iron from checking Tabel 7.2.1 A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.) volume and density of water and initial and final temperatures amount of energy stored The mass of water is \[ mass \; of \; H_{2}O=400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) = 3.99\times 10^{5}g\; H_{2}O \] The temperature change (Δ ) is 38.0°C − 22.0°C = +16.0°C. From Table \(\Page {1}\), the specific heat of water is 4.184 J/(g•°C). From Equation 7.3.3, the heat absorbed by the water is thus \[ q=mC_{s}\Delta T=\left ( 3.99X10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right ) \left ( 16.0 \; \cancel{^{o}C} \right ) = 2.67 \times 10^{7}J = 2.67 \times 10^{4}kJ \] Both and Δ are positive, consistent with the fact that the water has absorbed energy. Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO ) in Table \(\Page {1}\). 2.7 × 10 kJ (Even though the mass of sandstone is more than six times the mass of the water in Example \(\Page {1}\), the amount of thermal energy stored is the same to two significant figures.) When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process: \[q_{cold} + q_{hot} = 0 \label{7.3.4}\] The equation implies that the amount of heat that flows a warmer object is the same as the amount of heat that flows a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite: \[q_{cold} = −q_{hot} \label{7.3.5}\] Thus heat is conserved in any such process, consistent with the law of conservation of energy. The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object. Substituting for from Equation 7.3.2 gives \[ \left [ mC_{s} \Delta T \right ] _{cold} + \left [ mC_{s} \Delta T \right ] _{hot}=0 \label{7.3.6} \] which can be rearranged to give \[ \left [ mC_{s} \Delta T \right ] _{cold} = - \left [ mC_{s} \Delta T \right ] _{hot} \label{7.3.7} \] When two objects initially at different temperatures are placed in contact, we can use Equation 7.3.7 to calculate the final temperature if we know the chemical composition and mass of the objects. If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings. mass and initial temperature of two objects final temperature Using Equation 7.3.7 and writing Δ as − for both the copper and the water, substitute the appropriate values of , , and into the equation and solve for . \[ \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \] Substituting the data provided in the problem and Table \(\Page {1}\) gives \[ \left [ \left (30 \; g \right ) \left (0.385 \; J \right ) \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \] \[ T_{final}\left ( 11.6 \; J/ ^{o}C \right ) -924 \; J + T_{final}\left ( 418.4 \; J/ ^{o}C \right ) -11,300 \; J \] \[ T_{final}\left ( 430 \; J/\left ( g\cdot ^{o}C \right ) \right ) = -12,224 \; J \] \[ T_{final} = -28.4 \; ^{o}C \] If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings? 80.0°C A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.) 90.6°C Conservation of Energy: The Movement of Heat between Substances:   ).

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Samples containing only a single are described as enantiomerically pure. However, many processes give mixtures of stereoisomers, at least to some extent. A contains two enantiomers in equal amounts. As a result, a racemic mixture has no net optical activity. For non-racemic mixtures of enantiomers, one enantiomer is more abundant than the other. The composition of these mixtures is described by the , which is the difference between the relative abundance of the two enantiomers. Therefore, if a mixture contains 75% of the R enantiomer and 25% S, the enantiomeric excess if 50%. Similarly, a mixture that is 95% of one enantiomer, the enantiomeric excess is 90%, etc. Enantiomeric excess is useful because it reflects the of the mixture. The standard optical rotation by the mixture (\([\alpha]_{mix}\)) is equal to the product of the standard optical rotation of the major isomer (\([\alpha]_{major}\)) and the ic excess (\(EE\)): \[[\alpha]_{mix} = EE \times [\alpha]_{major}\] In the same way, the enantiomeric excess in a mixture can be measured if the optical rotation of the pure enantiomer is known. A similar approach can be used to describe mixtures of , resulting in the .

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The geometry of a complex is governed almost entirely by the coordination number. We will consider only the most common coordination numbers, namely, 2, 4, and 6. Complexes with two ligands are invariably . The best-known examples of such compounds are Ag(I) and Au(I) complexes such as Both of these complexes are important. The Au(CN) complex is used to extract minute gold particles from the rock in which they occur. The crushed ore is treated with KCN solution and air is blown through it: \[ 4 \text{Au} (s) + 8 \text{CN}^{-} (aq) + \text{O}_{2} (g) + 2 \text{H}_{2} \text{O} (l) \rightarrow 4 \text{[Au(CN)}_{2} \text{]}^{-} (aq) + 4 \text{OH}^{-} (aq) \label{1} \] The resultant complex is water soluble. The silver complex is also water soluble and affords a method for dissolving AgCl, which is otherwise very insoluble. \[ \text{AgCl} (s) + 2 \text{NH}_{3} (aq) \rightarrow \text{[Ag(NH}_{3} \text{)}_{2} \text{]}^{+} (aq) + \text{Cl}^{-} (aq) \nonumber \] This reaction is often used in the laboratory to be sure a precipitate is AgCl( ). Two geometries are possible for this coordination number. Some complexes, like the [Pt(NH ) ] ion shown in Figure \(\Page {1}\), are , while others, like Cd(NH ) , are tetrahedral. Most of the four-coordinated complexes of Zn(II), Cd(II), and Hg(II) are tetrahedral, while the square planar arrangement is preferred by Pd(II), Pt(II), and Cu(II) complexes. Because the square planar geometry is less symmetrical than the tetrahedral geometry, it offers more possibilities for isomerism. A well-known example of such isomerism is given by the two square planar complexes These two isomers are called . That isomer in which two identical ligands are next to each other is called the cis isomer, while that in which they are on opposite sides is called the trans isomer. Though these two isomers have some properties which are similar, no properties are identical and some are very different. For example, the cis isomer of the above complex is used as an anti-tumor drug to treat cancerous cells. The trans form, by contrast, shows no similar biological activity. It is worth noting that cis-trans isomerism is not possible in the case of tetrahedral complexes. As you can quickly verify by examining any three-dimensional tetrahedral shape, any given corner of a tetrahedron is adjacent to the other three. Since all the corners are cis to each other, none are trans. When there are six ligands, the geometry of the complex is almost always octahedral, like the geometry of SF , or of [Cr(H O) ] . All ligands are equidistant from the central atom, and all ligand-metal-ligand angles are 90°. An octahedral complex may also be thought of as being derived from a square planar structure by adding a fifth ligand above and a sixth below on a line through the central metal ion and perpendicular to the plane. The octahedral structure also gives rise to geometrical isomerism. For example, two different compounds, one violet and one green, have the formula [Co(NH ) Cl ]Cl. The violet complex turns out to have the cis structure and the green one trans, as shown in Figure \(\Page {2}\).

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Make sure you thoroughly understand the following essential concepts: One of the interesting things about thermodynamics is that although it deals with matter, it makes no assumptions about the microscopic nature of that matter. Thermodynamics deals with matter in a macroscopic sense; it would be valid even if the atomic theory of matter were wrong. This is an important quality, because it means that reasoning based on thermodynamics is unlikely to require alteration as new facts about atomic structure and atomic interactions come to light. In thermodynamics, we must be very precise in our use of certain words. The two most important of these are and . A is that part of the world to which we are directing our attention. Everything that is not a part of the system constitutes the . The system and surroundings are separated by a . If our system is one mole of a gas in a container, then the boundary is simply the inner wall of the container itself. The boundary need not be a physical barrier; for example, if our system is a factory or a forest, then the boundary can be wherever we wish to define it. We can even focus our attention on the dissolved ions in an aqueous solution of a salt, leaving the water molecules as part of the surroundings. The single property that the boundary must have is that it be clearly defined, so we can unambiguously say whether a given part of the world is in our system or in the surroundings. If matter is not able to pass across the boundary, then the system is said to be closed; otherwise, it is open. A closed system may still exchange energy with the surroundings unless the system is an isolated one, in which case neither matter nor energy can pass across the boundary. The tea in a closed Thermos bottle approximates a closed system over a short time interval. The of a system are those quantities such as the pressure, volume, temperature, and its composition, which are in principle measurable and capable of assuming definite values. There are of course many properties other than those mentioned above; the density and thermal conductivity are two examples. However, the pressure, volume, and temperature have special significance because they determine the values of all the other properties; they are therefore known as because if their values are known then the system is in a definite . In dealing with thermodynamics, we must be able to unambiguously define the change in the state of a system when it undergoes some process. This is done by specifying changes in the values of the different state properties using the symbol Δ ( ) as illustrated here for a change in the volume: \[ΔV = V_{final} – V_{initial} \label{1-1}\] We can compute similar delta-values for changes in (the number of moles of component ), and the other state properties we will meet later. is simply the totality of all forms of kinetic and potential energy of the system. Thermodynamics makes no distinction between these two forms of energy and it does not assume the existence of atoms and molecules. But since we are studying thermodynamics in the context of chemistry, we can allow ourselves to depart from “pure” thermodynamics enough to point out that the internal energy is the sum of the kinetic energy of motion of the molecules, and the potential energy represented by the chemical bonds between the atoms and any other intermolecular forces that may be operative. How can we know how much internal energy a system possesses? The answer is that we cannot, at least not on an absolute basis; all scales of energy are arbitrary. The best we can do is measure changes in energy. However, we are perfectly free to define zero energy as the energy of the system in some arbitrary , and then say that the internal energy of the system in any other state is the difference between the energies of the system in these two different states. This law is one of the most fundamental principles of the physical world. Also known as the , it states that energy can not be created or destroyed; it can only be redistributed or changed from one form to another. A way of expressing this law that is generally more useful in Chemistry is that any change in the internal energy of a system is given by the sum of the heat that flows across its boundaries and the work done on the system by the surroundings. \[ΔU = q + w \label{2-1}\] This says that there are two kinds of processes, heat and work, that can lead to a change in the internal energy of a system. Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the world outside the system- in other words, energy cannot be created or destroyed. There is an important for heat and work that you are expected to know. If heat flows into a system or the surroundings to do work on it, the internal energy increases and the sign of or is positive. Conversely, heat flow out of the system or work done by the system will be at the expense of the internal energy, and will therefore be negative. (Note that this is the opposite of the sign convention that was commonly used in much of the pre-1970 literature.) The full significance of Equation \(\ref{2-1}\) cannot be grasped without understanding that is a . This means that a given change in internal energy Δ can follow an infinite variety of corresponding to all the possible combinations of and that can add up to a given value of Δ . As a simple example of how this principle can simplify our understanding of change, consider two identical containers of water initially at the same temperature. We place a flame under one until its temperature has risen by 1°C. The water in the other container is stirred vigorously until its temperature has increased by the same amount. There is now no physical test by which you could determine which sample of water was warmed by performing work on it, by allowing heat to flow into it, or by some combination of the two processes. In other words, there is no basis for saying that one sample of water now contains more “work”, and the other more “heat”. The only thing we can know for certain is that both samples have undergone identical increases in internal energy, and we can determine the value of simply by measuring the increase in the temperature of the water. The kind of work most frequently associated with chemical change occurs when the volume of the system changes owing to the disappearance or formation of gaseous substances. This is sometimes called expansion work or -work, and it can most easily be understood by reference to the simplest form of matter we can deal with, the hypothetical . The figure shows a quantity of gas confined in a cylinder by means of a moveable piston. Weights placed on top of the piston exert a force f over the cross-section area , producing a pressure = / which is exactly countered by the pressure of the gas, so that the piston remains stationary. Now suppose that we heat the gas slightly; according to Charles’ law, this will cause the gas to expand, so the piston will be forced upward by a distance Δ . Since this motion is opposed by the force , a quantity of work Δ will be done by the gas on the piston. By convention, work done by the system (in this case, the gas) on the surroundings is negative, so the work is given by \[w = – f Δx \label{3-1}\] When dealing with a gas, it is convenient to think in terms of the more relevant quantities pressure and volume rather than force and distance. We can accomplish this by multiplying the second term by which of course leaves it unchanged: \[ w = -f Δx \dfrac{A}{A} \label{3-2}\] By grouping the terms differently, but still not changing anything, we obtain \[ w = -\dfrac{f}{A} Δx A \label{3-3}\] Since pressure is force per unit area and the product of the length and the area has the dimensions of volume, this expression becomes \[w = –P ΔV \label{3-4}\] It is important to note that although \(P\) and \(V\) are , the is not (that's why we denote it by a lower-case .) As is shown farther below, the quantity of work done will depend on whether the same net volume change is realized in a single step (by setting the external pressure to the final pressure ), or in multiple stages by adjusting the restraining pressure on the gas to successively smaller values approaching the final value of \(P\). Find the amount of work done on the surroundings when 1 liter of an ideal gas, initially at a pressure of 10 atm, is allowed to expand at constant temperature to 10 liters by: at \(ΔV\), which \[V_2 = (10/1) × (1 L) = 10 L\] \[ΔV = 9\; L\] in Figure \(\Page {1}\) so th When a gas expands, it does work on the surroundings; compression of a gas to a smaller volume similarly requires that the surroundings perform work on the gas. If the gas is thermally isolated from the surroundings, then the process is said to occur . In an adiabatic change, = 0, so the First Law becomes Δ = 0 + . Since the temperature of the gas changes with its internal energy, it follows that adiabatic compression of a gas will cause it to warm up, while adiabatic expansion will result in cooling. In contrast to this, consider a gas that is allowed to slowly escape from a container immersed in a constant-temperature bath. As the gas expands, it does work on the surroundings and therefore tends to cool, but the thermal gradient that results causes heat to pass into the gas from the surroundings to exactly compensate for this change. This is called an expansion. In an isothermal process the internal energy remains constant and we can write the First Law (Equation \(Ref{2-1}\)) as \[0 = q + w\] or \[q = –w\] This illustrates that the heat flow and work done exactly balance each other. Because no thermal insulation is perfect, truly adiabatic processes do not occur. However, heat flow does take time, so a compression or expansion that occurs more rapidly than thermal equilibration can be considered adiabatic for practical purposes. If you have ever used a hand pump to inflate a bicycle tire, you may have noticed that the bottom of the pump barrel can get quite warm. Although a small part of this warming may be due to friction, it is mostly a result of the work you (the surroundings) are doing on the system (the gas.) Adiabatic expansion and contractions are especially important in understanding the behavior of the . Although we commonly think of the atmosphere as homogeneous, it is really not, due largely to uneven heating and cooling over localized areas. Because mixing and heat transfer between adjoining parcels of air does not occur rapidly, many common atmospheric phenomena can be considered at least quasi-adiabatic. From Example \(\Page {1}\) we see that when a gas expands into a vacuum (\(P_{external} = 0\) the work done is zero. This is the the gas can do; what is the the gas can perform on the surroundings? To answer this, notice that more work is done when the process is carried out in two stages than in one stage; a simple calculation will show that even more work can be obtained by increasing the number of stages— that is, by allowing the gas to expand against a series of successively lower external pressures. In order to extract the maximum possible work from the process, the expansion would have to be carried out in an infinite sequence of infinitesimal steps. Each step yields an increment of work Δ which can be expressed as ( ) and integrated: \[ \begin{align} w &= \int_{V_1}^{V_2} \dfrac{RT}{V} dv \\[4pt] &= RT \ln \dfrac{V_2}{V_1} \label{3-5} \end{align}\] Although such a path (which corresponds to what is called a ) cannot be realized in practice, it can be approximated as closely as desired. Even though no real process can take place reversibly (it would take an infinitely long time!), reversible processes play an essential role in thermodynamics. The main reason for this is that and are which are important and are easily calculated. Moreover, many real processes take place sufficiently gradually that they can be treated as approximately reversible processes for easier calculation. Each expansion-compression cycle leaves the gas unchanged, but in all but the one in the bottom row, the surroundings are forever altered, having expended more work in compressing the gas than was performed on it when the gas expanded. For a chemical reaction that performs no work on the surroundings, the heat absorbed is the same as the change in internal energy: = Δ . But many chemical processes do involve work in one form or another: We will consider only pressure-volume work in this lesson. If the process takes place at a constant pressure, then the work is given by Δ and the change in internal energy will be \[ΔU = q – PΔV \label{4-1}\] Bear in mind why \(q\) is so important: the heat flow into or out of the system is directly measurable. Δ , being "internal" to the system, is not directly observable. Thus the amount of heat that passes between the system and the surroundings is given by \[q = ΔU + PΔV \label{4-2}\] This means that if an exothermic reaction is accompanied by a net increase in volume under conditons of constant pressure, some heat additional to Δ must be absorbed in order to supply the energy expended as work done on the surroundings if the temperature is to remain unchanged process.) For most practical purposes, changes in the volume of the system are only significant if the reaction is accompanied by a difference in the moles of gaseous reactants and products. For example, in the reaction H + O → ½H O , the total volume of the system decreases from that correponding to 2 moles of gaseous reactants to 0.5 mol of liquid water which occupies only 9 mL — a volume so small in comparison to that of the reactants that it can be neglected without significant error. So all we are really concerned with is the difference in the number of moles of gas Δ : Δ   = (0 – 2) mol = –2 mol This corresponds to a net contraction (negative expansion) of the system, meaning that the surroundings perform work on the system. The molar volume of an ideal gas at 25° C and 1 atm is (298/273) × (22.4 L mol ) = 24.5 L mol Remember the sign convention: a flow of heat or performance of work that supplies energy is positive; if it consumes energy, it is negative. Thus work performed by the surroundings diminishes the energy of the surroundings ( < 0) and increases the energy of the system ( > 0). and the work done (by the surroundings on the system) is (1 atm) (–2 mol)(24.5 L mol ) = –49.0 L-atm. Using the conversion factor 1 J = 101.3 J, and bearing in mind that work performed the system supplies energy the system, the work associated solely with the volume change of the system increases its energy by (101.3 J/L-atm)(–49.0 L-atm) = 4964 J = 4.06 kJ The above reaction H + ½ O → H O is carried out at a constant pressure of 1 atm and a constant temperature of 25° C. What quantity of heat will cross the system boundary (and in which direction?) For this reaction, the change in internal energy is Δ = –281.8 kJ/mol. = (–281.8 + 4.06) k J = –285.8 kJ (Eq. 4.2 above.) In order to maintain the constant 25° temperature, an equivalent quantity of heat must pass from the system to the surroundings. This terminology can be somewhat misleading unless you bear in mind that the conditions Δ and Δ refer to the differences between the and states of the system — that is, and the reaction. During the time the reaction is in progress, the temperature of the mixture will rise or fall, depending on whether the process is exothermic or endothermic. But because Δ is a state function, its value is independent of what happens "in between" the initial state (reactants) and final state (products). The same is true of Δ Enthalpy hides work and saves it too! Because most chemical changes we deal with take place at constant pressure, it would be tedious to have to explicitly deal with the pressure-volume work details that were described above. Fortunately, chemists have found a way around this; they have simply defined a new state function that incorporates and thus hides within itself any terms relating to incidental kinds of work ( , electrical, etc.). Since both Δ and Δ in Equation \(\ref{4-2}\) are state functions, then \(q_P\), the heat that is absorbed or released when a process takes place at constant pressure, must also be a state function and is known as the Δ \[ΔH ≡ q_P = ΔU + PΔV \label{4-3}\] Since most processes that occur in the laboratory, on the surface of the earth, and in organisms do so under a constant pressure of one atmosphere, Equation \(\ref{4-3}\) is the form of the First Law that is of greatest interest to most of us most of the time. Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find Δ for the system when one mole of HCl dissolves in water under these conditions. In this process the volume of liquid remains practically unchanged, so Δ = –24.5 L. The work done is \[ \begin{align*} w &= –PΔV \\[4pt] &= –(1\; atm)(–24.5\; L) \\[4pt] &= 24.6 \;L-atm \end{align*}\] (The work is positive because it is being done on the system as its volume decreases due to the dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm = 101.33 J mol and substituting in Equation \ref{4-3} we obtain \[ \begin{align*} ΔU &= q +PΔV \\[4pt] = –(75,300\; J) + [101.33\; J/L-atm) (24.5\; L-atm)] \\[4pt] &= –72.82\; kJ \end{align*}\] In other words, if the gaseous HCl could dissolve without volume change, the heat released by the process (75.3 kJ) would cause the system’s internal energy to diminish by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system by the pressure of the atmosphere performing work on it and consuming part of the energy that would otherwise be liberated, reducing the net value of Δ to J. For systems in which no change in composition (chemical reaction) occurs, things are even simpler: to a very good approximation, the enthalpy depends only on the temperature. This means that the temperature of such a system can serve as a direct measure of its enthalpy. The functional relation between the internal energy and the temperature is given by the measured at constant pressure: \[ c_p =\dfrac{dH}{dT} \label{5-1}\] (or Δ /Δ over a finite duration) An analogous quantity relates the heat capacity at constant to the internal energy: \[ c_v =\dfrac{dU}{dT} \label{5-2}\] The greater the heat capacity of a substance, the smaller will be the effect of a given absorption or loss of heat on its temperature. Heat capacity can be expressed in joules or calories per mole per degree (molar heat capacity), or in joules or calories per gram per degree; the latter is called the specific heat capacity or just the specific heat. The difference between \(c_p\) and \(c_v\) is of importance only when the volume of the system changes significantly— that is, when different numbers of moles of gases appear on either side of the chemical equation. For reactions involving only liquids and solids, \(c_p\) and \(c_v\) are for all practical purposes identical.

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Virtually all chemical processes involve the absorption or release of heat, and thus changes in the internal energy of the system. In this section, we survey some of the more common chemistry-related applications of enthalpy and the First Law. While the first two sections relate mainly to chemistry, the remaining ones impact the everyday lives of everyone. Comparison and interpretation of enthalpy changes is materially aided by a graphical construction in which the relative enthalpies of various substances are represented by horizontal lines on a vertical energy scale. The zero of the scale can be placed anywhere, since energies are always arbitrary; it is generally most useful to locate the elements at zero energy, which reflects the convention that their standard enthlapies of formation are zero. This very simple for carbon and oxygen and its two stable oxides (Figure \(\Page {1}\)) shows the changes in enthalpy associated with the various reactions this system can undergo. Notice how Hess’s law is implicit in this diagram; we can calculate the enthalpy change for the combustion of carbon monoxide to carbon dioxide, for example, by subtraction of the appropriate arrow lengths without writing out the thermochemical equations in a formal way. The zero-enthalpy reference states refer to graphite, the most stable form of carbon, and gaseous oxygen. All temperatures are 298 K. This enthalpy diagram for the hydrogen-oxygen system (Figure \(\Page {2}\)) shows the known stable configurations of these two elements. Reaction of gaseous H and O to yield one mole of liquid water releases 285 kJ of heat . If the H O is formed in the gaseous state, the energy release will be smaller. Notice also that... You may have heard the venerable urban legend, probably by now over 80 years old, that some obscure inventor discovered a process to do this, but the invention was secretly bought up by the oil companies in order to preserve their monopoly. The enthalpy diagram for the hydrogen-oxygen system shows why this cannot be true — there is simply no known compound of H and O that resides at a lower enthalpy level. Enthalpy diagrams are especially useful for comparing groups of substances having some common feature. This one shows the molar enthalpies of species relating to two hydrogen halides, with respect to those of the elements. From this diagram we can see at a glance that the formation of HF from the elements is considerably more exothermic than the corresponding formation of HCl. The upper part of this diagram shows the gaseous atoms at positive enthalpies with respect to the elements. The endothermic processes in which the H and the dihalogen are dissociated into atoms can be imagined as taking place in two stages, also shown. From the enthalpy change associated with the dissociation of H (218 kJ mol ), the dissociation enthalpies of F and Cl can be calculated and placed on the diagram. The enthalpy change associated with the reaction \[\ce{HI(g) → H(g) + I(g)}\] is the of the \(\ce{HI}\) molecule; it is also the of the hydrogen-iodine bond in this molecule. Under the usual standard conditions, it would be expressed either as the bond enthalpy °(HI,298K) or internal energy (HI,298); in this case the two quantities differ from each other by Δ = . Since this reaction cannot be studied directly, the H–I is calculated from the appropriate standard enthalpies of formation: Bond energies and enthalpies are important properties of chemical bonds, and it is very important to be able to estimate their values from other thermochemical data. The total bond enthalpy of a more complex molecule such as ethane can be found from the following combination of reactions: When a molecule in its ordinary state is broken up into gaseous atoms, the process is known as (Figure \(\Page {4}\)). The standard refers to the transformation of an element into gaseous atoms: \[\ce{ C_{(graphite)} → C(g)} \;\;\;\; ΔH^o = 716.7\; kJ\] Atomization is an endothermic process. Heats of atomization are most commonly used for calculating bond energies. They are usually measured spectroscopically. The total bond energy of a molecule can be thought of as the sum of the energies of the individual bonds. Pauling’s Rule is only an approximation, because the energy of a given type of bond is not really a constant, but depends somewhat on the particular chemical environment of the two atoms. In other words, all we can really talk about is the average energy of a particular kind of bond, such as C–O, for example, the average being taken over a representative sample of compounds containing this type of bond, such as CO, CO , COCl , (CH ) CO, CH COOH, etc. Despite the lack of strict additivity of bond energies, Pauling’s Rule is extremely useful because it allows one to estimate the heats of formation of compounds that have not been studied, or have not even been prepared. Thus in the foregoing example, if we know the enthalpies of the C–C and C–H bonds from other data, we could estimate the total bond enthalpy of ethane, and then work back to get some other quantity of interest, such as ethane’s enthalpy of formation. By assembling a large amount of experimental information of this kind, a consistent set of average bond energies can be obtained. The energies of double bonds are greater than those of single bonds, and those of triple bonds are higher still (Table \(\Page {1}\)). A is any substance capable of providing useful amounts of energy through a process that can be carried out in a controlled manner at economical cost. For most practical fuels, the process is combustion in air (in which the oxidizing agent O is available at zero cost.) The is obviously an important criterion for a substance’s suitability as a fuel, but it is not the only one; a useful fuel must also be easily ignited, and in the case of a fuel intended for self-powered vehicles, its in terms of both mass (kJ kg ) and volume (kJ m ) must be reasonably large. Thus substances such as methane and propane which are gases at 1 atm must be stored as pressurized liquids for transportation and portable applications. Notes on the above table What, exactly, is meant by the statement that a particular food “contains 1200 calories” per serving? This simply refers to the standard enthalpy of combustion of the foodstuff, as measured in a . Although this unit is still employed in the popular literature, the SI unit is now commonly used in the scientific and clinical literature, in which energy contents of foods are usually quoted in kJ per unit of weight. Although the mechanisms of oxidation of a carbohydrate such as glucose to carbon dioxide and water in a bomb calorimeter and in the body are complex and entirely different, : \[C_6H_{12}O_6 + 6 O_2 → 6 CO_2 + 6 H_2O \;\;\;\; ΔH^o = – 20.8\; kJ \;mol^{–1}\] Glucose is a , a breakdown product of , and is the most important energy source at the cellular level; fats, proteins, and other sugars are readily converted to glucose. By writing balanced equations for the combustion of sugars, fats, and proteins, a comparison of their relative energy contents can be made. The stoichiometry of each reaction gives the amounts of oxygen taken up and released when a given amount of each kind of food is oxidized; these gas volumes are often taken as indirect measures of energy consumption and metabolic activity; a commonly-accepted value that seems to apply to a variety of food sources is 20.1 J (4.8 kcal) per liter of O consumed. For some components of food, particularly proteins, oxidation may not always be complete in the body, so the energy that is actually available will be smaller than that given by the heat of combustion. Mammals, for example, are unable to break down cellulose (a polymer of sugar) at all; animals that derive a major part of their nutrition from grass and leaves must rely on the action of symbiotic bacteria which colonize their digestive tracts. The amount of energy available from a food can be found by measuring the heat of combustion of the waste products excreted by an organism that has been restricted to a controlled diet, and subtracting this from the heat of combustion of the food (Table \(\Page {3}\)). The amount of energy an animal requires depends on the age, sex, surface area of the body, and of course on the amount of physical activity. The rate at which energy is expended is expressed in watts: 1 W = 1 J sec . For humans, this value varies from about 200-800 W. This translates into daily food intakes having energy equivalents of about 10-15 MJ for most working adults. In order to just maintain weight in the absence of any physical activity, about 6 MJ per day is required. The above table is instructive in that although larger animals consume more energy, the energy consumption decreases with size. This reflects the fact the rate of heat loss to the environment depends largely on the surface area of an animal, which increases with mass at a greater rate than does an animal's volume ("size"). It is common knowledge that large bodies of water have a “moderating” effect on the local weather, reducing the extremes of temperature that occur in other areas. Water temperatures change much more slowly than do those of soil, rock, and vegetation, and this effect tends to affect nearby land masses. This is largely due to the high heat capacity of water in relation to that of land surfaces— and thus ultimately to the effects of . The lower efficiency of water as an absorber and radiator of infrared energy also plays a role. The specific heat capacity of water is about four times greater than that of soil. This has a direct consequence to anyone who lives near the ocean and is familiar with the daily variations in the direction of the winds between the land and the water. Even large lakes can exert a moderating influence on the local weather due to water's relative insensitivity to temperature change. During the daytime the land and sea receive approximately equal amounts of heat from the Sun, but the much smaller heat capacity of the land causes its temperature to rise more rapidly. This causes the air above the land to heat, reducing its density and causing it to rise. Cooler oceanic air is drawn in to fill the void, thus giving rise to the daytime . In the evening, both land and ocean lose heat by radiation to the sky, but the temperature of the water drops less than that of the land, continuing to supply heat to the oceanic air and causing it to rise, thus reversing the direction of air flow and producing the evening . The air receives its heat by absorbing far-infrared radiation from the earth, which of course receives its heat from the sun. The amount of heat radiated to the air immediately above the surface varies with what's on it (forest, fields, water, buildings) and of course on the time and season. When a parcel of air above a particular location happens to be warmed more than the air immediately surrounding it, this air expands and becomes less dense. It therefore rises up through the surrounding air and undergoes further expansion as it encounters lower pressures at greater altitudes. Whenever a gas expands against an opposing pressure, it does work on the surroundings. According to the First Law Δ = + , if this work is not accompanied by a compensating flow of heat into the system, its internal energy will fall, and so, therefore, will its temperature. It turns out that heat flow and mixing are rather slow processes in the atmosphere in comparison to the convective motion we are describing, so the First Law can be written as Δ = (recall that w is negative when a gas expands.) Thus as air rises above the surface of the earth it undergoes adiabatic expansion and cools. The actual rate of temperature decrease with altitude depends on the composition of the air (the main variable being its moisture content) and on its heat capacity. For dry air, this results in an of 9.8 C° per km of altitude. Just the opposite happens when winds develop in high-altitude areas and head downhill. As the air descends, it undergoes compression from the pressure of the air above it. The surroundings are now doing work on the system, and because the process occurs too rapidly for the increased internal energy to be removed as heat, the compression is approximately adiabatic. The resulting winds are warm (and therefore dry) and are often very irritating to mucous membranes. These are known generically as (which is the name given to those that originate in the Alps). In North America they are often called (or, in winter, "snow melters") when they originate along the Rocky Mountains. Among the most notorious are the winds of Southern California which pick up extra heat (and dust) as they pass over the Mohave Desert before plunging down into the Los Angeles basin (Figure 13.5.X). Their dryness and high velocities feed many of the disastrous wildfires that afflict the region.

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Make sure you thoroughly understand the following essential concepts: The heat that flows across the boundaries of a system undergoing a change is a fundamental property that characterizes the process. It is easily measured, and if the process is a chemical reaction carried out at constant pressure, it can also be predicted from the difference between the enthalpies of the products and reactants. The quantitative study and measurement of heat and enthalpy changes is known as . In order to define the thermochemical properties of a process, it is first necessary to write a that defines the actual change taking place, both in terms of the formulas of the substances involved and their physical states (temperature, pressure, and whether solid, liquid, or gaseous. To take a very simple example, here is the complete thermochemical equation for the vaporization of water at its normal boiling point: \[\ce{H2O(\ell,\, 373 \,K, \,1 \,atm) → H2O(g, \,373 \,K, \,1 \,atm)} \,\,\,ΔH = 40.7\, kJ\, mol^{-1} \nonumber\] The quantity 40.7 is known as the (often referred to as “heat of vaporization”) of liquid water. The following points should be kept in mind when writing thermochemical equations: Any thermodynamic quantity such as \(ΔH\) that is associated with a thermochemical equation refers to the number of moles of substances explicitly shown in the equation. Thus for the synthesis of water we can write \[\ce{2 H2(g) + O2(g)→ 2 H2O(l) }\,\,\,ΔH = -572 \,kJ \nonumber\] or \[\ce{H2(g) + 1/2 O2(g)→ H2O(l)} \,\,\, ΔH = -286 \,kJ \nonumber\] The thermochemical equations for reactions taking place in solution also specify the concentrations of the dissolved species. For example, the enthalpy of neutralization of a strong acid by a strong base is given by \[\ce{H^{+}(aq,\, 1M,\, 298\, K,\, 1\, atm) + OH^{-} (aq,\, 1M,\, 298 \,K,\, 1 \,atm) → H2O(\ell, 373\, K,\, 1 \,atm) } \,\,\, ΔH = -56.9\, kJ\, mol^{-1} \nonumber\] in which the abbreviation refers to the hydrated ions as they exist in aqueous solution. Since most thermochemical equations are written for the standard conditions of 298 K and 1 atm pressure, we can leave these quantities out if these conditions apply both before and after the reaction. If, under these same conditions, the substance is in its preferred (most stable) physical state, then the substance is said to be in its . Thus the standard state of water at 1 atm is the solid below 0°C, and the gas above 100°C. A thermochemical quantity such as \(ΔH\) that refers to reactants and products in their standard states is denoted by \(ΔH^°\). In the case of dissolved substances, the standard state of a solute is that in which the “effective concentration”, known as the activity, is unity. For non-ionic solutes the activity and molarity are usually about the same for concentrations up to about 1M, but for an ionic solute this approximation is generally valid only for solutions more dilute than 0.001-0.01M, depending on electric charge and size of the particular ion. The enthalpy change for a chemical reaction is the difference \[ΔH = H_{products} - H_{reactants}\] If the reaction in question represents the formation of one mole of the compound from its elements in their standard states, as in \[\ce{H2(g) + 1/2O2(g) -> H2O(l)} \;\;\; ΔH = -286\; kJ\] then we can arbitrarily set the enthalpy of the to zero and write \[\begin{align*} H_f^o &= \sum H_f^o (products) - \sum H_f^o (reactants) \\[4pt] &= -286\; kJ - 0 \\[4pt] &= -268\; kJ \,mol^{-1} \end{align*}\] which defines the of water at 298 K. The value ° = -268 kJ tells us that when hydrogen and oxygen, each at a pressure of 1 atm and at 298 K (25° C) react to form 1 mole of liquid water also at 25°C and 1 atm pressure, 268 kJ will have passed from the system (the reaction mixture) into the surroundings. The negative sign indicates that the reaction is exothermic: the enthalpy of the product is smaller than that of the reactants. The standard enthalpy of formation is a fundamental property of any compound. list ° values (usually alongside values of other thermodynamic properties) in their appendices. The standard enthalpy of formation of a compound is defined as the heat associated with the formation of one mole of the compound from its elements in their standard states. In general, the is given by the expression \[ΔH_f^o= \sum H_f^o (products) - \sum H_f^o (reactants) \label{2-1}\] in which the \(\sum H_f^o\) terms indicate the sums of the standard enthalpies of formations of all products and reactants. The above definition is one of the most important in chemistry because it allows The following examples illustrate some important aspects of the standard enthalpy of formation of substances. The thermochemical equation defining \(H_f^o\) is always written in terms of one mole of the substance in question> For example, the relavante thermodynamic equation for the heat of formation of ammonia (\(\ce{NH3}\)) is: \[\ce{ 1/2 N2(g) + 3/2 H2(g)→ NH3(g)}\,\,\, ΔH^o = -46.1\, kJ \,(\text{per mole of } \ce{NH3}) \nonumber\] The standard heat of formation of a compound is always taken in reference to the forms of the elements that are most stable at 25°C and 1 atm pressure. A number of elements, of which sulfur and carbon are common examples, can exist in more then one solid crystalline form (called ). For carbon dixoide (\(ce{CO2}\), one can construct two thermodynamics equations: \[\ce{C(graphite) + O2(g) → CO2(g)}\,\,\, ΔH^o ≡ H_f^o = -393.5\, kJ\, mol^{-1} \nonumber\] \[\ce{C(diamond) + O2(g) → CO2(g)}\,\,\, ΔH^{o} = -395.8\, kJ\, mol^{-1} \nonumber\] However for carbon, the graphite form is the more stable form and the correct thermodynamic equations for the heat of formation. The physical state of the product of the formation reaction must be indicated explicitly if it is not the most stable one at 25°C and 1 atm pressure: \[\ce{H2(g) + 1/2 O2(g) → H2O(aq)}\,\,\, ΔH^o ≡ H_f^o = -285.8\, kJ\, mol^{-1} \nonumber\] \[\ce{H2(g) + 1/2 O2(g) → H2O(g)} \,\,\, ΔH^o = -241.8\, kJ\, mol^{-1} \nonumber\] Notice that the difference between these two \(ΔH^o\) values is just the of water. Although the formation of most molecules from their elements is an exothermic process, the formation of some compounds is mildly endothermic: \[\ce{1/2 N2(g) + O2(g) -> NO2(g)}\,\,\, ΔH^o ≡ H_f^° = +33.2\, kJ\, mol^{-1} \nonumber\] A positive heat of formation is frequently associated with instability— the tendency of a molecule to decompose into its elements, although it is not in itself a sufficient cause. In many cases, however, the rate of this decomposition is essentially zero, so it is still possible for the substance to exist. In this connection, it is worth noting that all molecules will become unstable at higher temperatures. The thermochemical reactions that define the heats of formation of most compounds cannot actually take place. For example, the direct synthesis of methane from its elements \[\ce{C(graphite) + 2 H2(g) → CH4(g)} \nonumber\] cannot be observed directly owing to the large number of other possible reactions between these two elements. However, the standard enthalpy change for such a reaction be found indirectly from other data, as explained in the next section. The standard enthalpy of formation of gaseous atoms from the element is known as the . Heats of atomization are always positive, and are important in the calculation of bond energies. \[\ce{Fe(s) → Fe(g)}\,\, ΔH° = 417 \,kJ\, mol^{-1} \nonumber\] The standard enthalpy of formation of an ion dissolved in water is expressed on a separate scale in which that of \(\ce{H^{+}(aq)}\) is . The standard heat of formation of a such as \(\ce{Cl^{-}(aq)}\) (that is, formation of the ion from the element) cannot be measured because it is impossible to have a solution containing a single kind of ion. For this reason, ionic enthalpies are expressed on a separate scale on which \(\ce{H_f^o}\) of the hydrogen ion at (1 M effective concentration) is defined as zero: \[\ce{1/2 H2(g) → H^{+}(aq)}\,\,\, ΔH^o ≡ H_f^o = 0\, kJ\, mol^{-1} \nonumber\] Other ionic enthalpies (as they are commonly known) are found by combining appropriate thermochemical equations (as explained in Section 3 below). For example, ° of is found from the enthalpies of formation and solution of , yielding \[\ce{1/2 H2(g) + 1/2 Cl2(g) → HCl(aq)}\,\,\,\ ΔH^o ≡ H_f^o = -167\, kJ\, mol^{-1} \nonumber\] Because ° for is zero, this value establishes the standard enthalpy of the chloride ion. The standard enthalpy of formation of Ca , given by \[\ce{Ca(s) + Cl2(g) -> CaCl2(aq)} \nonumber\] could then be calculated by combining other measurable quantities such as the enthalpies of formation and solution of \(\ce{CaCl2(s)}\) to find ° for CaCl , from which ° of \(\ce{Ca^{2+}(aq)}\) is found by difference from that of \(\ce{Cl^{-}(aq)}\). Tables of the resulting ionic enthalpies are widely available ( ). Two or more chemical equations can be combined algebraically to give a new equation. Even before the science of thermodynamics developed in the late nineteenth century, it was observed that the heats associated with chemical reactions can be combined in the same way to yield the heat of another reaction. For example, the standard enthalpy changes for the oxidation of graphite and diamond can be combined to obtain ΔH° for the transformation between these two forms of solid carbon, a reaction that cannot be studied experimentally. \[\ce{C(graphite) + O2(g)→ CO2(g)}\,\,\, ΔH^o = -393.51\, kJ\, mol^{-1}\] \[\ce{C(diamond) + O2(g)→ CO2(g)} \,\,\, ΔH^o = -395.40\, kJ\, mol^{-1}\] Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding it to the first one) yields \[\ce{C(graphite) → C(diamond)}\,\,\, ΔH^{°} = 1.89\, kJ\, mol^{-1}\] This principle, known as is a direct consequence of the enthalpy being a state function. Hess’ law is one of the most powerful tools of chemistry, for it allows the change in the enthalpy (and in other thermodynamic functions) of huge numbers of chemical reactions to be predicted from a relatively small base of experimental data. Germain Henri Hess (1802-1850) was a Swiss-born professor of chemistry at St. Petersburg, Russia. He formulated his famous law, which he discovered empirically, in 1840. Very little appears to be known about his other work in chemistry. Because most substances cannot be prepared directly from their elements, heats of formation of compounds are seldom determined by direct measurement. Instead, Hess’ law is employed to calculate enthalpies of formation from more accessible data. The most important of these are the standard . Most elements and compounds combine with oxygen, and many of these oxidations are highly exothermic, making the measurement of their heats relatively easy. For example, by combining the heats of combustion of carbon, hydrogen, and methane, we can obtain the standard enthalpy of formation of methane, which as we noted above, cannot be determined directly (Example \(\Page {1}\)). Use the following heat of formation/combustion information to estimate the standard heat of formation of methane \(\ce{CH4}\). \[\begin{aligned} \ce{C(graphite)} + \ce{O2(g)} &→ \ce{CO2(g)} \quad &ΔH° = -393\, kJ\, mol^{-1} \label{P1-1} \\[4pt] \ce{H2(g)} + \ce{1/2 O2(g)} &→ \ce{H2O(g)} \quad &ΔH° = -286 kJ mol^{-1} \label{P1-2} \\[4pt] \ce{CH4(g)} + \ce{2O2(g)} &→ \ce{CO2(g)} + \ce{2H2O(g)} \quad &ΔH^o = -890\, kJ\, mol^{-1} \label{P1-3} \end{aligned}\] The standard heat of formation of methane is defined by the reaction \[\ce{C(graphite) + 2H2(g) → CH4(g)} \quad ΔH^o = ? \label{P1-4}\] Our task is thus to combine the top three equations in such a way that they add up to (4). Begin by noting that (3), the combustion of methane, is the only equation that contains the CH term, so we need to write it in reverse (not forgetting to reverse the sign of Δ °!) so that CH appears as the product. \[\ce{CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)} \quad ΔH^o = +890\, kJ\, mol^{-1} \label{P1-3Rev}\] Since H O does not appear in the net reaction (4), add two times (2) to cancel these out. Notice that this also cancels one of the oxygens in (3Rev): CO + → + O Δ = +890 kJ mol (P1-3Rev) → Δ = -484 kJ mol (P1-2) Finally, get rid of the remaining O and CO by adding (1); this also adds a needed C: Step 4 So our creative cancelling has eliminated all except the substances that appear in (4). Just add up the enthalpy changes and we are done: \[\ce{C(graphite) + 2H2(g) → CH4(g)} \] (The tablulated value is -74.6 kJ mol ) How are enthalpy changes determined experimentally? First, you must understand that Moreover, is equal to the standard enthalpy change only when the reactants and products are both at the same temperature, normally 25°C. The measurement of is generally known as . The most common types of calorimeters contain a known quantity of water which absorbs the heat released by the reaction. Because the specific heat capacity of water (4.184 J g K ) is known to high precision, a measurement of its temperature rise due to the reaction enables one to calculate the quantity of heat released. In all but the very simplest calorimeters, some of the heat released by the reaction is absorbed by the components of the calorimeter itself. It is therefore necessary to "calibrate" the calorimeter by measuring the temperature change that results from the introduction of a known quantity of heat. The resulting , expressed in J K , can be regarded as the “heat capacity of the calorimeter”. The known source of heat is usually produced by passing a known quantity of electric current through a resistor within the calorimeter, but it can be measured by other means as described in the following problem example. In determining the heat capacity of a calorimeter, a student mixes 100.0 g of water at 57.0 °C with 100.0 g of water, already in the calorimeter, at 24.2°C. (yhe specific heat of water is 4.184 J g K ). After mixing and thermal equilibration with the calorimeter, the temperature of the water stabilizes at 38.7°C. Calculate the heat capacity of the calorimeter in J/K. The hot water loses heat, the cold water gains heat, and the calorimter itself gains heat, so this is essentially a thermal balance problem. Conservation of energy requires that \[q_{hot} + q_{cold} + q_{cal} = 0 \nonumber\] We can evaluate the first two terms from the observed temperature changes: \[q_{hot} = (100\; g) (38.7\;K - 57.0\;K) (4.184\; J \;g^{-1}K^{-1}) = -7,657\; J \nonumber\] \[q_{cold} = (100\; g) (38.7 \;K - 24.2 \;K) (4.184\; J\; g^{-1} K^{-1}) = 6,067\; J \nonumber\] So \[q_{cal} = 7,657 \;J - ,6067\; J = 1,590;\ J \nonumber\] The \[\dfrac{1590\; J}{38.7\; K - 24.2\; K} = 110\; J\; K^{-1} \nonumber\] : Strictly speaking, there is a fourth thermal balance term that must be considered in a highly accurate calculation: the water in the calorimeter expands as it is heated, performing work on the atmosphere. For reactions that can be initiated by combining two solutions, the temperature rise of the solution itself can provide an approximate value of the reaction enthalpy if we assume that the heat capacity of the solution is close to that of the pure water — which will be nearly true if the solutions are dilute. For example, a very simple calorimetric determination of the standard enthalpy of the reaction \[\ce{H^{+}(aq) + OH^{-}(aq) → H2O(\ell)} \nonumber\] could be carried out by combining equal volumes of 0.1M solutions of HCl and of NaOH initially at 25°C. Since this reaction is exothermic, a quantity of heat will be released into the solution. From the temperature rise and the specific heat of water, we obtain the number of joules of heat released into each gram of the solution, and can then be calculated from the mass of the solution. Since the entire process is carried out at constant pressure, we have Δ ° = . For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonly placed within a larger insulated container of water. During the reaction, heat passes between the inner and outer containers until their temperatures become identical. Again, the temperature change of the water is observed, but in this case we need to know the value of the calorimeter constant described above. \(\Delta H_{combustion}\), since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, and this accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following: Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as . The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture. Another consequence of the constant-volume condition is that the heat released corresponds to  , and thus to the internal energy change Δ rather than to Δ . The enthalpy change is calculated according to the formula \[ΔH = q_v + Δn_gRT\] A sample of biphenyl (C H ) weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C H COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol (that is, Δ = -3226 kJ mol .) Use this information to determine the standard enthalpy of combustion of biphenyl. Begin by working out the calorimeter constant: \[\dfrac{(0.825 g}{122.1 \;g/mol} = 0.00676\; mol\nonumber\] \[(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ\nonumber\] \[\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K\nonumber\] Now determine \(ΔU_{combustion}\) of the biphenyl ("BP"): \[\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol\nonumber\] \[(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ\nonumber\] \[\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol\nonumber\] \[ΔU_{combustion} (BP) = -6,293\; kJ/mol\nonumber\] (This is the heat change at constant volume, \(q_v\); the negative sign indicates that the reaction is exothermic, as all combustion reactions are.) From the reaction equation \[(C_6H_5)_{2(s)} + \frac{19}{2} O_{2(g)} \rightarrow 12 CO_{2(g)} + 5 H_2O_{(l)} \nonumber\] we have \[Δn_g = 12 - \frac{19}{2} = \frac{-5}{2} \nonumber\] Thus the volume of the system when the reaction takes place. Converting to Δ , we can write the following equation. Additionally, recall that at constant volume, \(ΔU = q_V\). \[\begin{align*} ΔH &= q_V + Δn_gRT \\[4pt] &= ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K) \\[4pt] &=(-6,293 \; kJ/mol)-(6,194\; J/mol)=(-6,293-6.2)\;kJ/mol= -6299 \; kJ/mol \end{align*}\] A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in \(ΔH\) compared to \(ΔU\) reflects the work that the surroundings do on the system as the volume of gases decreases according to the reaction equation. The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the . The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents (\(q_{calorimeter}\)) through the combustion reaction. \[q_{rxn} = -q_{calorimeter} \label{2A}\] where \[q_{calorimeter} = q_{bomb} + q_{water} \label{3A}\] If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula: \[q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \label{4A}\] Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate \(q_{rxn}\) from \(q_{calorimeter}\) calculated by Equation 2. The heat capacity of the calorimeter can be determined by conducting an experiment. 1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.64°C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, \(C_{12}H_{22}O_{11}\), in kJ per mole of \(C_{12}H_{22}O_{11}\). Given: Using Equation 4 to calculate \(q_{calorimeter}\): \[q_{calorimeter} = (4.90\; kJ/°C) \times (27.64 - 23.42)°C = (4.90 \times 4.22) \;kJ = 20.7\; kJ \] Plug into Equation 2: \(q_{rxn} = -q_{calorimeter} = -20.7 \; kJ \;\) But the question asks for kJ/mol \(C_{12}H_{22}O_{11}\), so this needs to be converted: \(q_{rxn} = \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} = \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}}\) Per Mole \(C_{12}H_{22}O_{11}\): \(q_{rxn} = \dfrac{-18.0 \; kJ}{g \; C_{12}H_{22}O_{11}} \times \dfrac{342.3 \; g \; C_{12}H_{22}O_{11}}{1 \; mol \; C_{12}H_{22}O_{11}} = \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; C_{12}H_{22}O_{11}}\) Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories. Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms: The is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water. The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.04%3A_The_Ideal_Gas_Law

