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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/11%3A_Strategies_in_Steroids_Synthesis

The main skeletal features in steroid rings are depicted by the two figures shown in Figure 11.1. The same numbering system is maintained even while describing part structures obtained as synthetic intermediates. This convention helps us to follow the development of structural features through long synthetic schemes. Students should also become familiar with another convention followed by chemists to categorize synthetic schemes, originally evolved for steroids. Similar descriptions are also found in alkaloid chemistry. ‘An \(AB \rightarrow ABC \rightarrow ABCD\) Approach’ would mean that a naphthalene skeleton (either aromatic or suitable perhydro- skeleton) is chosen as SM. The C ring is then constructed on the AB rings. The D ring is then formed by ring closure. An example to this strategy is Bechmann’s synthesis (1940). Such descriptions do not indicate any details like the substitution patterns on the SM or synthetic intermediates nor do they spill light on stereochemical details. These details are discussed in the synthetic schemes. This classical synthesis exploits the chemistry of the naphthalene ring. 1-naphthylamine-2-sulphonic acid was chosen as the AB ring . The sulphonic acid moiety was converted to a phenol and protected as the methyl ether. The amine moiety was converted to iodide through diazotisation. The C and D rings were then built on the AB rings Woodward’s synthesis of Cholesterol (J. Am. Chem. Soc., 73, 2403, 3547, 3548 (1951); ibid, 74, 4223 (1952) could be described as \(C \rightarrow CD \rightarrow BCD \rightarrow ABCD\) Approach. Since the D ring remains D-homo until the last step of ring construction and the required 5-membered ring was obtained only after a ring contraction, it could also be termed as \(C \rightarrow BC \rightarrow ABC \rightarrow ABCD\) Approach. Remember that this molecule has 8 asymmetric centers and could therefore have 28 (256) optical isomers. This synthesis envisaged the synthesis of just one set of diastereomers. This stereospecific synthesis incorporated all the stereopoints in a stereo- and regiospecific manner. The D ring provides an anchor for variations in the chain. The double bond on the C ring allowed an opening for the synthesis of cortisone. The legends on the arrows show the reactions. Estrone has attracted several chemists as a target for executing new methodologies on this complex yet useful steroid. A very popular method for the introduction of the D ring, followed by cyclization of the C ring in steroid synthesis was introduced by Torgov (1950, 63). His synthesis of Estrone is shown in Figure 11.4. His procedure for \(AB \rightarrow ABD \rightarrow ABCD\) Approach became very popular. Several modifications were later developed to further improve this methodology. Bartlet used a mechanistic transform in his synthesis of Estrone. He applied a biogenetic-type cyclisation for his \(\ce{A -> AD -> ABCD}\) approach shown in Figure 11.5. A furan ring served as a masked 1,4-diketone needed for the D ring. Also note the rearrangement transform used for the stereospecific introduction of the angular methyl group at C13 position in the last step of the synthesis. Hughes (1960) relied on aldol transforms for his A  AD  ABCD Approach. Here he made use of the reliable reduction protocols developed by several workers for controlling the stereochemistry at ring junctions P. Hermann et.al., (J. Org. Chem., 73, 6202 (2008)) had attempted a sequential Media:Ring Closure Reactions (RCR) strategy for the stereospecific synthesis of Estrone. These workers relied on the C ZrB catalyst studied extensively in their laboratory and developed a short 9-step synthesis from known diene (A  AB  ABC  ABCD Approach) . Cyclisation of the B ring proceeded satisfactorily. Cylisation of the C ring was sensitive to the halogen. After several attempt, cyclisation proceeded well with the vinyl fluoride. The formation of the D ring using Zr catalyst was unsuccessful. The ring closure of the D ring was finally accomplished using the second-generation Media:Grubbs catalyst. The final modification of the D-ring has been already reported by Bartlett P.A et.al., ( J. Am. Chem. Soc., 95, 7501 (1973) thereby completing a formal synthesis of Estrone. Marko Weimar et.al., (J. Org. Chem., 75, 2718 (2010)) have reported a very successful DA Transform for the formation of CD ring from Dane’s diene as AB ring and a D ring as dienoplile (AB  ABCD Approach; see references cited for similar approaches). Key for the success of this strategy was the metal-free chiral catalyst, which they successfully tailored by the introduction of three H-bonding sites. The catalyst that worked efficiently was the Media:axially chiral amidine salt . They have suggested that this ligand formed a Host-Guest complex with the dienoplile as depicted in Under these conditions, the Diels-Alder reaction proceeded in excellent yield. This compound was converted to Estrone as shown in Figure 11.11 Cortisone attracted the attention of several synthetic chemists, because this wonder drug was available only in minute quantities from animal sources. Two challenging features in the structure of cortisone were the keto- group at C11 and the 1,2,3- oxygenation pattern at the two-carbon side chain at C17. Sarrett (J. Am. Chem. Soc., 74, 4974 (1952); J. Am. Chem. Soc., 76, 5031 (1954)) used an ABC  ABCD Approach. His starting molecule had all the features needed for the A, B and C rings of Cortisone. It had a C11 –OH group and a conveniently placed ketone for building the D ring. They exploited the known stereoelectronic constraints of anion reactions on a rigid 6-membered ring to construct the C/D trans ring junction . This cyclisation has two noteworthy features. The olefin served as a masked ketone. Unlike Woodward’s Cholesterol synthesis , which had a competing site for anion formation, in this scheme only one anion is feasible leading to predictable product. An interesting new approach in steroid synthesis came from Y. Horiguchi (J. Org. Chem., 51, 4325 (1986)). This could be described as CD  BCD  ABCD Approach. Note that the C11 oxygenation and the carbons needed for the A ring came through a single oxidation step, as envisaged in their retroanalysis shown in Figure 11.13. A detailed synthetic scheme of Horiguchi is shown in Figure 11.14. The strategy for formation of A as well as the D ring is of interest in this synthesis. Nemoto’s retroanalysis (J. Org. Chem., 55, 5625 (1990)) provided an interesting electrocyclic ring opening – cycloaddition strategy for a B  BCD  ABCD Approach . The synthetic scheme is shown in Figure 11.16. Note the versatile chiral auxiliary in the first step that serves several useful bond-forming purposes in addition to guiding three asymmetric center on C/D rings. The fascination for mastering the steroid skeleton still continues unabated.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/07%3A_Solids_and_Liquids/7.06%3A_Introduction_to_Crystals