In this module, the relationship between pressure, temperature, volume, and amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas. Any set of relationships between a single quantity (such as \(V\)) and several other variables (\(P\), \(T\), and \(n\)) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions were derived previously: \[V \propto \dfrac{1}{P} \;\; \text{@ constant n and T} \nonumber \] \[V \propto T \;\; \text{@ constant n and P} \nonumber \] \[V \propto n \;\; \text{@ constant T and P} \nonumber \] Combining these three expressions gives \[V \propto \dfrac{nT}{P} \label{10.4.1} \] which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as \[V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \label{10.4.2} \] By convention, the proportionality constant in Equation \(\ref{10.4.1}\) is called the gas constant, which is represented by the letter \(R\). Inserting R into Equation \(\ref{10.4.2}\) gives \[ V = \dfrac{RnT}{P} = \dfrac{nRT}{P} \label{10.4.3} \] Clearing the fractions by multiplying both sides of Equation \(\ref{10.4.4}\) by \(P\) gives \[PV = nRT \label{10.4.4} \] This equation is known as the . An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed. Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures. Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then \[R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \label{10.4.5} \] Because the product PV has the units of energy, R can also have units of J/(K•mol): \[R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\label{10.4.6} \] Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and \(\rm1\; bar = 100 \;kPa = 10^5\;Pa\) pressure, referred to as standard temperature and pressure ( ). \[\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa \nonumber \] Please note that STP was defined differently in the past. The old definition was based on a standard pressure of 1 atm. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation \(\ref{10.4.4}\): \[V=\dfrac{nRT}{P}\label{10.4.7} \] Thus the volume of 1 mol of an ideal gas is and , approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm​ are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm​. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed. The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables ( , , , and ). It also allows us to predict the of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters ( , , , and ) are specified for an Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample. The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft ), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? volume, temperature, and pressure amount of gas We are given values for , , and and asked to calculate . If we solve the ideal gas law (Equation \(\ref{10.4.4}\)) for \(n\), we obtain \[\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber \] and are given in units that are not compatible with the units of the gas constant [ = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: \[T=273+30=303{\rm K}\nonumber \] Substituting these values into the expression we derived for , we obtain \[\begin{align*} n &=\dfrac{PV}{RT} \\[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \\[4pt] &=1.23\times10^3\;mol \end{align*} \nonumber \] Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO . What is the pressure of the gas at 25°C? 1.5 atm In Example \(\Page {1}\), we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of in any of the specified conditions on any of the other parameters, as shown in Example \(\Page {5}\). When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is: \[\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \\ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array} \nonumber \] Both equations can be rearranged to give: \[R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f} \nonumber \] The two equations are equal to each other since each is equal to the same constant \(R\). Therefore, we have: \[\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8} \] The equation is called the . The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties. Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example \(\Page {1}\)? temperature, pressure, amount, and volume in August; temperature in January volume in January To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions: Both \(n\) and \(P\) are the same in both cases​ (\(n_i=n_f,P_i=P_f\)). Therefore, Equation \ref{10.4.8} can be simplified to: \[\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber \] This is the relationship first noted by Charles. ​Solving the equation for \(V_f\), we get: \[\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \\[4pt] &=2.70\times10^4\;L \end{align*} \nonumber \] It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change. At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon? 0.52 L Example \(\Page {1}\) illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle ( = constant) and the relationship between volume and amount observed by Avogadro ( / = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example \(\Page {1}\) can be applied in such case, as we demonstrate in Example \(\Page {2}\) (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion). Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise \(\Page {1}\) (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)? initial volume, amount, temperature, and pressure; final temperature final pressure Follow the strategy outlined in Example \(\Page {2}\). Prepare a table to determine which parameters change and which are held constant: Both \(V\) and \(n\) are the same in both cases​ (\(V_i=V_f,n_i=n_f\)). Therefore, Equation can be simplified to: \[ \dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \nonumber \] By solving the equation for \(P_f\), we get: \[\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \\[4pt] &=5.1\;atm \end{align*} \nonumber \] This pressure is more than enough to rupture a thin sheet metal container and cause an explosion! Suppose that a fire extinguisher, filled with CO to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher? 23.4 atm \(\Page {1}\) \(\Page {2}\) We saw in Example \(\Page {1}\) that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 10 mol of H gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude? initial pressure, temperature, amount, and volume; final pressure and temperature final volume Follow the strategy outlined in Example \(\Page {3}\). Begin by setting up a table of the two sets of conditions: By eliminating the constant property (\(n\)) of the gas, Equation \(\ref{10.4.8}\) is simplified to: \[\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f} \nonumber \] By solving the equation for \(V_f\), we get: \[\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \\[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \\[4pt] &=5.96\times10^4\;L \end{align*} \nonumber \] Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to the volume of the gas, while decreasing the temperature tends to the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both and 1/ , the variable that changes the most will have the greatest effect on . In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation. We could also have solved this problem by solving the ideal gas law for and then substituting the relevant parameters for an altitude of 23,000 ft: Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.) 4.07 × 10 The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain \[\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9} \] The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (\(m\), in grams) divided by its molar mass (\(M\), in grams per mole): \[n=\dfrac{m}{M}\label{10.4.10} \] Substituting this expression for \(n\) into Equation \(\ref{10.4.9}\) gives \[\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11} \] Because \(m/V\) is the density \(d\) of a substance, we can replace \(m/V\) by \(d\) and rearrange to give \[\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12} \] The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL). Calculate the density of butane at 25°C and a pressure of 750 mmHg. compound, temperature, and pressure density The molar mass of butane (C H ) is Using 0.08206 (L•atm)/(K•mol) for means that we need to convert the temperature from degrees Celsius to kelvins ( = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres: \[P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber \] Substituting these values into Equation \(\ref{10.4.12}\) gives \[\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber \] Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics. radon, 9.23 g/L; N , 1.17 g/L A common use of Equation \(\ref{10.4.12}\) is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example \(\Page {6}\). The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound. pressure, temperature, mass, and volume molar mass and chemical formula Solving Equation \(\ref{10.4.12}\) for the molar mass gives \[M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \nonumber \] Density is the mass of the gas divided by its volume: \[\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.157\rm L}=1.84 \rm g/L\nonumber \] We must convert the other quantities to the appropriate units before inserting them into the equation: \[T=18+273=291 K\nonumber \] \[P=727 \, mmHg \times \dfrac{1\rm atm} {760\rm mmHg} =0.957\rm atm \nonumber \] The molar mass of the unknown gas is thus \[M=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\nonumber \] \[M({\rm NO})=14 + 16=30 \rm\; g/mol\nonumber \] \[M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\nonumber \] \[M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\nonumber \] The most likely choice is NO which is in agreement with the data. The red-brown color of smog also results from the presence of NO gas. You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it. 44 g/mol; \(CO_2\) The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known. : \(PV = nRT\), where \(R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}\) : \(\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\) \(\rho=\dfrac{MP}{RT}\) The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the , = . The proportionality constant, , is called the and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an , a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the . All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity ( , , , or ) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of , , , and ) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured.

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1. Equation shows how [ ] is determined using a two-point fixed-time integral method in which the concentration of for the pseudo-first-order reaction \[A+R \longrightarrow P \nonumber\] is measured at times and . Derive a similar equation for the case where the product is monitored under pseudo-first order conditions. 2. The concentration of phenylacetate is determined from the kinetics of its pseudo-first order hydrolysis reaction in an ethylamine buffer. When a standard solution of 0.55 mM phenylacetate is analyzed, the concentration of phenylacetate after 60 s is 0.17 mM. When a sample is analyzed the concentration of phenylacetate that remains after 60 s is 0.23 mM. What is the concentration of phenylacetate in the sample? 3. In the presence of acid, iodide is oxidized by hydrogen peroxide \[2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{I}_{2}(a q) \nonumber\] When I and H O are present in excess, we can use the reaction’s kinetics of the reaction, which is pseudo-first order in H O , to determine the concentration of H O by following the production of I with time. In one analysis the solution’s absorbance at 348 nm was measured after 240 s. Analysis of a set of standard gives the results shown below. What is the concentration of H O in a sample if its absorbance is 0.669 after 240 s? 4. The concentration of chromic acid is determined by reducing it under conditions that are pseudo-first order in analyte. One approach is to monitor the reaction absorbance at a wavelength of 355 nm. A standard of \(5.1 \times 10^{-4}\) M chromic acid yields absorbances of 0.855 and 0.709 at 100 s and 300 s after the reaction’s initiation. When a sample is analyzed under identical conditions, the absorbances are 0.883 and 0.706. What is the concentration of chromic acid in the sample? 5. Malmstadt and Pardue developed a variable time method for the determination of glucose based on its oxidation by the enzyme glucose oxidase [Malmstadt, H. V.; Pardue, H. L. , 1040–1047]. To monitor the reaction’s progress, iodide is added to the samples and standards. The H O produced by the oxidation of glucose reacts with I , forming I as a product. The time required to produce a fixed amount of I is determined spectrophotometrically. The following data was reported for a set of calibration standards 149.6 66.0 34.0 22.6 To verify the method a standard solution of 20.0 ppm glucose was analyzed in the same way as the standards, requiring 34.6 s to produce the same extent of reaction. Determine the concentration of glucose in the standard and the percent error for the analysis. 6. Deming and Pardue studied the kinetics for the hydrolysis of -nitrophenyl phosphate by the enzyme alkaline phosphatase [Deming, S. N.; Pardue, H. L. , , 192–200]. The reaction’s progress was monitored by measuring the absorbance of -nitrophenol, which is one of the reaction’s products. A plot of the reaction’s rate (with units of μmol mL–1 sec–1) versus the volume, , in milliliters of a serum calibration standard that contained the enzyme, yielded a straight line with the following equation. \[\text { rate } = 2.7 \times 10^{-7} \mu \text{mol } \mathrm{mL}^{-1} \text{ s}^{-1}+\left(3.485 \times 10^{-5} \mu \text{mol } \mathrm{mL}^{-2} \text{ s}^{-1}\right) V \nonumber\] A 10.00-mL sample of serum is analyzed, yielding a rate of \(6.84 \times 10^{-5}\) μmol mL sec . How much more dilute is the enzyme in the serum sample than in the serum calibration standard? 7. The following data were collected for a reaction known to be pseudo-first order in analyte, , during the time in which the reaction is monitored. 1.36 1.24 1.12 1.02 What is the rate constant and the initial concentration of analyte in the sample? 8. The enzyme acetylcholinesterase catalyzes the decomposition of acetylcholine to choline and acetic acid. Under a given set of conditions the enzyme has a of \(9 \times 10^{-5}\) M and a of \(1.4 \times 10^4\) s . What is the concentration of acetylcholine in a sample if the reaction’s rate is 12.33 μM s in the presence of \(6.61 \times 10^{-7}\) M enzyme? You may assume the concentration of acetylcholine is significantly smaller than . 9. The enzyme fumarase catalyzes the stereospecific addition of water to fumarate to form l-malate. A standard 0.150 μM solution of fumarase has a rate of reaction of 2.00 μM min under conditions in which the substrate’s concentration is significantly greater than . The rate of reaction for a sample under identical condition is 1.15 mM min . What is the concentration of fumarase in the sample? 10. The enzyme urease catalyzes the hydrolysis of urea. The rate of this reaction is determined for a series of solutions in which the concentration of urea is changed while maintaining a fixed urease concentration of 5.0 mM. The following data are obtained. 0.100 6.25 Determine the values of , , and for urease. 11. To study the effect of an enzyme inhibitor and are measured for several concentrations of inhibitor. As the concentration of the inhibitor increases remains essentially constant, but the value of increases. Which mechanism for enzyme inhibition is in effect? 12. In the case of competitive inhibition, the equilibrium between the enzyme, , the inhibitor, , and the enzyme–inhibitor complex, , is described by the equilibrium constant . Show that for competitive inhibition the equation for the rate of reaction is \[\frac{d[P]}{d t}=\frac{V_{\max }[S]}{K_{m}\left\{1+\left([I] / K_{E l}\right)\right\}+[S]} \nonumber\] where is the formation constant for the complex \[E+I \rightleftharpoons E I \nonumber\] You may assume that << . 13. Analytes and react with a common reagent with first-order kinetics. If 99.9% of must react before 0.1% of has reacted, what is the minimum acceptable ratio for their respective rate constants? 14. A mixture of two analytes, and , is analyzed simultaneously by monitoring their combined concentration, = [ ] + [ ], as a function of time when they react with a common reagent. Both and are known to follow first-order kinetics with the reagent, and is known to react faster than . Given the data in the following table, determine the initial concentrations of and , and the first-order rate constants, and . 15. provides a list of several isotopes used as tracers. The half-lives for these isotopes also are listed. What is the rate constant for the radioactive decay of each isotope? 16. Co is a long-lived isotope ( = 5.3 yr) frequently used as a radiotracer. The activity in a 5.00-mL sample of a solution of Co is \(2.1 \times 10^7\) disintegrations/sec. What is the molar concentration of Co in the sample? 17. The concentration of Ni in a new alloy is determined by a neutron activation analysis. A 0.500-g sample of the alloy and a 1.000-g sample of a standard alloy that is 5.93% w/w Ni are irradiated with neutrons in a nuclear reactor. When irradiation is complete, the sample and the standard are allowed to cool and their gamma ray activities measured. Given that the activity is 1020 cpm for the sample and 3540 cpm for the standard, determine the %w/w Ni in the alloy. 18. The vitamin B content of a multivitamin tablet is determined by the following procedure. A sample of 10 tablets is dissolved in water and diluted to volume in a 100-mL volumetric flask. A 50.00-mL portion is removed and 0.500 mg of radioactive vitamin B having an activity of 572 cpm is added as a tracer. The sample and tracer are homogenized and the vitamin B isolated and purified, producing 18.6 mg with an activity of 361 cpm. Calculate the milligrams of vitamin B in a multivitamin tablet. 19. The oldest sample that can be dated by C is approximately 30 000 yr. What percentage of the C remains after this time span? 20. Potassium–argon dating is based on the nuclear decay of K to Ar ( = \(1.3 \times 10^9\) yr). If no Ar is originally present in the rock, and if Ar cannot escape to the atmosphere, then the relative amounts of K and Ar can be used to determine the age of the rock. When a 100.0-mg rock sample is analyzed it is found to contain \(4.63 \times 10^{-6}\) mol of K and \(2.09 \times 10^{-6}\) mol Ar. How old is the rock sample? 21. The steady state activity for C in a sample is 13 cpm per gram of carbon. If counting is limited to 1 hr, what mass of carbon is needed to give a percent relative standard deviation of 1% for the sample’s activity? How long must we monitor the radioactive decay from a 0.50-g sample of carbon to give a percent relative standard deviation of 1.0% for the activity? 22. To improve the sensitivity of a FIA analysis you might do any of the following: inject a larger volume of sample, increase the flow rate, decrease the length and the diameter of the manifold’s tubing, or merge separate channels before injecting the sample. For each action, explain why it leads to an improvement in sensitivity. 23. The figure below shows a fiagram for a solution of 50.0-ppm \(\text{PO}_4^{3-}\) using the method in Representative Method 13.4.1. Determine values for , , , ′, \(\Delta t\), and ′. What is the sensitivity of this FIA method, assuming a linear relationship between absorbance and concentration? How many samples can be analyzed per hour? 24. A sensitive method for the flow injection analysis of Cu is based on its ability to catalyze the oxidation of di-2-pyridyl ketone hydrazone (DPKH) [Lazaro, F.; Luque de Castro, M. D.; Valcárcel, M. Analyst, 1984, 109, 333–337]. The product of the reaction is fluorescent and is used to generate a signal when using a fluorimeter as a detector. The yield of the reaction is at a maximum when the solution is made basic with NaOH. The fluorescence, however, is greatest in the presence of HCl. Sketch an appropriate FIA manifold for this analysis. 25. The concentration of chloride in seawater is determined by a flow injection analysis. The analysis of a set of calibration standards gives the following results. 5.00 A 1.00-mL sample of seawater is placed in a 500-mL volumetric flask and diluted to volume with distilled water. When injected into the flow injection analyzer an absorbance of 0.317 is measured. What is the concentration of Cl in the sample? 26. Ramsing and co-workers developed an FIA method for acid–base titrations using a carrier stream that is \(2.0 \times 10^{-3}\) M NaOH and that contains the acid–base indicator bromothymol blue [Ramsing, A. U.; Ruzicka, J.; Hansen, E. H. , , 1–17]. Standard solutions of HCl were injected, and the following values of \(\Delta t\) were measured from the resulting fiagrams. 7.71 8.13 9.27 10.45 A sample with an unknown concentration of HCl is analyzed five times, giving values of 7.43, 7.28, 7.41, 7.37, and 7.33 s for \(\Delta t\). Determine the concentration of HCl in the sample. 27. Milardovíc and colleagues used a flow injection analysis method with an amperometric biosensor to determine the concentration of glucose in blood [MilardoviĆ, S.; Kruhak, I.; Ivekovic, D.; Rumenjak, V.; Tkalčec, M.; Grabaric, B. S. , , 91–96]. Given that a blood sample that is 6.93 mM in glucose has a signal of 7.13 nA, what is the concentration of glucose in a sample of blood if its signal is 11.50 nA? 28. Fernández-Abedul and Costa-García developed an FIA method to determine cocaine in samples using an amperometric detector [Fernández-Abedul, M; Costa-García, A. , , 67–71]. The following signals (arbitrary units) were collected for 12 replicate injections of a \(6.2 \times 10^{-6}\) M sample of cocaine, C H NO 24.5 24.1 24.1 23.8 23.9 25.1 23.9 (a) What is the relative standard deviation for this sample? (b) The following calibration data are available 0.18 0.8 0.36 2.1 0.60 2.4 0.81 3.2 1.0 4.5 2.0 8.1 4.0 14.4 6.0 8.0 In a typical analysis a 10.0-mg sample is dissolved in water and diluted to volume in a 25-mL volumetric flask. A 125-mL aliquot is transferred to a 25-mL volumetric flask and diluted to volume with a pH 9 buffer. When injected into the flow injection apparatus a signal of 21.4 (arb. units) is obtained. What is the %w/w cocaine in the sample? 29. Holman, Christian, and Ruzicka described an FIA method to determine the concentration of H SO in nonaqueous solvents [Holman, D. A.; Christian, G. D.; Ruzicka, J. , , 1763–1765]. Agarose beads (22–45 mm diameter) with a bonded acid–base indicator are soaked in NaOH and immobilized in the detector’s flow cell. Samples of H SO in -butanol are injected into the carrier stream. As a sample passes through the flow cell, an acid–base reaction takes place between H SO and NaOH. The endpoint of the neutralization reaction is signaled by a change in the bound indicator’s color and is detected spectrophotometrically. The elution volume needed to reach the titration’s endpoint is inversely proportional to the concentration of H SO ; thus, a plot of endpoint volume versus [H SO ] is linear. The following data is typical of that obtained using a set of external standards. 0.358 0.266 0.436 0.560 0.752 What is the concentration of H SO in a sample if its endpoint volume is 0.157 mL?

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It is physically impossible to measure the potential difference between a piece of metal and the solution in which it is immersed. We can, however, measure the between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction , and we are measuring the sum of the two . This arrangement is called a . A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through. If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions: \[Zn_{(s)} + Cu^{2+} \rightarrow Zn^{2+} + Cu_{(s)}\] but this time, the oxidation and reduction steps (half reactions) take place in separate locations: The reaction can be started and stopped by connecting or disconnecting the two electrodes. If we place a variable resistance in the circuit, we can even control the rate of the net cell reaction by simply turning a knob. By connecting a battery or other source of current to the two electrodes, we can force the reaction to proceed in its non-spontaneous, or reverse direction. By placing an ammeter in the external circuit, we can measure the amount of electric charge that passes through the electrodes, and thus the number of moles of reactants that get transformed into products in the cell reaction. Electric charge is measured in coulombs. The amount of charge carried by one mole of electrons is known as the , which we denote by . Careful experiments have determined that 1 = 96467 C. For most purposes, you can simply use 96,500 Coulombs as the value of the faraday. When we measure electric current, we are measuring the rate at which electric charge is transported through the circuit. A current of one ampere corresponds to the flow of one coulomb per second. For the cell to operate, not only must there be an external electrical circuit between the two electrodes, but the two electrolytes (the solutions) must be in contact. The need for this can be understood by considering what would happen if the two solutions were physically separated. Positive charge (in the form of Zn ) is added to the electrolyte in the left compartment, and removed (as Cu ) from the right side, causing the solution in contact with the zinc to acquire a net positive charge, while a net negative charge would build up in the solution on the copper side of the cell. These violations of would make it more difficult (require more work) to introduce additional Zn ions into the positively-charged electrolyte or for electrons to flow into right compartment where they are needed to reduce the Cu ions, thus effectively stopping the reaction after only a chemically insignificant amount has taken place. In order to sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells. This means that we must provide a path for ions to move directly from one cell to the other. This ionic transport involves not only the electroactive species Cu and Zn , but also the , which in this example are nitrate, NO . Thus an excess of Cu in the left compartment could be alleviated by the drift of these ions into the right side, or equally well by diffusion of nitrate ions to the left. More detailed studies reveal that both processes occur, and that the relative amounts of charge carried through the solution by positive and negative ions depends on their relative , which express the velocity with which the ions are able to make their way through the solution. Since negative ions tend to be larger than positive ions, the latter tend to have higher mobilities and carry the larger fraction of charge. In the simplest cells, the barrier between the two solutions can be a porous membrane, but for precise measurements, a more complicated arrangement, known as a , is used. The salt bridge consists of an intermediate compartment filled with a concentrated solution of KCl and fitted with porous barriers at each end. The purpose of the salt bridge is to minimize the natural potential difference, known as the , that develops (as mentioned in the previous section) when any two phases (such as the two solutions) are in contact. This potential difference would combine with the two half-cell potentials so as introduce a degree of uncertainty into any measurement of the cell potential. With the salt bridge, we have two liquid junction potentials instead of one, but they tend to cancel each other out. In order to make it easier to describe a given electrochemical cell, a special symbolic notation has been adopted. In this notation the cell we described above would be Zn | Zn || Cu | Cu There are several other conventions relating to cell notation and nomenclature that you are expected to know: An refers to the net oxidation or reduction process that takes place at an electrode. This reaction may take place in a single electron-transfer step, or as a succession of two or more steps. The substances that receive and lose electrons are called the electroactive species. This process takes place within the very thin interfacial region at the electrode surface, and involves quantum-mechanical tunneling of electrons between the electrode and the electroactive species. The work required to displace the H O molecules in the hydration spheres of the ions constitutes part of the of the process. In the example of the Zn/Cu cell we have been using, the electrode reaction involves a metal and its hydrated cation; we call such electrodes metal-metal ion electrodes. There are a number of other kinds of electrodes which are widely encountered in electrochemistry and analytical chemistry. Many electrode reactions involve only ionic species, such as \(Fe^{2+}\) and \(Fe^{3+}\). If neither of the electroactive species is a metal, some other metal must serve as a conduit for the supply or removal of electrons from the system. In order to avoid complications that would arise from electrode reactions involving this metal, a relatively inert substance such as platinum is commonly used. Such a half cell would be represented as Pt(s) | Fe , Fe || ... and the half-cell reaction would be \[Fe^{2+}(aq) \rightarrow Fe^{3+} (aq) + e^-\] The reaction occurs at the surface of the electrode (Fig 4 above). The electroactive ion diffuses to the electrode surface and adsorbs (attaches) to it by and . In doing so, the waters of hydration that are normally attached to any ionic species must be displaced. This process is always endothermic, sometimes to such an extent that only a small fraction of the ions be able to contact the surface closely enough to undergo electron transfer, and the reaction will be slow. The actual electron-transfer occurs by quantum-mechanical . Some electrode reactions involve a gaseous species such as \(H_2\), \(O_2\), or \(Cl_2\). Such reactions must also be carried out on the surface of an electrochemically inert conductor such as platinum. A typical reaction of considerable commercial importance is \[Cl^-(aq) \rightarrow ½ Cl_2(g) + e^- \] Similar reactions involving the oxidation of \(Br_2\) or \(I_2\) also take place at platinum surfaces. A typical electrode of this kind consists of a silver wire covered with a thin coating of silver chloride, which is insoluble in water. The electrode reaction consists in the oxidation and reduction of the silver: \[AgCl(s) + e^– → Ag(s) + Cl^–(aq)\] The half cell would be represented as \[ ... || Cl^– (aq) | AgCl (s) | Ag (s)\] Although the usefulness of such an electrode may not be immediately apparent, this kind of electrode finds very wide application in electrochemical measurements, as we shall see later. In most electrochemical experiments our interest is concentrated on only one of the electrode reactions. Since all measurements must be on a complete cell involving two electrode systems, it is common practice to employ a as the other half of the cell. The major requirements of a reference electrode are that it be easy to prepare and maintain, and that its potential be stable. The last requirement essentially means that the concentration of any ionic species involved in the electrode reaction must be held at a fixed value. The most common way of accomplishing this is to use an electrode reaction involving a saturated solution of an insoluble salt of the ion. One such system, the silver-silver chloride electrode has already been mentioned: \[Ag | AgCl(s) | Cl^–(aq) || ...\] \[Ag(s) + Cl^–(aq) →AgCl(s) + e^–\] This electrode usually takes the form of a piece of silver wire coated with AgCl. The coating is done by making the silver the anode in an electrolytic cell containing HCl; the Ag ions combine with Cl ions as fast as they are formed at the silver surface. The other common reference electrode is the ; calomel is the common name for mercury(I) chloride. Such a half cell would be represented as \[Hg | Hg^{2+}(aq) | KCl || ...\] and the half-cell reaction would be \[Hg(l) + Cl^– → ½ HgCl2(s) + e^–\] The potentials of both of these electrodes have been very accurately determined against the hydrogen electrode. The latter is seldom used in routine electrochemical measurements because it is more difficult to prepare; the platinum surface has to be specially treated by preliminary electrolysis. Also, there is need for a supply of hydrogen gas which makes it somewhat cumbersome and hazardous. Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. Zn + Cu → Zn + Cu we write Zn | Zn || Cu | Cu in which the single vertical bars represent . The double bar denotes a which in laboratory cells consists of a salt bridge or in ion-permeable barrier. If the net cell reaction were written in reverse, the cell notation would become Cu | Cu || Zn | Zn the process is always shown on the .

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Make sure you thoroughly understand the following essential ideas which have been presented above. In this section we deal mainly with a very small but important class of solids that are commonly regarded as composed of ions. We will see how the relative sizes of the ions determine the energetics of such compounds. And finally, we will point out that not all solids that are formally derived from ions can really be considered "ionic" at all. The idealized ionic solid consists of two interpenetrating lattices of oppositely-charged point charges that are held in place by a balance of coulombic forces. But because real ions occupy space, no such "perfect" ionic solid exists in nature. Nevertheless, this model serves as a useful starting point for understanding the structure and properties of a small group of compounds between elements having large differences in electronegativity. Chemists usually apply the term "ionic solid" to binary compounds of the metallic elements of Groups 1-2 with one of the halogen elements or oxygen. As can be seen from the diagram, the differences in electronegativity between the elements of Groups 1-2 and those of Group 17 (as well as oxygen in Group 16) are sufficiently great that the binding in these solids is usually dominated by Coulombic forces and the crystals can be regarded as built up by aggregation of oppositely-charged ions. Structurally, each ion in sodium chloride is surrounded and held in tension by six neighboring ions of opposite charge. The resulting crystal lattice is of a type known as , meaning that the lattice points are equally spaced in all three dimensions and all cell angles are 90°. In Figure \(\Page {2}\), we have drawn two imaginary octahedra centered on ions of different kinds and extending partially into regions outside of the diagram. (We could equally well have drawn them at any of the lattice points, but show only two in order to reduce clutter.) Our object in doing this is to show that each ion is surrounded by six other ions of opposite charge; this is known as . Another way of stating this is that each ion resides in an within the cubic lattice. How can one sodium ion surrounded by six chloride ions (or ) be consistent with the simplest formula NaCl? The answer is that each of those six chloride ions also sits at the center of its own octahedron defined by another six sodium ions. You might think that this corresponds to Na Cl , but note that the central sodium ion shown in the diagram can claim only a one-sixth share of each of its chloride ion neighbors, so the formula NaCl is not just the simplest formula, but correctly reflects the 1:1 stoichiometry of the compound. But of course, as in all ionic structures, there are no distinguishable "molecular" units that correspond to the NaCl simplest formula. Bear in mind that large amount of empty space in diagrams depicting a crystal lattice structure can be misleading, and that the ions are really in direct contact with each other to the extent that this is geometrically possible. Sodium chloride, like virtually all salts, is a more energetically favored configuration of sodium and chlorine than are these elements themselves; in other words, the reaction \[Na_{(s)} + ½Cl_{2(g)} \rightarrow NaCl_{(s)}\] is accompanied by a release of energy in the form of heat. How much heat, and why? To help us understand, we can imagine the formation of one mole of sodium chloride from its elements proceeding in these hypothetical steps in which we show the energies explicitly: Step 1: of sodium (breaking one mole of metallic sodium into isolated sodium atoms) \[\ce{ Na(s) + 108 kJ → Na(g)} \label{Step1}\] Same thing with chlorine. This requires more energy because it involves breaking a covalent bond. \[\ce{ ½Cl2(g) + 127\, kJ → Cl(g)} \label{Step2}\] Step 3: We strip an electron from one mole of sodium atoms (this costs a lot of energy!) \[\ce{ Na(g) + 496\, kJ → Na^{+}(g) + e^{–}} \label{Step3}\] Step 4: Feeding these electrons to the chlorine atoms gives most of this energy back. \[\ce{ Cl(g) + e^{–} → Cl^{–}(g) + 348\, kJ}\label{Step4}\] Step 5: Finally, we bring one mole of the ions together to make the crystal lattice — with a huge release of energy. \[\ce{ Na^{+}(g) + Cl^{–}(g) → NaCl(s) + 787\, kJ} \label{Step5}\] If we add all of these equations together, we get \[\ce{Na(s) + 1/2Cl2(g) → NaCl(s)} + 404\; kJ\] In other words, the formation of solid sodium chloride from its elements is highly . As this energy is released in the form of heat, it spreads out into the environment and will remain unavailable to push the reaction in reverse. We express this by saying that "sodium chloride is more than its elements". Looking at the equations above, you can see that Equation \ref{Step5} constitutes the big payoff in energy. The 787 kj/mol noted there is known as the NaCl . Its large magnitude should be no surprise, given the strength of the coulombic force between ions of opposite charge. Note that this lattice energy, while due principally to coulombic attraction between each ion and its eight nearest neighbors, is really the sum of the interactions with the crystal. Lattice energies cannot be measured directly, but they can be estimated fairly well from the energies of the other processes described in the table immediately above. The most energetically stable arrangement of solids made up of identical molecular units (as in the noble gas elements and pure metals) are generally those in which there is a minimum of empty space; these are known as structures, and there are several kinds. In the case of ionic solids of even the simplest 1:1 stoichiometry, the positive and negative ions usually differ so much in size that packing is often much less efficient. This may cause the solid to assume lattice geometries that differ from the one illustrated above for sodium chloride. By way of illustration, consider the structure of cesium chloride (the spelling is also used), CsCl. The radius of the Cs ion is 168 pm compared to 98 pm for Na and cannot possibly fit into the octahedral hole of a simple cubic lattice of chloride ions. The CsCl lattice therefore assumes a different arrangement. Figure \(\Page {3}\) focuses on two of these cubic lattice elements whose tops and bottoms are shaded for clarity. It should be easy to see that each cesium ion now has eight nearest-neighbor chloride ions. Each chloride ion is also surrounded by eight cesium ions, so all the lattice points are still geometrically equivalent. We therefore describe this structure as having . The two kinds of lattice arrangements exemplified by NaCl ("rock salt") and CsCl are found in a large number of other 1:1 ionic solids, and these names are used generically to describe the structures of these other compounds. There are of course many other fundamental lattice arrangements (not all of them cubic), but the two we have described here are sufficient to illustrate the point that the (the ratio of the radii of the positive to the negative ion) plays an important role in the structures of simple ionic solids. The interaction of the many atomic properties that influence ionic binding are nicely illustrated by looking at a series of alkali halides, especially those involving extreme differences in atomic radii. The latter are all drawn to the same scale. On the energetic plots at the right, the lattice energies are shown in green. We will start with the one you already know very well. Sodium chloride - NaCl ("rock-salt") mp/bp 801/1413 °C; coordination (6,6) - mp/bp 846/1676 °C, rock-salt lattice structure (6,6). Tiny-tiny makes strong-strong! This is the most "ionic" of the alkali halides, with the largest lattice energy and highest melting and boiling points. The small size of these ions (and consequent high charge densities) together with the large electronegativity difference between the two elements places a lot of electronic charge between the atoms. Even in this highly ionic solid, the electron that is "lost" by the lithium atom turns out to be closer to the Li nucleus than when it resides in the 2 shell of the neutral atom. - mp/bp 703/1231 °C, (8,8) coordination. With five shells of electrons shielding its nucleus, the Cs ion with its low charge density resembles a big puff-ball which can be distorted by the highly polarizing fluoride ion. The resulting ion-induced dipoles (blue arrows) account for much of the lattice energy here. The reverse of this would be a tiny metal ion trying to hold onto four relatively huge iodide ions like Lithium iodide. - mp/bp 745/1410 °C. Negative ions can make even bigger puff-balls. The tiny lithium ion can't get very close to any of the iodides to generate a very strong coulombic binding, but does polarize them to create an ion-induced dipole component. It does not help that the negative ions are in contact with each other. The structural geometry is the same (6,6) coordination as NaCl. - mp/bp 626/1280 °C. Even with the (8,8) coordination afforded by the CsCl structure, this is a pretty sorry combination owing to the low charge densities. The weakness of coulombic- compared to van der Waals interactions makes this the least-"ionic" of all the alkali halide solids. Conclusion: Many of the alkali halide solids are not all that "ionic" in the sense that coulombic forces are the predominant actors; in many, such as the CsI illustrated above, ion-induced dipole forces are more important. As noted above, ionic solids are generally hard and brittle. Both of these properties reflect the strength of the coulombic force. measures resistance to . Because the ions are tightly bound to their oppositely-charged neighbors and, a mechanical force exerted on one part of the solid is resisted by the electrostatic forces operating over an extended volume of the crystal. But by applying sufficient force, one layer of ions can be made to slip over another; this is the origin of . This slippage quickly propagates along a plane of the crystal (more readily in some directions than in others), weakening their attraction and leading to physical . Because the "ions" in ionic solids lack , the solids themselves are . Even within the alkali halides, the role of coulombic attraction diminishes as the ions become larger and more polarizable or differ greatly in radii. This is especially true of the anions, which tend to be larger and whose electron clouds are more easily distorted. In solids composed of polyatomic ions such as (NH ) SO , SrClO , NH CO , ion-dipole and ion-induced dipole forces may actually be stronger than the coulombic force. Higher ionic charges help, especially if the ions are relatively small. This is especially evident in the extremely high melting points of the and higher oxides: These substances are known as , meaning that they retain their essential properties at high temperatures. Magnesia, for example, is used to insulate electric heating elements and, in the form of fire bricks, to line high-temperature furnaces. No boiling points have been observed for these compounds; on further heating, they simply dissociate into their elements. Their crystal structures can be very complex, and some (notably Al O ) can have several solid forms. Even in the most highly ionic solids there is some electron sharing, so the idea of a “pure” ionic bond is an abstraction. Many solids that are formally derived from ions cannot really be said to form "ionic" solids at all. For example, anhydrous copper(II) chloride consists of layers of copper atoms surrounded by four chlorine atoms in a square arrangement. Neighboring chains are offset so as to create an octahedral coordination of each copper atom. Similar structures are commonly encountered for other salts of transition metals. Similarly, most oxides and sulfides of metals beyond tend to have structures dominated by other than ion-ion attractions. The trihalides of aluminum offer another example of the dangers of assuming ionic character of solids that are formally derived from ions. Aqueous solutions of what we assume to be AlF , AlCl , AlBr , and AlI all exhibit the normal properties ionic solutions (they are electrically conductive, for example), but the solids are quite different: the melting point of AlF is 1290°C, suggesting that it is indeed ionic. But AlCl melts at 192°C — hardly consistent with ionic bonding, and the other two halides are also rather low-melting. Structural studies show that when AlCl vaporizes or dissolves in a non-polar solvent it forms a dimer Al Cl . The two other halides exist only as dimers in all states. The structural formula of the Al Cl molecule shows that the aluminum atoms are bonded to four chlorines, two of which are shared between the two metal atoms. The arrows represent coordinate covalent bonds in which the bonding electrons both come from the same atom (chlorine in this case.) As shown at the right above, the aluminum atoms can be considered to be located at the centers of two tetrahedra that possess one edge in common.

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As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called . Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent. Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.

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In this section, we will learn about alcohols and ethers. Incorporation of an oxygen atom into carbon- and hydrogen-containing molecules leads to new functional groups and new families of compounds. When the oxygen atom is attached by single bonds, the molecule is either an alcohol or ether. are derivatives of hydrocarbons in which an –OH group has replaced a hydrogen atom. Although all alcohols have one or more hydroxyl (–OH) functional groups, they do not behave like bases such as NaOH and . NaOH and KOH are ionic compounds that contain OH ions. Alcohols are covalent molecules; the –OH group in an alcohol molecule is attached to a carbon atom by a covalent bond. Ethanol, CH CH OH, also called ethyl alcohol, is a particularly important alcohol for human use. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars: Large quantities of ethanol are synthesized from the addition reaction of water with ethylene using an acid as a catalyst:   Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines): The name of an alcohol comes from the hydrocarbon from which it was derived. The final in the name of the hydrocarbon is replaced by , and the carbon atom to which the –OH group is bonded is indicated by a number placed before the name. Consider the following example. How should it be named? The carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to . In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol. Name the following molecule:   2-methyl-2-pentanol are compounds that contain the functional group –O–. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named “methoxy.” The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by “ether.” The common name for the compound shown in below is ethylmethyl ether: Provide the IUPAC and common name for the ether shown here: Provide the IUPAC and common name for the ether shown:   IUPAC: 2-methoxypropane; common: isopropylmethyl ether Ethers can be obtained from alcohols by the elimination of a molecule of water from two molecules of the alcohol. For example, when ethanol is treated with a limited amount of sulfuric acid and heated to 140 °C, diethyl ether and water are formed: In the general formula for ethers, R— —R, the hydrocarbon groups (R) may be the same or different. Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. -butyl methyl ether, C H OCH (abbreviated —italicized portions of names are not counted when ranking the groups alphabetically—so butyl comes before methyl in the common name), is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline. Carbohydrates are large biomolecules made up of carbon, hydrogen, and oxygen. The dietary forms of carbohydrates are foods rich in these types of molecules, like pastas, bread, and candy. The name “carbohydrate” comes from the formula of the molecules, which can be described by the general formula C (H O) , which shows that they are in a sense “carbon and water” or “hydrates of carbon.” In many cases, and have the same value, but they can be different. The smaller carbohydrates are generally referred to as “sugars,” the biochemical term for this group of molecules is “saccharide” from the Greek word for sugar (Figure \(\Page {1}\)). Depending on the number of sugar units joined together, they may be classified as monosaccharides (one sugar unit), disaccharides (two sugar units), oligosaccharides (a few sugars), or polysaccharides (the polymeric version of sugars—polymers were described in the feature box earlier in this chapter on recycling plastics). The scientific names of sugars can be recognized by the suffix at the end of the name (for instance, fruit sugar is a monosaccharide called “fructose” and milk sugar is a disaccharide called lactose composed of two monosaccharides, glucose and galactose, connected together). Sugars contain some of the functional groups we have discussed: Note the alcohol groups present in the structures and how monosaccharide units are linked to form a disaccharide by formation of an ether. Organisms use carbohydrates for a variety of functions. Carbohydrates can store energy, such as the polysaccharides glycogen in animals or starch in plants. They also provide structural support, such as the polysaccharide cellulose in plants and the modified polysaccharide chitin in fungi and animals. The sugars ribose and deoxyribose are components of the backbones of and , respectively. Other sugars play key roles in the function of the immune system, in cell-cell recognition, and in many other biological roles. Diabetes is a group of metabolic diseases in which a person has a high sugar concentration in their blood (Figure \(\Page {2}\)). Diabetes may be caused by insufficient insulin production by the pancreas or by the body’s cells not responding properly to the insulin that is produced. In a healthy person, insulin is produced when it is needed and functions to transport glucose from the blood into the cells where it can be used for energy. The long-term complications of diabetes can include loss of eyesight, heart disease, and kidney failure. In 2013, it was estimated that approximately 3.3% of the world’s population (~380 million people) suffered from diabetes, resulting in over a million deaths annually. Prevention involves eating a healthy diet, getting plenty of exercise, and maintaining a normal body weight. Treatment involves all of these lifestyle practices and may require injections of insulin. Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.06%3A_Enthalpy

The factors influencing a reaction are complicated and varied. Since a catalyst affects activation energy, we might assume it would have some sort of impact on the amount of heat that is absorbed or released by the reaction—but it does not. The change in heat content of a reaction depends solely on the chemical compositions of the reactants and products, not on the path taken to get from one to the other. Heat changes in chemical reactions are often measured in the laboratory under conditions in which the reacting system is open to the atmosphere. In these cases, the system is at a constant pressure. is the heat content of a system at constant pressure. Chemists routinely measure changes in enthalpy of chemical systems as reactants are converted into products. The heat that is absorbed or released by a reaction at constant pressure is the same as the enthalpy change, and is given the symbol \(\Delta H\). Unless otherwise specified, all reactions in this material are assumed to take place at constant pressure. The change in enthalpy of a reaction is a measure of the differences in enthalpy of the reactants and products. The enthalpy of a system is determined by the energies needed to break chemical bonds and the energies needed to form chemical bonds. Energy needs to be put into the system in order to break chemical bonds—they do not come apart spontaneously in most cases. Bond formation to produce products will involve release of energy. The change in enthalpy shows the trade-offs made in these two processes. Does it take more energy to break bonds than that needed to form bonds? If so, the reaction is endothermic and the enthalpy change is positive. If more energy is produced in bond formation than that needed for bond breaking, the reaction is exothermic and the enthalpy is negative. Several factors influence the enthalpy of a system. Enthalpy is an extensive property, determined in part by the amount of material being dealt with. The state of reactants and products (solid, liquid, or gas) influences the enthalpy value for a system. The direction of the reaction affects the enthalpy value. A reaction that takes place in the opposite direction has the same numerical enthalpy value, but the opposite sign.

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The molecular units of a liquid, like those of solids, are in direct contact, but never for any length of time and in the same locations. Whereas the molecules or ions of a solid maintain the same average positions, those of liquids are continually jumping and sliding to new ones, giving liquids something of the mobility of gases. From the standpoint of chemistry, this represents the best of two worlds; rapid chemical change requires intimate contact between the agents undergoing reaction, but these agents, along with the reaction products, must be free to move away to allow new contacts and further reaction to take place. This is why so much of what we do with chemistry takes place in the liquid phase. Liquids occupy a rather peculiar place in the trinity of solid, liquid and gas. A liquid is the preferred state of a substance at temperatures intermediate between the realms of the solid and the gas. Howeve, if one look at the melting and boiling points of a variety of substances Figure \(\Page {1}\), you will notice that the temperature range within which many liquids can exist tends to be rather small. In this, and in a number of other ways, the liquid state appears to be somewhat tenuous and insecure, as if it had no clear right to exist at all, and only does so as an oversight of Nature. Certainly the liquid state is the most complicated of the three states of matter to analyze and to understand. But just as people whose personalities are more complicated and enigmatic are often the most interesting ones to know, it is these same features that make the liquid state of matter the most fascinating to study. Anyone can usually tell if a substance is a liquid simply by looking at it. What special physical properties do liquids possess that make them so easy to recognize? One obvious property is their , which refers to their ability to move around, to change their shape to conform to that of a container, to flow in response to a pressure gradient, and to be displaced by other objects. But these properties are shared by gases, the other member of the two of matter. The real giveaway is that a liquid occupies a fixed volume, with the consequence that a liquid possesses a definite . Gases, of course, do not; the volume and shape of a gas are simply those of the container in which it is confined. The higher density of a liquid also plays a role here; it is only because of the large density difference between a liquid and the space above it that we can see the surface at all. (What we are really seeing are the effects of reflection and refraction that occur when light passes across the boundary between two phases differing in density, or more precisely, in their .) The term is a measure of resistance to flow. It can be measured by observing the time required for a given volume of liquid to flow through the narrow part of a tube. The viscosity of a substance is related to the strength of the forces acting between its molecular units. In the case of water, these forces are primarily due to hydrogen bonding. Liquids such as syrups and honey are much more viscous because the sugars they contain are studded with hydroxyl groups (–OH) which can form multiple hydrogen bonds with water and with each other, producing a sticky disordered network. Even in the absence of hydrogen bonding, dispersion forces are universally present (as in mercury). Because these forces are additive, they can be very significant in long carbon-chain molecules such as those found in oils used in cooking and for lubrication. Most "straight-chain" molecules are really bent into complex shapes, and dispersion forces tend to preserve their spaghetti-like entanglements with their neighbors. The temperature dependence of the viscosity of liquids is well known to anyone who has tried to pour cold syrup on a pancake. Because the forces that give rise to viscosity are weak, they are easily overcome by thermal motions, so it is no surprise that viscosity decreases as the temperature rises. Automotive lubricating oils can be too viscous at low temperatures (making it harder for your car to operate on a cold day), while losing so much viscosity at engine operating temperatures that their lubricating properties become impaired. These engine oils are sold in a wide range of viscosities; the higher-viscosity oils are used in warmer weather and the lower-viscosity oils in colder weather. The idea is to achieve a fairly constant viscosity that is ideal for the particular application. By blending in certain ingredients, lubricant manufacturers are able to formulate “multigrade” oils whose viscosities are less sensitive to temperatures, thus making a single product useful over a much wider temperature range. The next time you pour a viscous liquid over a surface, notice how different parts of the liquid move at different rates and sometimes in different directions. To flow freely, the particles making up a fluid must be able to move independently. Intermolecular attractive forces work against this, making it difficult for one molecule to pull away from its neighbors and force its way in between new neighbors. The pressure drop that is observed when a liquid flows through a pipe is a direct consequence of viscosity. Those molecules that happen to find themselves near the inner walls of a tube tend to spend much of their time attached to the walls by intermolecular forces, and thus move forward very slowly. Movement of the next layer of molecules is impeded as they slip and slide over the slow-movers; this process continues across successive layers of molecules as we move toward the center of the tube, where the velocity is greatest. This effect is called , and is directly responsible for the pressure drop that can be quite noticeable when you are taking a shower bath and someone else in the house suddenly turns on the water in the kitchen. Liquids and gases are both and exhibit resistance to flow through a confined space. However, it is interesting (and not often appreciated) that their viscosities have entirely different origins, and that they vary with temperature in opposite ways. Why should the viscosity of a gas with temperature? A molecule within the bulk of a liquid experiences attractions to neighboring molecules in all directions, but since these average out to zero, there is no net force on the molecule because it is, on the average, as energetically comfortable in one location within the liquid as in another. Liquids ordinarily do have surfaces, however, and a molecule that finds itself in such a location is attracted to its neighbors below and to either side, but there is no attraction operating in the 180° solid angle above the surface. As a consequence, a molecule at the surface will tend to be drawn into the bulk of the liquid. Conversely, must be done in order to move a molecule within a liquid to its surface. Clearly there must always be some molecules at the surface, but the smaller the surface area, the lower the potential energy. Thus . The geometric shape that has the smallest ratio of surface area to volume is the , so very small quantities of liquids tend to form spherical drops. As the drops get bigger, their weight deforms them into the typical tear shape. Think of a bubble as a hollow drop. Surface tension acts to minimize the surface, and thus the radius of the spherical shell of liquid, but this is opposed by the pressure of vapor trapped within the bubble. The imbalance of forces near the upper surface of a liquid has the effect of an elastic film stretched across the surface. You have probably seen water striders and other insects take advantage of this when they walk across a pond. Similarly, you can carefully "float" a light object such as a steel paperclip on the surface of water in a cup. Surface tension is defined as the amount of work that must be done in order to create unit area of surface. The SI units are J m (or N m ), but values are more commonly expressed in mN m or in cgs units of dyn cm or erg cm . Table \(\Page {3}\) compares the surface tensions of several liquids at room temperature. Note especially that: Surface tension and viscosity are not directly related, as you can verify by noting the disparate values of these two quantities for mercury. Viscosity depends on intermolecular forces within the liquid, whereas surface tension arises from the difference in the magnitudes of these forces within the liquid and at the surface. Surface tension is also affected by the electrostatic charge of a body. This is most dramatically illustrated by the famous demonstration.. Surface tension always decreases with temperature as thermal motions reduce the effect of intermolecular attractions (Table \(\Page {4}\)). This is one reason why washing with warm water is more effective; the lower surface tension allows water to more readily penetrate a fabric. Why do "tears" form inside a wine glass? You have undoubtedly noticed this; pour some wine into a glass, and after a few minutes, droplets of clear liquid can be seen forming on the inside walls of the glass about a centimeter above the level of the wine. This happens even when the wine and the glass are at room temperature, so it has nothing to do with condensation. The explanation involves , hydrogen bonding, adsorption, and surface tension, so this phenomenon makes a good review of much you have learned about liquids and solutions. The tendency of a surface tension gradient to draw water into the region of higher surface tension is known as the Maringoni effect First, remember that both water and alcohol are hydrogen-bonding liquids; as such, they are both strongly attracted to the oxygen atoms and -OH groups on the surface of the glass. This causes the liquid film to creep up the walls of the glass. Alcohol, the more volatile of the two liquids, vaporizes more readily, causing the upper (and thinnest) part of the liquid film to become enriched in water. Because of its stronger hydrogen bonding, water has a larger surface tension than alcohol, so as the alcohol evaporates, the surface tension of the upper part of the liquid film increases. This that part of the film draw up more liquid and assume a spherical shape which gets distorted by gravity into a "tear", which eventually grows so large that gravity wins out over adsorption, and the drop falls back into the liquid, soon to be replaced by another. The surface tension discussed immediately above is an attribute of a liquid in contact with a gas (ordinarily the air or vapor) or a vacuum. But if you think about it, the molecules in the part of a liquid that is in contact with any other phase (liquid or solid) will experience a different balance of forces than the molecules within the bulk of the liquid. Thus surface tension is a special case of the more general which is defined by the work associated with moving a molecule from within the bulk liquid to the interface with any other phase. Take a plastic mixing bowl from your kitchen, and splash some water around in it. You will probably observe that the water does not cover the inside surface uniformly, but remains dispersed into drops. The same effect is seen on a dirty windshield; running the wipers simply breaks hundreds of drops into thousands. By contrast, water poured over a clean glass surface will wet it, leaving a uniform film. When a molecule of a liquid is in contact with another phase, its behavior depends on the relative attractive strengths of its neighbors on the two sides of the phase boundary. If the molecule is more strongly attracted to its own kind, then interfacial tension will act to minimize the area of contact by increasing the curvature of the surface. This is what happens at the interface between water and a hydrophobic surface such as a plastic mixing bowl or a windshield coated with oily material. A liquid will wet a surface if the angle at which it makes contact with the surface is less than 90°. The value of this can be predicted from the properties of the liquid and solid separately. A clean glass surface, by contrast, has –OH groups sticking out of it which readily attach to water molecules through hydrogen bonding; the lowest potential energy now occurs when the contact area between the glass and water is maximized. This causes the water to spread out evenly over the surface, or to it. The surface tension of water can be reduced to about one-third of its normal value by adding some soap or synthetic detergent. These substances, known collectively as , are generally hydrocarbon molecules having an ionic group on one end. The ionic group, being highly polar, is strongly attracted to water molecules; we say it is . The hydrocarbon ( ) portion is just the opposite; inserting it into water would break up the local hydrogen-bonding forces and is therefore energetically unfavorable. What happens, then, is that the surfactant molecules migrate to the surface with their hydrophobic ends sticking out, effectively creating a new surface. Because hydrocarbons interact only through very weak dispersion forces, this new surface has a greatly reduced surface tension. How do soaps and detergents help get things clean? There are two main mechanisms. First, by reducing water's surface tension, the water can more readily penetrate fabrics (see the illustration under "Water repellency" below.) Secondly, much of what we call "dirt" consists of non-water soluble oils and greasy materials which the hydrophobic ends of surfactant molecules can penetrate. When they do so in sufficient numbers and with their polar ends sticking out, the resulting aggregate can hydrogen-bond to water and becomes "solubilized". Washing is usually more effective in warm water; higher temperatures reduce the surface tension of the water and make it easier for the surfactant molecules to penetrate the material to be removed. The answer is no, but claims that they can are widely circulated in promotions of dubious products such as "magnetic laundry disks" which are supposed to reduce the need for detergents. In Gore-Tex, one of the more successful water-proof fabrics, the fibers are made non-wettable by coating them with a Teflon-like fluoropolymer. Water is quite strongly attracted to many natural fibers such as cotton and linen through hydrogen-bonding to their cellulosic hydroxyl groups. A droplet that falls on such a material will flatten out and be drawn through the fabric. One way to prevent this is to coat the fibers with a polymeric material that is not readily wetted. The water tends to curve away from the fibers so as to minimize the area of contact, so the droplets are supported on the gridwork of the fabric but tend not to fall through. If the walls of a narrow tube can be efficiently wetted by a liquid, then the the liquid will be drawn up into the tube by . This effect is only noticeable in narrow containers (such as burettes) and especially in small-diameter . The smaller the diameter of the tube, the higher will be the capillary rise. A clean glass surface is highly attractive to most molecules, so most liquids display a concave in a glass tube. To help you understand capillary rise, the above diagram shows a glass tube of small cross-section inserted into an open container of water. The attraction of the water to the inner wall of the tube pulls the edges of the water up, creating a curved whose surface area is smaller than the cross-section area of the tube. The surface tension of the water acts against this enlargement of its surface by attempting to reduce the curvature, stretching the surface into a flatter shape by pulling the liquid farther up into the tube. This process continues until the weight of the liquid column becomes equal to the surface tension force, and the system reaches mechanical equilibrium. Capillary rise results from a combination of two effects: the tendency of the liquid to wet (bind to) the surface of the tube (measured by the value of the contact angle), and the action of the liquid's surface tension to minimize its surface area. In the formula shown at the left (which you need not memorize!) = elevation of the liquid (m) γ = surface tension (N/m) θ = contact angle (radians) ρ = density of liquid (kg/m ) = acceleration of gravity (m/s ) = radius of tube (m) The contact angle between water and ordinary soda-lime glass is essentially zero; since the cosine of 0 radians is unity, its capillary rise is especially noticable. In general, water can be drawn very effectively into narrow openings such as the channels between fibers in a fabric and into porous materials such as soils. Note that if θ is greater than 90° (π/2 radians), the capillary "rise" will be negative — meaning that the molecules of the liquid are more strongly attracted to each other than to the surface. This is readily seen with mercury in a glass container, in which the meniscus is upwardly convex instead of concave. Capillary rise is the principal mechanism by which water is able to reach the highest parts of trees. Water strongly bonds to the narrow (25 μM) cellulose channels in the xylem. (Osmotic pressure and "suction" produced by loss of water vapor through the leaves also contribute to this effect, and are the main drivers of water flow in smaller plants.) Bubbles can be thought of as "negative drops" — spherical spaces within a liquid containing a gas, often just the vapor of the liquid. Bubbles within pure liquids such as water (which we see when water boils) are inherently unstable because the liquid's surface tension causes them to collapse. But in the presence of a surfactant, bubbles can be stabilized and given an independent if evanescent existence. The pressure of the gas inside a bubble must be sufficient to oppose the pressure outside of it ( , the atmospheric pressure plus the hydrostatic pressure of any other fluid in which the bubble is immersed. But the force caused by surface tension γ of the liquid boundary also tends to collapse the bubble, so must be than by the amount of this force, which is given by 4γ/ : \[ P_{in}=P_{out} + \dfrac{4\gamma}{r}\] The most important feature of this relationship (known as ) is the that the pressure required to maintain the bubble is inversely proportional to its radius. This means that the bubbles have the internal gas pressures! This might seem counterintuitive, but if you are an experienced soap-bubble blower, or have blown up a rubber balloon (in which the elastic of the rubber has an effect similar to the surface tension in a liquid), you will have noticed that you need to puff harder to begin the expansion. All of us at one time or another have enjoyed the fascination of creating soap bubbles and admiring their intense and varied colors as they drift around in the air, seemingly aloof from the constraints that govern the behavior of ordinary objects — but only for a while! Their life eventually comes to an abrupt end as they fall to the ground or pop in mid-flight. The walls of these bubbles consist of a thin layer of water molecules sandwiched between two layers of surfactant molecules. Their spherical shape is of course the result of water's surface tension. Although the surfactant (soap) initially reduces the surface tension, expansion of the bubble spreads the water into a thinner layer and spreads the surfactant molecules over a wider area, deceasing their concentration. This, in turn, allows the water molecules to interact more strongly, increasing its surface tension and stabilizing the bubble as it expands. The bright colors we see in bubbles arises from interference between light waves that are reflected back from the inner and outer surfaces, indicating that the thickness of the water layer is comparable the range of visible light (around 400-600 nm). Once the bubble is released, it can endure until it strikes a solid surface or collapses owing to loss of the water layer by evaporation. The latter process can be slowed by adding a bit of glycerine to the liquid. A variety of recipes and commercial "bubble-making solutions" are available; some of the latter employ special liquid polymers which slow evaporation and greatly extend the bubble lifetimes. Bubbles blown at very low temperatures can be frozen, but these eventually collapse as the gas diffuses out. The sites of gas exchange with the blood in mammalian lungs are tiny sacs known as . In humans there are about 150 million of these, having a total surface area about the size of a tennis court. The inner surface of each alveolus is about 0.25 mm in diameter and is coated with a film of water, whose high surface tension not only resists inflation, but would ordinarily cause the thin-walled alveoli to collapse. In order to counteract this effect, special cells in the alveolar wall secrete a phospholipid that reduces the surface tension of the water film to about 35% of its normal value. But there is another problem: the alveoli can be regarded physically as a huge collection of interconnected bubbles of varying sizes. As noted above, the surface tension of a surfactant-stabilized bubble increases with their size. So by making it easier for the smaller alveoli to expand while inhibiting the expansion of the larger ones, the surfactant helps to equalize the volume changes of all the alveoli as one inhales and exhales. Pulmonary surfactant is produced only in the later stages of fetal development, so premature infants often do not have enough and are subject to respiratory distress syndrome which can be fatal. You can think of a simple liquid such as argon or methane as a collection of loosely-packed marbles that can assume various shapes.Although the overall arrangement of the individual molecular units is entirely random, there is a certain amount of short-range order: the presence of one molecule at a given spot means that the neighboring molecules must be at least as far away as the sum of the two radii, and this in turn affects the possible locations of more distant concentric shells of molecules. An important consequence of the disordered arrangement of molecules in a liquid is the presence of void spaces. These, together with the increased kinetic energy of colliding molecules which helps push them apart, are responsible for the approximately 15-percent decrease in density that is observed when solids based on simple spherical molecules such as Ne and Hg melt into liquids. These void spaces are believed to be the key to the flow properties of liquids; the more “holes” there are in the liquid, the more easily the molecules can slip and slide over one another. As the temperature rises, thermal motions of the molecules increase and the local structure begins to deteriorate, as shown in the plots below. This plot shows the relative probability of finding a mercury atom at a given distance from another atom located at distance 0. You can see that as thermal motions increase, the probabilities even out at greater distances. It is very difficult to design experiments that yield the kind of information required to define the microscopic arrangement of molecules in the liquid state. Many of our current ideas on the subject come from computer simulations based on hypothetical models. In a typical experiment, the paths of about 1000 molecules in a volume of space are calculated. The molecules are initially given random kinetic energies whose distribution is consistent with the Boltzmann distribution for a given temperature. The trajectories of all the molecules are followed as they change with time due to collisions and other interactions; these interactions must be calculated according to an assumed potential energy-vs.-distance function that is part of the particular model being investigated. These computer experiments suggest that whatever structure simple liquids do possess is determined mainly by the repulsive forces between the molecules; the attractive forces act in a rather nondirectional, general way to hold the liquid together. It is also found that if spherical molecules are packed together as closely as geometry allows (in which each molecule would be in contact with twelve nearest neighbors), the collection will have a long-range order characteristic of a solid until the density is decreased by about ten percent, at which point the molecules can slide around and move past one another, thus preserving only short-range order. In recent years, experimental studies based on ultra-short laser flashes have revealed that local structures in liquids have extremely short lifetimes, of the order of picoseconds to nanoseconds. It has long been suspected that the region of a liquid that bounds a solid surface is more ordered than within the bulk liquid. This has been confirmed for the case of water in contact with silicon, in which the liquid's layers form layers, similar to what is found in liquid crystals.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/06%3A_Strategies_in_(-)-Menthol_Synthesis

(-)-Menthol is amongst the most important perfume / flavor chemical, extensively used in pharmaceuticals, cosmetics, toothpastes, chewing gums and toiletries. Out a the estimated total production of about 20,000 m.tons, natural menthol accounts to about 13 m.tons, the rest coming from synthetic sources. The natural source - oil of Mentha Arvensis - being erratic due to dependence on monsoon, the demand for synthetic menthol is on the increase. The manufacturing processes chosen for discussions here demonstrate three important methodologies used in industry for the synthesis of chiral compounds. A summary of some of the known processes is provided in Figure 6.1. (formerly know as Haarmann & Reimer process) (US Patent 3,943,181 (Mar 9 1976)) – In this process , thymol is synthesized from m-cresol. Catalytic hydrogenation gave a mixture of Menthols from which menthols were first obtained as a racemic mixture by careful fractional distillation. The residual mixture was epimerised to increase the content of racemic menthol using a patented catalytic process. The breakthrough in the process is the resolution of the benzoate ester of the racemate by recrystallization by a process of seeding the concentrate with one pure epimer. The mother liquor that was now rich in the (+) isomer was recycled by taking it back to the distillation cycle. In this process, overall yield of (-)-menthol is about 90%. In this process a (S)-DINAP catalysed isomerization is the key step . Addition of lithium amide to Myrcene gave an addition compound that was isomerised using a chiral ruthenium catalyst. Hydrolysis of the resulting enamine gave an aldehyde citronellal in high enantiomeric purity. This was cyclized by Lewis catalyst. Catalytic reduction of the olefin gave (-)-Menthol1. BASF has already set up processes for the synthesis of a series of terpenes starting from butene. In view of the high demand for (-)-Menthol, they extended the product chain to (-)-Menthol as well. The scheme for the synthesis of the product chain is shown in Figure 6.4. Fig 6.4 Extention of the Citral value chain to (+)-Citronellal, (-)-Isopulegol and finally to (-)-Menthol gave a range of value added products. Note that these processes have taken advantage of the developments in catalytic processes in recent years. Details of the catalysts are not made public. Limonene is abundantly available from peels of citrus fruits. On selective catalytic reduction with Ra-Ni, it could be reduced to (+)-1-Menthene, which on epoxidation and hydrolysis gave (+)-1-hydroxyneocarvomenthol. Acylation followed by pyrolysis gave (-)-trans-menth-2-ene-1-ol as the major product. The crude product was solvolysed to give a mixture of piperityl actates as the allylic migration products. The crude product was distilled at this stage to separate the cis- and trans piperitols. The minor ring contraction product was useful as perfume intermediate elsewhere. The final reduction was achieved by H2 / Pd-C to give 75% yield of (-)-Menthol after fractional distillation. The Starting material has the correct configuration at C1. The problem is to reduce the double bond enantioselectively. The double bond is first reduced by catalytic hydrogenation to give a mixture of (-)-Menthone and (+)-Isomenthone. The all-equatorial configuration of (-)-menthol is best attained by dissolving metal reduction. The enolate intermediate is protonated to the thermodynamically stable (-)-menthol. Look closely at the stereochemistry at the asymmetric center of (-)-Piperitone. The isopropyl is in the wrong configuration for (-)-Menthone. The challenge here is (1) isomerise this center and (2) enantioselectively reduce the double bond. All attempts to produce (-)-Menthol produce only mixtures as shown in Figure 6.7. Hence this process has not been very successful. (-)-β-Pinene offers a good route because the requisite structural features are present and is available in sufficient optical purity. Hydrogenation of (-)-β-Pinene gave cis-Pinane is a major product. On pyrolysis, the strained bridged ring system cleaved to give optically pure 2,6-dimethyl-2,7-octadiene. It was converted to (+)-Citronellol in good yields by direct oxidation. Alternately the more substituted olefin was first subjected to a Markovnikov addition of HCl followed by an anti-Markovnikov addition of HBr. Solvolysis reaction provided a mixture of citronellols. Catalytic oxidation of the alcohol provided (+)-Citronellal. This could be converted to (-)-Menthol by known procedures. The product was however contaminated with trace amounts of (+)-Menthol arising from trans-Pinane generated in the first step. δ-3-Carene is another chiral synthon that has the required structural features to serve as a starting material for (-)-Menthol. Catalytic isomerization of δ-3-Carene gave (+)-δ-2-Carene. Two different routes were investigated. In the first route (+)-δ-2-Carene was pyrolysed to cleave the cyclopropane ring. The resultant diene had the right stereochemistry at C1 and C4. The latter did not matter because this asymmetry is lost soon and regenerated in the process. Treatment of the unconjugated diene with HCl and dehydrohalogenation led to a conjugated diene. Addition of HCl led to an allyl chloride. Solvolysis with acetic acid-sodium acetate provided an S ’ displacement causing an allylic rearrangement. The resulting piperitol acetates gave (+)-cis and (-)-trans-piperitols which could be fractionally distilled. Pure(-)-trans-piperitols yielded (-)-menthol on hydrogenation. In the second route (+)-δ-2-Carene was epoxidised to yield (+)-cis-2,8-p-menthadienol directly. On buffering with formic acid – acetic acid mixture, allylic rearrangement occurred to give a mixture of formate and acetate of piperityl esters. The corresponding alcohols could be fractionally distilled. The cis-isomer could be isomerised to improve the yield of pure (-)-trans-piperitol. Hydrgenation gave (-)-Menthol. Conclusion: In this section we were focused on development of commercially viable routes for (-)-Menthol. Note the way modern reagents have influenced industrial processes.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Complex_Molecular_Synthesis_(Salomon)/03%3A_Fatty_Acids_and_Prostaglandins

In the previous chapter, two hypothetical biosynthetic strategies were presented that are not used biosynthetically. That the individual steps in each of these hypothetical strategies are reasonable, is indicated by the interesting fact that both strategies are exploited in reverse in Nature; one for the conversion of glucose into ribulose and the other for the conversion of acetyl CoA into malonyl CoA. The actual biosynthetic strategies for acetyl CoA and malonyl CoA were then presented. This format is intended not only to exemplify the numerous potential strategic options available, but also to provide foils that highlight the unique features of the actual biosynthetic strategies. These contrasting strategies encourage the reader to go beyond understanding the logic that governs the success of the biosynthetic sequence of reactions, and to ask: why does Nature choose this particular strategy? In the ensuing chapters, a variety of strategies will be presented, compared, and contrasted for each natural product. Besides unimplemented hypothetical biosynthetic strategies, fatally flawed strategies, and often several topologically unique successful strategies will be considered with the goal of familiarizing the student with the options and pitfalls presented by the challenge of designing and executing the total synthesis of structurally and functionally complex organic molecules.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/09%3A_Gases/9.4%3A_Effusion_and_Diffusion_of_Gases

If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called (shown in Figure \(\Page {1}\)). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure \(\Page {1}\). The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no transfer of molecules occurs). We are often interested in the , the amount of gas passing through some area per unit time: \[\textrm{rate of diffusion}=\dfrac{\textrm{amount of gas passing through an area}}{\textrm{unit of time}} \nonumber \] The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation. A process involving movement of gaseous species similar to diffusion is , the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure \(\Page {1}\)). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same. If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure \(\Page {2}\)). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated : : \[\textrm{rate of effusion}∝\dfrac{1}{\sqrt{ℳ}} \nonumber \] This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles: \[\dfrac{\textrm{rate of effusion of B}}{\textrm{rate of effusion of A}}=\dfrac{\sqrt{ℳ_\ce{A}}}{\sqrt{ℳ_\ce{B}}} \nonumber \] Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen. From Graham’s law, we have: \[\dfrac{\textrm{rate of effusion of hydrogen}}{\textrm{rate of effusion of oxygen}}=\mathrm{\dfrac{\sqrt{1.43\cancel{g\: L^{−1}}}}{\sqrt{0.0899\cancel{g\: L^{−1}}}}=\dfrac{1.20}{0.300}=\dfrac{4}{1}} \nonumber \] Using molar masses: \[\dfrac{\textrm{rate of effusion of hydrogen}}{\textrm{rate of effusion of oxygen}}=\mathrm{\dfrac{32\cancel{g\: mol^{−1}}}{2\cancel{g\: mol^{−1}}}=\dfrac{\sqrt{16}}{\sqrt{1}}=\dfrac{4}{1}} \nonumber \] Hydrogen effuses four times as rapidly as oxygen. At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide effuse? 52 mL/s Here’s another example, making the point about how determining times differs from determining rates. It takes 243 s for 4.46 × 10 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10 mol Ne to effuse? It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion: \[\textrm{rate of effusion}=\dfrac{\textrm{amount of gas transferred}}{\textrm{time}}\nonumber \] and combine it with Graham’s law: \[\dfrac{\textrm{rate of effusion of gas Xe}}{\textrm{rate of effusion of gas Ne}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}\nonumber \] To get: \[\dfrac{\dfrac{\textrm{amount of Xe transferred}}{\textrm{time for Xe}}}{\dfrac{\textrm{amount of Ne transferred}}{\textrm{time for Ne}}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}\nonumber \] Noting that = , and solving for : \[\dfrac{\dfrac{\cancel{\textrm{amount of Xe}}}{\textrm{time for Xe}}}{\dfrac{\cancel{\textrm{amount of Ne}}}{\textrm{time for Ne}}}=\dfrac{\textrm{time for Ne}}{\textrm{time for Xe}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}} \nonumber \] and substitute values: \[\mathrm{\dfrac{time\: for\: Ne}{243\:s}=\sqrt{\dfrac{20.2\cancel{g\: mol}}{131.3\cancel{g\: mol}}}=0.392}\nonumber \] Finally, solve for the desired quantity: \[\mathrm{time\: for\: Ne=0.392×243\:s=95.3\:s}\nonumber \] Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for \(\ce{Xe}\), which means the time of effusion for Ne will be smaller than that for Xe. A party balloon filled with helium deflates to \(\dfrac{2}{3}\) of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to \(\dfrac{1}{2}\) of its original volume? 32 h Finally, here is one more example showing how to calculate molar mass from effusion rate data. An unknown gas effuses 1.66 times more rapidly than CO . What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity? From Graham’s law, we have: \[\mathrm{\dfrac{rate\: of\: effusion\: of\: Unknown}{rate\: of\: effusion\: of\: CO_2}}=\dfrac{\sqrt{ℳ_\mathrm{CO_2}}}{\sqrt{ℳ_{Unknown}}} \nonumber \] Plug in known data: \[\dfrac{1.66}{1}=\dfrac{\sqrt{44.0\:\ce{g/mol}}}{\sqrt{ℳ_{Unknown}}} \nonumber \] Solve: \[ℳ_{Unknown}=\mathrm{\dfrac{44.0\:g/mol}{(1.66)^2}=16.0\:g/mol} \nonumber \] The gas could well be CH , the only gas with this molar mass. Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas. 163 g/mol Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF , the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The UF molecules have a higher average speed and diffuse through the barrier a little faster than the heavier UF molecules. The gas that has passed through the barrier is slightly enriched in UF and the residual gas is slightly depleted. The small difference in molecular weights between UF and UF only about 0.4% enrichment, is achieved in one diffuser (Figure \(\Page {4}\)). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained. The large scale separation of gaseous UF from UF was first done during the World War , at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF . Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.   Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.03%3A_Fermentation_Paper

You will write a research paper explaining the production of a fermented product not discussed in class or expanding on a covered topic. There must be significant chemistry/biochemistry in your paper. Additionally, there will be a comparison of the use or production in the US vs another country. Potential Topics for Review Article on Fermentation: • Meat preservation • Bletting of fruit (beyond ripening) • Olive Fermentation (effects on oleuropein) • Kimchee • Tempeh • Shalgam juice, hardaliye, or boza (Turkish fermented vegetable and grain beverages) • Injera (organisms, fermentation, and carbohydrates in t'eff) • Miso and Soy • Distilled alcoholic beverages • Impacts of Nitrogen/nutrients on fermentation in a specific product • Impacts of pH on fermentation in beer or wine • Effect of local water chemistry on brewing or distilling • Tannin and polyphenolics in beer production • Megasphaera cerevisiae effects on beer production (H2S formation) • Hop content on flavor profiles • Sulfur compounds in beers (production, regulation, flavor profiles) • 'Head' or foam on beers • Wheat ales • Barley wines • Cask conditioning of beers • Production of two short branched-chain fatty acids, 2-methylbutanoic acid and 3- methylbutanoic acid, imparting the “cheesy/sweaty” notes in many cheeses. • Propionic acid fermentation and the distinctive flavor of Swiss cheese • Mold Fermentations (e.g. roquefort cheese) • Buttermilk • Microbe variability in flavors for a specific fermented product • Lactic Acid Bacteria and the undesirable flavor products in cider such as 'piqûre acroléique’ • Phenolic variation in wine varietals and flavor profiles • Impact of oxygen on wine (what happens to chemical profile after you open the bottle?) • Effects of chemical aging on wine • Champagne and sparkling wines • Wine (broad topic -- will need a narrower focus) • Tej: ethiopian honey wine • Sulfur compounds in wine (production, regulation, flavor profiles) • Champagne and sparkling wines • Malolactic fermentation in wine. This secondary fermentation process is standard for most red wine production and common for some white grape varieties such as Chardonnay, where it can impart a "buttery" flavor from diacetyl, a byproduct of the reaction. • Use of additives in wine. Ascorbic Acid, lysozyme, fumaric acid, sorbic acid, DMDC, tannins, gum arabic, colors. How do these impact chemistry and flavor? • Biological aging of wines. Sherry. Use of 'flor'. Chemical byproducts and pathways involved. • Astringency. Astringency is an important factor in the sensory perception of beers, ciders, and wines. What compounds are responsible for this sensation and how do they interact with tastebuds on a molecular level? • Sake • Tea • Chocolate • Coffee • Kombucha • Bulk chemical production • Pharmaceuticals • Wood-Ljungdal pathway for biofuel production • ABE fermentation • Enzymes needed for Gluten free bread • FODMAPs (fermentable oligosaccharides disaccharides, monosaccharides and, polyols) cause IBS and gluten sensitivity -- diets, solutions? • Microbe variability in flavors for a specific fermented product • Propose your own topic Confirm your topic for your research paper that includes these three key ideas: 1. Thesis statement (Purdue Online Writing Lab ) 2. Biochemistry/chemistry content 3. Cultural Comparison Write a 1-2 page outline of the literature on your topic. It should be in a typical bulleted or numbered form. See for more details about writing an outline. This outline should contain an introduction and sufficient background biochemical pathway information, key experimental results, topics for discussion (applications/uses, variations), and a possible direction for cultural comparison essay. List in your bibliography at least 15 references, 10 of which must be primary references. For each reference, cite it in the appropriate format and write a 2-3 sentence summary of each reference. Complete the background and literature review of your fermentation topic. This section should cover the biochemical pathways involved in your topic. This should be a minimum of five pages. • Include drawings with structures (in ChemDraw) not clipped from a literature article. This section of the paper should address the applications or uses of your fermentation topic. It should be a complete story with current uses and modifications. This section of the paper should be at least 2-3 pages long. Some possible topics to cover: • What food or industrial applications are you exploring? • Why are people interested in this topic? • How is this technique or process or food used in US culture? • What are current concerns/problems with the process? • How are people attempting to improve this process? • Is climate change going to affect production? • Quality control issues? • Regulatory issues? • Are there different types of related fermentation products or processes? Outline or draft of the cultural comparison of your topic. This last section should be 2-3 pages that looks at cultural differences in either the production or process or use of your topic. This could include cultural differences in consumption or different regulatory processes or production. Compare and contrast differences between at least two countries or cultures. Please use citations to support your ideas. This is your final Fermentation Paper. There should be three parts: 1. Literature Review (with edits incorporated). 2. Application Section (with edits incorporated). 3. Cultural comparison of your topic (with edits and insights from Amsterdam and Belgium incorporated).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.3%3A_Phase_Transitions

We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored. When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called . When the rate of condensation becomes equal to the rate of , neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a , the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure \(\Page {1}\), and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase. The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces. Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs: Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two −OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest. At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols: All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed: \[P_{methanol} > P_{ethanol} > P_{propanol} > P_{butanol} \nonumber \] As temperature increases, the vapor pressure of a liquid also increases due to the increased average of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure \(\Page {2}\). The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure. When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure \(\Page {3}\) shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure. A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure \(\Page {3}\) to determine the boiling point of water at this elevation. The graph of the vapor pressure of water versus temperature in Figure \(\Page {3}\) indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil. The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure \(\Page {3}\) to determine the approximate atmospheric pressure at the camp. Approximately 40 kPa (0.4 atm)   The quantitative relation between a substance’s vapor pressure and its temperature is described by the equation: \[P=Ae^{−ΔH_{vap}/RT} \label{10.4.1} \] where Equation \(\ref{10.4.1}\) is often rearranged into logarithmic form to yield the linear equation: \[\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A \label{10.4.2} \] This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the examples and exercises that follow. If at temperature \(T_1\), the vapor pressure is \( P_1\), and at temperature \(T_2\), the vapor pressure is \(T_2\), the corresponding linear equations are: \[\ln P_1=−\dfrac{ΔH_\ce{vap}}{RT_1}+\ln A \nonumber \] and \[\ln P_2=−\dfrac{ΔH_\ce{vap}}{RT_2}+\ln A \label{10.4.3} \] Since the constant, ln A, is the same, these two equations may be rearranged to isolate \(\ln A\) and then set them equal to one another: \(\ln P_1+\dfrac{ΔH_\ce{vap}}{RT_1}=\ln P_2+\dfrac{ΔH_\ce{vap}}{RT_2}\label{10.4.5}\) which can be combined into: \[\ln \left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R} \left( \dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \label{10.4.6} \] Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane. The enthalpy of vaporization, \(ΔH_\ce{vap}\), can be determined by using the Clausius-Clapeyron equation (Equation \(\ref{10.4.6}\)): \[\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber \] Since we have two vapor pressure-temperature values we can substitute them into this equation and solve for \(ΔH_{vap}\). Rearranging the Clausius-Clapeyron equation and solving for \(ΔH_{vap}\) yields: \[ \begin{align*} ΔH_\ce{vap} &= \dfrac{R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)} \\[4pt] &= \dfrac{(8.3145\:J/mol⋅K)⋅\ln \left(\dfrac{100\: kPa}{10.0\: kPa}\right)}{\left(\dfrac{1}{307.2\:K}−\dfrac{1}{372.0\:K}\right)} \\[4pt] &=33,800\, J/mol =33.8\, kJ/mol \end{align*} \nonumber \] Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid. At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol. 47,782 J/mol = 47.8 kJ/mol For benzene (C H ), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, Δ then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation (Equation \(\ref{10.4.1}\)) : \[\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber \] Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (\(T_1\) = 80.1 °C = 353.3 K, \(P_1\) = 101.3 kPa, \(ΔH_{vap}\) = 30.8 kJ/mol) and want to find the temperature (\(T_2\)) that corresponds to vapor pressure = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for \(T_2\). Rearranging the Clausius-Clapeyron equation and solving for \(T_2\) yields: \[\begin{align*} T_2 &=\left(\dfrac{−R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{ΔH_\ce{vap}}+\dfrac{1}{T_1}\right)^{−1} \\[4pt] &=\mathrm{\left(\dfrac{−(8.3145\:J/mol⋅K)⋅\ln\left(\dfrac{83.4\:kPa}{101.3\:kPa}\right)}{30,800\: J/mol}+\dfrac{1}{353.3\:K}\right)^{−1}}\\[4pt] &=\mathrm{346.9\: K\: or\:73.8^\circ C} \end{align*} \nonumber \] For acetone \(\ce{(CH3)2CO}\), the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C? 30.1 kPa   Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, \(ΔH_{vap}\). For example, the vaporization of water at standard temperature is represented by: \[\ce{H2O}(l)⟶\ce{H2O}(g)\hspace{20px}ΔH_\ce{vap}=\mathrm{44.01\: kJ/mol} \nonumber \] As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat: \[\ce{H2O}(g)⟶\ce{H2O}(l)\hspace{20px}ΔH_\ce{con}=−ΔH_\ce{vap}=\mathrm{−44.01\:kJ/mol} \nonumber \] One way our body is cooled is by evaporation of the water in sweat (Figure \(\Page {4}\)). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at = 37 °C (normal body temperature); \(ΔH_{vap} = 43.46\, kJ/mol\) at 37 °C. Solution We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed: \[\mathrm{1.5\cancel{L}×\dfrac{1000\cancel{g}}{1\cancel{L}}×\dfrac{1\cancel{mol}}{18\cancel{g}}×\dfrac{43.46\:kJ}{1\cancel{mol}}=3.6×10^3\:kJ} \nonumber \] Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water. How much heat is required to evaporate 100.0 g of liquid ammonia, \(\ce{NH3}\), at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol? 28 kJ When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or . At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure \(\Page {5}\). If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing). The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures. The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process: \[\ce{H2O}_{(s)} \rightarrow \ce{H2O}_{(l)} \;\; ΔH_\ce{fus}=\mathrm{6.01\; kJ/mol} \label{10.4.9} \] The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C: \[\ce{H_2O}_{(l)} \rightarrow \ce{H_2O}_{(s)}\;\; ΔH_\ce{frz}=−ΔH_\ce{fus}=−6.01\;\mathrm{kJ/mol} \label{10.4.10} \] Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as . At room temperature and standard pressure, a piece of dry ice (solid CO ) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure \(\Page {6}\)). The reverse of sublimation is called , a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition. Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by: \[\ce{CO2}(s)⟶\ce{CO2}(g)\hspace{20px}ΔH_\ce{sub}=\mathrm{26.1\: kJ/mol} \nonumber \] Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation: \[\ce{CO2}(g)⟶\ce{CO2}(s)\hspace{20px}ΔH_\ce{dep}=−ΔH_\ce{sub}=\mathrm{−26.1\:kJ/mol} \nonumber \] Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law. \[\mathrm{solid⟶liquid}\hspace{20px}ΔH_\ce{fus}\\\underline{\mathrm{liquid⟶gas}\hspace{20px}ΔH_\ce{vap}}\\\mathrm{solid⟶gas}\hspace{20px}ΔH_\ce{sub}=ΔH_\ce{fus}+ΔH_\ce{vap} \nonumber \]   In the chapter on thermochemistry, the relation between the amount of heat absorbed or related by a substance, q, and its accompanying temperature change, ΔT, was introduced: \[q=mcΔT \nonumber \] where m is the mass of the substance and c is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure \(\Page {8}\) shows a typical heating curve. Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress. How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C? The transition described involves the following steps: The heat needed to change the temperature of a given substance (with no change in phase) is: = × × Δ (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by = × Δ . Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have: \[\begin{align*} q_\ce{total}&=(m⋅c⋅ΔT)_\ce{ice}+n⋅ΔH_\ce{fus}+(m⋅c⋅ΔT)_\ce{water}+n⋅ΔH_\ce{vap}+(m⋅c⋅ΔT)_\ce{steam}\\[7pt] &=\mathrm{(135\: g⋅2.09\: J/g⋅°C⋅15°C)+\left(135⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \right)}\\[7pt] &\mathrm{+(135\: g⋅4.18\: J/g⋅°C⋅100°C)+\left(135\: g⋅\dfrac{1\: mol}{18.02\:g}⋅40.67\: kJ/mol\right)}\\[7pt] &\mathrm{+(135\: g⋅1.84\: J/g⋅°C⋅20°C)}\\[7pt] &=\mathrm{4230\: J+45.0\: kJ+56,500\: J+305\: kJ+4970\: J} \end{align*} \nonumber \] Converting the quantities in J to kJ permits them to be summed, yielding the total heat required: \[\mathrm{=4.23\:kJ+45.0\: kJ+56.5\: kJ+305\: kJ+4.97\: kJ=416\: kJ} \nonumber \] What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C? 40.5 kJ Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.