The delicately faceted surfaces of large crystals that occur in nature have always been a source of fascination and delight. In some ways they seem to represent a degree of perfection that is not apparent in other forms of matter. But in the realm of pure solid substances, crystals are the rule rather than the exception, although this may not be apparent unless they are observed under a hand-lens or a microscope. It is remarkable that the visual examination of crystals was able to establish a fairly mature science of crystallography (applied mainly to the study of minerals) by the end of the 19 Century, even before the atomic theory of matter had been universally accepted. Today this aspect of crystallography is of importance not only to chemists and physicists, but also to geologists, amateur minerologists and "rock-hounds" who maintain some of the best Web resources on crystals. In this lesson we will see how the external shape of a crystal can reveal much about the underlying arrangement of its constituent atoms, ions, or molecules. The first thing we notice about a crystal is the presence of planes — called — which constitute the external boundaries of the solid. Of course, any solid, including non-crystalline glass, can be carved, molded or machined to display planar faces; examples of these can be found in any "dollar store" display of costume jewelry. What distinguishes and defines a true crystal is that these faces develop as the solid forms from a melt or from solution. The multiple faces invariably display certain geometrical relationships to one another, resulting in a symmetry that attracts our attention and delights the eye. One of the most apparent elements of this geometrical regularity are the sets of parallel faces that many crystals display. Nowhere is this more apparent than in the cubes that develop when sodium chloride crystallizes from solution. We usually think of a cubic shape in terms of the equality of its edge lengths and the 90° angles between its sides, but there is a more fundamental way of classifying shapes that chemists find very useful. This is to look at what (such as rotations around an axis) we can perform that leave the appearance unchanged. For example, you can rotate a cube 90° around an axis perpendicular to any pair of its six faces without making any apparent change to it. We say that the cube possesses three mutually perpendicular , abbreviated C axes. But if you think about it, a cube can also be rotated around an axis that extends between opposite corners; in this case, it takes three 120° rotations to go through a complete circle, so these axes (also four in number) are three-fold or C axes. And finally, there are two-fold (C ) axes that pass diagonally through the centers of the six pairs of opposite edges. In addition, there are imaginary symmetry planes that mirror the portions of the cube that lie on either side of them. Three of these are parallel to the three major axes of the crystal, and an additional six pass diagonally through opposite edges. All told, there are 13 rotational axes and 9 mirror planes (only a few of which are shown above) that define cubic symmetry. Why is this important? Although anyone can recognize a cube when they see one, it turns out that many crystals, both natural and synthetic, are for one reason or another unable to develop all of their faces equally. Thus a crystal that forms on the bottom of a container will be unable to grow any faces that project downward for the simple reason that there is no supply of ions or molecules from that direction. The same effect occurs when a mineral crystal tries to grow in contact with other solids. Finally, the presence of certain impurities that selectively adsorb to one or more faces can block the addition of more material to them, thus either completely inhibiting their formation or forcing them to grow at slower rates. These alternative shapes that can develop from a single basic crystal type are known as . Sodium chloride grown from pure aqueous solution forms simple cubes, but the addition of various impurities can result in habits that can be regarded as cubes that have been truncated along planes normal to some of the symmetry axes. (The same effects can sometimes be seen as a crystal slowly dissolves and material is released more rapidly from some directions than others.) In this example, the perfect cube develops triangular faces at the corners . If these enlarge beyond their maximum size , the triangular faces meet in a new set of edges that are hexagonal . Eventually we are left with the eight faces of what is obviously a regular octahedron . One might think that these five shapes bear no relationship to one another, but in fact they all possess the same set of set of symmetry elements as the simple cube and are thus various of the same underlying cubic structure and belong to the described further below. Even though a given crystal may be distorted or broken, the angles between corresponding faces remain the same. Thus you can crush a crystal underfoot or break it up with a hammer, but you will always find that the fragments possess a limited set of interfacial angles. This fundamental law, discovered by Nicholas Steno in 1669, was a major key development in crystallography. About 100 years later, a protractor-like device (the ) was invented to enable more accurate measurements than the rather crude ones that had formerly been traced out on paper. When a crystal is broken by applying a force in certain directions (as opposed to being pulverized by a hammer) it will often be seen to break cleanly into two pieces along what are known as cleavage planes. The new faces thus formed always correspond to the symmetry planes associated with a particular crystal type, and of course make constant angles with any other faces that may be present. Cleavage planes were first described in the late 17 century, but nothing much was thought about their significance until about a hundred years later when the Abbe Hauy accidently dropped a friend's sample of calcite and noticed how cleanly it broke. Further experimentation showed that other calcite crystals, even ones of different initial shapes (habits), displayed similar rhombohedral shapes upon cleavage, and that these in turn produced similar shapes when they were cleaved. This led Haüy to suggest that continued cleavages would ultimately lead to the smallest possible unit which would be the fundamental building block of the crystal. (Remember that the atomic theory of matter had not developed at this time.) Haüy's elaborately drawn figures (published in 1784) showed how external faces of a crystal could be produced by stacking the units in various ways. For example, by omitting rows from a cubic stack of primal cubelets, one could arrive at the various stages between the cube and the octahedra for sodium chloride that we saw earlier on this page. The modern interpretation of these observations replaces Haüy's primal shapes with atoms or molecules, or more generally with points in space that these define the possible locations of atoms or molecules. It is easy to see how plane faces can develop along some directions and not others if one assumes that the new faces must follow a linear sequence of points. Lattice: = , 90° angles lattice ≠ , angles < 90° lattice 0° or lattice: = , angles neither 60° or 90°; lattice (but unit cell is a rhombus with = and angles 60°) Although everyone has seen and admired the huge variety of patterns on printed fabrics or wallpapers, few are aware that these are all based on one of five types of two-dimensional "unit cells" that form the basis for these infinitely-extendable patterns. One of the most remarkable uses of this principle is in the work of the Dutch artist Maurits Escher (1888-1972). Shown below are two-dimensional views of the unit cells for two very common types of crystal lattices, one having cubic symmetry and the other being hexagonal. Although we could use a hexagon for the second of these lattices, the is preferred because it is simpler. Notice that in both of these lattices, the corners of the unit cells are centered on a lattice point. This means that an atom or molecule located on this point in a real crystal lattice is shared with its neighboring cells. As is shown more clearly here for a two-dimensional square-packed lattice, a single unit cell can claim "ownership" of only one-quarter of each molecule, and thus "contains" 4 × ¼ = 1 molecule. The unit cell of the graphite form of carbon is also a rhombus, in keeping with the hexagonal symmetry of this arrangement.Notice that to generate this structure from the unit cell, we need to shift the cell in both the - and - directions in order to leave empty spaces at the correct spots. We could alternatively use regular hexagons as the unit cells, but the + shifts would still be required, so the simpler rhombus is usually preferred. This image nicely illustrates the relations between the unit cell, the lattice structure, and the actual packing of atoms in a typical crystal. We saw above that five basic cell shapes can reproduce any design motif in two dimensions. If we go to the three-dimensional world of crystals, there are just seven possible basic lattice types, known as , that can produce an infinite lattice by successive translations in three-dimensional space so that each lattice point has an identical environment. Each system is defined by the relations between the axis lengths and angles of its unit cell. For example, if the three edge lengths are identical and all corner angles are 90°, a crystal belongs to the cubic system. The simplest possible cube is defined by the eight lattice points at its corner, but variants are also possible in which additional lattice points exist in the faces ("face-centered cubic") or in the center of the cube ("body-centered cubic"). If variants of this kind are taken into account, the total number of possible lattices is fourteen; these are known as the fourteen . = = α = β = γ = 90° The F cell corresponds to , a very common and important structure. = ≠ α = β = γ = 90° A cube that has been extended in one direction, creating a unique -axis. An F cell would simply be a network of joined I cells. ≠ ≠ α = β = γ = 90° = ≠ α = β = 60°, γ = 120° Just as in the 2-dimensional examples given above, the unit cell of the hexagonal lattice has a rhombic cross-section; the entire hexagonal unit is built from three of these rhombic prisms. (rhombohedral) = = α = β = γ ≠ 90°, ≠ ≠ α = γ = 90°, β > 90° Two 90° angles, one > 90°, with all sides of different lengths. A cell (also seen in the orthorhombic class) has additional points in the center of each end. Monoclinic I and F cells can be constructed from C cells. ≠ ≠ α ≠ β ≠ γ ≠ 90° This is the most generalized of the crystal systems, with all lengths and angles unequal, and no right angles. In any kind of repeating pattern, it is useful to have a convenient way of specifying the orientation of elements relative to the unit cell. This is done by assigning to each such element a set of integer numbers known as its . To understand indexing, it will be easier to begin with a unit cell plane that we are viewing from above, along the [invisible] -axis. The drawing shows such a plane with three lines crossing it at various slopes. The index of each line is found by first determining the points where it intersects the and axes as fractions of the unit cell parameters and . Thus in the above example: The of the lines are given by the of these values: Miller indices are written in parentheses with no spaces between numbers. Negative values are indicated by an overbar as in . We proceed in exactly the same way, except that we now have 3-digit Miller indices corresponding to the axes and . It is important to note that multiple parallel planes that repeat at the same interval have identical Miller indices. This simply reflects the fact that we can repeat the coordinate axes at any regular interval. We mentioned previously that the plane faces of crystals are their most important visually-distinctive property, so it is important to have a convenient way of referring to any given face. First, we define a set of reference directions ( ) which are known as the . In most cases these axes correspond to directions that are fairly apparent on visual examination of one or more crystals of a given kind. They are parallel to actual or possible edges of the crystal, and they are not necessarily orthogonal. We now know, as Haüy first suggested, that these directions correspond to rows of lattice points in the underlying structure of the crystal. We also define three ( ) which mark out the boundaries of the unit cell along the crystallographic axes. The index of a particular face is determined by the fractional values of (a,b,c) at which the face intersects the axes ( ). Study the examples shown below for three different habits of a cubic lattice. Below is a more complicated example of one particular habit of an orthorhombic crystal. The figure at the right shows how the (113) face is indexed. In this case, the plane at the top of the crystal is extended downward to the ( ) plane. This extended plane cuts the ) axes at (2 , 2 , 2/3 ). The corresponding inverses would be (½,½,3/2). In order to make them into proper Miller indices (which should always be integers) we multiply everything by 2, yielding (113). It is remarkable that the faces that bound real crystals generally have small Miller indices. The low values for the indices suggest that a given lattice plane has a high density of lattice points per unit area, a logical consequence of each molecule being surrounded and held by its closely-packed neighbors. In the 2-dimensional projection below, compare the facial lattice-point density in the (11) plane with that of the (31) plane. Crystals with a single long unit-cell axis tend to form planes with the long axis normal to the plane, so that the major faces of the crystal are planes containing the short-axis translations. Similarly, crystals with a single, short unit-cell axis tend to be needles. The main faces, on the sides of the needles, contain the short lattice translation — a high density of lattice points. In general, it is found that crystals have linear dimensions that mirror the reciprocals of the lattice parameters. Faces having a lower density of lattice points (as in the (31) face shown above) can acquire new layers more rapidly, and thus grow more rapidly than faces having a high lattice-point density. The faces that can potentially develop in a crystal are determined entirely by the symmetry properties of the underlying lattice. But the faces that actually develop under specific conditions — and thus the overall of the crystal — is determined by the relative of growth of the various faces. The slower the growth rate, the larger the face. This relation can be understood by noting that faces that grow normal to shorter unit cell axes (as in the needle-shaped crystal shown above) present a larger density of lattice points to the surface (that is, more points per unit surface area.) This means that more time is required for diffusion of enough new particles to build out a new layer on such a surface. is to grind a large crystal of salt into a spherical shape and immerse it in a saturated solution of sodium chloride. At first, the most disturbed and exposed parts on the surface dissolve, revealing a large variety of underlying plane faces. As growth resumes, the smaller of these are rapidly replaced by larger faces. Eventually, the fast-growing faces eliminate themselves and the high-lattice point density faces that correspond to the sides of the cube win out. In addition to these structural effects, the conditions under which a crystal is grown can affect its habit. Temperature, degree of supersaturation, nature of the solvent all have their effects, and these may affect the growth of different faces in different ways. The presence of impurities in the solution can radically alter the habit of a crystal, as seen in the following table for the growth of sodium chloride: These effects presumably come about because these substances preferentially adsorb to certain faces, impeding their growth.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Scientific_Method

You know what science is. A is a way to do something. The is how scientific knowledge is produced. You have probably learned science before as a set of knowledge that is just given to you, and you have to memorize facts and procedures to follow to get answers. This way of learning that you have probably experienced is completely different from how science professionals think of science and do science. Now it is time to learn science like a scientist. Because scientists create new knowledge, they have to be able to think independently. Even if this seems very hard or unfamiliar, you can learn to make new scientific conclusions yourself. The basic steps of the scientific method involve observations about what actually happens, about what the observations mean, making about what will be observed in the future, and then making to see if the guesses are right. The are evidence or , the results of experiments. The guesses are usually called . If the hypotheses fit a lot of data and seem to work well, then they are called . In science, a describes a pattern of consistent results. For instance, you know that when you drop things, they fall. That could be called the law of gravity. If we had a good explanation of why that happens, we would call that a theory of gravity. Gravity is easy to describe (in Newton's law, which describes how things move on Earth and how planets and moons and stars moves in space) but very hard to explain: how can objects that are very far apart feel a force towards each other? The real test of whether something is science is whether it works reliably. An experiment is good and useful if you can (if someone different can repeat it) and . There will be some small differences in the data because measurements are always a little imprecise and there might be some small differences in the way the experiment was done, but the data should be the same within the "error range." Likewise, a law is good if observations always follow it within the context it describes. And a theory is good if it explains all the available evidence. For theories, it's also important that it be possible to prove the theory wrong, that it makes predictions that can be tested. Otherwise, it might be true but it isn't useful. A useful skill for learning science is building "mental models" of how things work. A is a bigger or smaller version of something, depending on the size of the original object, that shows all its parts and how they are related. For instance, an architect builds a small 3-D model of a new building before people start constructing the building. The architect also has complex computer models of the building showing how it stays up, how the doors don't bump into each other, etc. These models help the architect predict whether their design will be successful or not. The is a structured approach to gathering and analyzing data, and drawing conclusions about various subjects or phenomena. consist of evidence or data that may verify or falsify ideas about what actually happens in a given event. These ideas are usually , educated guesses about how or why the particular event occurred which could lead to experiments to test the guesses. If a hypothesis seems reasonable and fits with a lot of data through rigorous testing, the hypothesis evolves into a . Good experiments have replicable results, meaning that other scientists should be able to come up with the same results or conclusions after repeating the experiment. A describes what will be observed in the future given a pattern of consistent results on the phenomenon. A is a visualization of something to show how all of its parts are related.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.08%3A_Electrochemical_Corrosion

Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. Corrosion can be defined as the deterioration of materials by chemical processes. Of these, the most important by far is electrochemical corrosion of metals, in which the oxidation process M → M + e is facilitated by the presence of a suitable electron acceptor, sometimes referred to in corrosion science as a In a sense, corrosion can be viewed as the spontaneous return of metals to their ores; the huge quantities of energy that were consumed in mining, refining, and manufacturing metals into useful objects is dissipated by a variety of different routes. The economic aspects of corrosion are far greater than most people realize; the estimated cost of corrosion in the U.S. alone was $276 billion per year. Of this, about $121 billion was spent to control corrosion, leaving the difference of $155 billion as the net loss to the economy. Utilities, especially drinking water and sewer systems, suffer the largest economic impact, with motor vehicles and transportation being a close second. The special characteristic of most corrosion processes is that the oxidation and reduction steps occur at separate locations on the metal. This is possible because metals are conductive, so the electrons can flow through the metal from the anodic to the cathodic regions (Figure \(\Page {1}\)). The presence of water is necessary in order to transport ions to and from the metal, but a thin film of adsorbed moisture can be sufficient. A corrosion system can be regarded as a short-circuited electrochemical cell in which the anodic process is something like \[\ce{Fe(s) \rightarrow Fe^{2+}(aq) + 2 e^{-}} \label{1.7.1}\] and the cathodic steps may invove the reduction of oxygen gas \[ \ce{O_2} + \ce{2 H_2O} + \ce{4e^{-}} \rightarrow \ce{4 OH^{-}} \label{1.7.2}\] or the reduction of protons \[ \ce{H^{+} + e^{-}} \rightarrow \ce{1/2 H2(g)} \label{1.7.2b} \] or the reduction of a metal ion \[\ce{M^{2+} + 2 e^{–}} \rightarrow \ce{M(s)} \label{1.7.2c}\] where \(\ce{M}\) is a metal. Which parts of the metal serve as anodes and cathodes can depend on many factors, as can be seen from the irregular corrosion patterns that are commonly observed. Atoms in regions that have undergone stress, as might be produced by forming or machining, often tend to have higher free energies, and thus tend to become anodic. If one part of a metallic object is protected from the atmosphere so that there is insufficient \(\ce{O2}\) to build or maintain the oxide film, this "protected" region will often be the site at which corrosion is most active. The fact that such sites are usually hidden from view accounts for much of the difficulty in detecting and controlling corrosion. In contrast to anodic sites, which tend to be localized to specific regions of the surface, the cathodic part of the process can occur almost anywhere. Because metallic oxides are usually semiconductors, most oxide coatings do not inhibit the flow of electrons to the surface, so almost any region that is exposed to \(\ce{O2}\) or to some other electron acceptor can act as a cathode. The tendency of oxygen-deprived locations to become anodic is the cause of many commonly-observed patterns of corrosion. Anyone who has owned an older car has seen corrosion occur at joints between body parts and under paint films. You will also have noticed that once corrosion starts, it tends to feed on itself. One reason for this is that one of the products of the O reduction reaction is hydroxide ion. The high pH produced in these cathodic regions tends to destroy the protective oxide film, and may even soften or weaken paint films, so that these sites can become anodic. The greater supply of electrons promotes more intense cathodic action, which spawns even more anodic sites, and so on. A very common cause of corrosion is having two dissimilar metals in contact, as might occur near a fastener or at a weld joint. Moisture collects at the junction point, acting as an electrolyte and forming a cell in which the two metals serve as electrodes. Moisture and conductive salts on the outside surfaces provide an external conductive path, effectively short-circuiting the cell and producing very rapid corrosion; this is why cars rust out so quickly in places where salt is placed on roads to melt ice. Dissimilar-metal corrosion can occur even if the two metals are not initially in direct contact. For example, in homes where copper tubing is used for plumbing, there is always a small amount of dissolved \(\ce{Cu^{2+}}\) in the water. When this water encounters steel piping or a chrome-plated bathroom sink drain, the more-noble copper will plate out on the other metal, producing a new metals-in-contact corrosion cell. In the case of chrome bathroom sink fittings, this leads to the formation of \(\ce{Cr^{3+}}\) salts which precipitate as greenish stains. Since both the cathodic and anodic steps must take place for corrosion to occur, prevention of either one will stop corrosion. The most obvious strategy is to stop both processes by coating the object with a paint or other protective coating. Even if this is done, there are likely to be places where the coating is broken or does not penetrate, particularly if there are holes or screw threads. A more sophisticated approach is to apply a slight negative charge to the metal, thus making it more difficult for the reaction to take place: \[\ce{M -> M^{2+} + 2 e^{-}}.\] One way of supplying this negative charge is to apply a coating of a more active metal. Thus a very common way of protecting steel from corrosion is to coat it with a thin layer of zinc; this process is known as galvanizing.The zinc coating, being less noble than iron, tends to corrode selectively. Dissolution of this sacrificial coating leaves behind electrons which concentrate in the iron, making it cathodic and thus inhibiting its dissolution. The effect of plating iron with a less active metal provides an interesting contrast. The common tin-plated can (on the right) is a good example. As long as the tin coating remains intact, all is well, but exposure of even a tiny part of the underlying iron to the moist atmosphere initiates corrosion. The electrons released from the iron flow into the tin, making the iron more anodic so now the tin is actively promoting corrosion of the iron! You have probably observed how tin cans disintegrate very rapidly when left outdoors. A more sophisticated strategy is to maintain a continual negative electrical charge on a metal, so that its dissolution as positive ions is inhibited. Since the entire surface is forced into the cathodic condition, this method is known as . The source of electrons can be an external direct current power supply (commonly used to protect oil pipelines and other buried structures), or it can be the corrosion of another, more active metal such as a piece of zinc or aluminum buried in the ground nearby, as is shown in the illustration of the buried propane storage tank below.