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Liquids and solids differ from gases in that they are held together by forces that act between the individual molecular units of which they are composed. In this lesson we will take a closer look at these forces so that you can more easily understand, and in many cases predict, the diverse physical properties of the many kinds of solids and liquids we encounter in the world. The very existence of condensed states of matter suggests that there are attractive forces acting between the basic molecular units of solids and liquids. The term refers to the smallest discrete structural unit that makes up the liquid or solid. In most of the over 15 million chemical substances that are presently known, these structural units are actual — that is, aggregates of atoms that have their own distinguishing properties, formulas, and molecular weights. But the molecular units can also be individual , and more extended units. As with most artificial classifications, these distinctions tend to break down in extreme cases: most artificial polymers ("plastics") are composed of molecules of various sizes and shapes, some metal alloys contain identifiable molecular units, and it is not too much of a stretch to regard a diamond or a crystal of NaCl as a "molecule" in itself. On the atomic or molecular scale, all particles exert both attractive and repulsive forces on each other. If the attractive forces between two or more atoms are strong enough to make them into an enduring unit with its own observable properties, we call the result a "molecule" and refer to the force as a "chemical bond". The two diatomic molecules depicted in Figure \(\Page {1}\) have come into close contact with each other, but the attractive force that acts between them is not strong enough to bind them into a new molecular unit, so we call this force a . In the absence of these non-bonding attractions, all matter would exist in the gaseous state only; there would be no condensed phases. The distinction between bonding- and non-bonding attractions can be seen by comparing the potential energy plot for a pair of hydrogen atoms with that for two argon atoms (Figure \(\Page {2}\)). As two hydrogen atoms are brought together, the potential energy falls to a minimum and then rises rapidly as the two electron clouds begin to repel each other. The potential energy minimum defines the and the of the H–H bond — two of its unique measurable properties. The potential energy of a pair of argon atoms also falls as they are brought together, but not enough to hold them together. (e.g., the laws of quantum mechanics do not allow this noble gas element to form stable \(Ar_2\) molecules.) However, these non-bonding attractions enable argon to exist as a liquid and solid at low temperatures, but are unable to withstand disruptions caused by thermal energy at ordinary temperatures, so we commonly know argon as a gas. From a classic pictures, at temperatures above absolute zero, all molecular-scale particles possess that keeps them in constant motion (and from a quantum picture, motion does not stop even at absolute zero due to Heisenberg Uncertainly principle). The average thermal energy is given by the product of the gas constant and the absolute temperature. At 25°C, this works out to \[RT = (8.314 \,J \,K^{–1} mol^{–1}) (298\, K) = 2,480\, J\, mol^{–1} \approx 2.5\, kJ\, mol^{–1}\] A functional chemical bond is much stronger than this (typically over 100 kJ/mol), so the effect of thermal motion is simply to cause the bond to vibrate; only at higher temperatures (where the value of is larger) will most bonds begin to break. Non-bonding attractive forces between pairs of atoms are generally too weak to sustain even a single vibration. In addition to unique distinguishing properties such as bond energy, bond length and stretching frequencies, covalent bonds usually have that depend on the orbital structures of the component atoms. The much-weaker non-bonding attractions possess none of these properties. The shape of a potential energy curve (often approximated as a "Morse" curve) shows how repulsive and attractive forces affect the potential energy in opposite ways: repulsions always raise this energy, and attractions reduce it. The curve passes through a minimum when the attractive and repulsive forces are exactly in balance. As we stated above, all particles exert both kinds of forces on one another; these forces are all basically electrical in nature and they manifest themselves in various ways and with different strengths. The distance corresponding to the minimum potential energy is known as the . This is the average distance that will be maintained by the two particles if there are no other forces acting on them, such as might arise from the presence of other particles nearby. A general empirical expression for the interaction potential energy curve between two particles can be written as \[ E = Ar^{-n} + Br^{-m} \label{7.2.1}\] \(A\) and \(B\) are proportionality constants and \(n\) and \(m\) are integers. This expression is sometimes referred to as the . The first term, \(A\), corresponds to repulsion is always positive, and \(n\) must be larger than \(m\), reflecting the fact that repulsion always dominates at small separations. The \(B\) coefficient is negative for attractive forces, but it will become positive for electrostatic repulsion between like charges. The larger the value of one of these exponents, the closer the particles must come before the force becomes significant. Table \(\Page {1}\) lists the exponents for the types of interactions we will describe in this lesson. The value of \(n\) for the repulsive force in Figure \(\Page {3}\) is 9; this may be the highest inverse-power law to be found in nature. The magnitude of such a force is negligible until the particles are almost in direct contact, but once it kicks in, it becomes very strong; if you want to get a feel for it, try banging your head into a concrete wall. Because the repulsive force is what prevents two atoms from occupying the same space, this is just what you would expect. If the repulsive force did not always win out against all attractive forces, all matter would collapse into one huge glob! The universal repulsive force arises directly from two main aspects of quantum theory. In a wonderful article ( ), the physicist Victor Weiskopf showed how these considerations, combined with a few fundamental constants, leads to realistic estimates of such things as the hardness and compressibility of solids, the heights of mountains, the lengths of ocean waves, and the sizes of stars. Electrostatic attraction between electrically-charged particles is the strongest of all the intermolecular forces. These Coulombic forces (as they are often called) cause opposite charges to attract and like charges to repel. Coulombic forces are involved in all forms of chemical bonding; when they act between separate charged particles (ion-ion interactions) they are especially strong. Thus the energy required to pull a mole of Na and Cl ions apart in the sodium chloride crystal is greater than that needed to break the covalent bond in \(H_2\) (Figure \(\Page {1}\)). The effects of ion-ion attraction are seen most directly in solids such as NaCl which consist of oppositely-charged ions arranged in two inter-penetrating crystal lattices. According to Coulomb's Law the force between two charged particles is given by \[ F= \dfrac{q_1q_2}{4\pi\epsilon_0 r^2} \label{7.2.2}\] Instead of using SI units, chemists often prefer to express atomic-scale distances in picometers and charges as electron charge (±1, ±2, etc.) Using these units, the proportionality constant \(1/4\pi\epsilon\) works out to \(2.31 \times 10^{16}\; J\; pm\). The sign of \(F\) determines whether the force will be attractive (–) or repulsive (+); notice that the latter is the case whenever the two 's have the same sign. Equation \(\ref{7.2.2}\) is an example of an ; the force falls off as the square of the distance. A similar law governs the manner in which the illumination falls off as you move away from a point light source; recall this the next time you walk away from a street light at night, and you will have some feeling for what an inverse square law means. The stronger the attractive force acting between two particles, the greater the amount of work required to separate them. Work represents a flow of energy, so the foregoing statement is another way of saying that when two particles move in response to a force, their potential energy is lowered. This work, as you may recall if you have studied elementary mechanics, is found by integrating the negative force with respect to distance over the distance moved. Thus the energy that must be supplied in order to completely separate two oppositely-charged particles initially at a distance is given by \[ w= - \int _{r_o} ^{\infty} \dfrac{q_1q_2}{4\pi\epsilon_0 r}dr =- \dfrac{q_1q_2}{4\pi\epsilon_0 r_o} \label{7.2.3}\] When sodium chloride is melted, some of the ion pairs vaporize and form neutral NaCl molecules. How much energy would be released when one mole of Na and Cl ions are brought together in this way? The energy released will be the same as the work required to separate \[ \begin{align*} E &= \dfrac{(2.31 \times 10^{16} J pm) (+1) (–1)}{276\; pm} \\[4pt] &= –8.37 \times 10^{–19}\; J \end{align*} \] The ion-ion interaction is the simplest of electrostatic interactions and other higher order interactions exists as discussed below. According to Coulomb's law (Equation \(\ref{7.2.1}\)), the electrostatic force between an ion and an uncharged particle having = 0 should be zero. Bear in mind, however, that this formula assumes that the two particles are point charges having zero radii. A real particle such as an atom or a molecule occupies a certain volume of space. Even if the electric charges of the protons and electrons cancel out (as they will in any neutral atom or molecule), it is possible that the spatial distribution of the electron cloud representing the most loosely-bound [valence] electrons might be asymmetrical, giving rise to an . There are two kinds of dipole moments: An refers to a separation of electric charge. An idealized electric dipole consists of two point charges of magnitude + and – separated by a distance . Even though the overall system is electrically neutral, the charge separation gives rise to an electrostatic effect whose strength is expressed by the given by \[μ = q \times r \label{\(\Page {5}\)}\] Dipole moments possess both magnitude and direction, and are thus vectorial quantities; they are conventionally represented by arrows whose heads are at the negative end. These are commonly referred to simply as "dipole moments". The most well-known molecule having a dipole moment is ordinary water. The charge imbalance arises because oxygen, with its nuclear charge of 8, pulls the electron cloud that comprises each O–H bond toward itself. These two "bond moments" add vectorially to produce the permanent dipole moment denoted by the red arrow. Note the use of the δ (Greek ) symbol to denote the positive and negative ends of the dipoles. When an electric dipole is subjected to an external electric field, it will tend to orient itself so as to minimize the potential energy; that is, its negative end will tend to point toward the higher (more positive) electric potential. In liquids, thermal motions will act to disrupt this ordering, so the overall effect depends on the temperature. In condensed phases the local fields due to nearby ions or dipoles in a substance play an important role in determining the physical properties of the substance, and it is in this context that dipolar interactions are of interest to us here. We will discuss each kind of interaction in order of decreasing strength. Even if a molecule is electrically neutral and possesses no permanent dipole moment, it can still be affected by an external electric field. Because all atoms and molecules are composed of charged particles (nuclei and electrons), the electric field of a nearby ion will cause the centers of positive and negative charges to shift in opposite directions. This effect, which is called , results in the creation of a temporary, or induced dipole moment. The induced dipole then interacts with the species that produced it, resulting in a net attraction between the two particles. The larger an atom or ion, the more loosely held are its outer electrons, and the more readily will the electron cloud by distorted by an external field. A quantity known as the expresses the magnitude of the temporary dipole that can be induced in it by a nearby charge. A dipole that is close to a positive or negative ion will orient itself so that the end whose partial charge is opposite to the ion charge will point toward the ion. This kind of interaction is very important in aqueous solutions of ionic substances; H O is a highly polar molecule, so that in a solution of sodium chloride, for example, the Na ions will be enveloped by a shell of water molecules with their oxygen-ends pointing toward these ions, while H O molecules surrounding the Cl ions will have their hydrogen ends directed inward. As a consequence of ion-dipole interactions, all ionic species in aqueous solution are hydrated; this is what is denoted by the suffix in formulas such as K (aq), etc. The strength of ion-dipole attraction depends on the magnitude of the dipole moment and on the charge density of the ion. This latter quantity is just the charge of the ion divided by its volume. Owing to their smaller sizes, positive ions tend to have larger charge densities than negative ions, and they should be more strongly hydrated in aqueous solution. The hydrogen ion, being nothing more than a bare proton of extremely small volume, has the highest charge density of any ion; it is for this reason that it exists entirely in its hydrated form H O in water. As two dipoles approach each other, they will tend to orient themselves so that their oppositely-charged ends are adjacent. Two such arrangements are possible: the dipoles can be side by side but pointing in opposite directions, or they can be end to end. It can be shown that the end-to-end arrangement gives a lower potential energy. Dipole-dipole attraction is weaker than ion-dipole attraction, but it can still have significant effects if the dipole moments are large. The most important example of dipole-dipole attraction is hydrogen bonding. The most significant induced dipole effects result from nearby ions, particularly cations (positive ions). Nearby ions can distort the electron clouds even in polar molecules, thus temporarily changing their dipole moments. The larger ions (especially negative ones such as SO and ClO ) are highly polarizable, and the dipole moments induced in them by a cation can play a dominant role in compound formation. A permanent dipole can induce a temporary one in a species that is normally nonpolar, and thus produce a net attractive force between the two particles (Figure \(\Page {9}\)). This attraction is usually rather weak, but in a few cases it can lead to the formation of loosely-bound compounds. This effect explains the otherwise surprising observation that a wide variety of neutral molecules such as hydrocarbons, and even some of the noble gas elements, form stable hydrate compounds with water. The fact that noble gas elements and completely non-polar molecules such as H and N can be condensed to liquids or solids tells us that there must be yet another source of attraction between particles that does not depend on the existence of permanent dipole moments in either particle (Figure \(\Page {10}\)). To understand the origin of this effect, it is necessary to realize that when we say a molecule is “nonpolar”, we really mean that the dipole moment is zero. This is the same kind of averaging we do when we draw a picture of an orbital, which represents all the locations in space in which an electron can be found with a certain minimum probability. On a very short time scale, however, the electron must be increasingly localized; not even quantum mechanics allows it to be in more than one place at any given instant. As a consequence, there is no guarantee that the distribution of negative charge around the center of an atom will be perfectly symmetrical at every instant; every atom therefore has a weak, fluctuating dipole moment that is continually disappearing and reappearing in another direction. Dispersion or London forces can be considered to be "spontaneous dipole - induced dipole" interactions. Although these extremely short-lived fluctuations quickly average out to zero, they can still induce new dipoles in a neighboring atom or molecule, which helps sustain the original dipole and gives rise to a weak attractive force known as the or . Although dispersion forces are the weakest of all the intermolecular attractions, they are . Their strength depends to a large measure on the number of electrons in a molecule. This can clearly be seen by looking at the noble gas elements in Table \(\Page {2}\), whose ability to condense to liquids and freeze to solids is entirely dependent on dispersion forces. It is important to note that dispersion forces are ; if two elongated molecules find themselves side by side, dispersion force attractions will exist all along the regions where the two molecules are close. This can produce quite strong attractions between large polymeric molecules even in the absence of any stronger attractive forces. Although nonpolar molecules are by no means uncommon, many kinds of molecules possess permanent dipole moments, so liquids and solids composed of these species will be held together by a combination of dipole-dipole, dipole-induced dipole, and dispersion forces. These weaker forces (that is, those other than Coulombic attractions) are known collectively as These include attraction and repulsions between atoms, molecules, and surfaces, as well as other intermolecular forces. The term includes: Table \(\Page {3}\) shows some estimates of the contributions of the various types of van der Waals forces that act between several different types of molecules. Note particularly how important dispersion forces are in all of these examples, and how this, in turn, depends on the .

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Make sure you thoroughly understand the following essential ideas: The previous Module emphasized the dynamic character of equilibrium as expressed by the Law of Mass Action. This law serves as a model explaining how the composition of the equilibrium state is affected by the "active masses" (concentrations) of reactants and products. In this lesson, we develop the consequences of this law to answer the very practical question of how an existing equilibrium composition is affected by the addition or withdrawal of one of the components. If a reaction is at equilibrium and we alter the conditions so as to create a new , then the composition of the system will tend to change until that new equilibrium state is attained. (We say "tend to change" because if the reaction is , the change may be too slow to observe or it may never take place.) In 1884, the French chemical engineer and teacher (1850-1936) showed that in every such case, the new equilibrium state is one that partially reduces the effect of the change that brought it about. This law is known to every Chemistry student as the . His original formulation was somewhat complicated, but a reasonably useful paraphrase of it reads as follows: To see how this works (and you do so, as this is of such fundamental importance that you simply cannot do any meaningful chemistry without a thorough working understanding of this principle), look again at the hydrogen iodide dissociation reaction \[2 HI \rightarrow H_2 + I_2\] Consider an arbitrary mixture of these three components at equilibrium, and assume that we inject more hydrogen gas into the container. Because the H concentration now exceeds its new equilibrium value, the system is no longer in its equilibrium state, so a net reaction now ensues as the system moves to the new state. The Le Chatelier principle states that the net reaction will be in a direction that tends to reduce the effect of the added H . This can occur if some of the H is consumed by reacting with I to form more HI; in other words, a net reaction occurs in the reverse direction. Chemists usually simply say that "the equilibrium shifts to the left". To get a better idea of how this works, carefully examine the diagram below which follows the concentrations of the three components of this reaction as they might change in time (the time scale here will typically be about an hour): At the left, the concentrations of the three components do not change with time because the system is at equilibrium. We then add more hydrogen to the system, disrupting the equilibrium. A net reaction then ensues that moves the system to a new equilibrium state (right) in which the quantity of hydrogen iodide has increased; in the process, some of the I and H are consumed. Notice that the new equilibrium state contains more hydrogen than did the initial state, but not as much as was added; as the Le Chatelier principle predicts, the change we made (addition of H ) has been partially counteracted by the "shift to the right". Table \(\Page {1}\) Contains several examples showing how changing the quantity of a reaction component can shift an established equilibrium. Virtually all chemical reactions are accompanied by the liberation or uptake of heat. If we regard heat as a "reactant" or "product" in an endothermic or exothermic reaction respectively, we can use the Le Chatelier principle to predict the direction in which an increase or decrease in temperature will shift the equilibrium state. Thus for the oxidation of nitrogen, an endothermic process, we can write \[\text{[heat]} + N_2 + O_2 \rightleftharpoons 2 NO\] Suppose this reaction is at equilibrium at some temperature \(T_1\) and we raise the temperature to \(T_2\). The Le Chatelier principle tells us that a net reaction will occur in the direction that will partially counteract this change. Since the reaction is endothermic, a shift of the equilibrium to the right will take place. Nitric oxide, the product of this reaction, is a major air pollutant which initiates a sequence of steps leading to the formation of atmospheric smog. Its formation is an unwanted side reaction which occurs when the air (which is introduced into the combustion chamber of an engine to supply oxygen) gets heated to a high temperature. Designers of internal combustion engines now try, by various means, to limit the temperature in the combustion region, or to restrict its highest-temperature part to a small volume within the combustion chamber. You will recall that if the pressure of a gas is reduced, its volume will increase; pressure and volume are inversely proportional. With this in mind, suppose that the reaction \[2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}\] is in equilibrium at some arbitrary temperature and pressure, and that we double the pressure, perhaps by compressing the mixture to a smaller volume. From the Le Chatelier principle we know that the equilibrium state will change to one that tends to counteract the increase in pressure. This can occur if some of the NO reacts to form more of the dinitrogen tetroxide, since two moles of gas are being removed from the system for every mole of N O formed, thereby decreasing the total volume of the system. Thus increasing the pressure will shift this equilibrium to the right. \[Δn_g = (n_{products} – n_{reactants}) = 1 – 2 = –1.\] In the case of the nitrogen oxidation reaction \[N_2 + O_2 \rightleftharpoons 2 NO\] Δ = 0 and changing pressure will have no effect on the equilibrium. The volumes of solids and liquids are hardly affected by the pressure at all, so for reactions that do not involve gaseous substances, the effects of pressure changes are ordinarily negligible. Exceptions arise under conditions of very high pressure such as exist in the interior of the Earth or near the bottom of the ocean. A good example is the dissolution of calcium carbonate \[CaCO_{3(s)} \rightleftharpoons Ca^{2+} + CO_3^{2–}\] There is a slight decrease in the volume when this reaction takes place, so an increase in the pressure will shift the equilibrium to the right, with the results that calcium carbonate becomes more soluble at higher pressures. The skeletons of several varieties of microscopic organisms that inhabit the top of the ocean are made of CaCO , so there is a continual rain of this substance toward the bottom of the ocean as these organisms die. As a consequence, the floor of the Atlantic ocean is covered with a blanket of calcium carbonate. This is not true for the Pacific ocean, which is deeper; once the skeletons fall below a certain depth, the higher pressure causes them to dissolve. Some of the seamounts (undersea mountains) in the Pacific extend above the solubility boundary so that their upper parts are covered with CaCO sediments. The effect of pressure on a reaction involving substances whose boiling points fall within the range of commonly encountered temperature will be sensitive to the states of these substances at the temperature of interest. For reactions involving gases, changes in the partial pressures of those gases directly involved in the reaction are important; the presence of other gases has no effect. The commercial production of hydrogen is carried out by treating natural gas with steam at high temperatures and in the presence of a catalyst (“steam reforming of methane”): \[CH_4 + H_2O \rightleftharpoons CH_3OH + H_2 \nonumber\] Given the following boiling points: CH (methane) = –161°C, H O = 100°C, CH OH = 65°, H = –253°C, predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C. Calculate the change in the moles of gas for each process: 50° The Haber process for the synthesis of ammonia is based on the exothermic reaction N + 3 H → 2 NH Δ = –92 kJ/mol The Le Chatelier principle tells us that in order to maximize the amount of product in the reaction mixture, it should be carried out at high pressure and low temperature. However, the lower the temperature, the slower the reaction (this is true of virtually all chemical reactions.) As long as the choice had to be made between a low yield of ammonia quickly or a high yield over a long period of time, this reaction was infeasible economically. Nitrogen is available for free, being the major component of air, but the strong triple bond in N makes it extremely difficult to incorporate this element into species such as NO and NH which serve as the starting points for the wide variety of nitrogen-containing compounds that are essential for modern industry. This conversion is known as , and because nitrogen is an essential plant nutrient, modern intensive agriculture is utterly dependent on huge amounts of fixed nitrogen in the form of fertilizer. Until around 1900, the major source of fixed nitrogen was the NaNO found in extensive deposits in South America. Several chemical processes for obtaining nitrogen compounds were developed in the early 1900's, but they proved too inefficient to meet the increasing demand. Although the direct synthesis of ammonia from its elements had been known for some time, the yield of product was found to be negligible. In 1905, Fritz Haber (1868-1934) began to study this reaction, employing the thinking initiated by Le Chatelier and others, and the newly-developing field of thermodynamics that served as the basis of these principles. From the Le Chatelier law alone, it is apparent that this exothermic reaction is favored by low temperature and high pressure. However, it was not as simple as that: the rate of any reaction increases with the temperature, so working with temperature alone, one has the choice between a high product yield achieved only very slowly, or a very low yield quickly. Further, the equipment and the high-strength alloy steels need to build it did not exist at the time. Haber solved the first problem by developing a that would greatly speed up the reaction at lower temperatures. The second problem, and the development of an efficient way of producing hydrogen, would delay the practical implementation of the process until 1913, when the first plant based on the Haber-Bosch process (as it is more properly known, Carl Bosch being the person who solved the major engineering problems) came into operation. The timing could not have been better for Germany, since this country was about to enter the First World War, and the Allies had established a naval blockade of South America, cutting off the supply of nitrate for the the German munitions industry. Bosch's plant operated the ammonia reactor at 200 atm and 550°C. Later, when stronger alloy steels had been developed, pressures of 800-1000 atm became common. The source of hydrogen in modern plants is usually natural gas, which is mostly methane: The Haber-Bosch process is considered the most important chemical synthesis developed in the 20th century. Besides its scientific importance as the first large-scale application of the laws of chemical equilibrium, it has had tremendous economic and social impact; without an inexpensive source of fixed nitrogen, the intensive crop production required to feed the world's growing population would have been impossible. Haber was awarded the 1918 Nobel Prize in Chemistry in recognition of his work. Carl Bosch, who improved the process, won the Nobel Prize in 1931. Many of the chemical reactions that occur in living organisms are regulated through the Le Chatelier principle. Few of these are more important to warm-blooded organisms than those that relate to aerobic respiration, in which oxygen is transported to the cells where it is combined with glucose and metabolized to carbon dioxide, which then moves back to the lungs from which it is expelled. hemoglobin + O oxyhemoglobin The partial pressure of O in the air is 0.2 atm, sufficient to allow these molecules to be taken up by hemoglobin (the red pigment of blood) in which it becomes loosely bound in a complex known as oxyhemoglobin. At the ends of the capillaries which deliver the blood to the tissues, the O concentration is reduced by about 50% owing to its consumption by the cells. This shifts the equilibrium to the left, releasing the oxygen so it can diffuse into the cells. Carbon dioxide reacts with water to form a weak acid H CO which would cause the blood pH to fall to dangerous levels if it were not promptly removed as it is excreted by the cells. This is accomplished by combining it with carbonate ion through the reaction \[H_2CO_3 + CO_3^{2–} \rightleftharpoons 2 HCO_3^– \nonumber\] which is forced to the right by the high local CO concentration within the tissues. Once the hydrogen carbonate (bicarbonate) ions reach the lung tissues where the CO partial pressure is much smaller, the reaction reverses and the CO is expelled. Carbon monoxide, a product of incomplete combustion that is present in automotive exhaust and cigarette smoke, binds to hemoglobin 200 times more tightly than does O . This blocks the uptake and transport of oxygen by setting up a competing equilibrium O -hemoglobin hemoglobin CO-hemoglobin Air that contains as little as 0.1 percent carbon monoxide can tie up about half of the hemoglobin binding sites, reducing the amount of O reaching the tissues to fatal levels. Carbon monoxide poisoning is treated by administration of pure O which promotes the shift of the above equilibrium to the left. This can be made even more effective by placing the victim in a hyperbaric chamber in which the pressure of O can be made greater than 1 atm.

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Gases, liquids, and especially solids surround us and give form to our world. Chemistry at its most fundamental level is about atoms and the forces that act between them to form larger structural units. But the matter that we experience with our senses is far removed from this level. This unit will help you see how these properties of matter depend on the particles of which it is composed. What distinguishes solids, liquids, and gases– the three major — from each other? Let us begin at the microscopic level, by reviewing what we know about gases, the simplest state in which matter can exist. At ordinary pressures, the molecules of a gas are so far apart that intermolecular forces have an insignificant effect on the random thermal motions of the individual particles. As the temperature decreases and the pressure increases, intermolecular attractions become more important, and there will be an increasing tendency for molecules to form temporary clusters. These are so short-lived, however, that even under extreme conditions, gases cannot be said to possess “structure” in the usual sense. The contrast at the level between solids, liquids and gases is most clearly seen in the simplified schematic views above. The molecular units of crystalline solids tend to be highly ordered, with each unit occupying a fixed position with respect to the others. In liquids, the molecules are able to slip around each other, introducing an element of disorder and creating some void spaces that decrease the density. Gases present a picture of almost total disorder, with practically no restrictions on where any one molecule can be. Having lived our lives in a world composed of solids, liquids, and gases, few of us ever have any difficulty deciding into which of these categories a given sample of matter falls. Our decision is most commonly based on purely visual cues: Our experience also tells us that these categories are quite distinct; a , which you will recall is a region of matter having uniform intensive properties, is either a gas, a liquid, or a solid. Thus the three states of matter are not simply three points on a continuum; when an ordinary solid melts, it usually does so at a definite temperature, without apparently passing through any states that are intermediate between a solid and a liquid. Although these common-sense perceptions are usually correct, they are not infallible, and in fact there are solids such as glasses and many plastics that do not have sharp melting points, but instead undergo a gradual transition from solid to liquid known as , and when subject to enough pressure, solids can exhibit something of the flow properties of liquids (glacial ice, for example). A more scientific approach would be to compare the macroscopic physical properties of the three states of matter, but even here we run into difficulty. It is true, for example, that the density of a gas is usually about a thousandth of that of the liquid or solid at the same temperature and pressure; thus one gram of water vapor at 100°C and 1 atm pressure occupies a volume of 1671 mL; when it condenses to liquid water at the same temperature, it occupies only 1.043 mL. Table \(\Page {1}\) compares the molar volumes of in its three states. For the gaseous state, P = 1 atm and T = 0°C. The excluded volume is the volume actually taken up by the neon atoms according to the equation of state model. It is this extreme contrast with the gaseous states that leads to the appellation “ ” for liquids and solids. However, gases at very high pressures can have densities that exceed those of other solid and liquid substances, so density alone is not a sufficiently comprehensive criterion for distinguishing between the gaseous and condensed states of matter. Similarly, the density of a is usually greater than that of the corresponding liquid at the same temperature and pressure, but not always: you have certainly seen ice floating on water! Compare the density of gaseous xenon (molar mass 131 g) at 100 atm and 0°C with that of a hydrocarbon liquid for which \(ρ = 0.104\, g/mL\) at the same temperature. For simplicity, we will assume that xenon approximates an ideal gas under these conditions, which it really does not. The ideal molar volume at 0° C and 1 atm is 22.4 L; at 100 atm, this would be reduced to 0.22 L or 220 mL, giving a density \[ρ = \dfrac{131\, g}{224\, mL} = 0.58\, g/mL. \nonumber\] In his autobiographical , the physician/author Oliver Sacks describes his experience with xenon-filled balloons of [imagined]. Other physical properties, such as the compressibility, surface tension, and viscosity, are somewhat more useful for distinguishing between the different states of matter. Even these, however, provide no well-defined dividing lines between the various states. Rather than try to develop a strict scheme for classifying the three states of matter, it will be more useful to simply present a few generalizations. Some of these deal with macroscopic properties (that is, properties such as the that relate to ), and others with microscopic properties that refer to the individual molecular units. Even the most casual inspection of the above table shows that solids and liquids possess an important commonality that distinguishes them from gases: . As a consequence, these generally possess much higher than gases. In our study of gases, we showed that the macroscopic properties of a gas (the pressure, volume, and temperature) are related through an , and that for the limiting case of an ideal gas, this equation of state can be derived from the relatively small set of assumptions of the kinetic molecular theory. To the extent that a volume of gas consists mostly of empty space, all gases have very similar properties. Equations of state work for gases because gases consist mostly of empty space, so intermolecular interactions can be largely neglected. In condensed matter, these interactions dominate, and they tend to be unique to each particular substance, so there is no such thing as a generally useful equation of state of liquids and solids. Is there a somewhat more elaborate theory that can predict the behavior of the other two principal states of matter, liquids and solids? Very simply, the answer is "no"; despite much effort, no one has yet been able to derive a general equation of state for condensed states of matter. The best one can do is to construct models based on the imagined interplay of attractive and repulsive forces, and then test these models by computer simulation. Nevertheless, the very factors that would seem to make an equation of state for liquids and solids impossibly complicated also give rise to new effects that are easily observed, and which ultimately define the distinguishing characteristics of the gaseous, liquid, and solid states of matter. In this unit, we will try to learn something about these distinctions, and how they are affected by the chemical constitution of a substance. Crystalline solids and gases stand at the two extremes of the spectrum of perfect order and complete chaos. Liquids display elements of both qualities, and both in limited and imperfect ways. Liquids and solids share most of the properties of having their molecular units in direct contact as discussed in the previous section on condensed states of matter. At they same time, liquids, like gases, are fluids, meaning that their molecular units can move more or less independently of each other. But whereas the of a gas depends entirely on the pressure (and thus generally on the volume within which it is confined), the volume of a liquid is largely independent of the pressure. Here we offer just enough to help you see how they relate to the other major states of matter. Of the four ancient elements of "fire, air, earth and water", it is the many forms of solids ("earths") that we encounter in daily life and which give form, color and variety to our visual world. The solid state, being the form of any substance that prevails at lower temperatures, is one in which thermal motion plays an even smaller role than in liquids. The thermal kinetic energy that the individual molecular units do have at temperatures below their melting points allows them to oscillate around a fixed center whose location is determined by the balance between local forces of attraction and repulsion due to neighboring units, but only very rarely will a molecule jump out of the fixed space allotted to it in the lattice. Thus solids, unlike liquids, exhibit , and , and possess definite . Most people who have lived in the world long enough to read this have already developed a rough way of categorizing solids on the basis of macroscopic properties they can easily observe; everyone knows that a piece of metal is fundamentally different from a rock or a chunk of wood. Unfortunately, nature's ingenuity is far too versatile to fit into any simple system of classifying solids, especially those composed of more than a single chemical substance. The most commonly used classification is based on the kinds of forces that join the molecular units of a solid together. We can usually distinguish four major categories on the basis of properties such as general appearance, hardness, and melting point. It's important to understand that these four categories are in a sense idealizations that fail to reflect the diversity found in nature. The triangular diagram shown here illustrates this very nicely by displaying examples of binary compounds whose properties suggest that they fall somewhere other than at a vertex of the triangle. The triangle shown above excludes what is probably the largest category: molecular solids that are bound by van der Waals forces. One way of including these is to expand the triangle to a tetrahedron (the so-called ). Although this illustrates the concept, it is visually awkward to include many examples of the intermediate cases. Solids, like the other states of matter, can be classified according to whether their fundamental molecular units are atoms, electrically-neutral molecules, or ions. But solids possess an additional property that gases and liquids do not: an enduring structural arrangement of their molecular units. Over-simplifying only a bit, we can draw up a rough classification of solids according to the following scheme: Notice how the boiling points in the following selected examples reflect the major type of attractive force that binds the molecular units together. Bear in mind, however, that more than one type of attractive force can be operative in many substances. In a solid comprised of identical molecular units, the most favored (lowest potential energy) locations occur at regular intervals in space. If each of these locations is actually occupied, the solid is known as a crystal. What really defines a crystalline solid is that its structure is composed of repeating each containing a small number of molecular units bearing a fixed geometric relation to one another. The resulting long-range order defines a three-dimensional geometric framework known as a Geometric theory shows that only fourteen different types of lattices are possible in three dimensions, and that just six different unit cell arrangements can generate these lattices. The regularity of the external faces of crystals, which in fact correspond to lattice planes, reflects the long-range order inherent in the underlying structure. Perfection is no more attainable in a crystal than in anything else; real crystals contain of various kinds, such as lattice positions that are either vacant or occupied by impurities, or by abrupt displacements or dislocations of the lattice structure. Most pure substances, including the metallic elements, form crystalline solids. But there are some important exceptions. In metals the valence electrons are free to wander throughout the solid, instead of being localized on one atom and shared with a neighboring one. The valence electrons behave very much like a mobile fluid in which the fixed lattice of atoms is immersed. This provides the ultimate in electron sharing, and creates a very strong binding effect in solids composed of elements that have the requisite number of electrons in their valence shells. The characteristic physical properties of metals such as their ability to bend and deform without breaking, their high thermal and electrical conductivities and their metallic sheen are all due to the fluid-like behavior of the valence electrons. Recall that a "molecule" is defined as a discrete aggregate of atoms bound together sufficiently tightly (that is, by forces) to allow it to retain its individuality when the substance is dissolved, melted, or vaporized. The two words italicized in the preceding sentence are important; implies that the forces acting between atoms the molecule are much stronger than those acting molecules, and the of covalent bonding confers on each molecule a distinctive shape which affects a number of its properties. Most compounds of carbon — and therefore, most chemical substances, fall into this category. Many simpler compounds also form molecules; H O, NH , CO , and PCl are familiar examples. Some of the elements, such as H , O , O , P and S also occur as discrete molecules. Liquids and solids that are composed of molecules are held together by , and many of their properties reflect this weak binding. Thus molecular solids tend to be soft or deformable, have low melting points, and are often sufficiently volatile to evaporate (sublime) directly into the gas phase; the latter property often gives such solids a distinctive odor. is a good example of a volatile molecular crystal. The solid (mp 114° C , bp 184°) consists of I molecules bound together only by dispersion forces. If you have ever worked with solid iodine in the laboratory, you will probably recall the smell and sight of its purple vapor which is easily seen in a closed container. Because dispersion forces and the other van der Waals forces increase with the number of atoms, larger molecules are generally less volatile, and have higher melting points, than do the smaller ones. Also, as one moves down a column in the periodic table, the outer electrons are more loosely bound to the nucleus, increasing the polarizibility of the atom and thus its susceptibility to van der Waals-type interactions. This effect is particularly apparent in the progression of the boiling points of the successively heavier noble gas elements. These are a class of extended-lattice compounds (see Section 6 below) in which each atom is covalently bonded to its nearest neighbors. This means that the entire crystal is in effect one super-giant “molecule”. The extraordinarily strong binding forces that join all adjacent atoms account for the extreme hardness of such substances; these solids cannot be broken or abraded without cleaving a large number of covalent chemical bonds. Similarly, a covalent solid cannot “melt” in the usual sense, since the entire crystal is its own giant molecule. When heated to very high temperatures, these solids usually decompose into their elements. Diamond is the hardest material known, defining the upper end of the 1-10 scale known as . Diamond cannot be melted; above 1700°C it is converted to graphite the more stable form of carbon. The diamond unit cell is face-centered cubic and contains 8 carbon atoms. The four darkly shaded ones are contained within the cell and are completely bonded to other members of the cell. The other carbon atoms (6 in faces and 4 at corners) have some bonds that extend to atoms in other cells. (Two of the carbons nearest the viewer are shown as open circles in order to more clearly reveal the bonding arrangement.) BN is similar to carbon in that it exists as a diamond-like cubic polymorph as well as in a hexagonal form analogous to graphite. Cubic BN is the second hardest material after diamond, and finds use in industrial abrasives and cutting tools. Recent interest in BN has centered on its carbon-like ability to form nanotubes and related nanostructures. SiC is also known as . Its structure is very much like that of diamond with every other carbon replaced by silicon. On heating at atmospheric pressure, it decomposes at 2700°C, but has never been observed to melt. Structurally, it is very complex; at least 70 crystalline forms have been identified. Its extreme hardness and ease of synthesis have led to a diversity of applications — in cutting tools and abrasives, high-temperature semiconductors, and other high-temperature applications, manufacture of specialty steels, jewelry, and many more. Silicon carbide is an extremely rare mineral on the earth, and comes mostly from meteorites which are believed to have their origins in carbonaceous stars. The first synthetic SiC was made accidently by E.G. Acheson in 1891 who immediately recognized its industrial prospects and founded the Carborundum Co. WC is probably the most widely-encountered covalent solid owing to its use in "carbide" cutting tools and as the material used to make the rotating balls in ball-point pens. It's high-melting (2870°C) form has a structure similar to that of diamond and is only slightly less hard. In many of its applications it is embedded in a softer matrix of cobalt or coated with titanium compounds. In some solids there is so little long-range order that the substance cannot be considered crystalline at all; such a solid is said to be . Amorphous solids possess short-range order but are devoid of any organized structure over longer distances; in this respect they resemble liquids. However, their rigidity and cohesiveness allow them to retain a definite shape, so for most practical purposes they can be considered to be solids. Glasses refers generally to solids formed from their melts that do not return to their crystalline forms on cooling, but instead form hard, and often transparent amorphous solids. Although some organic substances such as sugar can form glasses ("rock candy"), the term more commonly describes inorganic compounds, especially those based on , SiO . Natural silica-based glasses, known as , are formed when certain volcanic magmas cool rapidly. is composed mostly of SiO , which usually exists in nature in a crystalline form known as . If quartz (in the form of sand) is melted and allowed to cool, it becomes so viscous that the molecules are unable to move to the low potential energy positions they would occupy in the crystal lattice, so that the disorder present in the liquid gets “frozen into” the solid. In a sense, glass can be regarded as a supercooled liquid. Glasses are transparent because the distances over which disorder appears are small compared to the wavelength of visible light, so there is nothing to scatter the light and produce cloudiness. Ordinary glass is made by melting silica sand to which has been added some calcium and sodium carbonates. These additives reduce the melting point and make it more difficult for the SiO2 molecules to arrange themselves into crystalline order as the mixture cools. Glass is believed to have first been made in the Middle East at least as early as 3000 BCE. Its workability and ease of coloring has made it one of mankind's most important and versatile materials. Molecules, not surprisingly, are the most common building blocks of pure substances. Most of the 15-million-plus chemical substances presently known exist as distinct molecules. Chemists commonly divide molecular compounds into "small" and "large-molecule" types, the latter usually falling into the class of (see below.) The dividing line between the two categories is not very well defined, and tends to be based more on the properties of the substance and how it is isolated and purified. We usually think of atoms as the building blocks of molecules, so the only pure substances that consist of plain atoms are those of some of the elements — mostly the metallic elements, and also the noble-gas elements. The latter do form liquids and crystalline solids, but only at very low temperatures. Although the form crystalline solids that are essentially atomic in nature, the special properties that give rise to their "metallic" nature puts them into a category of their own. Most of the non-metallic elements exist under ordinary conditions as small molecules such as O or S , or as extended structures that can have a somewhat polymeric nature. Many of these elements can form more than one kind of structure, each one stable under different ranges of temperature and pressure. Multiple structures of the same element are known as , although the more general term is now preferred. Ions, you will recall, are atoms or molecules that have one or more electrons missing (positive ions) or in excess (negative ions), and therefore possess an electric charge. A basic law of nature, the , states that bulk matter cannot acquire more than a trifling (and chemically insignificant) net electric charge. So one important thing to know about ions is that in ordinary matter, whether in the solid, liquid, or gaseous state, any positive ions must be accompanied by a compensating number of negative ions. Ionic substances such as sodium chloride form crystalline solids that can be regarded as made of ions. These solids tend to be quite hard and have high melting points, reflecting the strong forces between oppositely-charged ions. Solid metal oxides, such as CaO and MgO which are composed of doubly-charged ions don't melt at all, but simply dissociate into the elements at very high temperatures. Plastics and natural materials such as rubber or cellulose are composed of very large molecules called ; many important biomolecules are also polymeric in nature. Owing to their great length, these molecules tend to become entangled in the liquid state, and are unable to separate to form a crystal lattice on cooling. In general, it is very difficult to get such substances to form anything other than amorphous solids. actually exist in their solid forms as linked assemblies of these basic units arranged in chains or layers that extend indefinitely in one, two, or three dimensions. Thus the very simple models of chemical bonding that apply to the isolated molecules in gaseous form must be modified to account for bonding in some of these solids. The terms "one-dimensional" and "two-dimensional", commonly employed in this context, should more accurately be prefixed by " "; after all, even a single atom occupies three-dimensional space! Atoms of some elements such as sulfur and selenium can bond together in long chains of indefinite length, thus forming polymeric, amorphous solids. The most well known of these is the amorphous "plastic sulfur" formed when molten is cooled rapidly by pouring it into water.These are never the most common (or stable) forms of these elements, which prefer to form discrete molecules. Rubber-like strands of plastic sulfur formed by pouring hot molten sulfur into cold water. After a few days, it will revert to ordinary crystalline sulfur. But small molecules can also form extended chains. Sulfur trioxide is a gas above room temperature, but when it freezes at 17°C the solid forms long chains in which each S atom is coordinated to four oxygen atoms. Many inorganic substances form crystalline solids which are built up from parallel chains in which the basic formula units are linked by weak bonds involving dipole-dipole and dipole-induced dipole interactions. Neighboring chains are bound mainly by dispersion forces. Solid cadmium chloride is a good example of a structure. The Cd and Cl atoms occupy separate layers; each of these layers extends out in a third dimension to form a . The CdCl crystal is built up from stacks of these layers held together by van der Waals forces. It's worth pointing out that although salts such as CuCl and CdCl are dissociated into ions when in aqueous solution, the solids themselves should not be regarded as "ionic solids". See also this section of the lesson on ionic solids. Graphite is a polymorph of carbon and its most stable form. It consists of sheets of fused benzene rings stacked in layers. The spacing between layers is sufficient to admit molecules of water vapor and other atmospheric gases which become absorbed in the interlamellar spaces and act as lubricants, allowing the layers to slip along each other. Thus graphite itself often has a flake-like character and is commonly used as a solid lubricant, although it loses this property in a vacuum. As would be expected from its anistropic structure, the electric and thermal conductivity of graphite are much greater in directions parallel to the layers than across the layers. The melting point of 4700-5000°C makes graphite useful as a high-temperature refractory material. Graphite is the most common form of relatively pure carbon found in nature. Its name comes from the same root as the Greek word for "write" or "draw", reflecting its use as pencil "lead" since the 16th century. (The misnomer, which survives in common use, is due to its misidentification as an ore of the metallic element of the same name at a time long before modern chemistry had developed.) is a two-dimensional material consisting of a single layer of graphite — essentially "chicken wire made of carbon" that was discovered in 2004. Small fragments of graphene can be obtained by several methods; one is to attach a piece of Scotch Tape™ to a piece of graphite and then carefully pull it off (a process known as .) Fragments of graphene are probably produced whenever one writes with a pencil. Graphene has properties that are uniquely different from all other solids. It is the strongest known material, and it exhibits extremely high electrical conductivity due to its massless electrons which are apparently able to travel at relativistic velocities through the layer.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.03%3A_Colligative_Properties-_Raoult's_Law