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

A large number of important and useful polymeric materials are not formed by chain-growth processes involving reactive species such as radicals, but proceed instead by conventional functional group transformations of polyfunctional reactants. These polymerizations often (but not always) occur with loss of a small byproduct, such as water, and generally (but not always) combine two different components in an alternating structure. The polyester Dacron and the polyamide Nylon 66, shown here, are two examples of synthetic condensation polymers, also known as step-growth polymers. In contrast to chain-growth polymers, most of which grow by carbon-carbon bond formation, step-growth polymers generally grow by carbon-heteroatom bond formation (C-O & C-N in Dacron & Nylon respectively). Although polymers of this kind might be considered to be alternating copolymers, the repeating monomeric unit is usually defined as a combined moiety. Examples of naturally occurring condensation polymers are cellulose, the polypeptide chains of proteins, and poly(β-hydroxybutyric acid), a polyester synthesized in large quantity by certain soil and water bacteria. Formulas for these will be displayed below by clicking on the diagram. Condensation polymers form more slowly than addition polymers, often requiring heat, and they are generally lower in molecular weight. The terminal functional groups on a chain remain active, so that groups of shorter chains combine into longer chains in the late stages of polymerization. The presence of polar functional groups on the chains often enhances chain-chain attractions, particularly if these involve hydrogen bonding, and thereby crystallinity and tensile strength. The following examples of condensation polymers are illustrative. Note that for commercial synthesis the carboxylic acid components may actually be employed in the form of derivatives such as simple esters. Also, the polymerization reactions for Nylon 6 and Spandex do not proceed by elimination of water or other small molecules. Nevertheless, the polymer clearly forms by a step-growth process. Some Condensation Polymers The difference in Tg and Tm between the first polyester (completely aliphatic) and the two nylon polyamides (5th & 6th entries) shows the effect of intra-chain hydrogen bonding on crystallinity. The replacement of flexible alkylidene links with rigid benzene rings also stiffens the polymer chain, leading to increased crystalline character, as demonstrated for polyesters (entries 1, 2 &3) and polyamides (entries 5, 6, 7 & 8). The high Tg and Tm values for the amorphous polymer Lexan are consistent with its brilliant transparency and glass-like rigidity. Kevlar and Nomex are extremely tough and resistant materials, which find use in bullet-proof vests and fire resistant clothing. Many polymers, both addition and condensation, are used as fibers The chief methods of spinning synthetic polymers into fibers are from melts or viscous solutions. Polyesters, polyamides and polyolefins are usually spun from melts, provided the Tm is not too high. Polyacrylates suffer thermal degradation and are therefore spun from solution in a volatile solvent. Cold-drawing is an important physical treatment that improves the strength and appearance of these polymer fibers. At temperatures above T , a thicker than desired fiber can be forcibly stretched to many times its length; and in so doing the polymer chains become untangled, and tend to align in a parallel fashion. This cold-drawing procedure organizes randomly oriented crystalline domains, and also aligns amorphous domains so they become more crystalline. In these cases, the physically oriented morphology is stabilized and retained in the final product. This contrasts with elastomeric polymers, for which the stretched or aligned morphology is unstable relative to the amorphous random coil morphology. This cold-drawing treatment may also be used to treat polymer films (e.g. Mylar & Saran) as well as fibers. Step-growth polymerization is also used for preparing a class of adhesives and amorphous solids called epoxy resins. Here the covalent bonding occurs by an S 2 reaction between a nucleophile, usually an amine, and a terminal epoxide. In the following example, the same bisphenol A intermediate used as a monomer for Lexan serves as a difunctional scaffold to which the epoxide rings are attached. Bisphenol A is prepared by the acid-catalyzed condensation of acetone with phenol. ),

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.02%3A_Heat

Blacksmiths heat solid iron in order to shape it into a variety of different objects. Iron is a rigid, solid metal. At room temperature, it is extremely difficult to bend iron. However, when heated to a high enough temperature, iron can be easily worked. The heat energy in the forge is transferred to the metal, making the iron atoms vibrate more and move around more readily. is energy that is transferred from one object or substance to another because of a difference in temperature between the two. Heat always flows from an object at a higher temperature to an object at a lower temperature (see figure below). The flow of heat will continue until the two objects are at the same temperature. is the study of energy changes that occur during chemical reactions and during changes of state. When chemical reactions occur, some chemical bonds are broken, while new chemical bonds form. As a result of the rearrangement of atoms, the total chemical potential energy of the system either increases or decreases.

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

Metalloproteases ( metal) are members of a clan of proteases that contain a metal ion at their active site which acts as a catalyst in the hydrolysis peptide binds. The metallic core of each enzyme is the location of the specific reaction performed by the enzyme, in the case of metalloproteases, the cleavage of peptide bonds within proteins. The most common metal ion metalloproteases utilize is a zinc ion (Zn ). Other transition metals have been found at active sites, such as Co and Mn , and some have been used to restore function in zinc-metalloproteases in which the Zn core has been removed. Generally, metal ions are bound in a nearly tetrahedral conformation at the active site. Three amino acid ligands, usually charged residues, associate with the metal core along with one water molecule which is used for hydrolysis. There are two major divisions of metalloproteases: metalloendopeptidases and metalloexopeptidases. Each division is named for the region of the hydrolysis-targeted protein at which the reaction takes place. Within these divisions are nested more highly specified target sites and conserved catalytic residues wholly dependent on the nature of the enzyme. Thermolysin (TLN; , Figure 1) is a 34.6 kDa Zn -endopeptidase secreted by the bacterium . TLN and TLN-like proteins are used by bacteria to break down exogenous proteins for nutrition and as virulence factors aiding in host colonization and tissue degradation. TLN is active in the hydrolysis of internal peptide bonds on the N-terminal side of large hydrophobic amino acids, between leucine, isoleucine, or phenylalanine. Thermolysin was the first metalloproteases to be completely sequenced. Commercially, TLN is used as a nonspecific protease (within its cleavage site specificity) in peptide sequencing and is used in the production of the artificial sweetener aspartame. Figure 1. General structure of thermolysin (PDBID ). Alpha-helices in blue, beta-sheets in red, Zn ion in yellow, active site side chains in magenta, Ca ions in green. The overall structure of the protein consists of 316 amino acid residues organized into two domains separated by the active site. The N-terminal domain is predominately composed of β-sheets, while the C-terminal domain is primarily composed of α-helices. The two large domains are separated by a central α-helix. One Zn is bound via three residues in the active site within a conserved binding motif of HE H (His 142, His 146, Glu 166) located on the C-terminal side (Figure 2). The tetrahedral binding of Zn is rounded out by a nucleophilic water molecule. This water molecule is capable of alternating between two distinct association positions, W1 and W2, which are in turn stabilized by His 231 and Glu 143, respectively. Four Ca2+ ions are associated with the enzyme that aide in thermostability. Figure 2. Active site of thermolysin. The bound Zn is responsible for catalyzing peptide hydrolysis and stabilizing the various intermediates of the reaction. Although normally bound in a tetrahedral structure, during catalysis Zn assumes a pentacoordinate geometry between the original three residues (His142, His146, and Glu166), the oxygen of the nucleophilic water, and the carbonyl oxygen of the substrate. The formation of a -diolate intermediate is stabilized by Zn . Removal of Zn yields an inactive enzyme. Exogenous addition of other divalent transition metals, specifically Zn Co , Fe , and Mn , results in the regaining of 100%, 200%, 60%, and 10% enzymatic activity. Zn is also responsible for the polarization of the carbonyl bond of the substrate and the enhancement of nucleophilicity of the catalytic water molecule. Glu143 is responsible for the polarization of the catalytic water molecule leading to an enhancement of nucleophilicity. Additionally, Glu143 abstracts a proton from the water molecule and transfers it to the amide leaving group. In site-directed mutagenesis experiments involving the neutral protease of , a protease with an amino acid sequence identical to that of thermolysin, Glu143Asp and Glu143Gln substitutions resulted in no catalytic activity of the enzyme. The mechanism proposed by Mock and colleagues suggests a diminished catalytic role of Glu143, which is instead used solely for charge stabilization with no association with the nucleophilic water molecule. Mutagenesis studies show that Glu143Asp substitutions result in inactive enzymes, despite the same side chain charge that would provide electrostatic stability. His231 is responsible for substrate stabilization during hydrolysis. The carbonyl oxygen of the peptide is hydrogen bonded to N of His231 in the intermediate step of hydrolysis. Mutagenesis experiments involving His231Phe and His231Ala show 430- and 500-fold reductions in catalytic activity with no significant change in . These findings led to a proposed TLN-mechanism with an emphasis on the role of His231 as a general base, but this mechanism is less favored due to the residual activities of His231-substituted mutants compared to inactive mutants generated by Glu143 mutagenesis studies. The importance of Tyr157 has been debated in several mechanisms. Site-directed mutagenesis studies show an 80% drop in catalytic activity of Tyr157Trp mutants, and it has been suggested that the hydroxyl-H of Ty157 stabilizes the carboxylate-O of the peptide substrate during hydrolysis. Energetics modeling suggests a primary role for Tyr157 in transition state stabilization and substrate binding with an increase in the activation barrier energy of 2.7 kcal/mol and a decrease of binding affinity of 0.5 kcal/mol. Asp226 has been suggested to stabilize the catalytic His231 through H-bonding of the Asp226 carboxylate to N of His231. Mutagenesis of Asp226Ala shows a relatively small decrease in catalytic activity of 40-fold and energetics modeling suggest an increase in overall ΔG‡ 2.2 kcal/mol. Scheme 1. Mechanism of peptide hydrolysis by thermolysin. Several mechanisms for TLN mediated peptide cleavage have been proposed with varying emphasis on the importance of the roles of the catalytic residues, Glu143 and His231, as well as several catalysis-associated residues, Tyr157 and Asp226. The generally accepted mechanism for TLN-mediated hydrolysis proceeds via the two-step process depicted in Scheme 1. Briefly: The catalytic activity of TLN is dependent on both temperature and pH. The s of TLN are 5.0 and 8.25, and maximum catalytic activity has been measured at pH 7.2. The thermostability of TLN conferred by the four Ca ions allows an increase in catalytic activity at temperatures approaching 40°C, with no significant loss in activity or conformation alteration until temperatures exceed 70°C. Chelating agents such as EDTA have been shown to completely inactivate TLN and other metalloproteases by removal of the metal ion. In addition to direct removal of the catalytic metal ion, substrate and transition state analogs have been synthesized that greatly decrease the catalytic activity of TLN and TLN-like proteins. Transition state analogs that resemble the mechanistic transition states of the normal hydrolysis reaction catalyzed by TLN have been both isolated and synthesize. Phosphoramidates, a group of phosphoryl group containing amino acid or peptide compounds have been shown to mimic the tetrahedral intermediate of the TLN mechanism. Overall inhibition has been shown to vary from five to 1000-fold decrease in catalytic activity through use of transition state analogs. Holmquist and Vallee combined substrate analog inhibition with anionic ligands that tightly bind active site metals such as R-S and R-P-O . By combining two approaches, catalytic activity of TLN was decreased 10,000-fold. This compared to ~800-fold decrease when transition state analogs were combined with metal-binding ligands, suggesting a stronger inhibition effect of the substrate analog. Carboxypeptidase A (carboxypolypeptidase; CPA; ) is a 35 kD metalloenzyme within the zinc hydrolase family and as such contains a Zn2+ ion cofactor located within its active site. Originally isolated from bovine pancreas tissue in 1929, CPA is an exopeptidase that catalyzes the hydrolysis of C-terminal esters and peptides with large hydrophobic side chains. Biologically, CPA facilitates the breakdown of proteins during metabolism, while proposed commercial applications include its use the hydrolysis of cheese whey protein and the production of phenylalanine-free protein hydrolysates for use by individuals with phenylketonuria. Figure 3: General structure of carboxypeptidase A (PDBID ). Alpha-helices in blue, beta-sheets in red, Zn ion in yellow, active site side chains in cyan. Figure 4: Carboxypeptidase A active site. The structure of CPA was first determined in 1967 using x-ray diffraction, making it one earliest proteins to have been characterized using the technique (Figure 3). CPA consists of a single chain containing 307 amino acids and a single Zn ion in its active site. Zn is stabilized within the active site through interactions with His69, Glu72, and His196 and is additionally bound by a catalytic water molecule that has been shown to interact with Glu270 (Figure 4). Mutagenesis studies have indicated that the additional residues Arg127, Tyr248, Arg71, Asn144, and Arg145 form an outer shell, which includes Glu270, around the active site and contribute to the catalytic function of CPA through the stabilization of substrate molecules. In both proposed mechanisms for the catalytic activity of CPA, Glu270 plays an important role, either acting as a general base-general acid or as a nucleophile. Two mechanisms have been suggested for CPA catalyzed hydrolysis—the promoted-water pathway, also known as the general base-general acid pathway, and the nucleophilic, or anhydride, pathway—with experimental evidence existing for both mechanisms. In the enzyme substrate (ES) complex, the Zn tetrahedron consists of a water molecule, His69, Glu72, and His196 bound to the ion while the water molecule in the near attack position is maintained through hydrogen bonding with Glu270 and Ser197. The substrate position is maintained through interactions with Arg127, Asn144, Arg145, Tyr248, and Arg71. The water molecule attacks the scissile carbonyl carbon of the substrate molecule through nucleophilic addition with Glu270 acting as a general base, leading to the first transition state (TS1) and the formation of the tetrahedral intermediate (TI). Following the formation of the TI, the leaving group is protonated and the peptide bond is cleaved with Glu270 now severing as a general acid, forming the second transition state (TI2) and finally the enzyme product (EP) complex. An oxyanion hole, created by the polarization of the scissile carbonyl carbon of the substrate in the ES complex by Arg127 and the presence of Zn , helps stabilize the charge generated on the carbonyl oxygen during TS1 and TI. In the enzyme substrate (ES) complex, the Zn tetrahedron consists of His69, Glu72, His196, and the scissile carbonyl oxygen of the substrate molecule. This direct binding by Zn polarizes the carbonyl oxygen, facilitating the nucleophilic attack by the carboxylate side chain of Glu270, which is in the near attack position. The substrate position is maintained through interactions with Arg127, Asn144, Arg145, Tyr248, and Arg71. The nucleophilic attack by Glu270 on the scissile carbonyl carbon leads to the first transition state (TS1) and the formation of the acyl enzyme intermediate (AE). A water molecule present in the active site nucleophilically attacks the carboxylate carbon of Glu270, resulting in deacylation and the transition through the second transition state (TS2) and finally the enzyme product (EP) complex. Computational evidence suggests that in regards to proteolysis, the promoted-water pathway is the only feasible pathway of the two. However, both pathways are feasible in esterolysis reactions, with the promoted-water pathway having the lower kinetic barrier. This suggests that under normal conditions the promoted-water pathway is favored, whereas at low temperatures, the formation of the tetrahedral intermediate by the promoter-water pathway and the formation of the acyl enzyme intermediate by the nucleophilic pathways are comparable. The second deacylation step in the nucleophilic pathway presents too high of a barrier however to be viable versus the promoted-water pathway. This then suggests that in regards to both proteolysis and esterolysis, the promoted-water pathway is the dominant pathway of CPA. Ultraviolet-visible radiation (400 W, λ=250-750 nm) has been shown to cause uncompetitive inhibition of CPA with the decrease in enzymatic activity indirectly proportional to the irradiation time, with total enzymatic inactivation after 20 minutes of exposure. Additionally, exposure times of greater than 24 minutes are suspected to adversely affect the structure of CPA, resulting in the formation of protein aggregates. Active site-directed inhibitors of CPA, which are characterized by the presence of a terminal carboxylate, a hydrophobic side chain, and a zinc-binding group, have been identified, among which include the enantiomers of 2-benzyl-5-hydroxy-4-oxopentanoic acid. ( )-2-benzyl-5-hydroxy-4-oxopentanoic acid interacts through its carboxylate by forming hydrogen bonds with Arg145, Arg127, and Tyr248 and through its terminal hydroxyl group by forming a hydrogen bond with Ser197. Transition state analog inhibitors of CPA have also been found, which when bound with the enzyme active site create a pseudo-transition state complex. Both ( )-2-benzyl-3-nitropropanoic acid and ( )-2-benzyl-5-nitro-4-oxopentanoic acid are capable of inhibiting CPA by forming complexes with their respective nitro groups, Glu270, Arg127, and Zn .