Make sure you thoroughly understand the following essential ideas: The tendency of molecules to escape from a liquid phase into the gas phase depends in part on how much of an increase in entropy can be achieved in doing so. Evaporation of solvent molecules from the liquid always leads to a large increase in entropy because of the greater volume occupied by the molecules in the gaseous state. But if the liquid solvent is initially “diluted“ with solute, its entropy is already larger to start with, so the amount by which it can increase on entering the gas phase will be less. There will accordingly be less tendency for the solvent molecules to enter the gas phase, and so the vapor pressure of the solution diminishes as the concentration of solute increases and that of solvent decreases. The number 55.5 mol L (= 1000 g L ÷ 18 g mol ) is a useful one to remember if you are dealing a lot with aqueous solutions; this represents the concentration of water in pure water. (Strictly speaking, this is the concentration of H O; it is only the molar concentration at temperatures around 4° C, where the density of water is closest to 1.000 g cm .) Diagram 1 (above left) represents pure water whose concentration in the liquid is 55.5 M. A tiny fraction of the H O molecules will escape into the vapor space, and if the top of the container is closed, the pressure of water vapor builds up until equilibrium is achieved. Once this happens, water molecules continue to pass between the liquid and vapor in both directions, but at equal rates, so the partial pressure of H O in the vapor remains constant at a value known as the of water at the particular temperature. In Figure \(\Page {1}\), we have replaced a fraction of the water molecules with a substance that has zero or negligible vapor pressure — a such as salt or sugar. This has the effect of diluting the water, reducing its escaping tendency and thus its vapor pressure. What's important to remember is that the reduction in the vapor pressure of a solution of this kind is directly proportional to the fraction of the [volatile] solute molecules in the liquid — that is, to the mole fraction of the solvent. The reduced vapor pressure is given by (1886): \[\chi_{solvent} = 1–\chi_{solute}.\] Estimate the vapor pressure of a 40 % (W/W) solution of ordinary cane sugar (C O H , 342 g mol ) in water. The vapor pressure of pure water at this particular temperature is 26.0 torr. 100 g of solution contains (40 g) ÷ (342 g mol ) = 0.12 mol of sugar and (60 g) ÷ (18 g mol ) = 3.3 mol of water. The mole fraction of water in the solution is \[ \dfrac{3.3}{3.3 + 12} = 0.96\] and its vapor pressure will be 0.96 × 26.0 torr = 25.1 torr. The vapor pressure of water at 10° C is 9.2 torr. Estimate the vapor pressure at this temperature of a solution prepared by dissolving 1 mole of CaCl in 1 L of water. Each mole of CaCl dissociates into one mole of Ca and two moles of Cl , giving a total of three moles of solute particles. The mole fraction of water in the solution will be \[ \dfrac{55.5}{3 + 55.5} = 0.95\] The vapor pressure will be 0.95 × 9.2 torr = 8.7 torr. Since the sum of all mole fractions in a mixture must be unity, it follows that the more moles of solute, the smaller will be the mole fraction of the solvent. Also, if the solute is a salt that dissociates into ions, then the proportion of solvent molecules will be even smaller.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.05%3A_Reduction_of_Metals

The ease with which a metal may be obtained from its ore varies considerably from one metal to another. Since the majority of ores are oxides or can be converted to oxides by roasting, the free-energy change accompanying the decomposition of the oxide forms a convenient measure of how readily a metal may be obtained from its ore. Values of the free energy change per mol O produced are given in the table for a representative sample of metals at 298 and 2000 K. A high positive value of Δ ° in this table indicates a very stable oxide from which it is difficult to remove the oxygen and obtain the metal, while a negative value of Δ ° indicates an oxide which will spontaneously decompose into its elements. Note how the value of Δ ° decreases with temperature in each case. This is because a gas (oxygen) is produced by the decomposition, and ΔS is accordingly positive. The two metals in the table which are easiest to obtain from their oxide ores are Hg and Ag. Since the Δ ° value for the decomposition of these oxides becomes negative when the temperature is raised, simple heating will cause them to break up into O and the metal. The next easiest metals to obtain are Sn and Fe. These can be reduced by coke, an impure form of C obtained by heating coal. Coke is the cheapest readily obtainable reducing agent which can be used in metallurgy. When C is oxidized to CO , the free-energy change is close to – 395 kJ mol over a wide range of temperatures. This fall in free energy is not quite enough to offset the free-energy rise when Fe O and SnO are decomposed at 298 K, but is more than enough if the temperature is 2000 K. Thus, for example, if Fe O is reduced by C at 2000 K, we have, from Hess’ law, \({}_{\text{3}}^{\text{2}}\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{(}s\text{) }\to \text{ }{}_{\text{3}}^{\text{4}}\text{Fe(}l\text{) + O}_{\text{2}}\text{(}g\text{)}\) Δ ° = +314 kJ mol \(\text{C(}s\text{) + O}_{\text{2}}\text{(}g\text{)}\to \text{ CO}_{\text{2}}\text{(}g\text{)}\) Δ ° = –394 kJ mol \({}_{\text{3}}^{\text{2}}\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{(}s\text{) + C(}s\text{) }\to \text{ }{}_{\text{3}}^{\text{4}}\text{Fe(}l\text{) + CO}_{\text{2}}\) Δ ° = –82 kJ mol Thus Δ ° for the reduction is negative, and the reaction is spontaneous. The two metals in the table which are most difficult to obtain from their ores are Mg and Al. Since they cannot be reduced by C or any other readily available cheap reducing agent, they must be reduced electrolytically. The electrolytic reduction of bauxite to yield Al (the Hall process) is used to . Since iron is the most important metal in our industrial civilization, its reduction from iron ore in a blast furnace (Figure \(\Page {1}\) ) deserves a detailed description. The oxides present in most iron ores are Fe O and Fe O . These oxides are reduced stepwise: first to FeO and then to Fe. Ore, coke, and limestone are charged to the furnace through an airlock-type pair of valves at the top. Near the bottom a blast of air, preheated to 900 to 1000 K, enters through blowpipes called tuyères. Oxygen from the air blast reacts with carbon in the coke to form carbon monoxide and carbon dioxide, releasing considerable heat. The blast carries these gases up through the ore, coke, and limestone, and they exit from the top of the furnace. By the time the ore works its way to the lower part of the furnace, most of the Fe O has already been reduced to FeO. In this region, temperatures reach 1600 to 2000 K, high enough to melt FeO and bring it into close contact with the coke. Most of the FeO is reduced by direct reaction with carbon, the latter being oxidized to carbon monoxide: \[\text{2C}(s) + \text{2FeO}(l) \rightarrow \text{2Fe}(l) + \text{2CO} \triangle G_m^o(2000 k) = -280 \frac{kJ}{mol} \nonumber \] The molten iron produced by this reaction drips to the bottom of the furnace where it is collected and occasionally tapped off. Higher in the furnace, temperatures fall below the melting points of the iron oxides. Because there is little contact between solid chunks of ore and of coke, direct reduction by solid carbon is rather slow. Gaseous carbon monoxide contacts all parts of the ore, however, and reacts much more rapidly: \[\text{CO}(g) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{CO}_2 (g) \text{2FeO}(s) \nonumber \] \(\text{CO}(g) + \text{FeO}(s) \rightarrow \text{CO}_2(g) + \text{Fe}(s)\) Thus much of the “carbon reduction” in ironmaking is actually carried out by carbon monoxide. The gangue in iron ore consists mainly of silicates and silica, SiO . These impurities are removed in slag. Limestone added with coke and ore is calcined (decomposed to the oxide) by the high temperatures of the blast furnace: \[\text{CaCO}_3(s) \underset{\text{1100 K}}{\mathop{\rightarrow}}\, \text{CaO}(s) + \text{CO}_2(g) \nonumber \] Lime (CaO) serves as a flux, reducing the melting points (mp) of silica (SiO ) and silicates: \[\underset{\text{mp = 2853 K}}{\mathop{\text{CaO(}s\text{)}}}\,\text{ + }\underset{\text{mp = 1986 K}}{\mathop{\text{SiO}_{\text{2}}\text{(}s\text{)}}}\,\text{ }\to \text{ }\underset{\text{mp = 1813 K}}{\mathop{\text{CaSiO}_{\text{3}}\text{(}l\text{)}}} \nonumber \] The liquid silicates flow rapidly down through the hottest part of the furnace. This helps to prevent reduction of silica to silicon, hence yielding purer iron. The slag is less dense than molten iron and immiscible with it. Therefore the slag floats on the surface of the iron and can easily be tapped off. Although most blast-furnace iron now goes directly to a steelmaking furnace in molten form, much of it used to be run into molds where it hardened into small ingots called pigs because of their shape. Consequently blast-furnace iron is still referred to as pig iron. A single large blast furnace may produce more than 10 kg iron per day. For each kilogram of iron, 2 kg iron ore, 1 kg coke, 0.3 kg limestone, 4 kg air, 63 kg water, and 19 MJ of fossil-fuel energy are required. The furnace produces 0.6 kg slag and 5.7 kg, flue gas per kg iron. Nearly 5 percent of the iron ore is lost in the form of small particles suspended in the flue gas unless, as in the furnace shown in Figure 1, air-pollution controls are installed. The latter trap FeO particles for recycling to the furnace and also make the flue gas (which contains about 12% CO and 1% H ) suitable as a fuel for preheating air fed to the tuyères. Thus control of blast-furnace air pollution (a major contributor to the one-time “smoky city” reputations of Pittsburgh, Pennsylvania and Gary, Indiana) also conserves ore supplies and energy resources.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/29%3A_Polymers

Polymers are substances made up of recurring structural units, each of which can be regarded as derived from a specific compound called a . The number of monomeric units usually is large and variable, each sample of a given polymer being characteristically a mixture of molecules with different molecular weights. The range of molecular weights is sometimes quite narrow, but is more often very broad. The concept of polymers being mixtures of molecules with long chains of atoms connected to one another seems simple and logical today, but was not accepted until the 1930's when the results of the extensive work of H. Staudinger, who received the Nobel Prize in Chemistry in 1953, finally became appreciated. Prior to Staudinger's work, polymers were believed to be colloidal aggregates of small molecules with quite nonspecific chemical structures. and (1977)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.05%3A_Equilibrium_Calculations

The five Examples presented above were carefully selected to span the range of problem types that students enrolled in first-year college chemistry courses are expected to be able to deal with. If you are able to reproduce these solutions on your own, you should be well prepared on this topic. This page presents examples that cover most of the kinds of equilibrium problems you are likely to encounter in a first-year university course. Reading this page will teach you how to work equilibrium problems! The only one who can teach you how to interpret, understand, and solve problems is . So don't just "read" this and think you are finished. You need to find and solve similar problems on your own. Look over the problems in your homework assignment or at the end of the appropriate chapter in a textbook, and see how they fit into the general types described below. When you can solve them without looking at the examples below, you will be well on your way! Clearly, if the concentrations or pressures of all the components of a reaction are known, then the value of \(K\) can be found by simple substitution. Observing individual concentrations or partial pressures directly may be not always be practical, however. If one of the components is colored, the extent to which it absorbs light of an appropriate wavelength may serve as an index of its concentration. Pressure measurements are ordinarily able to measure only the total pressure of a gaseous mixture, so if two or more gaseous products are present in the equilibrium mixture, the partial pressure of one may need to be inferred from that of the other, taking into account the stoichiometry of the reaction. In an experiment carried out by Taylor and Krist ( ), hydrogen iodide was found to be 22.3% dissociated at 730.8 K. Calculate \(K_c\) for \[\ce{ 2 HI(g) <=> H2(g) + I2} \nonumber \] No explicit molar concentrations are given, but we do know that for every \(n\) moles of \(\ce{HI}\), \(0.223n\) moles of each product is formed and \((1–0.223)n = 0.777n\) moles of \(\ce{HI}\) remains. For simplicity, we assume that \(n=1\) and that the reaction is carried out in a 1.00-L vessel, so that we can substitute the required concentration terms directly into the equilibrium expression for \(K_c\). \[\begin{align*} K_c &= \dfrac{[\ce{H2},\ce{I2}]}{[\ce{HI}]^2} \\[4pt] &= \dfrac{(0.223)(0.223)}{(0.777)^2} \\[4pt] &= 0.082 \end{align*}\] Ordinary white phosphorus, \(\ce{P4}\), forms a vapor which dissociates into diatomic molecules at high temperatures: \[\ce{P4(g) <=> 2 P2(g)} \nonumber\] A sample of white phosphorus, when heated to 1000°C, formed a vapor having a total pressure of 0.20 atm and a density of 0.152 g L . Use this information to evaluate the equilibrium constant \(K_p\) for this reaction. Before worrying about what the density of the gas mixture has to do with \(K_p\), start out in the usual way by laying out the information required to express \(K_p\) in terms of an unknown \(x\) The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure: \[P_i = \chi_i P_{tot} \nonumber\] with the term in the denominator of each mole fraction is the total number of moles of gas present at equilibrium: \[P_{tot} = 1-x + 2x = 1+x \nonumber\] Expressing the equilibrium constant in terms of \(x\) gives \[ \begin{align*} K_p &= \dfrac{p^2_{P_2}}{p_{P_4}} \\[4pt] &= \dfrac{\left(\dfrac{2x}{1+x}\right)^2 0.2^2} {\left(\dfrac{1-x}{1+x}\right) 0.2} \\[4pt] &= \left(\dfrac{4x^2}{(1-x)(1+x)}\right) 0.2 \\[4pt] &= \left(\dfrac{4x^2}{1+x^2}\right) 0.2 \end{align*}\] Now we need to find the dissociation fraction \(x\) of \(P_4\), and at this point we hope you remember those gas laws that you were told you would be needing later in the course! The density of a gas is directly proportional to its molecular weight, so you need to calculate the densities of pure \(P_4\) and pure \(P_2\) vapors under the conditions of the experiment. One of these densities will be greater than 0.152 gL and the other will be smaller; all you need to do is to find where the measured density falls in between the two limits, and you will have the dissociation fraction. The molecular weight of phosphorus is 31.97, giving a molar mass of 127.9 g for \(\ce{P4}\). This mass must be divided by the volume to find the density; assuming ideal gas behavior, the volume of 127.9 g (1 mole) of \(\ce{P4}\) is given by , which works out to 522 L (remember to use the absolute temperature here.) The density of pure \(\ce{P4}\) vapor under the conditions of the experiment is then \[\rho = \dfrac{m}{V} = (128\; g \;mol^{–1}) \times x = (522\; L mol^{–1}) = 0.245\; g\; L^{–1} \nonumber\] The density of pure \(P_2\) would be half this, or 0.122 g L . The difference between these two limiting densities is 0.123 g L , and the difference between the density of pure \(\ce{P4}\) and that of the equilibrium mixture is (0.245 – 0.152) g L or 0.093 g L . The ratio 0.093/0.123 = 0.76 is therefore the fraction of \(\ce{P4}\) that remains and its fractional dissociation is (1 – 0.76) = 0.24. Substituting into the equilibrium expression above gives \(K_p = 1.2\). Solve Example \(\Page {2}\) by using a different set of initial conditions to demonstrated that the initial conditions indeed have no effect on determining the Equilibrium state and \(K_p\). This is by far the most common kind of equilibrium problem you will encounter: starting with an arbitrary number of moles of each component, how many moles of each will be present when the system comes to equilibrium? The principal source of confusion and error for beginners relates to the need to determine the values of several unknowns (a concentration or pressure for each component) from a single equation, the equilibrium expression. The key to this is to make use of the stoichiometric relationships between the various components, which usually allow us to express the equilibrium composition in terms of a single variable. The easiest and most error-free way of doing this is adopt a systematic approach in which you create and fill in a small table as shown in the following problem example. You then substitute the equilibrium values into the equilibrium constant expression, and solve it for the unknown. This very often involves solving a quadratic or higher-order equation. Quadratics can of course be solved by using the familiar quadratic formula, but it is often easier to use an algebraic or graphical approximation, and for higher-order equations this is the only practical approach. There is almost never any need to get an exact answer, since the equilibrium constants you start with are rarely known all that precisely anyway. Phosgene (\(\ce{COCl2}\)) is a poisonous gas that dissociates at high temperature into two other poisonous gases, carbon monoxide and chlorine. The equilibrium constant = 0.0041 at 600°K. Find the equilibrium composition of the system after 0.124 atm of \(\ce{COCl2}\) is allowed to reach equilibrium at this temperature. First we need a balanced chemical reaction \[\ce{COCl_2 <=> CO(g) + Cl2(g)} \nonumber\] Start by drawing up a table showing the relationships between the components: Substitution of the equilibrium pressures into the equilibrium expression gives \[ \dfrac{x^2}{0.124 - x} = 0.0041 \nonumber\] This expression can be rearranged into standard polynomial form \[x^2 +0.0041 x – 0.00054 = 0 \nonumber\] and solved by the quadratic formula, but we will simply obtain an approximate solution by iteration. Because the equilibrium constant is small, we know that will be rather small compared to 0.124, so the above relation can be approximated by \[ \dfrac{x^2}{0.124 - x} \approx \dfrac{x^2}{0.124}= 0.0041 \nonumber\] which gives = 0.0225. To see how good this is, substitute this value of x into the denominator of the original equation and solve again: \[ \dfrac{x^2}{0.124 - 0.0225} = \dfrac{x^2}{0.102}= 0.0041 \nonumber\] This time, solving for \(x\) gives 0.0204. Iterating once more, we get \[ \dfrac{x^2}{0.124 - 0.0204} = \dfrac{x^2}{0.104}= 0.0041 \nonumber\] and = 0.0206 which is sufficiently close to the previous to be considered the final result. The final partial pressures are then 0.104 atm for COCl , and 0.0206 atm each for CO and Cl . : Using the quadratic formula to find the exact solution yields the two roots –0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good. The gas-phase dissociation of phosphorus pentachloride to the trichloride has \(K_p = 3.60\) at 540°C: \[\ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \nonumber\] What will be the partial pressures of all three components if 0.200 mole of \(\ce{PCl5}\) and 3.00 moles of \(\ce{PCl3}\) are combined and brought to equilibrium at this temperature and at a total pressure of 1.00 atm? As always, set up a table showing what you know (first two rows) and then expressing the equilibrium quantities: The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure: \[P_i = \chi_i P_{tot} \nonumber\] with the term in the denominator of each mole fraction is the total number of moles of gas present at equilibrium: \[P_{tot} = (0.200 – x) + (3.00 + x) + x = 3.20 + x \nonumber\] Substituting the equilibrium partial pressures into the equilibrium expression, we have \[ \dfrac{ (3.00 +x)(x)}{(0.200 -x)(3.20 +x)} = 3.60 \nonumber\] whose polynomial form is \[4.60x^2 + 13.80x – 2.304 = 0. \nonumber\] You can use the quadratic question to solve this or you can do it graphically (more useful for higher order equations). Plotting this on a graphical calculator yields \(x = 0.159\) as the positive root: Substitution of this root into the expressions for the equilibrium partial pressures in the table yields the following values: In the section that introduced the Le Chatelier principle, it was mentioned that diluting a weak acid such as acetic acid \(\ce{CH3COOH}\) (“\(\ce{HAc}\)”) will shift the dissociation equilibrium to the right: \[\ce{HAc + H_2O \rightleftharpoons H_3O^{+} + Ac^{–}} \nonumber\] Thus a \(0.10\,M\) solution of acetic acid is 1.3% ionized, while in a 0.01 M solution, 4.3% of the \(\ce{HAc}\) molecules will be dissociated. This is because as the solution becomes more dilute, the product [H O ,Ac ] decreases more rapidly than does the \(\ce{[HAc]}\) term. At the same time the concentration of \(\ce{H2O}\) becomes greater, but because it is so large to start with (about 55.5 M), any effect this might have is negligible, which is why no \(\ce{[H2O]}\) term appears in the equilibrium expression. For a reaction such as \[\ce{CH_3COOH (l) + C_2H_5OH (l) \rightleftharpoons CH_3COOC_2H_5 (l) + H_2O (l) }\] (in which the water concentration does change), dilution will have no effect on the equilibrium; the situation is analogous to the way the pressure dependence of a gas-phase reaction depends on the number of moles of gaseous components on either side of the equation (i.e., \(\Delta n_g\)). The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction \[\text{fructose} (aq) + \text{glucose-6-phosphate} (aq) → \text{sucrose-6-phosphate} (aq) + \ce{H2O} (l)\] (Sucrose is ordinary table sugar.) To what volume should a solution containing 0.050 mol of each monosaccharide be diluted in order to bring 5% conversion to sucrose phosphate? The equilibrium constant for this reaction is \(K_{c} = 7.1 \times 10^{-6}\) at room temperature. The initial and final numbers of moles in this equation are as follows: What is the value of \(x\)? That is when 5% of the reaction has proceeded or when 5% of the fructose (or glucose-6-P) is consumed: \[\dfrac{x}{0.05} = 0.05 \nonumber\] so \(x = 0.0025 \nonumber\). The equilibrated concentrations are then Substituting into the values in for the expression of \(K_c\) (in which the solution volume is the unknown), we have \[\begin{align*} K_{c} &= \dfrac{[\text{suc6P}]_{equil} }{[\text{fruc}]_{equil} [\text{gluc6P}]_{equil} } \\[4pt] &= \dfrac{\left(\dfrac{0.0025}{V}\right)}{\left(\dfrac{0.0475}{V}\right)^2} = 7.1 \times 10^{-6} \end{align*}\] \[V = (7.1 \times 10^{-6}) \dfrac{(0.0475)^2}{0.0025} \nonumber\] Solving for \(V\) gives a final solution volume of \(6.4 \times 10^{-4}\,L\) or \(640 \mu L\). Why so small? The reaction is not favored and to push it forward, large concentrations of reactants are needed (Le Chatelier principle in action).

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/12%3A_Woodwards_Synthesis_of_Chlorophyll

Chlorophyll, the most conspicuous of natural products, has held the fancy of organic chemists for more than two century, not only for its complex structure but also for its photochemical roll in the production of food that sustains all living things on this planet. Isolation and structure elucidation work began during the end of nineteenth century. This monumental activity that started with Willstätter, culminated in elucidation of the complete structure for Chlorophyll ‘a’ only by the middle of twentieth century. Most of these outstanding works, stretched over half a century, took place when modern organic chemistry was at its infancy. The synthesis of Chlorophyll ‘a’ by R. B. Woodward is acclaimed was an outstanding achievement in organic synthesis and ranks amongst the shinning gems in synthesis. The preliminary analysis for the synthesis, the persistent planned attach on this complex, delicate chemistry by his school and the logic of the famous Woodwardian approach are all good lessons for any discerning student of Organic Synthesis. In the cited lecture, Woodward starts his discussions on synthesis with a series of questions that delineates not only the critical features of the structure under attach, but also arrives at a plausible target and approach for synthesis of this complex molecule. The structure of Chlorophyll ‘α’ is shown in . Earlier work had established that the magnesium atom and the phytyl group were easy to remove and put in place through well established chemistry. Chlorophyll ‘α’ belongs to the green coloured pigment family . Simple chlorins are readily oxidized (lost their ‘extra’ hydrogens) to give the more conjugated porphyrins . But some chlorins like chlorophyll do so only under drastic conditions. The second feature that drew their attention is the carbocyclic five membered fused ring attached to ring III. A close study of the molecular models suggested that a porphyrin, which had a string of substituents on C5, C6, Cγ, C7 and C8 positions, would be highly crowded as shown in and . They noted that chlorins and porphyrins that had a substitution at Cγ and a carboxyl group at C6, loose C with ease, while absence of Cγ subsitution endeared the C6 carbonyl increased stability. All these features led them to conclude that in porphyrins, the space at C7 / Cγ and Cγ / C6 are most crowded. Decarboxylation reduces this strain partially . Presence of substituent at Cγ increases the strain for C7 / Cγ space as well. This crowding is perhaps partly relieved by formation of the carbocyclic ring at Cγ / C6. Furthermore, the strain at C7 / Cγ could be relieved by introduction of the ‘extra’ hydrogen at C7 and C8. The same strain factors would influence the trans- arrangement for substituents at C7 and C8. Such detailed analysis of the given structure not only helped them to understand the given structure but also suggested that the ‘Target’ for synthesis could be a porphylin like . Once it is synthesized, such a structure could be coerced to move on to take up the ‘extra’ hydrogens and move to . Some questions remained. The vinyl group on the first ring was considered as very sensitive to withstand the rigour of the projected synthesis. Hence, it was replaced with an equivalent aminopropane chain. As shown in structure , the residue needed at Cγ is an acetic acid moiety. However, mechanistic analysis of such a porphyrin suggested that such a unit would readily eliminate due to an expected ‘electronic factor’ (actually, we would now say ‘enamine-like activity’). They decided to make this unit a propionic acid residue to avoid this instability. We have now arrived at a target structure . Based on the known pyrrole chemistry, they decided to make the of the molecule independently and condense these units into a tetramer. Thus, they arrived at the following four monomer units . The pyrrole units II and III were combined to give the expected dimer unit in good yields. Acylation with β-carbomethoxypropionyl chloride gave the RHS unit . With this RHS unit on hand, they tried the crucial cyclisation with a readily available model LHS unit . On condensation under acid conditions, followed by oxidation with iodine, the required porphyrin could be obtained in acceptable 25% yield. Encouraged with this result, they went ahead to synthesize the actual LHS unit . With the LHS unit ready at hand, they looked at the cyclisation of these two units. They realized that this condensation could give two products due to two different orientations of the reactants. Though the yield of this condensation product was comparable to previous condensation, they saw some drawbacks. The yield was not acceptable for their projected purpose. Furthermore, this route resulted in the formation of isomeric products, whose structure assignment posed problems. Such ‘inelegance’ (an expression used by RBW in his lecture) was not acceptable for their group. Hence, they decided to freeze the mobility of the two units by linking the amine and the aldehyde units into a Schiff base . This masterly defined craftsmanship in molecular engineering was, however, not easy to achieve in practice. Pyrrole aldehydes were generally unreactive and could be converted to Schiff bases only under acid catalysis or buffered conditions. The problem arose from the LHS unit that proved to be very sensitive to such reaction conditions. No trace of the expected condensation product was observed. After several experiments, activation at aldehyde unit via Schiff bases followed by condensation with LHS unit via base exchange became a viable approach. But the discovery that such Schiff bases could be smoothly converted to thioaldehyde provided a breakthrough. This thioaldehyde smoothly condensed with the LHS unit under neutral conditions in good yield. This ‘extraordinarily sensitive’ Schiff base, on acid treatment gave the cation , which was isolated as the dibromide. This dication was immediately oxidized with iodine to and isolated as the acetamide on acylation with acetic anhydride. In spite of the sensitivity of the intermediates and quick subsequent steps, the porphyrin could be obtained in good yield and could be scaled up to several gram scale. When the porphyrin was heated in acetic acid in air, migration of the first hydrogen occurred to give . On heating in acetic acid at 110 0C, the chain at Cγ cyclised at C7 to give the reduced ring unit IV . Note the orientation of the two chains on ring IV is trans as expected. At this stage, attention was shifted to the vinyl group on ring I . This was readily achieved by Hoffmann exhaustive methylation-elimination sequence shown above . On exposure to light and air, the extra ring readily cleaved to give a pyruvic ester and a formyl group . The extra pyruvyl moiety was readily cleaved in methanolic KOH, which led to a methoxy lactone . Dry HCl gave the hamiacetal that could be resolved using quinine salt method to give (+)-Chlorin-5 . Diazomethane esterification exposed the aldehyde unit . Treatment of the formyl compound with HCN in triethyl amine gave the cyanolactone , which on reduction, esterification and hydrolysis with methanolic HCl gave , the precursor for a Dieckmann cyclisation product. From here to (-)-Clorophyll α was already known. A Diekmann followed by introduction of the phytyl group by enzymztic methods completed the synthesis (Fortchr. Chem. Forsch., 2, 538 (1952)). To this day, this masterpiece in organic synthesis has several feature that hold the interest for lovers of Art and Logic in organic synthesis.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/07%3A_Strategies_in_Longfolene_Synthesis

The synthesis of Longifolene has held the fascination of synthetic organic chemists for several decades. Since the compound was available in a pure form from natural sources in sufficient quantities, the fascination was purely academic. During structure elucidation studies, it was observed that this bridged structure underwent a host of migration reactions. These rearrangements were of interest both from theoretical and practical points of view. The concept of ‘disconnection’ and ‘retrosynthesis’ that was evolving around 1960’ led to the development of Logic in Organic Synthesis during this period. The structure of Longifolene was a happy exploration ground for those grand masters who were involved in these developments. For this reason synthesis of Longifolene has been closely associated with these developments. While analyzing such structures, Corey had suggested a few ‘strategic bond disconnection’ for logical approaches towards synthesis (Fig 7.1). Please note that these suggestions were meant to provide guidelines and therefore need not restrict any further innovations. It was further pointed out that any one of these disconnections could lead to several paths for the construction of synthetic trees. For example, let us consider one such disconnections at bond ‘a’ shown in Fig 7.2. This could throw open four branches on the synthetic tree. Out of these, Corey first selected the Michael strategy because one such cyclisation was already known in the chemistry of Santonin. Execution of this concept by Corey ( J. Am. Chem. Soc., 83, 2151 (1961); ibid.,86, 478 (1964)) is shown in Fig 7.3. L.W. Oppolzer et.al., set up the longifolene ring system through a rearrangement transform. He made use of De Mayo reaction, which is a photocyclisation – retroaldol sequence shown below . As shown in the synthetic scheme Figure 7.5, a [2+2] cycloaddition reaction on C gave D, which exposed the retroaldol components after an hydrogenolysis. Another useful feature of this synthesis is the utilization hydrogenolysis reaction on cyclopropane to expose a gem-dimethyl group. A chiral starting material A yielded (+)-Longifolene in 25% overall yield. A carbene insertion strategy was reported by A.G. Schutz et.al., (J. Org. Chem., 50, 915(1985)). The retroanalysis is shown in Figure 7.6. The synthetic scheme is shown in Figure 7.7. The Diels-Alder strategy indicated in the introduction to this section on Longifolene was demonstrated by Fallis et.al., Another Diels-Alder strategy came from Ho and Liu . Note the utilization of the exo- cyclic olefin by converting this unit to the required seven membered ring moiety. A cationic cyclisation strategy for Longifolene skeleton reported by Johnson’s school has some interesting features. Figure 7.10 depicts the cyclisation reaction that forms the key step in this scheme. The detailed synthetic plan is shown in Figure 7.11. Note that this cyclization still leaves the methyl group on the wrong carbon. The unwanted –OH group was removed via a Lewis acid complexed hydride transfer reaction. The acid catalyzed isomerisation of the double bond is followed by a hydroxylation- oxidation sequence to expose a carbonyl group. The quaternary methyl group was then introduced via., an enolate. The exocyclic methylene was reintroduced to complete the synthesis.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Introduction_to_Polymers

Prior to the early 1920's, chemists doubted the existence of molecules having molecular weights greater than a few thousand. This limiting view was challenged by , a German chemist with experience in studying natural compounds such as rubber and cellulose. In contrast to the prevailing rationalization of these substances as aggregates of small molecules, Staudinger proposed they were made up of composed of 10,000 or more atoms. He formulated a structure for , based on a repeating isoprene unit (referred to as a monomer). For his contributions to chemistry, Staudinger received the 1953 Nobel Prize. The terms and were derived from the Greek roots (many), (one) and (part). Recognition that polymeric macromolecules make up many important natural materials was followed by the creation of synthetic analogs having a variety of properties. Indeed, applications of these materials as fibers, flexible films, adhesives, resistant paints and tough but light solids have transformed modern society. Some important examples of these substances are discussed in the following sections. ),

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Electron_Paramagnetic_Resonance_(Jenschke)

Electron Paramagnetic Resonance (EPR) spectroscopy is less well known and less widely applied than NMR spectroscopy. The reason is that EPR spectroscopy requires unpaired electrons and electron pairing is usually energetically favorable. Hence, only a small fraction of pure substances exhibit EPR signals, whereas NMR spectroscopy is applicable to almost any compound one can think of. On the other hand, as electron pairing underlies the chemical bond, unpaired electrons are associated with reactivity. Accordingly, EPR spectroscopy is a very important technique for understanding radical reactions, electron transfer processes, and transition metal catalysis, which are all related to the ’reactivity of the unpaired electron’. Some species with unpaired electrons are chemically stable and can be used as spin probes to study systems where NMR spectroscopy runs into resolution limits or cannot provide sufficient information for complete characterization of structure and dynamics. These notes introduce the basics for applying EPR spectroscopy on reactive or catalytically active species as well as on spin probes. Thumbnail: HYSCORE spectrum of a Ti(III) surface species (CC BY-NC 4.0; Junnar Jeschke in collaboration with C. Copéret, F. Allouche, V. Kalendra)

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Stoichiometry

is a general term for relationships between amounts of substances in chemical reactions. It also describes calculations done to determine how much of a substance will be used in a reaction, left over after a reaction, produced by a reaction, etc. The calculations of theoretical yield in the previous section are simple stoichiometry calculations. To do stoichiometry calculations, you'll need: Doing stoichiometry calculations isn't just a procedure to solve a problem in a book. It is a way to describe things that actually happen, and it is related to concepts and understanding of chemistry. Stoichiometry calculations are based on the conservation of mass (see the Lavoisier page) and the idea that particles like nuclei and electrons aren't created or destroyed during reactions, just rearranged. These understandings allow us to make these calculations. Stoichiometry calculations are also related to the concept of chemical equilibrium. means a stable state in which opposing forces are balanced. For instance, when you stand on one foot, to maintain your balance and not fall over, if you move a little to one side, you will need to correct that and move back a little in the other direction so you don't lose your balance. In chemistry, equilibrium means a state in which 2 opposite processes are occuring, but at the same rate. For instance, the reaction may go in the forward direction (to the right), and at the same time some molecules of product are turning into reactants, going in the reverse direction (to the left). In many chemical reactions, both directions are possible; when they are happening at the same rate, that is called , which means "moving equilibrium", because the individual molecules are moving back and forth between "reactant" and "product" (which are really just defined by how you write the equation), but the total amounts of reactant and product aren't changing. When we do stoichiometry calculations, we assume that the reaction will be complete, meaning that the limiting reactant will react completely, so none is left, forming as much product as possible. However, this is only sometimes what happens. There are several reason why this might not happen. For your calculations, you don't usually need to really worry about rate, equilibrium or side reactions, because we haven't learned strategies for dealing with them yet. However, to keep your mental models of chemical reactions matching real chemical systems, it's important to know that real reactions are more complicated than we make them seem in the first few weeks of chem class.

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Spices are aromatic substances obtained from the dried parts of plants such as the roots, shoots, fruits, bark, and leaves. They are sold as seeds, blends of spices, whole or ground spices, and seasonings. The aromatic substances that give a spice its particular aroma and flavor are the essential oils. The flavor of the essential oil or flavoring compound will vary depending on the quality and freshness of the spice. The aromas of ground spices are volatile. This means they lose their odor or flavoring when left exposed to the air for extended periods. They should be stored in sealed containers when not in use. Whole beans or unground seeds have a longer shelf life but should also be stored in sealed containers. Allspice is only one spice, yet it has a flavor resembling a blend of cloves, nutmeg, and cinnamon. At harvest time, the mature (but still green) berries from the allspice trees (a small tropical evergreen) are dried in the sun. During drying they turn reddish-brown and become small berries. The berries are about 0.6 cm (1/4 in.) in diameter and contain dark brown seeds. Allspice is grown principally in Jamaica and to a lesser degree in Mexico. Allspice is available whole or ground. Bakers usually use ground allspice in cakes, cookies, spices, and pies. Anise is the small, green-grey fruit or seed of a plant of the parsley family. The plant grows to a height of 45 cm (18 in.) and has fine leaves with clusters of small white flowers. It is native to Mexico and Spain, with the latter being the principal producer. Anise seeds are added to pastries, breads, cookies, and candies. Caraway is the dried fruit or seed of a biennial plant of the parsley family, harvested every second year, primarily in the Netherlands. It is also produced in Poland and Russia. The many-branched, hollow-stemmed herb grows up to 60 cm (24 in.) high and has small white flowers. Caraway is a small crescent-shaped brown seed with a pleasant aroma but somewhat sharp taste. Although it is most familiar in rye bread, caraway is also used in cookies and cakes. Native to India, Sri Lanka, and Guatemala, cardamom is the fruit or seed of a plant of the ginger family. The three-sided, creamy-white, flavorless pod holds the tiny aromatic, dark brown seeds. It is available in whole and ground (pod removed). Cardamom in ground form flavors Danish pastries and coffee cakes, Christmas baking, and Easter baking such as hot cross buns. Cinnamon comes from the bark of an aromatic evergreen tree. It is native to China, Indonesia, and Indochina. Cinnamon may be purchased in ground form or as cinnamon sticks. Ground cinnamon is used in pastries, breads, puddings, cakes, candy, and cookies. Cinnamon sticks are used for preserved fruits and flavoring puddings. Cinnamon sugar is made with approximately 50 g (2 oz.) of cinnamon to 1 kg (2.2 lb.) of granulated sugar. Cassia, sometimes known as Chinese cinnamon, is native to Assam and Myanmar. It is similar to cinnamon but a little darker with a sharper taste. It is considered better for savory rather than sweet foods. It is prized in Germany and some other countries as a flavor in chocolate. Cloves are the dried, unopened buds of a tropical evergreen tree, native to Indonesia. The flavor is characterized by a sweet, pungent spiciness. The nail-shaped whole cloves are mainly used in cooking, but the ground version of this spice heightens the flavor of mincemeat, baked goods, fruit pies, and plum pudding. Ginger is one of the few spices that grow below the ground. It is native to southern Asia but is now imported from Jamaica, India, and Africa. The part of the ginger plant used is obtained from the root. Ground ginger is the most commonly used form in baking — in fruitcakes, cookies, fruit pies, and gingerbread. Candied ginger is used in pastries and confectionery. Originating in the East and West Indies, mace is the fleshy growth between the nutmeg shell and outer husk, yellow-orange in color. It is usually sold ground, but sometimes whole mace (blades of mace) is available. Mace is used in pound cakes, breads, puddings, and pastries. Nutmeg is the kernel or seed of the nutmeg fruit. The fruit is similar to the peach. The fleshy husk, grooved on one side, splits, releasing the deep-brown aromatic nutmeg. It is available whole or ground. Ground nutmeg is used extensively in custards, cream puddings, spice cakes, gingerbread, and doughnuts. Poppy seed comes from the Netherlands and Asia. The minute, blue-grey, kidney-shaped seeds are so small they seem to be round. Poppy seeds are used in breads and rolls, cakes and cookies, and fillings for pastries. Sesame or benne seeds are the seeds of the fruit of a tropical annual herb grown in India, China, and Turkey. The seeds are tiny, shiny, and creamy-white with a rich almond-like flavor and aroma. Bakers use sesame seeds in breads, buns, coffee cakes, and cookies. The Spaniards named vanilla. The word derives from vaina, meaning pod. Vanilla is produced from an orchid-type plant native to Central America. The vanilla beans are cured by a complicated process, which helps explain the high cost of genuine vanilla. The cured pods should be black in color and packed in airtight boxes. Imitation vanilla extracts are made from a colorless crystalline synthetic compound called vanillin. Pure vanilla extract is superior to imitation vanilla. Artificial vanilla is more intense than real vanilla by a factor of 3 to 4 and must be used sparingly. To use vanilla beans, split the pod down the middle to scrape out the seeds. The seeds are the flavoring agents. Alternatively, the split pod can be simmered in the milk or cream used in dessert preparation. Its flavoring power is not spent in one cooking and it can be drained, kept frozen, and reused. A vanilla bean kept in a container of icing sugar imparts the flavor to the sugar, all ready for use in cookies and cakes. Vanilla extract is volatile at temperatures starting at 138°C (280°F) and is therefore not ideal for flat products such as cookies. It is suitable for cakes, where the interior temperature does not get so high. Vanilla beans and vanilla extract are used extensively by bakers to flavor a wide range of desserts and other items.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.09%3A_Cheese_Production

Cheese making is essentially a dehydration process in which milk casein, fat and minerals are concentrated 6 to 12-fold, depending on the variety. The basic steps common to most varieties are acidification, coagulation, dehydration, and salting. Cheese Production Process: Milk is primarily composed of water with four biological macromolecules; carbohydrate (lactose), fats, casein phosphoproteins, and whey protein. Caseins are phosphoproteins. These proteins are mostly random coils will little secondary or tertiary structure. They are highly heat stable. Casein exists in the milk as micelles that consist of hundreds of casein molecules coordinated with Ca ions. Although the casein micelle is fairly stable, here are two major ways in which aggregation can be induced. Aggregation is a key step of cheese production. During the rennet cleaves the Phe(105)-Met(106) linkage of kappa-casein forming a soluble protein which diffuses away from the micelle and para-kappa-casein. During the , the micelles aggregate. This is due to the loss of steric repulsion of the kappa-casein. Calcium assists coagulation by acting as a bridge between micelles. During the of coagulation, a gel forms, the milk curd firms, and the liquid separates. 2. . Acidification causes the casein micelles to destabilize or aggregate. Acid coagulated fresh cheeses may include Cottage cheese, Quark, and Cream cheese. Acid coagulation can be achieved naturally with the starter culture of . Acid curd is more fragile than rennet curd due to the loss of calcium. Whey proteins include \(\beta\) -lactoglobulin, alpha-lactalbumin, bovine serum albumin (BSA), and immunoglobulins (Ig). These proteins have well defined tertiary and quaternary structures. They are soluble in water at lower pH but do not coagulate after proteolysis or acid treatment. When the casein is coagulated with enzymes or acid treatment, there is usually a straining step whereby the water is separated from the curd. There is a third process for casein coagulation, . In this process, heat causes denaturation of the whey proteins which can interact with the caseins. With the addition of acid, the caseins precipitate the whey proteins. In heat-acid coagulation, 90% of protein can be recovered. Examples of cheeses made by this method include Paneer, Ricotta and Queso Blanco. When are added to milk, the bacterium uses enzymes to produce energy (ATP) from lactose. The lactic acid curdles the milk that then separates to form curds, which are used to produce cheese and whey. We previously covered the pathway for bacteria to convert glucose to lactic acid. However, we haven’t talked about how this bacterium can convert lactose to glucose. Lactose is hydrolyzed to glucose and \(\beta\)-galactose. Glucose can be converted to lactic acid as discussed before. Galactose is converted into glucose 6-phosphate in four steps in the Leloir pathway. 1. The first reaction is the phosphorylation of galactose to galactose 1-phosphate. 2. Galactose 1-phosphate reacts with uridine diphosphate glucose (UDP-glucose) to form UDP-galactose and glucose 1-phosphate are formed. 3. The galactose moiety of UDP-galactose is then epimerized to glucose-1-phosphate. The configuration of the hydroxyl group at carbon 4 is inverted by UDP-galactose 4-epimerase. This enzyme utilizes NAD in the first step. And then regenerates the NAD in the second step. 4. Glucose 1-phosphate, formed from galactose, is isomerized to glucose 6-phosphate by phosphoglucomutase. In this pathway, UDP-glucose and UDP-galactose fulfill catalytic roles but are not subject to any net turnover. It might therefore be said that they form a tiny metabolic cycle between the two of them. After the whey is removed the curds, there is a wide variety of curd handling dependent upon the type of cheese being prepared. Some cheese varieties, such as Colby or Gouda require a curd washing to increases the moisture content and reduce the acidity. Salt is added to some cheeses through different methods: Gouda is soaked in brine, while Feta has surface salt added. The curd is then ripened until the desired flavors and textures are produced. This ripening process includes further fermentation by bacteria, added yeasts or molds, and enzymatic reactions from added lipases or rennet. These processes develop distinctive characteristics for each cheese. The table below shows a sample of flavor molecules derived from the breakdown of milk components. More examples: Simon Cotton, Education in Chemistry, Royal Society of Chemistry, Really Cheesy Chemistry Lipolysis is a critical step is the lipolysis of triglycerides to esters and acids which yield many flavorful molecules. Fatty acid metabolism (b-oxidation) removes to carbons at a time to each of these fatty acids. Esterases are often present that can turn these shorter chain fatty acids into methyl esters. species are facultative anaerobes that can ferment sugars (glucose or lactose) into propionic acid. This process creates aroma and flavors found in Swiss cheeses. A is an organism that: (choose the correct definition) This process hijacks a part of the TCA Cycle. A key step in the Wood-Werkman Pathway is to transfer a carboxyl group from methylmalonyl CoA to pyruvate to form propionyl CoA and oxaloacetate. This mechanism utilizes vitamin B12 (biotin). Draw the arrows for the decarboxylation of methylmalonyl CoA: Process continues with the carboxylated biotin and the enolate of pyruvate. Draw the enolate anion of pyruvate. Is this a nucleophile or an electrophile? As this step continues with the carboxylated biotin and the enolate of pyruvate to form oxaloacetate. Draw the arrows for this conversion. Now that you have made propionyl CoA. How is it converted to propionic acid? Draw arrows for this trans-thioesterification process. Be sure to include a tetrahedral intermediate. D. H. Hettinga and G. W. Reinbold, THE PROPIONIC-ACID BACTERIA–A REVIEW. (6), 358-372.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/05%3A_Strategies_in_Disparlure_Synthesis

The gypsy moth (Porthytria dispar) is a serious pest of the forests. In 1976 B.A. Bierl et.al., (Science, 170,88 (1970)) isolated the sex pheromone from extracts of 78,000 tips of the last two abdominal segments of female moths. The structure was assigned as Later, the precursor molecule – the cis-olefin was also isolated from the same source. A laboratory bioassay from synthetic materials showed that just 2 pg of 5.1 was enough to elicit bioactivity. Since the availability of the molecule from natural sources was very minute even for structure elucidation problems and study its anticipated role as pest control molecule, there was intense interest in an efficient synthesis of this molecule. Some disconnections for this simple molecule are depicted in Figure 5.2. Epoxides could be made from corresponding olefins. In this case, the olefin should be Z-olefin. When synthesis of such olefins are not stereospecific, direct epoxidation using peroxides would yield a mixture of α- and β-epoxide from both isomeric olefins. To avoid such mixtures at the last stage, one should introduce selectivity at an early stage of the synthesis. The first attempt was directed towards synthesis of the appropriate olefin and epoxidation (B.A. Bierl et.al., (Science, 170, 88 (1970)). The stereoselectivity was unsatisfactory (Fig 5.3). This necessitated extensive purification. The ratio of cis- / trans- isomers in Wittig Olefination reaction could be altered by modification of reagents and reaction parameters. H.T. Bestmann et.al., (Chem. Ber., 109, 3375 (1976)) were able to improve the synthesis by modifying the Wittig reaction conditions . Pure cis- olefins could be obtained by catalytic reduction of acetylenes (Angew. Chem., Int. Ed., 11, 60 (1972). Klunenberg et.al., (Angew. Chem., Int. Ed., 17, 47 (1978) took advantage of the cis- olefin moieties in 1,5-cyclooctadiene by selective oxidation of one double bond. The chains were introduced by sequential Kolbe electrolysis . Epoxidation of olefins yield only racemates unless the epoxidation step involves an asymmetric synthesis. Synthesis of pure (+)- and (-)- isomers could be achieved in three ways. Mori et.al.,( Tet. Lett., 3953 (1976); Tetrahedron, 53, 833 (1979)) soon followed with a synthesis of (+)- and (-)- Disparlures starting from L-(+)-tartaric acid . This synthesis had the merit that some of the chiral intermediated were crystallisable and therefore amenable for easy purification to very pure intermediates and pure final products. Their intermediates were subjected to critical spectral analysis to assess their purity. Thus, their bio-assays gave more reliable data. A synthesis of (+)- and (-)- Disparlure from another chiral synthon - isopropylidene D- and L- erythroses - was reported by Alexandros E. Koumbis et.al., (Tetrahedron Letters, 46, 4353 (2005)) . A successful synthesis of (+)- Disparlure by the application of Sharpless epoxidation was reported by Kossier B. E.. et.al., (J. Am. Chem. Soc., 103, 464 (1981)) . Synthesis of all four isomers in a very pure form came from the school of Sharpless E. B. ( Tet. Lett., 6411 (1992)). Using both the chiral hydroxylation agents, they reported an efficient synthesis of all enantiomers . The efficiency of the asymmetric synthesis was as high as 95% and gave 100% pure intermediated by crystallization. The overall process was very efficient.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.16%3A_Distilled_Spirits

Distilled spirits are all alcoholic beverages in which the concentration of ethanol has been increased above that of the original fermented mixture by a method called distillation. More Information about Distilling: , A Guide for Small Distilleries, Kris Berglund Distilled Spirits Production Steps: Any sugar containing fruit or syrup can be used for fermentation and then distilled to prepare spirits. Similarly, grains and potatoes are fermentable and can be used for whiskey or vodka production. Like wine and cider production, the fruits are harvested and mashed to release enzymes and simple mono- and di-saccharides. Review: Describe the steps and any necessary additives (like pectinases and sulfite) For grain spirits, the process involves malting of the grain, milling, boiling a mash to release the complex carbohydrates. Review: Describe the steps and any necessary additives (like amylases) As you remember from the ‘Cider’ unit, many fruits contain a large amount of pectin. Pectin is a polymer of the sugar galacturonic acid. This pectin can form a gel that is undesirable in ciders or fruit beverages, so it is necessary to allow native pectolytic enzymes to hydrolyze this polysaccharide. In addition, some producers add extra pectolytic enzymes. Pectin methylesterase is an enzyme found in cherries, pears, and apples that hydrolyzes the esters that are on the side chains of pectin. Show the product of this reaction. In cider or wines, the small amounts of methanol formed in this process are not a concern. However, when the wine is distilled the methanol is also concentrated and can have toxic impacts on consumers. One way to limit the formation of methanol is by heating of the mash to a temperature of 80- 85 °C. What will this do to the enzyme? Fermentation is the same process as seen in the previous discussions of Bread, Beer, Cider, and Wine. 5% 10% 15% 20% 25% 30% 35% 45% 50% 60% 75% Distillation in the concentration of ethanol content in an alcoholic beverage through boiling. Ethanol boils at a lower temperature ( 78.4 °C or 173.12 °F) than water ( 100 °C or 212 °F). When the fermentation mixture is heated, the ethanol is evaporated in a higher concentration in the steam. This is condensed and collected resulting in a product that is approximately 25- 35% alcohol. If a distillery desires a higher concentration of alcohol, then what will they need to do? The still vessel is filled with mash, wine, or beer up to 50-75 % full and then closed. More viscous mashes are diluted with 20 % water. Pomaces which yield a low alcohol content are mixed preferentially with 20 % coarse spirit. Most distilleries use copper stills as they produce cleaner and aromatic because copper reacts with the sulfur side-products found in mashes to form non-volatile compounds. What is the problem with sulfur side-products in mash (and then the final product)? Boiling points of different alcohols present in mashes: Most distillers will collect three fractions from the distillation process: fore-run (head), middlerun (heart), and after-run (tail). What is the primary component(s) in each fraction? Which fraction will be sold as a distilled spirit? With direct heating of the fermentation product in the pot stills, the highly viscous mashes/fruit pulps can lead to burning. The decomposition products of sugar leads to ___________________. The products formed in this process can lend a bitter or burnt flavor to the final distilled spirits. Wood fires directly below the pot are problematic due to leads to concerns about burning the mash and possible explosions. Why is distillation prone to fires? Hint: consider flammability of the product. Some whisky distillers choose to use the wood fired heating because they like the flavors. To keep the mash from burning, they use a ‘rummager’ to continuously stir the mash. The fire also requires careful tending, making sure it’s not burning too hot or too cold. To prevent burning the mash, other distillers have moved to steam, hot water baths around the pot, or electrical heating. With column distillation, the mash enters near the top of the still and begins flowing downward. This brings it closer to the heating source, and once it’s heated enough to evaporate, the vapor rises up through a series of partitions known as plates or stripping plates. At each plate along the way, the vapor ends up leaving behind some of the higher boiling compounds. It is important to note that pot stills operate on a batch by batch basis, while column stills may be operated continuously allowing higher throughput. Is scorching a problem with this method? The aging process is similar to wines. The aging process allows tannins, terpenes, lignins, polyphenols, and minerals from the wood of the barrel to dissolve into the spirits. Many of the barrels have been charred so there are oxidized lignin and wood sugars also available. As these compounds are dissolved into the spirits, new condensation and oxidation reactions can occur during this process. Some barrels have been previously used for wines so they will also release flavors from the polyphenols of wines that were absorbed into the wood. Each of the distilled spirits have a slightly different aging process. You will notice that the more northern the climate in which the distilled spirit is produced, the longer it is aged. Most distillates have greater than 40-45 % alcohol content. In order to be drinkable, they have to be watered down. The distilled spirits still contain a variety of flavor and aroma compounds from the original mash, the fermentation process, the Maillard reaction in the still, or from the wood barrels in the aging process. Some of these compounds can cause a cloudy or hazy appearance to the distilled spirits. Distillers will often cool the spirits to between 0 and -10 °C. After cold storage, the distilled spirits are filtered to remove any precipitates. The bottling of the distilled spirits is straightforward. There are many flavors in distilled spirits. It is highly dependent upon the original raw materials, yeast fermentation process, presence of any microbial contaminants, aging, etc. However, distillation can intensify flavors that are found in the middle-run, but many other flavors do not get transferred from the pot to the distillate. It is important to note that the addition of flavorings, sugars or other sweetening products after distillation is forbidden for distilled beverages such as rum, whisky, fruit distillates or wine brandy. The addition of caramel in fruit distillates is not allowed, while whiskey is allowed plain caramel coloring only. The most common spirits are those derived from grains (whiskey, vodkas), grapes (cognac, brandy), molasses (rum), and agave (tequila). Whisky is a distilled beverage from cereal grains and matured in barrels. There are different regional variations on this drink. The malt from corn, barley, rye, or wheat is mashed in a process similar to beer. The wort is then directly distilled. Brandy is a distilled wine beverage. Rum is a distilled beverage from sugar cane. Tequila is a distilled beverage from agave. An eau de vie is a clear fruit brandy that is produced by means of fermentation and double distillation. For example, Framboise is a double distilled raspberry brandy. Unlike liqueurs, are not sweetened. Although is a French term, similar beverages are produced in other countries (e.g. German Schnapps, German Kirschwasser, Turkish rakı, Hungarian pálinka, and Sri Lankan coconut arrack). Liqueurs are drinks made by adding fruit, herbs or nuts to neutral distilled spirits. Usually a distilled beverage like vodka is used as it is mostly alcohol and little flavoring. They are often also heavily sweetened. They are often served with dessert. You might drink it straight, with coffee, used in cocktails, or in cooking. Coldea, Mudura & Socaciu, , In Ideas and Applications Toward Sample Preparation for Food and Beverage Analysis, M. Stauffer, Ed., IntechOpen, 2017. S. Canas, . A Review, Beverages, 2017, 3(4), 55-77. Schaller, Structure and Reactivity, Purification of Molecular Compounds, PM3: Distillation N. Spaho, , In Distillation – Innovative Applications and Modeling, M. Mendes, Ed., IntechOpen, 2017.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.04%3A_Equilibrium_Expressions

Make sure you thoroughly understand the following essential ideas: You know that an equilibrium constant expression looks something like K = [products] / [reactants]. But how do you translate this into a format that relates to the actual chemical system you are interested in? This lesson will show you how to write the equilibrium constant expressions that you will need to use when dealing with the equilibrium calculation problems in the chapter that follows this one. Although we commonly write equilibrium quotients and equilibrium constants in terms of molar concentrations, any concentration-like term can be used, including mole fraction and molality. Sometimes the symbols \(K_c\) , \(K_x\) , and \(K_m\) are used to denote these forms of the equilibrium constant. Most of the equilibria we deal with in this course occur in liquid solutions and gaseous mixtures. We can express \(K_c\) values in terms of moles per liter for both, but when dealing with gases it is often more convenient to use partial pressures. These two measures of concentration are of course directly proportional: \[c=\dfrac{n}{V}=\dfrac{\dfrac{PV}{RT}}{V}=\dfrac{P}{RT} \label{Eq1}\] so for a reaction \(A_{(g)} \rightarrow B_{(g)}\) we can write the equilibrium constant as \[K_p =\dfrac{P_B}{P_A} \label{Eq2}\] Owing to interactions between molecules, especially when ions are involved, all of these forms of the equilibrium constant are only approximately correct, working best at low concentrations or pressures. The only equilibrium constant that is truly “constant” (except that it still varies with the temperature!) is expressed in terms of , which you can think of as “effective concentrations” that allow for interactions between molecules. In practice, this distinction only becomes important for equilibria involving gases at very high pressures (such as are often encountered in chemical engineering) and in ionic solutions more concentrated than about 0.001 . We will not deal much with activities in this course. For a reaction such as \[CO_{2\, (g)} + OH^–_{(aq)} \rightleftharpoons HCO_{3\, (aq)}^- \label{Eq3}\] that involves both gaseous and dissolved components, a “hybrid” equilibrium constant is commonly used: \[ K =\dfrac{[HCO_3^-]}{P_{CO_2}[OH^-]} \label{Eq4}\] Clearly, it is essential to be sure of the units when you see an equilibrium constant represented simply by "\(K\)". In this lesson (and in most of the others in this set,) we express concentrations in mol L and pressures in atmospheres. Although this reflects common usage among chemists (older ones, especially!), these units are not part of the SI system which has been the international standard since the latter part of the 20th Century. are now more properly expressed in mol dm and the "standard " corresponds to a pressure of 101.325 kPa. Until 1990, 1 atm was the "standard pressure" employed in calculations involving the gas laws, and also in thermodynamics. Since that date, "standard pressure" has been 100.000 kPa, also expressed as 1 bar. For most practical purposes, the differences between these values are so small that they can be neglected. It is sometimes necessary to convert between equilibrium constants expressed in different units. The most common case involves pressure- and concentration equilibrium constants. Note that when is expressed in liters and in atmospheres, must have the value 0.08206 L-atm/mol K.). The ideal gas law relates the partial pressure of a gas to the number of moles and its volume: \[PV = nRT \] Concentrations are expressed in moles/unit volume , so by rearranging the above equation we obtain the explicit relation of pressure to concentration: \[P = \left(\dfrac{n}{V} \right)RT \label{Eq5}\] Conversely, \[c = \dfrac{n}{V} = \dfrac{P}{RT}\] so a concentration of a gas [A] can be expressed as \(\dfrac{P_A}{RT}\). For a reaction of the form \(A + 3 B\rightleftharpoons 2C\), we can write Assuming that all of the components are gases, the difference \[(\text{moles of gas in products}) – (\text{moles of gas in reactants}) = \Delta{n_g} \nonumber\] is given by \[ \color{red} {K_p = K_c (RT)^{\Delta{n_g}} \label{Eq6}}\] Substances whose concentrations undergo no significant change in a chemical reaction do not appear in equilibrium constant expressions. How can the concentration of a reactant or product change when a reaction involving that substance takes place? There are two general cases to consider. This happens all the time in acid-base chemistry. Thus for the hydrolysis of the cyanide ion \[\ce{CN^{-} + H2O <=> HCN + OH^{–}} \nonumber\] we write \[K_c= \dfrac{[\ce{HCN},\ce{OH^-}]}{[\ce{CN^-}]} \nonumber\] in which no \([H_2O]\) term appears. The justification for this omission is that water is both the solvent and reactant, but only the tiny portion that acts as a reactant would ordinarily go in the equilibrium expression. The amount of water consumed in the reaction is so minute (because \(K\) is very small) that any change in the concentration of \(H_2O\) from that of pure water (55.6 mol L ) will be negligible. Similarly, for the "autodissociation" of water \[H_2O = H^+ + OH^– \nonumber\] the equilibrium constant is expressed as the "ion product" \[K_w = [H^+,OH^–] \nonumber\] Be careful about throwing away H O whenever you see it. In the esterification reaction \[\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O} \nonumber\] that we discussed in a previous section, a [H O] term must be present in the equilibrium expression if the reaction is assumed to be between the two liquids acetic acid and ethanol. If, on the other hand, the reaction takes place between a dilute aqueous solution of the acid and the alcohol, then the [H O] term would not be included. This is most frequently seen in solubility equilibria, but there are many other reactions in which solids are directly involved: \[CaF_{2\, (s)} \rightarrow Ca^{2+}_{(aq)} + 2F^-_{(aq)} \label{Eq11}\] \[Fe_3O_4(s) + 4 H_{2\, (g)} \rightarrow 4 H_2O_{(g)} + 3Fe_{(s)} \label{Eq12}\] These are (meaning reactions in which some components are in different phases), and the argument here is that concentration is only meaningful when applied to a substance within a single phase. Thus the term \([CaF_2]\) would refer to the “concentration of calcium fluoride within the solid \(CaF_2\)", which is a constant depending on the molar mass of \(CaF_2\) and the density of that solid. The concentrations of the two ions will be independent of the quantity of solid \(CaF_2\) in contact with the water; in other words, the system can be in equilibrium as long as any \(CaF_2\) at all is present. Throwing out the constant-concentration terms can lead to some rather sparse-looking equilibrium expressions. For example, the equilibrium expression for each of the processes shown in the following table consists solely of a single term involving the partial pressure of a gas: \(K_p = P_{CO_2}\) \(K_p = P_{H_2O}^{10}\) Sodium sulfate decahydrate is a solid in which H2O molecules (“waters of hydration") are incorporated into the crystal structure.) \(K_p = P_{I_2}\) \(K_p = P_{H_2O}\) The last two processes represent changes of state ( ) which can be treated exactly the same as chemical reactions. In each of the heterogeneous processes shown in Table \(\Page {1}\), the reactants and products can be in equilibrium (that is, permanently coexist) only when the partial pressure of the gaseous product has the value consistent with the indicated \(K_p\). Bear in mind also that these \(K_p\)'s all increase with the temperature. What are the values of \(K_p\) for the equilibrium between liquid water and its vapor at 25°C, 100°C, and 120°C? The vapor pressure of water at these three temperatures is 23.8 torr, 760 torr (1 atm), and 1489 torr, respectively. e vapor pressures are the partial pressures of water vapor in equilibrium with the liquid, so they are identical with the \(K_p\)'s when expressed in units of atmospheres. or “ ”. As an equilibrium constant approaches the limits of zero or infinity, the reaction can be increasingly characterized as a one-way process; we say it is “ ” or “ ”. The latter term must of course not be taken literally; the Le Chatelier principle still applies (especially insofar as temperature is concerned), but addition or removal of reactants or products will have less effect. The examples in the following table are intended to show that numbers (values of ), no matter how dull they may look, do have practical consequences! The equilibrium expression for the synthesis of ammonia \[3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)} \label{Eq13}\] can be expressed as \[ K_p =\dfrac{P^2_{NH_3}}{P_{N_2}P^3_{H_2}} \label{Eq14}\] or \[ K_c = \dfrac{[NH_3]^2}{[N_2] [H_2]^3} \label{Eq15}\] so \(K_p\) for this process would appear to have units of atm , and \(K_c\) would be expressed in mol L . And yet these quantities are often represented as being dimensionless. Which is correct? The answer is that both forms are acceptable. There are some situations (which you will encounter later) in which ’s must be considered dimensionless, but in simply quoting the value of an equilibrium constant it is permissible to include the units, and this may even be useful in order to remove any doubt about the units of the individual terms in equilibrium expressions containing both pressure and concentration terms. In carrying out your own calculations, however, there is rarely any real need to show the units. It is important to remember that an equilibrium quotient or constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of \(Q\) or \(K\) will change. The rules are very simple: Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements: Example 1: \(\ce{2 H2 + O2 <=> 2 H2O} \) with equilibrium expression \[K_p = \dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}} \nonumber\] Example 2: \(\ce{10 H2 + 5 O2 <=> 10 H2O}\) with equilibrium expression \[\begin{align*} K_p &= \dfrac{P_{H_2O}^{10}}{P_{H_2}^{10}P_{O_2}^5} \\[4pt] &= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{5}\end{align*}\] Example 3: \(\ce{H2 + 1/2 O2 <=> H2O} \) with equilibrium expression \[\begin{align*} K_p &= \dfrac{P_{H_2O}}{P_{H_2}P_{O_2}^{1/2}} \\[4pt] &= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{1/2} \end{align*}\] Example 4: \(\ce{H2O <=> H2 + 1/2 O2 } \) with equilibrium expression \[\begin{align*} K_p &= \dfrac{P_{H_2}P_{O_2}^{1/2}}{P_{H_2O}} \\[4pt] &= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{-1/2} \end{align*}\] Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to the following rule. T Calculate the value of \(K\) for the reaction \[\ce{CaCO3(s) + H^{+}(aq) <=> Ca^{2+}(aq) + HCO^{–}3(aq)} \nonumber\] given the following equilibrium constants: \(CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}\) \(K_1 = 10^{–6.3}\) \(HCO^–_{3(aq)} \rightleftharpoons H^+_{(aq)} + CO^{2–}_{3(aq)}\) \(K_2 = 10^{–10.3}\) The net reaction is the sum of reaction 1 and the reverse of reaction 2: \(CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}\) \(K_1 = 10^{–6.3}\) \( H^+_{(aq)} + CO^{2–}_{3(aq)} \rightleftharpoons HCO^–_{3(aq)} \) \(CaCO_{3(s)} + H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + HCO^–_{3(aq)}\) t \(K_p = 4.5 \times 10^{15}\) a Heterogeneous reactions are those involving more than one phase. Some examples: \(NaHCO_3(s) + H^+(aq) \rightleftharpoons CO_2(g) + Na^+(aq) + H_2O(g)\) A particularly interesting type of heterogeneous reaction is one in which a solid is in equilibrium with a gas. The sublimation of ice illustrated in the above table is a very common example. The equilibrium constant for this process is simply the partial pressure of water vapor in equilibrium with the solid— the of the ice. Many common inorganic salts form solids which incorporate water molecules into their crystal structures. These water molecules are usually held rather loosely and can escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which four of the water molecules are coordinated to the Cu ion while the fifth is hydrogen-bonded to SO . This latter water is more tightly bound, so that the pentahydrate loses water in two stages on heating: \[\ce{CuSO4 \cdot 5H2O ->[140^oC] CuSO4 \cdot 5H2O ->[400^oC] CuSO4} \nonumber\] These dehydration steps are carried out at the temperatures indicated above, but at any temperature, some moisture can escape from a hydrate. For the complete dehydration of the pentahydrate we can define an equilibrium constant: \[\ce{CuSO4 \cdot 5H2O(s) <=> CuSO4(s) + 5 H2O(g)} \quad K_p = 1.14 \times 10^{10} \nonumber\] The vapor pressure of the hydrate (for this reaction) is the partial pressure of water vapor at which the two solids can coexist indefinitely; its value is atm. If a hydrate is exposed to air in which the partial pressure of water vapor is less than its vapor pressure, the reaction will proceed to the right and the hydrate will lose moisture. Vapor pressures always increase with temperature, so any of these compounds can be dehydrated by heating. ate, \(K_p = \sqrt{p_{H_2O)}}\), so the part \[\left(\dfrac{12.5}{760}\right)^5 = 1.20 \times 10^{-9}. \nonumber\] One of the first hydrates investigated in detail was calcium sulfate hemihydrate (CaSO ·½ H O) which Le Chatelier (he of the “principle”) showed to be the hardened form of CaSO known as plaster of Paris. Anhydrous CaSO forms compact, powdery crystals, whereas the elongated crystals of the hemihydrate bind themselves into a cement-like mass that makes this material useful for making art objects, casts for immobilizing damaged limbs, and as a construction material (fireproofing, drywall).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/21%3A_Nuclear_Chemistry/21.2%3A_Nuclear_Equations

Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are . To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction. Many entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure \(\Page {1}\). Protons \( (\ce{^{1}_{1}p}\), also represented by the symbol \(\ce{^1_1H})\) and neutrons \( (\ce{^1_0n})\) are the constituents of atomic nuclei, and have been described previously. Alpha particles \( (\ce{^4_2He}\), also represented by the symbol \(\ce{^{4}_{2}\alpha})\) are high-energy helium nuclei. Beta particles \( (\ce{^{0}_{−1}\beta}\), also represented by the symbol \(\ce{^0_{-1}e})\) are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons \( (\ce{^0_{+1}e}\), also represented by the symbol \(\ce{^0_{+1}β})\) are positively charged electrons (“anti-electrons”). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4, so it is symbolized \(\ce{^4_2He}\). This works because, in general, the ion charge is not important in the balancing of nuclear equations. Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of , particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays (γ)—and other much smaller subnuclear particles, which are beyond the scope of this chapter—according to the mass-energy equivalence equation \(E = mc^2\), seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created: \[\ce{^0_{−1}e + ^0_{+1}e } \rightarrow \gamma + \gamma \label{21.3.1} \] Gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays. Gamma rays are produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions. A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways: If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that \(\ce{^{17}_8O}\) is a product of the nuclear reaction of \(\ce{^{14}_7N}\) and \(\ce{^4_2He}\) if we knew that a proton, \(\ce{^1_1H}\), was one of the two products. Example \(\Page {1}\) shows how we can identify a nuclide by balancing the nuclear reaction. The reaction of an \(α\) particle with magnesium-25 \( (\ce{^{25}_{12}Mg})\) produces a proton and a nuclide of another element. Identify the new nuclide produced. The nuclear reaction can be written as: \[\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber \] where Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products: \[\mathrm{25+4=A+1} \nonumber \] so \[ \mathrm{A=28} \nonumber \] Similarly, the charges must balance, so: \[\mathrm{12+2=Z+1} \nonumber \] so \[\mathrm{Z=13} \nonumber \] Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is \(\ce{^{28}_{13}Al}\). The nuclide \(\ce{^{125}_{53}I}\) combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction? \[\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber \] Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry: Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Penicillin

The penicillins were the first antibiotics discovered as natural products from the mold Penicillium. In 1928, Sir Alexander Fleming, professor of bacteriology at St. Mary's Hospital in London, was culturing . He noticed zones of inhibition where mold spores were growing. He named the mold Penicillium rubrum. It was determined that a secretion of the mold was effective against Gram-positive bacteria. Penicillins as well as cephalosporins are called beta-lactam antibiotics and are characterized by three fundamental structural requirements: the fused beta-lactam structure (shown in the blue and red rings, a free carboxyl acid group (shown in red bottom right), and one or more substituted amino acid side chains (shown in black). The lactam structure can also be viewed as the covalent bonding of pieces of two amino acids - cysteine (blue) and valine (red). Penicillin-G where R = an ethyl pheny group, is the most potent of all penicillin derivatives. It has several shortcomings and is effective only against gram-positive bacteria. It may be broken down in the stomach by gastric acids and is poorly and irregularly absorbed into the blood stream. In addition many disease producing staphylococci are able to produce an enzyme capable of inactivating penicillin-G. Various semisynthetic derivatives have been produced which overcome these shortcomings. Powerful electron-attracting groups attached to the amino acid side chain such as in phenethicillin prevent acid attack. A bulky group attached to the amino acid side chain provides steric hindrance which interferes with the enzyme attachment which would deactivate the pencillins i.e. methicillin. Refer to Table 2 for the structures. Finally if the polar character is increased as in ampicillin or carbenicillin, there is a greater activity against Gram-negative bacteria. All penicillin derivatives produce their bacteriocidal effects by inhibition of bacterial cell wall synthesis. Specifically, the cross linking of peptides on the mucosaccharide chains is prevented. If cell walls are improperly made cell walls allow water to flow into the cell causing it to burst. Resemblances between a segment of penicillin structure and the backbone of a peptide chain have been used to explain the mechanism of action of beta-lactam antibiotics. The structures of a beta-lactam antibiotic and a peptide are shown on the left for comparison. Follow the trace of the red oxygens and blue nitrogen atoms. Gram-positive bacteria possess a thick cell wall composed of a cellulose-like structural sugar polymer covalently bound to short peptide units in layers.The polysaccharide portion of the peptidoglycan structure is made of repeating units of N-acetylglucosamine linked b-1,4 to N-acetylmuramic acid (NAG-NAM). The peptide varies, but begins with L-Ala and ends with D-Ala. In the middle is a dibasic amino acid, diaminopimelate (DAP). DAP (orange) provides a linkage to the D-Ala (gray) residue on an adjacent peptide. The bacterial cell wall synthesis is completed when a cross link between two peptide chains attached to polysaccharide backbones is formed. The cross linking is catalyzed by the enzyme transpeptidase. First the terminal alanine from each peptide is hydrolyzed and secondly one alanine is joined to lysine through an amide bond. Peptidoglycan image courtesy of the University of Texas-Houston Medical School Penicillin binds at the active site of the transpeptidase enzyme that cross-links the peptidoglycan strands. It does this by mimicking the D-alanyl-D-alanine residues that would normally bind to this site. Penicillin irreversibly inhibits the enzyme transpeptidase by reacting with a serine residue in the transpeptidase. This reaction is irreversible and so the growth of the bacterial cell wall is inhibited. Since mammal cells do not have the same type of cell walls, penicillin specifically inhibits only bacterial cell wall synthesis. As early as the 1940s, bacteria began to combat the effectiveness of penicillin. Penicillinases (or beta-lactamases) are enzymes produced by structurally susceptable bacteria which renders penicillin useless by hydrolysing the peptide bond in the beta-lactam ring of the nucleus. Penicillinase is a response of bacterial adaptation to its adverse environment, namely the presence of a substance which inhibits its growth. Many other antibiotics are also rendered ineffective because of this same type of resistance. It is estimated that between 300-500 people die each year from penicillin-induced anaphylaxis, a severe allergic shock reaction to penicillin. In afflicted individuals, the beta-lactam ring binds to serum proteins, initiating an IgE-mediated inflammatory response. Penicillin and ala-ala peptide - Cephalosporins are the second major group of beta-lactam antibiotics. They differ from penicillins by having the beta-lactam ring as a 6 member ring. The other difference, which is more significant from a medicinal chemistry stand point, is the existence of a functional group (R) at position 3 of the fused ring system. This now allows for molecular variations to effect changes in properties by diversifying the groups at position 3. The first member of the newer series of beta-lactams was isolated in 1956 from extracts of Cephalosporium acremonium, a sewer fungus. Like penicillin, cephalosporins are valuable because of their low toxicity and their broad spectrum of action against various diseases. In this way, cephalosporin is very similar to penicillin. Cephalosporins are one of the most widely used antibiotics, and economically speaking, has about 29% of the antibiotic market. The cephalosporins are possibly the single most important group of antibiotics today and are equal in importance to penicillin. The structure and mode of action of the cephalosporins are similar to that of penicillin. They affect bacterial growth by inhibiting cell wall synthesis, in Gram-positive and negative bacteria. Some brand names include: cefachlor, cefadroxil, cefoxitin, ceftriaxone. Cephalexin

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/21%3A_Nuclear_Chemistry/21.3%3A_Radioactive_Decay

Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed. The spontaneous change of an unstable nuclide into another is . The unstable nuclide is called the ; the nuclide that results from the decay is known as the . The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure \(\Page {1}\)). Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab. Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure \(\Page {2}\)) helped him determine that one type of radiation consisted of positively charged and relatively massive \(α\) particles; a second type was made up of negatively charged and much less massive \(β\) particles; and a third was uncharged electromagnetic waves, \(γ\) rays. We now know that \(α\) particles are high-energy helium nuclei, \(β\) particles are high-energy electrons, and \(γ\) radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced. is the emission of an α particle from the nucleus. For example, polonium-210 undergoes α decay: \[\ce{^{210}_{84}Po⟶ ^4_2He + ^{206}_{82}Pb} \hspace{40px}\ce{or}\hspace{40px} \ce{^{210}_{84}Po ⟶ ^4_2α + ^{206}_{82}Pb}\nonumber \] Alpha decay occurs primarily in heavy nuclei (A > 200, Z > 83). Because the loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing α decay lies below , the daughter nuclide will lie closer to the band. is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes β decay: \[\ce{^{131}_{53}I ⟶ ^0_{-1}e + ^{131}_{54}X} \hspace{40px}\ce{or}\hspace{40px} \ce{^{131}_{53}I ⟶ ^0_{-1}β + ^{131}_{54}Xe}\nonumber \] Beta decay, which can be thought of as the conversion of a neutron into a proton and a β particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a γ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits γ radiation and is used in many applications including cancer treatment: \[\mathrm{^{60}_{27}Co^* ⟶\, ^0_0γ +\, ^{60}_{27}Co}\nonumber \] There is no change in mass number or atomic number during the emission of a γ ray unless the γ emission accompanies one of the other modes of decay. (\(β^+\) decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission: \[\ce{^{15}_8O ⟶ ^0_{+1}e + ^{15}_7N} \hspace{40px}\ce{or}\hspace{40px} \ce{^{15}_8O ⟶ ^0_{+1}β + ^{15}_7N}\nonumber \] is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture: \[\ce{^{40}_{19}K + ^0_{-1}e ⟶ ^{40}_{18}Ar}\nonumber \] Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur. Figure \(\Page {3}\) summarizes these types of decay, along with their equations and changes in atomic and mass numbers. Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure \(\Page {4}\)). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions. For example, F-18 is produced by proton bombardment of O \( (\ce{^{18}_8O + ^1_1p⟶ ^{18}_9F + ^1_0n})\) and incorporated into a glucose analog called fludeoxyglucose (FDG). How is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The F emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan. The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or . Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure \(\Page {5}\)). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205. Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant ( ), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem. For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure \(\Page {6}\)). In a given cobalt-60 source, since half of the \(\ce{^{60}_{27}Co}\) nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective. Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, , for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is: \[\text{decay rate} = \lambda N\nonumber \] with \(\lambda\) is the for the particular radioisotope. The decay constant, \(\lambda\), which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, : \[λ=\dfrac{\ln 2}{t_{1/2}}=\dfrac{0.693}{t_{1/2}} \hspace{40px}\ce{or}\hspace{40px} t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}\nonumber \] The first-order equations relating amount, , and time are: \[N_t=N_0e^{−kt} \hspace{40px}\ce{or}\hspace{40px} t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber \] where is the initial number of nuclei or moles of the isotope, and is the number of nuclei/moles remaining at time . Example \(\Page {1}\) applies these calculations to find the rates of radioactive decay for specific nuclides. \(\ce{^{60}_{27}Co}\) decays with a half-life of 5.27 years to produce \(\ce{^{60}_{28}Ni}\). (a) The value of the rate constant is given by: \[λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5.27\:y}=0.132\:y^{−1}} \nonumber \] (b) The fraction of \(\ce{^{60}_{27}Co}\) that is left after time is given by \(\dfrac{N_t}{N_0}\). Rearranging the first-order relationship = to solve for this ratio yields: \[\dfrac{N_t}{N_0}=e^{-λt}=e^\mathrm{-(0.132/y)(15.0/y)}=0.138 \nonumber \] The fraction of \(\ce{^{60}_{27}Co}\) that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the \(\ce{^{60}_{27}Co}\) originally present will remain after 15 years. (c) 2.00% of the original amount of \(\ce{^{60}_{27}Co}\) is equal to 0.0200 × . Substituting this into the equation for time for first-order kinetics, we have: \[t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)=−\dfrac{1}{0.132\:\ce y^{−1}}\ln\left(\dfrac{0.0200×N_0}{N_0}\right)=29.6\:\ce y \nonumber \] Radon-222, \(\ce{^{222}_{86}Rn}\), has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222? 11.1 days Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of \(\ce{^{209}_{83}Bi}\) is 1.9 × 10 years; \(\ce{^{239}_{94}Ra}\) is 24,000 years; \(\ce{^{222}_{86}Rn}\) is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 × 10 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table \(\Page {1}\), and others are listed in Appendix N1. Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type. The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old. Naturally occurring carbon consists of three isotopes: \(\ce{^{12}_6C}\), which constitutes about 99% of the carbon on earth; \(\ce{^{13}_6C}\), about 1% of the total; and trace amounts of \(\ce{^{14}_6C}\). Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space: \[\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C + ^1_1H}\nonumber \] All isotopes of carbon react with oxygen to produce CO molecules. The ratio of \(\ce{^{14}_6CO2}\) to \(\ce{^{12}_6CO2}\) depends on the ratio of \(\ce{^{14}_6CO}\) to \(\ce{^{12}_6CO}\) in the atmosphere. The natural abundance of \(\ce{^{14}_6CO}\) in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of \(\ce{^{14}_6C ^{14}_6CO2}\) and \(\ce{^{12}_6CO2}\) into plants is a regular part of the photosynthesis process, which means that the \(\ce{^{14}_6C: ^{12}_6C}\) ratio found in a living plant is the same as the \(\ce{^{14}_6C: ^{12}_6C}\) ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because \(\ce{^{12}_6C}\) is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years: \[\ce{^{14}_6C⟶ ^{14}_7N + ^0_{-1}e}\nonumber \] Thus, the \(\ce{^{14}_6C: ^{12}_6C}\) ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure \(\Page {7}\) visually depicts this process. For example, with the half-life of \(\ce{^{14}_6C}\) being 5730 years, if the \(\ce{^{14}_6C : ^{12}_6C}\) ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of \(\ce{^{14}_6C : ^{12}_6C}\) ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer. A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls. The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, , in the relationship: \[t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)⟶t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right) \nonumber \] where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript represents the current time. The decay constant can be determined from the half-life of C-14, 5730 years: \[λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5730\: y}=1.21×10^{−4}\:y^{−1}} \nonumber \] Substituting and solving, we have: \[t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right)=\mathrm{−\dfrac{1}{1.21×10^{−4}\:y^{−1}}\ln\left(\dfrac{10.8\:dis/min/g\: C}{13.6\:dis/min/g\: C}\right)=1910\: y}\nonumber \] Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure \(\Page {8}\)). More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end? about 3350 years ago, or approximately 1340 BC There have been some significant, well-documented changes to the \(\ce{^{14}_6C : ^{12}_6C}\) ratio. The accuracy of a straightforward application of this technique depends on the \(\ce{^{14}_6C : ^{12}_6C}\) ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of CO molecules (largely \(\ce{^{12}_6CO2}\)) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the \(\ce{^{14}_6C}\) has decayed), the ratio of \(\ce{^{14}_6C : ^{12}_6C}\) in the atmosphere may be changing. This manmade increase in \(\ce{^{12}_6CO2}\) in the atmosphere causes the \(\ce{^{14}_6C : ^{12}_6C}\) ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years. Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old. An igneous rock contains 9.58 × 10 g of U-238 and 2.51 × 10 g of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed. The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay. The amount of U-238 currently in the rock is: \[\mathrm{9.58×10^{−5}\cancel{g\: U}×\left( \dfrac{1\: mol\: U}{238\cancel{g\: U}}\right )=4.03×10^{−7}\:mol\: U}\nonumber \] Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is: \[\mathrm{2.51×10^{-5}\cancel{g\: Pb}×\left( \dfrac{1\cancel{mol\: Pb}}{206\cancel{g\: Pb}}\right )×\left(\dfrac{1\: mol\: U}{1\cancel{mol\: Pb}}\right)=1.22×10^{-7}\:mol\: U}\nonumber \] The total amount of U-238 originally present in the rock is therefore: The amount of time that has passed since the formation of the rock is given by: with representing the original amount of U-238 and representing the present amount of U-238. U-238 decays into Pb-206 with a half-life of 4.5 × 10 y, so the decay constant is: \[t=\mathrm{−\dfrac{1}{1.54×10^{−10}\:y^{−1}}\ln\left(\dfrac{4.03×10^{−7}\cancel{mol\: U}}{5.25×10^{−7}\cancel{mol\: U}}\right)=1.7×10^9\:y}\nonumber \] Therefore, the rock is approximately 1.7 billion years old. A sample of rock contains 6.14 × 10 g of Rb-87 and 3.51 × 10 g of Sr-87. Calculate the age of the rock. (The half-life of the β decay of Rb-87 is 4.7 × 10 y.) 3.7 × 10 y Nuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some substances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.

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When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called . Substances that do not yield ions when dissolved are called . If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a . If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a . Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure \(\Page {1}\)). Water and other polar molecules are attracted to ions, as shown in Figure \(\Page {2}\). The electrostatic attraction between an ion and a molecule with a dipole is called an . These attractions play an important role in the dissolution of ionic compounds in water. When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as . Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Let us consider what happens at the microscopic level when we add solid KCl to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules penetrate between individual K and Cl ions and surround them, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure shows. The reduction of the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution, resulting in an increase in the disorder of the system as the ions change from their fixed and ordered positions in the crystal to mobile and much more disordered states in solution. This increased disorder is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat. In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in disorder cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust). Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton to another molecule of water, yielding hydronium and hydroxide ions. \[\ce{H_2O (l)+ H_2O (l) \rightleftharpoons H_3O^{+} (aq) + OH^{−} (aq)} \label{11.3.2} \] In some cases, we find that solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, when we dissolve hydrogen chloride in water, we find that the solution is a very good conductor. The water molecules play an essential part in forming ions: Solutions of hydrogen chloride in many other solvents, such as benzene, do not conduct electricity and do not contain ions. Hydrogen chloride is an , and so its molecules react with water, transferring H ions to form hydronium ions (\(H_3O^+\)) and chloride ions (Cl ):   This reaction is essentially 100% complete for HCl (i.e., it is a and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry. Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes, while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water.

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After you have completed Chapter 6, you should be able to This chapter is designed to provide a gentle introduction to the subject of reaction mechanisms. Two types of reactions are introduced—polar reactions and radical reactions. The chapter briefly reviews a number of topics you should be familiar with, including rates and equilibria, elementary thermodynamics and bond dissociation energies. You must have a working knowledge of these topics to obtain a thorough understanding of organic reaction mechanisms. Reaction energy diagrams are used to illustrate the energy changes that take place during chemical reactions, and to emphasize the difference between a reaction intermediate and a transition state.

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Most students of chemistry quickly learn to relate the structure of a molecule to its general properties. Thus we generally expect small molecules to form gases or liquids, and large ones to exist as solids under ordinary conditions. And then we come to H O, and are shocked to find that many of the predictions are way off, and that water (and by implication, life itself) should not even exist on our planet! In this section we will learn why this tiny combination of three nuclei and ten electrons possesses special properties that make it unique among the more than 15 million chemical species we presently know. In water, each hydrogen nucleus is covalently bound to the central oxygen atom by a pair of electrons that are shared between them. In H O, only two of the six outer-shell electrons of oxygen are used for this purpose, leaving four electrons which are organized into two non-bonding pairs. The four electron pairs surrounding the oxygen tend to arrange themselves as far from each other as possible in order to minimize repulsions between these clouds of negative charge. This would ordinarily result in a tetrahedral geometry in which the angle between electron pairs (and therefore the H-O-H ) is 109.5°. However, because the two non-bonding pairs remain closer to the oxygen atom, these exert a stronger repulsion against the two covalent bonding pairs, effectively pushing the two hydrogen atoms closer together. The result is a distorted tetrahedral arrangement in which the H—O—H angle is 104.5°. The H O molecule is electrically neutral, but the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an , represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule. Opposite charges attract, so it is not surprising that the negative end of one water molecule will tend to orient itself so as to be close to the positive end of another molecule that happens to be nearby. The strength of this is less than that of a normal chemical bond, and so it is completely overwhelmed by ordinary thermal motions in the gas phase. However, when the H O molecules are crowded together in the liquid, these attractive forces exert a very noticeable effect, which we call (somewhat misleadingly) . And at temperatures low enough to turn off the disruptive effects of thermal motions, water freezes into ice in which the hydrogen bonds form a rigid and stable network. Notice that the hydrogen bond (shown by the dashed green line) is somewhat longer than the covalent O—H bond. It is also , about 23 kJ mol compared to the O–H covalent bond strength of 492 kJ mol . Water has long been known to exhibit many physical properties that distinguish it from other small molecules of comparable mass. Although chemists refer to these as the "anomalous" properties of water, they are by no means mysterious; all are entirely predictable consequences of the way the size and nuclear charge of the oxygen atom conspire to distort the electronic charge clouds of the atoms of other elements when these are chemically bonded to the oxygen. The most apparent peculiarity of water is its for such a light molecule. Liquid methane CH (molecular weight 16) boils at –161°C. As you can see from this diagram, extrapolation of the boiling points of the various Group 16 hydrogen compounds to H O suggests that this substance should be a gas under normal conditions. Compared to most other liquids, water also has a high . Have you ever watched an insect walk across the surface of a pond? The takes advantage of the fact that the water surface acts like an elastic film that resists deformation when a small weight is placed on it. (If you are careful, you can also "float" a small paper clip or steel staple on the surface of water in a cup.) This is all due to the of the water. A molecule within the bulk of a liquid experiences attractions to neighboring molecules in all directions, but since these average out to zero, there is no net force on the molecule. For a molecule that finds itself the surface, the situation is quite different; it experiences forces only sideways and downward, and this is what creates the stretched-membrane effect. The distinction between molecules located at the surface and those deep inside is especially prominent in H O, owing to the strong hydrogen-bonding forces. The difference between the forces experienced by a molecule at the surface and one in the bulk liquid gives rise to the liquid's surface tension. This drawing highlights two H O molecules, one at the surface, and the other in the bulk of the liquid. The surface molecule is attracted to its neighbors below and to either side, but there are no attractions pointing in the 180° solid angle angle above the surface. As a consequence, a molecule at the surface will tend to be drawn into the bulk of the liquid. But since there must always be some surface, the overall effect is to minimize the surface area of a liquid. The geometric shape that has the smallest ratio of surface area to volume is the , so very small quantities of liquids tend to form spherical drops. As the drops get bigger, their weight deforms them into the typical tear shape. The most energetically favorable configuration of H O molecules is one in which each molecule is hydrogen-bonded to four neighboring molecules. Owing to the thermal motions described above, this ideal is never achieved in the liquid, but when water freezes to ice, the molecules settle into exactly this kind of an arrangement in the ice crystal. This arrangement requires that the molecules be somewhat farther apart then would otherwise be the case; as a consequence, ice, in which hydrogen bonding is at its maximum, has a more open structure, and thus a lower density than water. Here are three-dimensional views of a typical local structure of water (left) and ice (right.) Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The more crowded and jumbled arrangement in liquid water can be sustained only by the greater amount of thermal energy available above the freezing point. When ice melts, the more vigorous thermal motion disrupts much of the hydrogen-bonded structure, allowing the molecules to pack more closely. Water is thus one of the very few substances whose solid form has a lower density than the liquid at the freezing point. Localized clusters of hydrogen bonds still remain, however; these are continually breaking and reforming as the thermal motions jiggle and shove the individual molecules. As the temperature of the water is raised above freezing, the extent and lifetimes of these clusters diminish, so the density of the water increases. At higher temperatures, another effect, common to all substances, begins to dominate: as the temperature increases, so does the amplitude of thermal motions. This more vigorous jostling causes the average distance between the molecules to increase, reducing the density of the liquid; this is ordinary thermal expansion. Because the two competing effects (hydrogen bonding at low temperatures and thermal expansion at higher temperatures) both lead to a decrease in density, it follows that there must be some temperature at which the density of water passes through a maximum. This temperature is 4° C; this is the temperature of the water you will find at the bottom of an ice-covered lake in which this most dense of all water has displaced the colder water and pushed it nearer to the surface. The nature of liquid water and how the H O molecules within it are organized and interact are questions that have attracted the interest of chemists for many years. There is probably no liquid that has received more intensive study, and there is now a huge literature on this subject. The following facts are well established: A variety of techniques including infrared absorption, neutron scattering, and nuclear magnetic resonance have been used to probe the microscopic structure of water. The information garnered from these experiments and from theoretical calculations has led to the development of around twenty "models" that attempt to explain the structure and behavior of water. More recently, computer simulations of various kinds have been employed to explore how well these models are able to predict the observed physical properties of water. This work has led to a gradual refinement of our views about the structure of liquid water, but it has not produced any definitive answer. There are several reasons for this, but the principal one is that the very concept of "structure" (and of water "clusters") depends on both the time frame and volume under consideration. Thus, questions of the following kinds are still open: In the 1950's it was assumed that liquid water consists of a mixture of hydrogen-bonded clusters (H O) in which can have a variety of values, but little evidence for the existence of such aggregates was ever found. The present view, supported by computer-modeling and spectroscopy, is that on a very short time scale, water is more like a "gel" consisting of a single, huge hydrogen-bonded cluster. On a 10 -10 sec time scale, rotations and other thermal motions cause individual hydrogen bonds to break and re-form in new configurations, inducing ever-changing local discontinuities whose extent and influence depends on the temperature and pressure. , like all solids, has a well-defined structure; each water molecule is surrounded by four neighboring H Os. two of these are hydrogen-bonded to the oxygen atom on the central H O molecule, and each of the two hydrogen atoms is similarly bonded to another neighboring H O. Ice forms crystals having a hexagonal lattice structure, which in their full development would tend to form hexagonal prisms very similar to those sometimes seen in quartz. This does occasionally happen, and anyone who has done much winter mountaineering has likely seen needle-shaped prisms of ice crystals floating in the air. Under most conditions, however, the snowflake crystals we see are flattened into the beautiful fractal-like hexagonal structures that are commonly observed. The H O molecules that make up the top and bottom plane faces of the prism are packed very closely and linked (through hydrogen bonding) to the molecules inside. In contrast to this, the molecules that make up the sides of the prism, and especially those at the hexagonal corners, are much more exposed, so that atmospheric H O molecules that come into contact with most places on the crystal surface attach very loosely and migrate along it until they are able to form hydrogen-bonded attachments to these corners, thus becoming part of the solid and extending the structure along these six directions. This process perpetuates itself as the new extensions themselves acquire a hexagonal structure. At temperatures as low as 200 K, the surface of ice is highly disordered and water-like. As the temperature approaches the freezing point, this region of disorder extends farther down from the surface and acts as a lubricant. The illustration is taken from from an article in the April 7, 2008 issue of C&EN honoring the physical chemist Gabor Somorjai who pioneered modern methods of studying surfaces. To a chemist, the term "pure" has meaning only in the context of a particular application or process. The distilled or de-ionized water we use in the laboratory contains dissolved atmospheric gases and occasionally some silica, but their small amounts and relative inertness make these impurities insignificant for most purposes. When water of the highest obtainable purity is required for certain types of exacting measurements, it is commonly filtered, de-ionized, and triple-vacuum distilled. But even this "chemically pure" water is a mixture of isotopic species: there are two stable isotopes of both hydrogen (H and H , the latter often denoted by D) and oxygen (O and O ) which give rise to combinations such as H O , HDO , etc., all of which are readily identifiable in the infrared spectra of water vapor. And to top this off, the two hydrogen atoms in water contain protons whose magnetic moments can be parallel or antiparallel, giving rise to and water, respectively. The two forms are normally present in a ratio of 3:1. The amount of the varies enough from place to place that it is now possible to determine the age and source of a particular water sample with some precision. These differences are reflected in the H and O isotopic profiles of organisms. Thus the isotopic analysis of human hair can be a useful tool for crime investigations and anthropology research. Hydrogen bonds form when the electron cloud of a hydrogen atom that is attached to one of the more electronegative atoms is distorted by that atom, leaving a partial positive charge on the hydrogen. Owing to the very small size of the hydrogen atom, the density of this partial charge is large enough to allow it to interact with the lone-pair electrons on a nearby electronegative atom. Although hydrogen bonding is commonly described as a form of dipole-dipole attraction, it is now clear that it involves a certain measure of electron-sharing (between the external non-bonding electrons and the hydrogen) as well, so these bonds possess some covalent character. Hydrogen bonds are longer than ordinary covalent bonds, and they are also weaker. The experimental evidence for hydrogen bonding usually comes from X-ray diffraction studies on solids that reveal shorter-than-normal distances between hydrogen and other atoms. The following examples show something of the wide scope of hydrogen bonding in molecules. (mp –92, bp 33°C) is another common substance that is strongly hydrogen-bonded in its condensed phases. Hydrogen bonding plays an essential role in natural polymers of biological origin in two ways: The examples that follow are representative of several types of biopolymers. is a linear polymer of glucose (see above), containing 300 to over 10,000 units, depending on the source. As the principal structural component of plants (along with lignin in trees), cellulose is the most abundant organic substance on the earth. The role of hydrogen bonding is to cross-link individual molecules to build up sheets as shown here. These sheets than stack up in a staggered array held together by van der Waals forces. Further hydrogen-bonding of adjacent stacks bundles them together into a stronger and more rigid structure. These polymers made from amino acids R—CH(NH )COOH depend on intramolecular hydrogen bonding to maintain their shape (secondary and tertiary structure) which is essential for their important function as biological catalysts (enzymes). Hydrogen-bonded water molecules embedded in the protein are also important for their structural integrity. The principal hydrogen bonding in proteins is between the -N—H groups of the "amino" parts with the -C=O groups of the "acid" parts. These interactions give rise to the two major types of the secondary structure which refers to the arrangement of the amino acid polymer chain: [images] Although carbon is not usually considered particularly electronegative, C—H----X hydrogen bonds are also now known to be significant in proteins. DNA, as you probably know, is the most famous of the biopolymers owing to its central role in defining the structure and function of all living organisms. Each strand of DNA is built from a sequence of four different consisting of a , and a conventionally identified by the letters A,T, C and G. DNA itself consists of two of these that are coiled around a common axis in a configuration something like the protein alpha helix depicted above. The sugar-and-phosphate backbones are on the outside so that the nucleotide bases are on the inside and facing each other. The two strands are held together by hydrogen bonds that link a nitrogen atom of a nucleotide in one chain with a nitrogen or oxygen on the nucleotide that is across from it on the other chain. Efficient hydrogen bonding within this configuration can only occur between the pairs A-T and C-G, so these two complementary pairs constitute the "alphabet" that encodes the genetic information that gets transcribed whenever new protein molecules are built. Water molecules, hydrogen-bonded to the outer parts of the DNA helix, help stabilize it.

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are molecules that contain carbon-nitrogen bonds. The nitrogen atom in an amine has a lone pair of electrons and three bonds to other atoms, either carbon or hydrogen. Various nomenclatures are used to derive names for amines, but all involve the class-identifying suffix as illustrated here for a few simple examples: The genetic material for all living things is a polymer of four different molecules, which are themselves a combination of three subunits. The genetic information, the code for developing an organism, is contained in the specific sequence of the four molecules, similar to the way the letters of the alphabet can be sequenced to form words that convey information. The information in a DNA sequence is used to form two other types of polymers, one of which are proteins. The proteins interact to form a specific type of organism with individual characteristics. A genetic molecule is called DNA, which stands for deoxyribonucleic acid. The four molecules that make up DNA are called nucleotides. Each nucleotide consists of a single- or double-ringed molecule containing nitrogen, carbon, oxygen, and hydrogen called a nitrogenous base. Each base is bonded to a five-carbon sugar called deoxyribose. The sugar is in turn bonded to a phosphate group \(\ce{(−PO4^3- )}\) When new DNA is made, a polymerization reaction occurs that binds the phosphate group of one nucleotide to the sugar group of a second nucleotide. The nitrogenous bases of each nucleotide stick out from this sugar-phosphate backbone. DNA is actually formed from two such polymers coiled around each other and held together by hydrogen bonds between the nitrogenous bases. Thus, the two backbones are on the outside of the coiled pair of strands, and the bases are on the inside. The shape of the two strands wound around each other is called a double helix (Figure \(\Page {2}\)). It probably makes sense that the sequence of nucleotides in the DNA of a cat differs from those of a dog. But it is also true that the sequences of the DNA in the cells of two individual pugs differ. Likewise, the sequences of DNA in you and a sibling differ (unless your sibling is an identical twin), as do those between you and an unrelated individual. However, the DNA sequences of two related individuals are more similar than the sequences of two unrelated individuals, and these similarities in sequence can be observed in various ways. This is the principle behind DNA fingerprinting, which is a method used to determine whether two DNA samples came from related (or the same) individuals or unrelated individuals. Using similarities in sequences, technicians can determine whether a man is the father of a child (the identity of the mother is rarely in doubt, except in the case of an adopted child and a potential birth mother). Likewise, forensic geneticists can determine whether a crime scene sample of human tissue, such as blood or skin cells, contains DNA that matches exactly the DNA of a suspect. Like ammonia, amines are weak bases due to the lone pair of electrons on their nitrogen atoms: The basicity of an amine’s nitrogen atom plays an important role in much of the compound’s chemistry. Amine functional groups are found in a wide variety of compounds, including natural and synthetic dyes, polymers, vitamins, and medications such as penicillin and codeine. They are also found in many molecules essential to life, such as amino acids, hormones, neurotransmitters, and DNA. Since ancient times, plants have been used for medicinal purposes. One class of substances, called , found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with H O in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant: \[\ce{R3N + H3O+ + Cl- ⟶[R3NH+]Cl- + H2O} \nonumber \] The name alkaloid means “like an alkali.” Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base: \[\ce{[R3NH+]Cl- + OH- ⟶R3N + H2O + Cl-} \nonumber \] The structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant: are molecules that contain nitrogen atoms connected to the carbon atom of a carbonyl group. Like amines, various nomenclature rules may be used to name amides, but all include use of the class-specific suffix : Amides can be produced when carboxylic acids react with amines or ammonia in a process called amidation. A water molecule is eliminated from the reaction, and the amide is formed from the remaining pieces of the carboxylic acid and the amine (note the similarity to formation of an ester from a carboxylic acid and an alcohol discussed in the previous section): The reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both amine and carboxylic acid substituents) link together in a polymer to form proteins. Proteins are large biological molecules made up of long chains of smaller molecules called amino acids. Organisms rely on proteins for a variety of functions—proteins transport molecules across cell membranes, replicate DNA, and catalyze metabolic reactions, to name only a few of their functions. The properties of proteins are functions of the combination of amino acids that compose them and can vary greatly. Interactions between amino acid sequences in the chains of proteins result in the folding of the chain into specific, three-dimensional structures that determine the protein’s activity. Amino acids are organic molecules that contain an amine functional group (–NH ), a carboxylic acid functional group (–COOH), and a side chain (that is specific to each individual amino acid). Most living things build proteins from the same 20 different amino acids. Amino acids connect by the formation of a peptide bond, which is a covalent bond formed between two amino acids when the carboxylic acid group of one amino acid reacts with the amine group of the other amino acid. The formation of the bond results in the production of a molecule of water (in general, reactions that result in the production of water when two other molecules combine are referred to as condensation reactions). The resulting bond—between the carbonyl group carbon atom and the amine nitrogen atom is called a peptide link or peptide bond. Since each of the original amino acids has an unreacted group (one has an unreacted amine and the other an unreacted carboxylic acid), more peptide bonds can form to other amino acids, extending the structure. (Figure \(\Page {4}\)) A chain of connected amino acids is called a polypeptide. Proteins contain at least one long polypeptide chain. Enzymes are large biological molecules, mostly composed of proteins, which are responsible for the thousands of metabolic processes that occur in living organisms. Enzymes are highly specific catalysts; they speed up the rates of certain reactions. Enzymes function by lowering the activation energy of the reaction they are catalyzing, which can dramatically increase the rate of the reaction. Most reactions catalyzed by enzymes have rates that are millions of times faster than the noncatalyzed version. Like all catalysts, enzymes are not consumed during the reactions that they catalyze. Enzymes do differ from other catalysts in how specific they are for their substrates (the molecules that an enzyme will convert into a different product). Each enzyme is only capable of speeding up one or a few very specific reactions or types of reactions. Since the function of enzymes is so specific, the lack or malfunctioning of an enzyme can lead to serious health consequences. One disease that is the result of an enzyme malfunction is phenylketonuria. In this disease, the enzyme that catalyzes the first step in the degradation of the amino acid phenylalanine is not functional (Figure \(\Page {5}\)). Untreated, this can lead to an accumulation of phenylalanine, which can lead to intellectual disabilities. Kevlar (Figure \(\Page {6}\)) is a synthetic polymer made from two monomers 1,4-phenylene-diamine and terephthaloyl chloride (Kevlar is a registered trademark of DuPont). Kevlar’s first commercial use was as a replacement for steel in racing tires. Kevlar is typically spun into ropes or fibers. The material has a high tensile strength-to-weight ratio (it is about 5 times stronger than an equal weight of steel), making it useful for many applications from bicycle tires to sails to body armor. The material owes much of its strength to hydrogen bonds between polymer chains (refer back to the chapter on intermolecular interactions). These bonds form between the carbonyl group oxygen atom (which has a partial negative charge due to oxygen’s electronegativity) on one monomer and the partially positively charged hydrogen atom in the N–H bond of an adjacent monomer in the polymer structure (dashed lines in Figure \(\Page {7}\)). There is additional strength derived from the interaction between the unhybridized orbitals in the six-membered rings, called aromatic stacking. Kevlar may be best known as a component of body armor, combat helmets, and face masks. Since the 1980s, the military has used Kevlar as a component of the (personal armor system for ground troops) helmet and vest. Kevlar is also used to protect armored fighting vehicles and aircraft carriers. Civilian applications include protective gear for emergency service personnel such as body armor for police officers and heat-resistant clothing for fire fighters. Kevlar based clothing is considerably lighter and thinner than equivalent gear made from other materials (Figure \(\Page {8}\)). In addition to its better-known uses, Kevlar is also often used in cryogenics for its very low thermal conductivity (along with its high strength). Kevlar maintains its high strength when cooled to the temperature of liquid nitrogen (–196 °C). The table here summarizes the structures discussed in this chapter: The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides. Amines are a basic functional group. Amines and carboxylic acids can combine in a condensation reaction to form amides.

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Make sure you thoroughly understand the following essential ideas The picture of the atom that Niels Bohr developed in 1913 served as the starting point for modern atomic theory, but it was not long before Bohr himself recognized that the advances in quantum theory that occurred through the 1920's required an even more revolutionary change in the way we view the electron as it exists in the atom. This lesson will attempt to show you this view— or at least the portion of it that can be appreciated without the aid of more than a small amount of mathematics. About ten years after Bohr had developed his theory, de Broglie showed that the electron should have wavelike properties of its own, thus making the analogy with the mechanical theory of standing waves somewhat less artificial. One serious difficulty with the Bohr model still remained, however: it was unable to explain the spectrum of any atom more complicated than hydrogen. A refinement suggested by Sommerfeld assumed that some of the orbits are elliptical instead of circular, and invoked a second quantum number, , that indicated the degree of ellipticity. This concept proved useful, and it also began to offer some correlation with the placement of the elements in the periodic table. By 1926, de Broglie's theory of the wave nature of the electron had been experimentally confirmed, and the stage was set for its extension to all matter. At about the same time, three apparently very different theories that attempted to treat matter in general terms were developed. These were Schrödinger's wave mechanics, Heisenberg's matrix mechanics, and a more abstract theory of P.A.M. Dirac. These eventually were seen to be mathematically equivalent, and all continue to be useful. Of these alternative treatments, the one developed by Schrödinger is the most easily visualized. Schrödinger started with the simple requirement that the total energy of the electron is the sum of its kinetic and potential energies: \[ E = \underbrace{\dfrac{mv^2}{2}}_{\text{kinetic energy}} + \underbrace{\dfrac{-e^2}{r}}_{\text{potential energy}} \label{5.6.1}\] The second term represents the of an electron (whose charge is denoted by ) at a distance from a proton (the nucleus of the hydrogen atom). In quantum mechanics it is generally easier to deal with equations that use momentum (\(p = mv\)) rather than velocity, so the next step is to make this substitution: \[E = \dfrac{p^2}{2m} - \dfrac{e^2}{r} \label{5.6.2}\] This is still an entirely classical relation, as valid for the waves on a guitar string as for those of the electron in a hydrogen atom. The third step is the big one: in order to take into account the wavelike character of the hydrogen atom, a mathematical expression that describes the position and momentum of the electron at all points in space is applied to both sides of the equation. The function, denoted by \(\Psi\), "modulates" the equation of motion of the electron so as to reflect the fact that the electron manifests itself with greater probability in some locations that at others. This yields the celebrated Schrödinger equation \[ \left( \dfrac{mv^2}{2} - \dfrac{e^2}{r} \right) \Psi = E\Psi \label{5.6.3}\] How can such a simple-looking expression contain within it the quantum-mechanical description of an electron in an atom— and thus, by extension, of all matter? The catch, as you may well suspect, lies in discovering the correct form of Ψ , which is known as the . As this names suggests, the value of Ψ is a function of location in space relative to that of the proton which is the source of the binding force acting on the electron. As in any system composed of standing waves, certain boundary conditions must be applied, and these are also contained in Ψ; the major ones are that the value of must approach zero as the distance from the nucleus approaches infinity, and that the function be continuous. When the functional form of has been worked out, the Schrödinger equation is said to have been solved for a particular atomic system. The details of how this is done are beyond the scope of this course, but the consequences of doing so are extremely important to us. Once the form of is known, the allowed energies E of an atom can be predicted from the above equation. Soon after Schrödinger's proposal, his equation was solved for several atoms, and in each case the predicted energy levels agreed exactly with the observed spectra. There is another very useful kind of information contained in Ψ. Recalling that its value depends on the location in space with respect to the nucleus of the atom, the square of this function Ψ , evaluated at any given point, represents the probability of finding the electron at that particular point. The significance of this cannot be overemphasized; although the electron remains a particle having a definite charge and mass, and the question of "where" it is located is no longer meaningful. Any single experimental observation will reveal a definite location for the electron, but this will in itself have little significance; only a large number of such observations (similar to a series of multiple exposures of a photographic film) will yield meaningful results which will show that the electron can "be" anywhere with at least some degree of probability. This does not mean that the electron is "moving around" to all of these places, but that (in accord with the uncertainty principle) the concept of location has limited meaning for a particle as small as the electron. If we count only those locations in space at which the probability of the electron manifesting itself exceeds some arbitrary value, we find that the Ψ function defines a definite three-dimensional region which we call an . We can now return to the question which Bohr was unable to answer in 1912. Even the subsequent discovery of the wavelike nature of the electron and the analogy with standing waves in mechanical systems did not really answer the question; the electron is still a particle having a negative charge and is attracted to the nucleus. The answer comes from the Heisenberg Uncertainty Principle, which says that a quantum particle such as the electron cannot simultaneously have sharply-defined values of location and of momentum (and thus kinetic energy). To understand the implications of this restriction, suppose that we place the electron in a small box. The walls of the box define the precision δ to which the location is known; the smaller the box, the more exactly will we know the location of the electron. But as the box gets smaller, the uncertainty in the electron's kinetic energy will increase. As a consequence of this uncertainty, the electron will at times possess so much kinetic energy (the "confinement energy") that it may be able to penetrate the wall and escape the confines of the box. This process is known as ; the is exploited in various kinds of semiconductor devices, and it is the mechanism whereby electrons jump between dissolved ions and the electrode in batteries and other electrochemical devices. The region near the nucleus can be thought of as an extremely small funnel-shaped box, the walls of which correspond to the electrostatic attraction that must be overcome if an electron confined within this region is to escape. As an electron is drawn toward the nucleus by electrostatic attraction, the volume to which it is confined diminishes rapidly. Because its location is now more precisely known, its kinetic energy must become more uncertain; the electron's kinetic energy rises more rapidly than its potential energy falls, so that it gets ejected back into one of its allowed values of . We can also dispose of the question of why the orbiting electron does not radiate its kinetic energy away as it revolves around the nucleus. The Schrödinger equation completely discards any concept of a definite path or trajectory of a particle; what was formerly known as an "orbit" is now an "orbital", defined as the locations in space at which the probability of finding the electrons exceeds some arbitrary value. It should be noted that this wavelike character of the electron coexists with its possession of a momentum, and thus of an effective velocity, even though its motion does not imply the existence of a definite path or trajectory that we associate with a more massive particle. The modern view of atomic structure dismisses entirely the old but comfortable planetary view of electrons circling around the nucleus in fixed orbits. As so often happens in science, however, the old outmoded theory contains some elements of truth that are retained in the new theory. In particular, the old Bohr orbits still remain, albeit as spherical shells rather than as two-dimensional circles, but their physical significance is different: instead of defining the "paths" of the electrons, they merely indicate the locations in the space around the nucleus at which the probability of finding the electron has higher values. The electron retains its particle-like mass and momentum, but because the mass is so small, its wavelike properties dominate. The latter give rise to patterns of standing waves that define the possible states of the electron in the atom. Modern quantum theory tells us that the various allowed states of existence of the electron in the hydrogen atom correspond to different standing wave patterns. In the preceding lesson we showed examples of standing waves that occur on a vibrating guitar string. The wave patterns of electrons in an atom are different in two important ways: Aside from this, the similarities are striking. Each wave pattern is identified by an integer number , which in the case of the atom is known as the . The value of \(n\) tells how many peaks of amplitude (antinodes) exist in that particular standing wave pattern; the more peaks there are, the higher the energy of the state. The three simplest orbitals of the hydrogen atom are depicted above in pseudo-3D, in cross-section, and as plots of probability (of finding the electron) as a function of distance from the nucleus. The average radius of the electron probability is shown by the blue circles or plots in the two columns on the right. These radii correspond exactly to those predicted by the Bohr model. The principal quantum number. The value of \(n\) tells how many peaks of amplitude (antinodes) exist in that particular standing wave pattern; the more peaks there are, the higher the energy of the state. The potential energy of the electron is given by the formula \[ E =\dfrac{-4 \pi^2 e^4 m}{h^2n^2} \label{5.6.4}\] with The negative sign ensures that the potential energy is negative. Notice that this energy in inversely proportional to the of \(n\), so that the energy rises toward zero as \(n\) becomes very large, but it can never exceed zero. Equation \(\ref{5.6.4}\) was actually part of Bohr's original theory and is still applicable to the hydrogen atom, although to atoms with two or more electrons. In the Bohr model, each value of \(n\) corresponded to an orbit of a different radius. The larger the orbital radius, the higher the potential energy of the electron; the inverse square relationship between electrostatic potential energy and distance is reflected in the inverse square relation between the energy and n in the above formula. Although the concept of a definite trajectory or orbit of the electron is no longer tenable, the same orbital radii that relate to the different values of n in Bohr's theory now have a new significance: they give the average distance of the electron from the nucleus. The averaging process must encompass several probability peaks in the case of higher values of \(n\). The spatial distribution of these probability maxima defines the particular orbital. This physical interpretation of the principal quantum number as an index of the from the nucleus turns out to be extremely useful from a chemical standpoint, because it relates directly to the tendency of an atom to lose or gain electrons in chemical reactions. The electron wave functions that are derived from Schrödinger's theory are characterized by several quantum numbers. The first one, , describes the nodal behavior of the probability distribution of the electron, and correlates with its potential energy and average distance from the nucleus as we have just described. The theory also predicts that orbitals having the same value of can differ in shape and in their orientation in space. The quantum number , known as the , determines the shape of the orbital. (More precisely, determines the number of angular nodes, that is, the number of regions of zero probability encountered in a 360° rotation around the center.) When = 0, the orbital is spherical in shape. If = 1, the orbital is elongated into something resembling a figure-8 shape, and higher values of correspond to still more complicated shapes— but note that the number of peaks in the radial probability distributions (below) decreases with increasing l. The possible values that can take are strictly limited by the value of the principal quantum number; can be no greater than – 1. This means that for = 1, can only have the single value zero which corresponds to a spherical orbital. For historical reasons, the orbitals corresponding to different values of are designated by letters, starting with for = 0, for = 1, for = 2, and for = 3. The function relationship is given by \[ \bar{r} = (5.29\; pm) \dfrac{n^2}{Z} \left[ \dfrac{3}{2} - \dfrac{l(l-1)}{2n^2}\right]\] in which \(Z\) is the nuclear charge of the atom, which of course is unity for hydrogen. An -orbital, corresponding to = 0, is spherical in shape and therefore has no special directional properties. The probability cloud of a orbital is aligned principally along an axis extending along any of the three directions of space. The additional quantum number is required to specify the particular direction along which the orbital is aligned. "Direction in space" has no meaning in the absence of a force field that serves to establish a reference direction. For an isolated atom there is no such external field, and for this reason there is no distinction between the orbitals having different values of \(m\). If the atom is placed in an external magnetic or electrostatic field, a coordinate system is established, and the orbitals having different values of m will split into slightly different energy levels. This effect was first seen in the case of a magnetic field, and this is the origin of the term magnetic quantum number. In chemistry, however, electrostatic fields are much more important for defining directions at the atomic level because it is through such fields that nearby atoms in a molecule interact with each other. The electrostatic field created when other atoms or ions come close to an atom can cause the energies of orbitals having different direction properties to split up into different energy levels; this is the origin of the colors seen in many inorganic salts of transition elements, such as the blue color of copper sulfate. The quantum number m can assume 2 + 1 values for each value of , from – through 0 to + . When = 0 the only possible value of will also be zero, and for the orbital (l = 1), can be –1, 0, and +1. Higher values of introduce more complicated orbital shapes which give rise to more possible orientations in space, and thus to more values of . Certain fundamental particles have associated with them a that can align itself in either of two directions with respect to an external magnetic field. The electron is one such particle, and the direction of its magnetic moment is called its . A basic principle of modern physics states that for particles such as electrons that possess half-integral values of spin, no two of them can be in identical quantum states within the same system. The quantum state of a particle is defined by the values of its quantum numbers, so what this means is that This is known as the Pauli exclusion principle, named after the German physicist Wolfgang Pauli (1900-1958, Nobel Prize 1945). The mechanical analogy implied by the term spin is easy to visualize, but . Physical rotation of an electron is meaningless. However, the coordinates of the electron's wave function can be rotated mathematically; when this is done, it is found that a rotation of 720° is required to restore the function to its initial value— rather weird, considering that a 360° rotation will leave any extended body unchanged! Electron spin is basically a relativistic effect in which the electron's momentum distorts local space and time. It has no classical counterpart and thus cannot be visualized other than through mathematics. The exclusion principle was discovered empirically and was placed on a firm theoretical foundation by Pauli in 1925. A complete explanation requires some familiarity with quantum mechanics, so all we will say here is that if two electrons possess the same quantum numbers and (defined below), the wave function that describes the state of existence of the two electrons together becomes zero, which means that this is an "impossible" situation. A given is characterized by a fixed set of the quantum numbers , and . The electron spin itself constitutes a fourth quantum number , which can take the two values +1 and –1. Thus , which "cancel out" to produce zero magnetic moment. Two such electrons in a single orbital are often referred to as an . If it were not for the exclusion principle, the atoms of all elements would behave in the same way, and there would be no need for a science of Chemistry! As we have seen, the lowest-energy standing wave pattern the electron can assume in an atom corresponds to =1, which describes the state of the single electron in hydrogen, and of the two electrons in helium. Since the quantum numbers and are zero for =1, the pair of electrons in the helium orbital have the values ( ) = (1,0,0,+1) and (1,0,0,–1)— that is, they differ only in spin. These two sets of quantum numbers are the only ones that are possible for a =1 orbital. The additional electrons in atoms beyond helium must go into higher-energy ( >1) orbitals. Electron wave patterns corresponding to these greater values of are concentrated farther from the nucleus, with the result that these electrons are less tightly bound to the atom and are more accessible to interaction with the electrons of neighboring atoms, thus influencing their . If it were not for the Pauli principle, all the electrons of every element would be in the lowest-energy =1 state, and the differences in the chemical behavior the different elements would be minimal. Chemistry would certainly be a simpler subject, but it would not be very interesting!

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.06%3A_Refining_of_Metals

Once a metal is reduced, it is still not necessarily pure enough for all uses to which it might be put. An obvious example is the brittleness and low tensile strength of pig iron, characteristics which make it suitable for casting, but little else. These adverse properties are due to the presence of impurities, a typical analysis of blast-furnace iron showing about 4% C, 2% P, 2.5%, Si, 2.5%Mn, and 0.1% S by weight. Further refining to remove these impurities (especially carbon) produces , a much stronger and consequently more useful material. Steelmaking involves oxidation of the impurities in basic oxygen, open hearth, or electric furnaces. Some oxidation products (CO, CO , and SO ) are volatile and easily separated. The others end up, along with some iron oxides, in a slag which floats on the surface of the molten steel. In the oxidation is due to air at the surface of the molten metal. This method of steel refining was developed in the mid 1800s, contemporaneous with the industrial revolution. This method requires up to 12 hours—and natural gas or other fuel must be burned to keep the metal liquid. Thus the open hearth wastes large amounts of free energy. The use of fossil fuel does make it possible to recycle as much as 50 percent scrap iron, however, and the longer melting time allows somewhat greater control over the composition of a batch of steel. Developed in the 1950s, the has replaced the open-hearth furnace as the primary steelmaking method. In this process, pure oxygen is directed onto the surface of molten pig iron in a large crucible. Some of the iron is oxidized to Fe O and Fe O , forming an oxidizing slag. The impurities, namely, C, P, Si, Mn, and S, are all oxidized at the same time. Since all these reactions are spontaneous and exothermic, they provide enough heat so that up to 25 percent solid scrap iron may be melted in the crucible without cooling it to the point where solid iron would remain. Oxidation of one batch of pig iron and scrap normally takes slightly more than half an hour. Thus, this method is far quicker than the open-hearth furnace, is three times more efficient. Computers are now used to interpret spectroscopic analyses of steel in basic oxygen furnaces, indicating in a few minutes what metals must be added to obtain the desired composition. This has largely eliminated the last advantage of the open hearth and speeded up changeovers to basic oxygen. It has also decreased recycling of iron because the latter furnace cannot handle as much scrap. Much recycling of iron is now done in which can melt a charge of 100 percent scrap. In addition to the chemical oxidations used in steelmaking, electrolytic oxidation and reduction is quite important in refining metals. The electrolytic , and has already been described.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/04%3A_Alternative_Solvents

In a very general sense, solvents are a class of chemical compounds that allow chemistry to occur. The concept of a solvent has significant ramifications because they serve as the matrix, medium, or carrier for solutes. They are necessary in a number of processes, reactions, and systems. We tend to think of a substance like “water” as a universal solvent because it is so useful in so many disciplines. Water cleans up everything, allows biochemical reactions to occur, is used in paints, coatings, and films, allows cooking to occur (or else everything would catch fire), and provides lubrication and ease of movement for a great many devices.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Acid_Rain/Acid_Snow

The impact of acid precipitation on aquatic ecosystems may be intensified by melting snow. When snow melts rapidly in the spring, the stream or lake may be "shocked" with an excessive amount of acid. In the spring, at the time of acid snow melting, the various aquatic organisms are reproducing and are the most sensitive increases in acid. During the winter of 1976-77, there was no significant snow melt from mid-December through February at Little Moose Lake, New York. The highest depth was 130 cm. (top graph). The snow had an average pH of 4.4 in Feb. (middle graph). The snow started to melt in early March.(bottom graph). The first major thaw in early March resulted in the release of 80% of the stored acid in a one week period. During the first couple of days, the snow melt water pH was 3.4-3.6. The pH gradually rose over the eight day period to pH 5.0. The pH levels in Little Moose lake are normally about 7.0. During the snow melting, in early March, the lake pH dropped to 6.0. An outlet stream from the lake reached a low pH of 4.8. A small brook nearby hit a low pH of 4.6 during the snow melt period. The average pH in this brook during the rest of the year is about 5.4. There was a definite relationship of the amount of aluminum in the brook vs. pH. As the pH decreased, the aluminum concentration increased. Captive populations of adult, year old, and larval brook trout had been maintained over the winter in Little Moose Lake water without any problems. During the snow melt in March, 3 adult brook trout died, 25 one year old trout died, and about 50,000 recently hatched trout died. Various other abnormal behaviors were also observed.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Properties_of_Polymers

A comparison of the properties of polyethylene (both LDPE & HDPE) with the natural polymers rubber and cellulose is instructive. As noted above, synthetic HDPE macromolecules have masses ranging from 105 to 106 amu (LDPE molecules are more than a hundred times smaller). Rubber and cellulose molecules have similar mass ranges, but fewer monomer units because of the monomer's larger size. The physical properties of these three polymeric substances differ from each other, and of course from their monomers. To account for the differences noted here we need to consider the nature of the aggregate macromolecular structure, or morphology, of each substance. Because polymer molecules are so large, they generally pack together in a non-uniform fashion, with ordered or crystalline-like regions mixed together with disordered or amorphous domains. In some cases the entire solid may be amorphous, composed entirely of coiled and tangled macromolecular chains. Crystallinity occurs when linear polymer chains are structurally oriented in a uniform three-dimensional matrix. In the diagram on the right, crystalline domains are colored blue. Increased crystallinity is associated with an increase in rigidity, tensile strength and opacity (due to light scattering). Amorphous polymers are usually less rigid, weaker and more easily deformed. They are often transparent. Three factors that influence the degree of crystallinity are: i) Chain length ii) Chain branching iii) Interchain bonding The importance of the first two factors is nicely illustrated by the differences between LDPE and HDPE. As noted earlier, HDPE is composed of very long unbranched hydrocarbon chains. These pack together easily in crystalline domains that alternate with amorphous segments, and the resulting material, while relatively strong and stiff, retains a degree of flexibility. In contrast, LDPE is composed of smaller and more highly branched chains which do not easily adopt crystalline structures. This material is therefore softer, weaker, less dense and more easily deformed than HDPE. As a rule, mechanical properties such as ductility, tensile strength, and hardness rise and eventually level off with increasing chain length. The nature of cellulose supports the above analysis and demonstrates the importance of the third factor (iii). To begin with, cellulose chains easily adopt a stable rod-like conformation. These molecules align themselves side by side into fibers that are stabilized by inter-chain hydrogen bonding between the three hydroxyl groups on each monomer unit. Consequently, crystallinity is high and the cellulose molecules do not move or slip relative to each other. The high concentration of hydroxyl groups also accounts for the facile absorption of water that is characteristic of cotton. Natural rubber is a completely amorphous polymer. Unfortunately, the potentially useful properties of raw latex rubber are limited by temperature dependence; however, these properties can be modified by chemical change. The cis-double bonds in the hydrocarbon chain provide planar segments that stiffen, but do not straighten the chain. If these rigid segments are completely removed by hydrogenation (H2 & Pt catalyst), the chains lose all constrainment, and the product is a low melting paraffin-like semisolid of little value. If instead, the chains of rubber molecules are slightly cross-linked by sulfur atoms, a process called vulcanization which was discovered by Charles Goodyear in 1839, the desirable elastomeric properties of rubber are substantially improved. At 2 to 3% crosslinking a useful soft rubber, that no longer suffers stickiness and brittleness problems on heating and cooling, is obtained. At 25 to 35% crosslinking a rigid hard rubber product is formed. The following illustration shows a cross-linked section of amorphous rubber. By clicking on the diagram it will change to a display of the corresponding stretched section. The more highly-ordered chains in the stretched conformation are entropically unstable and return to their original coiled state when allowed to relax ),

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.10%3A_Yeast_Metabolism

Yeasts are ubiquitous unicellular fungi widespread in natural environments. Yeast have a broad set of carbon sources (e.g., polyols, alcohols, organic acids and amino acids) that they can metabolize but they prefer sugars. Yeast are capable of metabolizing hexoses (glucose, fructose, galactose or mannose) and disaccharides (maltose or sucrose) as well as compounds with two carbons (ethanol or acetate). The metabolic pathways utilized by yeast are Embden-Meyerhof glycolysis, tricarboxylic acid cycle (TCA), the pentose phosphate pathway, and oxidative phosphorylation. Embden-Meyerhof Glycolysis is the pathway utilized by most eukaryotes. Ethanol fermentation reaction occurs in two steps, decarboxylation and then hydride reduction. Show the curved arrows for this mechanism. 2. The second reaction, catalyzed by the enzyme alcohol dehydrogenase, regenerates NAD by reducing the acetaldehyde to ethanol. Ethanol has the added benefit of being toxic to competing organisms. However, it will also start to kill the yeast that is producing the ethanol. at the accumulation of alcohol will become toxic when it reaches a concentration between 14-18%, thereby killing the yeast cells Ethanol fermentation utilizes the pyruvate from glycolysis to regenerate NAD . This is an alternative pathway to metabolize glucose. The pathway is operated by and other yeast fermenters that ultimately produces ethanol and CO . When would you expect that an organism would choose to operate each pathway? Pasteur observed that yeast produce alcohol only as the product of a “starvation process” once they run out of oxygen. This observation has been shown to be incorrect! The Crabtree effect is the occurrence of alcoholic fermentation under conditions. The most common yeasts used in fermentation processes ( genus) will produce alcohol in both a beer wort and in bread dough immediately regardless of aeration. While you might expect the cell would perform aerobic respiration (full conversion of sugar and oxygen to carbon dioxide and water) as long as oxygen is present, while reverting to alcoholic fermentation, when there is no oxygen as it produces less energy. However, if a yeast finds itself in a sugar environment, it will immediately start producing ethanol, shunting sugar into the anaerobic respiration pathway while still running the aerobic process in parallel. This phenomenon is known as the . People have speculated that yeast use the ability to produce ethanol to kill competing organisms in the high-sugar environment. Summarize:

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/Nucleotides

Nucleotides are the basic monomer building block units in the nucleic acids. A nucleotide consists of a phosphate, pentose sugar, and a heterocyclic amine. The phosphoric acid forms a phosphate-ester bond with the alcohol on carbon #5 in the pentose. A nitrogen in the heterocyclic amines displaces the -OH group on carbon #1 of the pentose. The reaction is shown in the graphic below. If the sugar is ribose, the general name is ribonucleotide and deoxyribonucleotide if the sugar is deoxyribose. The other four nucleotides are synthesized in a similar fashion. Just as the exact amino acid sequence is important in proteins, the sequence of heterocyclic amine bases determines the function of the DNA and RNA. This sequence of bases on DNA determines the genetic information carried in each cell. Currently, much research is under way to determine the heterocyclic amine sequences in a variety of RNA and DNA molecules. The Genome Project as already succeeded in determining the DNA sequences in humans and other organisms. Future research will be to determine the exact functions of each DNA segment as these contain the codes for protein synthesis.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/Enzyme_Inhibitors

This page looks at the effect of inhibitors on reactions involving enzymes. This is the third and final page talking about how enzymes function as catalysts. This is the most straightforward and obvious form of enzyme inhibition - and the name tells you exactly what happens. The inhibitor has a similar shape to the usual substrate for the enzyme, and competes with it for the active site. However, once it is attached to the active site, nothing happens to it. It doesn't react - essentially, it just gets in the way. Remember the general equation for an enzyme reacting with a substrate? \[\ce{E + X <=> E-S}\, \text{complex} \ce{-> E + P}\] The equivalent equation for a competitive inhibitor looks like this: \[ \ce{E + I_{c} <=> E-I_{c}}\, \text{complex}\] The complex does not react any further to form products - but its formation is still reversible. It breaks up again to form the enzyme and the inhibitor molecule. That means that if you increase the concentration of the substrate, the substrate can out-compete the inhibitor, and so the normal reaction can take place at a reasonable rate. A simple example of this involves malonate ions inhibiting the enzyme succinate dehydrogenase. This enzyme catalyses the conversion of succinate ions to fumarate ions. The modern names are: The conversion that succinic dehydrogenase carries out is: The reaction is inhibited by malonate ions which have a very similar shape to succinate ions. The similar shape lets the malonate ions bind to the active site, but the lack of the CH -CH bond in the centre of the ion stops any further reaction taking place. The malonate ions therefore block the active site - but remember that this is reversible. The malonate ions will break away and free up the enzyme again. The malonate ions are in competition for the site - they aren't destroying it. If the succinate ions have a greater concentration than the malonate ions, by chance they will get access to the site more often than the malonate ions. That means that you can overcome the effect of a competitive inhibitor by increasing the concentration of the substrate. A non-competitive inhibitor doesn't attach itself to the active site, but attaches somewhere else on the enzyme. By attaching somewhere else it affects the structure of the enzyme and so the way the enzyme works. Because there isn't any competition involved between the inhibitor and the substrate, increasing the substrate concentration won't help. You will find two explanations for this. We'll look at the simple, fairly obvious one in some detail in a minute. I want to have a brief word about the other one first. This explanation says that the inhibitor doesn't affect the ability of the substrate to bond with the active site, but stops it reacting once it is there. I found a couple of biochemistry sites which said that inhibitors working like this (which they describe as pure non-competitive inhibitors) are virtually unknown. As a non-biochemist, I don't know what the truth is about this - if you want to find out, you will probably have to do a biochemistry degree! The straightforward explanation (which would seem to apply to most enzymes) is that reaction with the inhibitor causes the shape of the active site to change. Remember that non-competitive inhibitors aren't attaching directly to the active site, but elsewhere on the enzyme. The inhibitor attachs to a side group in the protein chain, and affects the way the protein folds into its tertiary structure. That in turn changes the shape of the active site. If the shape of the active site changes, then the substrate can't attach to it any more. Some non-competitive inhibitors attach irreversibly to the enzyme, and therefore stop it working permanently. Others attach reversibly. A relatively uncomplicated example of non-competitive inhibitors in a reasonably familiar situation is heavy metal poisoning. You are probably aware that compounds containing heavy metals such as lead, mercury, copper or silver are poisonous. This is because ions of these metals are non-competitive inhibitors for several enzymes. I'm going to take silver as a simple example. Silver ions react with -SH groups in the side groups of cysteine residues in the protein chain: There is not enough electronegativity difference between silver and sulfur for a full ionic bond and so the bond can be considered as covalent. If the cysteine residue is somewhere on the protein chain which affects the way it folds into its tertiary structure, then altering this group could have an effect on the shape of the active site, and so stop the enzyme from working. The 2+ ions from, for example, mercury, copper or lead can behave similarly - also attaching themselves to the sulphur in place of the hydrogen.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.01%3A_Solutions_and_their_Concentrations

Make sure you thoroughly understand the following essential ideas: Solutions are (single-phase) mixtures of two or more . For convenience, we often refer to the majority component as the ; minority components are ; there is really no fundamental distinction between them. Solutions play a very important role in Chemistry because they allow intimate and varied encounters between molecules of different kinds, a condition that is essential for rapid chemical reactions to occur. Several more explicit reasons can be cited for devoting a significant amount of effort to the subject of solutions: We usually think of a solution as a liquid made by adding a gas, a solid or another liquid in a liquid . Actually, solutions can exist as gases and solids as well. Solid solutions are very common; most natural minerals and many metallic alloys are solid solutions. Still, it is liquid solutions that we most frequently encounter and must deal with. Experience has taught us that sugar and salt dissolve readily in water, but that “oil and water don’t mix”. Actually, this is not strictly correct, since all substances have at least a slight tendency to dissolve in each other. This raises two important and related questions: why do solutions tend to form in the first place, and what factors limit their mutual solubilities? is a general term that expresses the quantity of solute contained in a given amount of solution. Various ways of expressing concentration are in use; the choice is usually a matter of convenience in a particular application. You should become familiar with all of them. In the consumer and industrial world, the most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The “quantities” referred to here can be expressed in weight, in volume, or both (i.e., the of solute in a given of solution.) In order to distinguish among these possibilities, the abbreviations (w/w), (v/v) and (w/v) are used. In most applied fields of Chemistry, (w/w) measure is often used, and is commonly expressed as weight-percent concentration, or simply "percent concentration". For example, a solution made by dissolving 10 g of salt with 200 g of water contains "1 part of salt per 20 g of water". The normal saline solution used in medicine for nasal irrigation, wound cleaning and intravenous drips is a 0.91% (w/v) solution of sodium chloride in water. How would you prepare 1.5 L of this solution? The solution will contain 0.91 g of NaCl in 100 mL of water, or 9.1 g in 1 L. Thus you will add (1.5 × 9.1g) = 13.6 g of NaCl to 1.5 L of water. Percent means parts per 100; we can also use parts per thousand (ppt) for expressing concentrations in grams of solute per kilogram of solution. For more dilute solutions, parts per million (ppm) and parts per billion (10 ; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment. Describe how you would prepare 30 g of a 20 percent (w/w) solution of KCl in water. The weight of potassium chloride required is 20% of the total weight of the solution, or 0.2 × (3 0 g) = 6.0 g of KCl. The remainder of the solution (30 – 6 = 24) g consists of water. Thus you would dissolve 6.0 g of KCl in 24 g of water. It is sometimes convenient to base concentration on a fixed volume, either of the solution itself, or of the solvent alone. In most instances, a 5% by volume solution of a solid will mean 5 g of the solute dissolved in 100 ml of the solvent. Fish, like all animals, need a supply of oxygen, which they obtain from oxygen dissolved in the water. The minimum oxygen concentration needed to support most fish is around 5 ppm (w/v). How many moles of O per liter of water does this correspond to? 5 ppm (w/v) means 5 grams of oxygen in one million mL (1000 L) of water, or 5 mg per liter. This is equivalent to (0.005 g) / (32.0 g mol ) = 1.6 × 10 mol. These kinds of concentration measure are mostly used in commercial and industrial applications. The "proof" of an alcoholic beverage is the (v/v)-percent, multiplied by two; thus a 100-proof vodka has the same alcohol concentration as a solution made by adding sufficient water to 50 ml of alcohol to give 100 ml of solution. This is the method most used by chemists to express concentration, and . Molar concentration (molarity) is the number of moles of solute per liter of solution. The important point to remember is that the volume of the is different from the volume of the ; the latter quantity can be found from the molarity only if the densities of both the solution and of the pure solvent are known. Similarly, calculation of the weight-percentage concentration from the molarity requires density information; you are expected to be able to carry out these kinds of calculations, which are covered in most texts. How would you make 120 mL of a 0.10 M solution of potassium hydroxide in water? The amount of KOH required is (0.120 L) × (0.10 mol L ) = 0.012 mol. The molar mass of KOH is 56.1 g, so the weight of KOH required is \[(0.012\; mol) \times (56.1\; g \;mol^{-1}) = 0.67\; g\] We would dissolve this weight of KOH in a volume of water that is less than 120 mL, and then add sufficient water to bring the volume of the solution up to 120 mL. Note if we had simply added the KOH to 120 mL of water, the molarity of the resulting solution would not be the same. This is because volumes of different substances are not strictly additive when they are mixed. Without actually measuring the volume of the resulting solution, its molarity would not be known. Calculate the molarity of a 60-% (w/w) solution of ethanol (C H OH) in water whose density is 0.8937 g mL . One liter of this solution has a mass of 893.7 g, of which \[0.60 \times (893.7\; g) = 536.2\; g\] consists of ethanol. The molecular weight of C H OH is 46.0, so the number of moles of ethanol present in one liter (that is, the molarity) will be \[ \dfrac{\dfrac{536.2\;g}{46.0\;g\;mol^{-1}}}{1 L} =11.6\; mol\,L^{-1}\] is a now-obsolete concentration measure based on the number of per liter of solution. Although the latter term is now also officially obsolete, it still finds some use in clinical- and environmental chemistry and in electrochemistry. Both terms are widely encountered in pre-1970 textbooks and articles. The of an is its molecular weight divided by the number of titratable hydrogens it carries. Thus for sulfuric acid H SO , one mole has a mass of 98 g, but because both hydrogens can be neutralized by strong base, its equivalent weight is 98/2 = 49 g. A solution of 49 g of H SO per liter of water is 0.5 molar, but also "1 normal" (1 = 1 eq/L). Such a solution is "equivalent" to a 1 solution of HCl in the sense that each can be neutralized by 1 mol of strong base. Although molar concentration is widely employed, it suffers from one serious defect: since volumes are temperature-dependent (substances expand on heating), so are molarities; a 0.100 M solution at 0° C will have a smaller concentration at 50° C. For this reason, molarity is not the preferred concentration measure in applications where physical properties of solutions and the effect of temperature on these properties is of importance. This is the most fundamental of all methods of concentration measure, since it makes no assumptions at all about volumes. The mole fraction of substance in a mixture is defined as \[ X_i= \dfrac{n_i}{\sum_j n_j}\] in which is the number of moles of substance , and the summation is over all substances in the solution. Mole fractions run from zero (substance not present) to unity (the pure substance). The sum of all mole fractions in a solution is, by definition, unity: \[\sum_i X_i=1\] What fraction of the molecules in a 60-% (w/w) solution of ethanol in water consist of H O? From the previous problem, we know that one liter of this solution contains 536.2 g (11.6 mol) of C H OH. The number of moles of H O is ( (893.7 – 536.2) g) / (18.0 g mol ) = 19.9 mol. The mole fraction of water is thus \[\dfrac{19.9}{19.9+11.6} = 0.63\] Thus 63% of the molecules in this solution consist of water, and 37% are ethanol. In the case of ionic solutions, each kind of ion acts as a separate component. Find the mole fraction of water in a solution prepared by dissolving 4.5 g of CaBr in 84.0 mL of water. The molar mass of CaBr is 200 g, and 84.0 mL of H O has a mass of very close to 84.0 g at its assumed density of 1.00 g mL . Thus the number of moles of CaBr in the solution is \[\dfrac{4.50\; g}{200\; g/mol} = 0.0225 \;mol\] Because this salt is completely dissociated in solution, the solution will contain 0.268 mol of Ca and (2 × .268) = 0.536 of Br . The number of moles of water is (84 g) / (18 g mol ) = 4.67 mol. The mole fraction of water is then \[\dfrac{0.467\; \cancel{mol}}{0.268 + 0.536 + 4.67\; \cancel{mol}} = \dfrac{0.467}{5.47} = 0.854\] Thus H O constitutes 85 out of every 100 molecules in the solution. A 1-molal solution contains one mole of solute per 1 kg of solvent. Molality is a hybrid concentration unit, retaining the convenience of mole measure for the solute, but expressing it in relation to a temperature-independent mass rather than a volume. Molality, like mole fraction, is used in applications dealing with certain physical properties of solutions; we will see some of these in the next lesson. Calculate the molality of a 60-% (w/w) solution of ethanol in water. From the above problems, we know that one liter of this solution contains 11.6 mol of ethanol in (893.7 – 536.2) = 357.5 g of water. The molarity of ethanol in the solution is therefore (11.6 mol) / (0.3575 kg) = 32.4 mol kg . Anyone doing practical chemistry must be able to convert one kind of concentration measure into another. The important point to remember is that any conversion involving molarity requires a knowledge of the of the solution. A solution prepared by dissolving 66.0 g of urea (NH ) CO in 950 g of water had a density of 1.018 g mL . Express the concentration of urea in The weight-percent of solute is (100%) (66.0 g) / (950 g) = 6.9% The molar mass of urea is 60, so the number of moles is (66 g) /(60 g mol ) = 1.1 mol. The number of moles of H O is (950 g) / (18 g mol ) = 52.8 mol. b) Mole fraction of urea: (1.1 mol) / (1.1 + 52.8 mol) = 0.020 c) molarity of urea: the volume of 1 L of solution is (66 + 950)g / (1018 g L )= 998 mL. The number of moles of urea (from a) is 1.1 mol. Its molarity is then (1.1 mol) / (0.998 L) = 1.1 mol L . d) The molality of urea is (1.1 mol) / (.066 + .950) kg = 1.08 mol kg . Ordinary dry air contains 21% (v/v) oxygen. About many moles of O can be inhaled into the lungs of a typical adult woman with a lung capacity of 4.0 L? The number of molecules (and thus the number of moles) in a gas is directly proportional to its volume ( ), so the mole fraction of O is 0.21. The molar volume of a gas at 25° C is (298/271) × 22.4 L mol = 24.4 L mol so the moles of O in 4 L of air will be (4 / 24.4) × (0.21 mol) × (24.4 L mol ) = 0.84 mol O . These kinds of calculations arise frequently in both laboratory and practical applications. If you have a thorough understanding of concentration definitions, they are easily tackled. The most important things to bear in mind are Commercial hydrochloric acid is available as a 10.17 molar solution. How would you use this to prepare 500 mL of a 4.00 molar solution? The desired solution requires (0.50 L) × (4.00 M L = 2.0 mol of HCl. This quantity of HCl is contained in (2.0 mol) / (10.17 M L ) = 0.197 L of the concentrated acid. So one would measure out 197 mL of the concentrated acid, and then add water to make the total volume of 500 mL. Calculate the molarity of the solution produced by adding 120 mL of 6.0 M HCl to 150 mL of 0.15 M HCl. What important assumption must be made here? The assumption, of course, is that the density of HCl within this concentration range is constant, meaning that their volumes will be additive. Moles of HCl in first solution: (0.120 L) × (6.0 mol L ) = 0.72 mol HCl Moles of HCl in second solution: (0.150 L) × (0.15 mol L ) = 0.02 mol HCl Molarity of mixture: (0.72 + 0.02) mol / (.120 + .150) L = 4.3 mol L .

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All four solution effects ( , , , ) result from “dilution” of the solvent by the added solute. Because of this commonality they are referred to as colligative properties (Lat. , connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of these phenomena occur. Basically, these all result from the effect of on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at equal densities. The temperatures at which this occurs are depicted by the shading. When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised. The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase. Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase. This phenomenon can explain . Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to osmotic flow of solvent into the solution. The effect of this pressure \(\Pi\) is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium. Osmotic pressure does not drive osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to osmotic flow of solvent into the solution.

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The structure of Reserpine was solved by 1953. R.B. Woodward’s group reported the first synthesis of Reserpine in 1956 (J. Am. Chem. Soc., 78, 2023, 2657 (1956); Tetrahedron, 2, 1 (1958)). His scholarly analysis clearly displayed aspects of retroanalysis, which was just evolving at that time. This synthesis commands admiration for the way he used conformational analysis and stereoelectronic effects to precisely develop the stereopoints in this exceedingly complex problem for that time. He recognized that the E-ring has a dense array of 5 asymmetric centers in a six membered E ring. His disconnection of reserpine led him to the key intermediate C. We could formalize his retroanalysis as shown in Figure 9.1. Another brilliant piece of conformational analysis could be seen in the way he converted Isoreserpine to Reserpine by introducing conformational strain in an otherwise comfortable molecule. In a cleaver execution by Woodward’s group , all the required carbons for the D/E rings and three of the five asymmetric centers were created by one Diels-Alder reaction . Note the simple dissymmetry in one component – methyl acrylate – could precisely place three asymmetric centers in a row in a correct fashion. This cycloaddition reaction developed a concave phase and a convex phase1 in the product that guided further modifications on this intermediate. The hydride reagent in the next step delivered the hydride from the less hindered convex phase, placing the –OH group in the concave phase. This facilitated the formation of a five-membered lactone ring . The alternate six-membered lactone ring was presumably more strained. Bromination on with molecular bromine formed the bromonium ion complex from the convex phase, while the –OH group could enter from the concave phase to form an ether . Treatment with sodium methoxide displaced the bromine from the convex phase, possibly via elimination – addition route . The next bromonium ion complex again proceeded from the convex phase, with the water molecule entering from the concave phase to enable a trans-diaxial opening of the bromonium complex . After one oxidation step, the molecule was now set for a complex double elimination reaction, with the zinc attacking two centers. An attack at the bromide center opened the ether ring while another attack at the carbonyl group opened the lactone ring. This complex Tandem Reaction placed all the five asymmetric centers. Opening the unsaturated ring gave the key intermediate . Condensation of with o-methoxytryptamine followed by Pictet-Spengler condensation led to the formation of Isoreserpine and not Reserpine. Woodward reasoned that this was due to the fact that the C, D and E rings in all-trans geometry, had all substituents in the stable equatorial orientation in Isoreserpine. Woodward executed the isomerisation at C3 in an ingenious way. Aqueous alkali hydrolyzed the acetate – ester functions, which was than lactonised under DCC conditions. Thus, an unstable all-axial E ring was locked in as a lactone . This forced the molecule in a crowded unstable state. On acid catalyzed isomerisation with the high boiling carboxylic acid, the C3 position isomerized to the reserpine configuration. On transesterification, the correct isomer was formed with a free –OH ready for final acylation. G. Metha et.al., (J. Chem. Soc., Perkin Trans. 1, 1319 (2000)) controlled the stereochemistry on the E ring through a bicyclo[2.2.1]heptane system as shown in Figure 9.3. Gilbert Stork (J. Am. Chem. Soc., 127, 16255 (2005) reasoned that the stereochemistry at C3 in the (AB  ABDE  ABCDE) cyclisation route of Woodward could be controlled if an iminium intermediate could be formed first. This would then orient a chair-like folding of the tether chain (potential C ring) for an axial attack on the iminium ion to give a stereoselective C/D ring closure to form reserpine. They reasoned that the intermediate would serve as the key intermediate. After some unsuccessful attempts, Stork’s school completed the synthesis of using a route shown in Figure 9.6. Condensation of the hexynal with the lithium enolate of methyl methoxy acetate followed by benzenesulphonyl chloride and further heating with DBU gave the conjugated methoxy ester with the (Z) isomer as the dominant product. Trimethylsilyl chloride trapped the diene ketene acetal without purification of the intermediate product. D.A reaction with maleic anhydride followed with aqueous THF gave the decarboxylated acid that was esterified to give . A free-radical cyclisation with tributylstannane in refluxing t-butanol and subsequent workup led to an epimeric mixture that was equilibrated with base to the desired isomer . Reduction of the keto- group with L-Selectride gave the axial alcohol. This alcohol was the inverted via mesylation and treatment with cesium acetate to give . Reduction with LAH, selective tosylation of the primary alcohol and silylation of the secondary alcohol and ozonolysis gave via . Opening of the 5-membered ring was achieved by trapping the kinetic enolate as TMS derivative and ozonolysis and final esterification gave . This key intermediate was also synthesized via an alternate more efficient route. Unfortunately, condensation of with the indole precursor gave isoreserpine as the major product . The researchers reasoned that this unexpected result was most probably due to formation of the C ring prior to cyclisation. In order to drive the reaction towards a DE ring formation, they decided to trap the intermediate imine as the cyanoamine . This was achieved by adding an excess of potassium cyanide to the reaction mixture The expected product with an axial cyano- group was the sole product. Refluxing the aminonitrile in acetonitril as solvent again gave (±) methyl isoreserpine precursor as the major product . the authors reasoned that the cyanide anion on the α- phase of the molecule formed a tight ion pair with the immonium ion under these conditions.This anion prevented axial attack on a chair-like folding of the chain, forcing a boat like conformation prior to cyclisation followed by an axial attack to give isoreserpine as a major product. In such an event, allowing the cyanide counter ion to escape from the influence of the iminium ion by the use of a polar solvent should clean the reaction site for a chair-axial attack. When the nitrilamine was treated 10% solution of 1 N HCl in THF solvent, the precursor for (±) methyl reserpine was the product in 90% yield. This was eventually converted to reserpine by known procedures. Based on these results, another very short chiral route has been reported by the authors. Thus, persistence and critical mechanistic reasoning at every point of failure led to a successful direct synthesis of the correct isomer. S. Hanessean et.al., (J.Org. Chem.,62, 465 (1997) applied the Chiron approach to the key intermediate suitable for the CD ring formation in a E  ABE  ABCDE approach discussed so far. Quinic acid had most of the chiral centers in the proper orientation. However it needed modification at the potential C20, C15 and C16 centres. Retroanalysis of Hanessian is shown in Figure 9.10. The actual synthesis is shown in Figure 9.11. Lactonisation, followed by selective benzyl protection and methylation of the remaining OH groups gave the compound B. Oxidation was followed by methanolysis of lactone ring led to elimination to give the conjugaten ketone D. Protection of the C18 – OH as TBDMS followed by vinylation using Grignard reagent placed the potential carboxylic acid function. The t-alcohol thus generated provided an anchor the stereospecific introduction of the needed two carbon chain. A reductive free radical cyclisation placed the carbon chain in the correct stereochemistry. The correct isomer at C20 (I) was then moved to the next step. This key intermediate L led to the reserpine and isoreserpine precursors and ’ in the ratio 1.4 : 1 respectively. These steps are shown in Figure 9.11.

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Herbs tend to be the leaves of fragrant plants that do not have a woody stem. Herbs are available fresh or dried, with fresh herbs having a more subtle flavor than dried. You need to add a larger quantity of fresh herbs (up to 50% more) than dry herbs to get the same desired flavor. Conversely, if a recipe calls for a certain amount of fresh herb, you would use about one-half of that amount of dry herb. The most common fresh herbs are basil, coriander, marjoram, oregano, parsley, rosemary, sage, tarragon, and thyme. Fresh herbs should have a clean, fresh fragrance and be free of wilted or brown leaves. They can be kept for about five days if sealed inside an airtight plastic bag. Fresh herbs are usually added near the completion of the cooking process so flavors are not lost due to heat exposure. Dried herbs lose their power rather quickly if not properly stored in airtight containers. They can last up to six months if properly stored. Dried herbs are usually added at the start of the cooking process as their flavor takes longer to develop than fresh herbs.

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Approximately three-quarters of the known elements display the macroscopic properties characteristic of metals. They conduct both heat and electricity very well; they have shiny surfaces; they are capable of being shaped by hammering (malleable) and also of being drawn into wires (ductile). These properties can be understood in terms of in which valence electrons are delocalized over an entire metallic crystal. Positive metal ions formed by loss of valence electrons are held together by an electron sea. The strength of metallic bonding varies roughly as the number of electrons available in this sea. Chemical properties of the metals include a tendency to lose electrons and form positive ions, and the ability of their oxides to function as bases. The extent of these characteristics varies from one metal to another. Several borderline cases such as B, Si, Ge, As, Sb, and Te are difficult to classify as metals or nonmetals. These elements are usually referred to as the or . As you will recall from the discussion of metals on the periodic table, one can draw a zigzag line across the periodic table from B to At which separates the metals from the nonmetals and semimetals. This line is clearly indicated in most periodic tables. Periodic Table Live allows you to select metals, semi-metals, or non-metals and see how they are divided on the periodic table. A great many metals and alloys are of commercial importance, but metals occur naturally in oxide, carbonate, or sulfide ores. Such ores must be concentrated (beneficiated) before they can be reduced to the metal, and usually the raw metal must be purified (refined) in a third step. An excellent example of these processes involves iron which can be readily beneficiated because its ore is ferromagnetic. Iron ore is then reduced in a blast furnace and purified in a steelmaking furnace. Since ore-reduction is a nonspontaneous process, its reverse, oxidation or corrosion of a metal, is often a problem. This is especially true in the case of iron because the oxide coating which forms on the metal surface does not protect the remaining metal from atmospheric oxidation. Here we will be concerned mainly with the transition metals. We have already covered metals which are , such as and . A discussion of the lanthanoid and actinoid metals is beyond the scope of general chemistry. Since transition metals contain electrons in their valence shell, their chemistry is somewhat different from that of the representative elements. In particular they form a family of compounds called complex compounds or coordination compounds which are very different from those we have encountered up to this point. In these complexes, several ligands which can serve as Lewis bases are bonded to a metal ion which serves as a Lewis acid. The number of ligands is called the coordination number, and defines the possible geometries of a complex. For a coordination number of 2 the complex is usually linear. Both square planar and tetrahedral structures occur for coordination number 4, and coordination, number 6 usually involves an octahedral structure. Square planar and octahedral structures give rise to cis-trans isomerism. Some ligands, called can coordinate-covalent bond to metal ions at more than one site. Chelate complexes are often important in biological systems because they can disguise the charge of a metal ion, stabilizing the ion in a hydrophobic environment. In aqueous solution transition-metal ions are usually octahedrally coordinated by water molecules, but often other ligands which are stronger Lewis bases replace water. Such reactions often produce color changes, and they are usually rapid. A few metal ions, such as Cr(III), Co(III), Pt(IV), and Pt(II), undergo ligand substitution rather slowly and are said to be inert. Metal ions whose reactions are rapid are said to be labile.

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When you have completed Chapter 23, you should be able to In this chapter, we consider the fourth and final general type of reaction that carbonyl compounds undergo—the carbonyl condensation reaction. Carbonyl condensation reactions take place between two carbonyl‑containing reactants, one of which must possess an alpha‑hydrogen atom. The first step of the reaction involves the removal of an alpha‑hydrogen atom by a base. In the second step, the enolate anion that results from this removal attacks the carbonyl‑carbon of the second reacting molecule. In the final step of the reaction, a proton is transferred to the tetrahedral intermediate formed in the second step, although in some cases the product that results may subsequently be dehydrated.

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The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships. The (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.) The test of the and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law. Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows: The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample. In a gas sample, individual molecules have widely varying speeds; however, because of the number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure \(\Page {2}\)). The kinetic energy (KE) of a particle of mass ( ) and speed ( ) is given by: \[\ce{KE}=\dfrac{1}{2}mu^2 \nonumber \] Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m s ). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the of a particle, , is defined as the square root of the average of the squares of the velocities with = the number of particles: \[u_\ce{rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u^2_1+u^2_2+u^2_3+u^2_4+…}{n}} \nonumber \] The average kinetic energy, KE , is then equal to: \[\mathrm{KE_{avg}}=\dfrac{1}{2}mu^2_\ce{rms} \nonumber \] The KE of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation: \[\mathrm{KE_{avg}}=\dfrac{3}{2}RT \nonumber \] where is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J mol K (8.314 kg m s mol K ). These two separate equations for KE may be combined and rearranged to yield a relation between molecular speed and temperature: \[\dfrac{1}{2}mu^2_\ce{rms}=\dfrac{3}{2}RT \nonumber \] \[u_\ce{rms}=\sqrt{\dfrac{3RT}{m}} \label{RMS} \] Calculate the root-mean-square velocity for a nitrogen molecule at 30 °C. Convert the temperature into Kelvin: \[30°C+273=303\: K \nonumber \] Determine the mass of a nitrogen molecule in kilograms: \[\mathrm{\dfrac{28.0\cancel{g}}{1\: mol}×\dfrac{1\: kg}{1000\cancel{g}}=0.028\:kg/mol} \nonumber \] Replace the variables and constants in the root-mean-square velocity formula (Equation \ref{RMS}), replacing Joules with the equivalent kg m s : \[ \begin{align*} u_\ce{rms} &= \sqrt{\dfrac{3RT}{m}} \\ u_\ce{rms} &=\sqrt{\dfrac{3(8.314\:J/mol\: K)(303\: K)}{(0.028\:kg/mol)}} \\ &=\sqrt{2.70 \times 10^5\:m^2s^{−2}} \\ &= 519\:m/s \end{align*} \nonumber \] Calculate the root-mean-square velocity for an oxygen molecule at –23 °C. 441 m/s If the temperature of a gas increases, its KE increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KE decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure \(\Page {3}\). At a given temperature, all gases have the same KE for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher , with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower , and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure \(\Page {4}\). The gas simulator may be used to examine the effect of temperature on molecular velocities. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures. According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates. The rate of effusion of a gas depends directly on the (average) speed of its molecules: \[\textrm{effusion rate} ∝ u_\ce{rms} \nonumber \] Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here: \[u_\ce{rms}=\sqrt{\dfrac{3RT}{m}} \nonumber \] \[m=\dfrac{3RT}{u^2_\ce{rms}}=\dfrac{3RT}{\overline{u}^2} \nonumber \] \[\mathrm{\dfrac{effusion\: rate\: A}{effusion\: rate\: B}}=\dfrac{u_\mathrm{rms\:A}}{u_\mathrm{rms\:B}}=\dfrac{\sqrt{\dfrac{3RT}{m_\ce{A}}}}{\sqrt{\dfrac{3RT}{m_\ce{B}}}}=\sqrt{\dfrac{m_\ce{B}}{m_\ce{A}}} \nonumber \] The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law. The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.

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In the previous module, the variation of a liquid’s equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance’s melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in Figure \(\Page {1}\). To illustrate the utility of these plots, consider the phase diagram for water shown in Figure \(\Page {2}\). We can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of −10 °C correspond to the region of the diagram labeled “ice.” Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state. Note that on the H O phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here. The curve in Figure \(\Page {2}\) is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This “liquid-vapor” curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm, the boiling point is 100 °C. Notice that the liquid-vapor curve terminates at a temperature of 374 °C and a pressure of 218 atm, indicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module. The solid-vapor curve, labeled in Figure \(\Page {2}\), indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in Figure \(\Page {2}\), we would see that ice has a vapor pressure of about 0.20 kPa at −10 °C. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa, ice will sublime. This is the basis for the “freeze-drying” process often used to preserve foods, such as the ice cream shown in Figure \(\Page {3}\). The solid-liquid curve labeled shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in Figure \(\Page {4}\). The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide. The point of intersection of all three curves is labeled B in Figure \(\Page {2}\). At the pressure and temperature represented by this point, all three phases of water coexist in equilibrium. This temperature-pressure data pair is called the . At pressures lower than the triple point, water cannot exist as a liquid, regardless of the temperature. Using the phase diagram for water given in Figure 10.4.2, determine the state of water at the following temperatures and pressures: Using the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f) gas. What phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa? If the pressure is held at 50 kPa? At 0.3 kPa: s⟶ g at −58 °C. At 50 kPa: s⟶ l at 0 °C, l ⟶ g at 78 °C Consider the phase diagram for carbon dioxide shown in Figure \(\Page {5}\) as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for CO increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions. Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous CO . Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water. Using the phase diagram for carbon dioxide shown in Figure 10.4.5, determine the state of CO at the following temperatures and pressures: Using the phase diagram for carbon dioxide provided, we can determine that the state of CO at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f) gas. Determine the phase changes carbon dioxide undergoes when its temperature is varied, thus holding its pressure constant at 1500 kPa? At 500 kPa? At what approximate temperatures do these phase changes occur? at 1500 kPa: s⟶ l at −45 °C, l⟶ g at −10 °C; at 500 kPa: s⟶ g at −58 °C If we place a sample of water in a sealed container at 25 °C, remove the air, and let the vaporization-condensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm. A distinct boundary between the more dense liquid and the less dense gas is clearly observed. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (Figure \(\Page {2}\)), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of 374 °C, the vapor pressure has risen to 218 atm, and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a , and the temperature and pressure above which this phase exists is the (Figure \(\Page {5}\)). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in Table \(\Page {1}\). Like a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus oils. It is nontoxic, relatively inexpensive, and not considered to be a pollutant. After use, the CO can be easily recovered by reducing the pressure and collecting the resulting gas. If we shake a carbon dioxide fire extinguisher on a cool day (18 °C), we can hear liquid CO sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day (35 °C). Explain these observations. On the cool day, the temperature of the CO is below the critical temperature of CO , 304 K or 31 °C (Table \(\Page {1}\)), so liquid CO is present in the cylinder. On the hot day, the temperature of the CO is greater than its critical temperature of 31 °C. Above this temperature no amount of pressure can liquefy CO so no liquid CO exists in the fire extinguisher. Ammonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior? The critical temperature of ammonia is 405.5 K, which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature.   Coffee is the world’s second most widely traded commodity, following only petroleum. Across the globe, people love coffee’s aroma and taste. Many of us also depend on one component of coffee—caffeine—to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee’s stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening. Since the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee’s taste and aroma also dissolve in H O, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and taste of the decaffeinated coffee. Dichloromethane (CH Cl ) and ethyl acetate (CH CO C H ) have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee. Supercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (Figure \(\Page {7}\)). At temperatures above 304.2 K and pressures above 7376 kPa, CO is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes 97−99% of the caffeine, leaving coffee’s flavor and aroma compounds intact. Because CO is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs. The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phase-transition temperatures and pressures. The point of intersection of all three curves represents the substance’s triple point—the temperature and pressure at which all three phases are in equilibrium. At pressures below the triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance’s critical point, the pressure and temperature above which a liquid phase cannot exist.