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.10%3A_Chelating_Agents

Although we have confined our discussion so far to simple ligands such as Cl , NH , or H O, much larger and more complicated molecules can also donate electron pairs to metal ions. An important and interesting example of this is the —ligands which are able to form two or more coordinate covalent bonds with a metal ion. One of the most common of these is 1,2-diaminoethane (usually called and abbreviated .) When both nitrogens coordinate to a metal ion, a stable five-member ring is formed. The word , derived from the Greek , “claw,” describes the pincerlike way in which such a ligand can grab a metal ion. A chelating agent which forms several bonds to a metal without unduly straining its own structure is usually able to replace a similar simpler ligand. For example, although both form coordinate covalent bonds via groups, ethylenediamine can readily replace ammonia from most complexes: For metals which display a coordination number of 6, an especially potent ligand is thylene iaminete ra cetate ion (abbreviated EDTA): All six electron pairs marked in color are capable of coordinating to a metal ion, in which case the EDTA ion wraps completely around the metal and is very difficult to dislodge. is used to treat lead and mercury poisoning because of its ability to chelate these metals and aid their removal from the body. Chelate complexes are often important in living systems. The coordination of iron in proteins such as myoglobin or hemoglobin involves four nitrogen of the heme group and one from a histidine side chain. Since iron normally has a coordination number of 6, this leaves one open site, to which oxygen can bond. The presence of carbon monoxide, a stronger ligand than oxygen, causes displacement of oxygen from hemoglobin. This prevents transport of oxygen from the lungs to the brain, causing drowsiness, loss of consciousness, and even death upon long exposure to carbon monoxide. Consequently operating an automobile in a closed garage, a cookstove in a tent, or burning any fossil fuel incompletely in an enclosed space may be hazardous to one’s health. Another important application of chelates is transport of metal ions across membranes. The interior of a biological membranes contain the nonpolar, hydrophobic tails of lipid molecules. This makes it quite difficult for ionic species such as K and Na to travel from one side of a membrane to the other. One way in which this barrier may be circumvented is by carrier molecules, called ionophores. Ionophores are able to chelate an ion, but also have a hydrophobic exterior. One such ionophore is the antibiotic , a medium-sized organic molecule with the formula This molecule is able to transport K ions but not Na ions. Apparently the Na ion is too small to fit in among the eight coordinating O’s, while the K ion can (Figure 1). Other than these O’s, most of the nonactin molecule is a hydrocarbon chain. Therefore once K is chelated, the outer part of the complex is quite hydrophobic. It can easily pass through the interior of a membrane, releasing K on the other side. The toxic effect of nonactin and several related antibiotics is the result of their ability to transport alkali-metal ions to regions of a cell where they should not be. This breaks down ion gradients the cell has created to perform tasks and store energy. Consequently the cell wastes energy pumping K and other ions out again.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.8%3A_Colloids

A colloid is one of the three primary types of mixtures, with the other two being a and suspension. A colloid is a mixture that has particles ranging between 1 and 1000 nanometers in diameter, yet are still able to remain evenly distributed throughout the solution. These are also known as colloidal dispersions because the substances remain dispersed and do not settle to the bottom of the container. In colloids, one substance is evenly dispersed in another. The substance being dispersed is referred to as being in the dispersed phase, while the substance in which it is dispersed is in the continuous phase. To be classified as a colloid, the substance in the dispersed phase must be larger than the size of a molecule but smaller than what can be seen with the naked eye. This can be more precisely quantified as one or more of the substance's dimensions must be between 1 and 1000 nanometers. If the dimensions are smaller than this the substance is considered a solution and if they are larger than the substance is a suspension. A sol When the dispersion medium is water, the collodial system is often referred to as a . The particles in the dispersed phase can take place in different phases depending on how much water is available. For example, Jello powder mixed in with water creates a hydrocolloid. A common use of hydrocolloids is in the creation of medical dressings. An easy way of determining whether a mixture is colloidal or not is through use of the . When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen. An example of this is shining a flashlight into fog. The beam of light can be easily seen because the fog is a colloid. Another method of determining whether a mixture is a colloid is by passing it through a semipermeable membrane. The larger dispersed particles in a colloid would be unable to pass through the membrane, while the surrounding liquid molecules can. Dialysis takes advantage of the fact that colloids cannot diffuse through semipermeable membranes to filter them out of a medium.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.01%3A_Chemistry_and_Electricity

Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. The connection between chemistry and electricity is a very old one, going back to ALESSANDRO VOLTA'S discovery, in 1793, that electricity could be produced by placing two dissimilar metals on opposite sides of a moistened paper. Nature seems to strongly discourage any process that would lead to an excess of positive or negative charge in matter. Suppose, for example, that we immerse a piece of zinc metal in pure water. A small number of zinc atoms go into solution as \(Zn^{2+}\) ions, leaving their electrons behind in the metal: \[Zn(s) \rightarrow Zn^{2+} + 2e^– \label{1.1.1}\] As this process goes on, the electrons which remain in the zinc cause a negative charge to build up within the metal which makes it increasingly difficult for additional positive ions to leave the metallic phase. A similar buildup of positive charge in the liquid phase adds to this inhibition. Very soon, therefore, the process comes to a halt, resulting in a solution in which the concentration of \(Zn^{2+}\) is still too low (around 10 ) to be detected by ordinary chemical means. There would be no build-up of this opposing charge in the two phases if the excess electrons could be removed from the metal or the positive ions consumed as the electrode reaction proceeds. For example, we could drain off the electrons left behind in the zinc through an external circuit that forms part of a complete ; this we will describe later. Another way to remove these same electrons is to bring a good electron acceptor (that is, an ) into contact with the electrode. A suitable acceptor would be hydrogen ions; this is why acids attack many metals. For the very active metals such as sodium, water itself is a sufficiently good electron acceptor. The degree of charge unbalance that is allowed produces differences in electric potential of no more than a few volts, and corresponds to unbalances in the concentrations of oppositely charged particles that are not chemically significant. There is nothing mysterious about this prohibition, known as the electroneutrality principle; it is a simple consequence of the thermodynamic work required to separate opposite charges, or to bring like charges into closer contact. The additional work raises the free energy change of the process, making it less spontaneous. The only way we can get the oxidation of the metal to continue is to couple it with some other process that restores electroneutrality to the two phases. A simple way to accomplish this would be to immerse the zinc in a solution of copper sulfate instead of pure water. The zinc metal quickly becomes covered with a black coating of finely-divided metallic copper. The reaction is a simple involving a transfer of two electrons from the zinc to the copper: \[ Zn(s) \rightarrow Zn^{2+} + 2e^– \label{1.1.2} \] and \[ Cu^{2+} + 2e^– \rightarrow Cu(s) \label{1.1.3}\] The dissolution of the zinc is no longer inhibited by a buildup of negative charge in the metal, because the excess electrons are removed from the zinc by copper ions that come into contact with it. At the same time, the solution remains electrically neutral, since for each \(Zn\) ion introduced to the solution, one \(Cu\) ion is removed. The net reaction \[Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s) \label{1.1.4}\] quickly goes to completion. The transition region between two phases consists of a region of charge unbalance known as the . As its name implies, this consists of an inner monomolecular layer of adsorbed water molecules and ions, and an outer diffuse region that compensates for any local charge unbalance that gradually merges into the completely random arrangement of the bulk solution. In the case of a metal immersed in pure water, the electron fluid within the metal causes the polar water molecules to adsorb to the surface and orient themselves so as to create two thin planes of positive and negative charge. If the water contains dissolved ions, some of the larger (and more polarizable) anions will loosely bond ( ) to the metal, creating a negative inner layer which is compensated by an excess of cations in the outer layer. A process of this kind can always be represented as a chemical reaction and is known generally as an . Electrode processes (also called ) take place within the double layer and produce a slight unbalance in the electric charges of the electrode and the solution. Much of the importance of electrochemistry lies in the ways that these potential differences can be related to the thermodynamics and kinetics of electrode reactions. The interfacial potential differences which develop in electrode-solution systems are limited to only a few volts at most. This may not seem like very much until you consider that this potential difference spans a very small distance. In the case of an electrode immersed in a solution, this distance corresponds to the thin layer of water molecules and ions that attach themselves to the electrode surface, normally only a few atomic diameters. Thus a very small voltage can produce a very large potential gradient. For example, a potential difference of one volt across a typical 10 cm interfacial boundary amounts to a potential gradient of 100 million volts per centimeter— a very significant value indeed! Interfacial potentials are not confined to metallic electrodes immersed in solutions; they can in fact exist between any two phases in contact, even in the absence of chemical reactions. In many forms of matter, they are the result of adsorption or ordered alignment of molecules caused by non-uniform forces in the interfacial region. Thus colloidal particles in aqueous suspensions selectively adsorb a given kind of ion, positive for some colloids, and negative for others. The resulting net electric charge prevents the particles from coming together and coalescing, which they would otherwise tend to do under the influence of ordinary van der Waals attractions. The usual way of measuring a potential difference between two points is to bring the two leads of a voltmeter into contact with them. It's simple enough to touch one lead of the meter to a metallic electrode, but there is no way you can connect the other lead to the solution side of the interfacial region without introducing a second electrode with its own interfacial potential, so you would be measuring the sum of two potential differences. Thus , as they are commonly known, are not directly observable. What we observe, and make much use of, are potential differences between of electrodes in . This is the topic of the next page in this series. 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.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.02%3A_Viscosity

Polymers have come to occupy a very important niche in the materials we use every day. What makes them unique? Why do they have properties that aren't easily replicated by other materials? Let's start with a property that you are probably familiar with. is a term we use to describe the "thickness" of different liquids. For example, we say that honey is more viscous than water. Motor oil is more viscous than gasoline. When we say that, we mean that water is much easier to stir or to pour than honey. The honey moves more slowly. It resists the movement of the spoon when we stir it. Viscosity is often described in very general terms as "resistance to flow". The honey doesn't flow very easily, especially compared to something like water. Now, honey isn't a polymer. It's a very concentrated solution. It contains a little bit of water and a whole lot of sugars, plus other small molecules produced by the plants from which the bees gathered the nectar to make the honey. The sugars aren't polymers, either; they are simple monosaccharides such as glucose and fructose. Nevertheless, this high viscosity is something honey has in common with polymer solutions, as well as with some , which are short-chain polymers that can be liquids instead of solids. Why is honey so viscous? Partly it's just that the sugar molecules in the water are much larger than the water molecules, so they experience a lot more drag as they move through the solution compared to just plain water. The strong intermolecular attractions between the individual sugar molecules are also a major factor. Sugars are covered in OH or hydroxyl groups that are capable of strong hydrogen bonds. As these molecules move past each other in the very concentrated solution, they cling to each other, slowing down the flow of the liquid. The motor oil example works in a similar way. Motor oil and gasoline are composed of the same class of compounds, hydrocarbons, composed of carbon chains covered in hydrogen atoms. They're very similar to each other. The intermolecular attractions between the molecules are much weaker than those between sugar molecules; they're just London dispersion forces. The major difference between motor oil and gasoline is that the motor oil contains much longer hydrocarbon chains. The molecules are bigger, and so they experience more drag as they move through the liquid. The amount of intermolecular interactions is also crucial in determining how tightly two molecules hold onto each other, and this factor is especially important in very weak London dispersion forces, where a small advantage goes a long way. Those longer hydrocarbon chains in the motor oil cling to each other much more strongly than the shorter ones in the gasoline, so gasoline flows much more easily than motor oil. For similar reasons, oligomers and polymers, if they are liquids, also display high viscosity. The extended contact between their long molecular chains leads to enhanced intermolecular attractions that contribute to a resistance to flow. Increased drag is also a factor; even in solution, these very large molecules encounter more resistance as they move past solvent molecules compared to the resistance smaller molecules would experience. In fact, viscosity measurements of polymer solutions are another way to determine the size of the polymer -- leading to the chain length and the molecular weight. The larger the polymer, the more drag and also the more intermolecular attraction, and so the higher the viscosity. So, the viscosity increases with the molecular weight, but not necessarily in a linear way. That can make it more difficult to use a graph like this to predict molecular weight from a viscosity measurement. One common thing to do in this situation is to take the log of the values. That approach will often give a straight line when two things are related. Often in nature one physical property that depends on another will follow a "power law". That means that one quantity depends on the other raised to some exponent. In this case, the relationship is described by the : \[ [η] = KM^α\] In this equation, [η] is the intrinsic viscosity; that's the viscosity attributed to the solute, rather than the viscosity of the solvent itself. is the molecular weight. and α are constants for a particular polymer. If you take the log of both sides of the equation -- that's OK to do mathematically, just like you could multiply both sides of an equation by four or subtract three from both sides and still have an equivalent expression -- you get: \[\log[η] = log(KM^a)\] If you take the log of two things multiplied together, it's the same as adding the individual logs of both things. \[\log[η] = \log K + \log(M^a)\] And if you take the log of something raised to an exponent, the exponent just comes down to become a coefficient. \[\log[η] = \log K + α\log M\] or \[\log[η] = α \log M + \log K\] That looks like an equation for a straight line: \[y = mx + b.\] Here, you plot \(\log[η]\) on the y axis, \(\log M\) on the x axis, and you get a straight line with those constants, \(α\) and \(K\), as the slope and y-intercept, respectively. This approach does not yield an absolute measurement. You couldn't take two completely different polymers, take viscosity measurements, and make conclusions about which one had a higher average molecular weight. However, if you took a series of polymers of the same kind and measured their viscosities in solution, you would be able to deduce which ones were longer and which ones were shorter. In fact, the dependence of viscosity on the molecular weight of polymers is more complicated than this graph suggests. If you deal with large enough polymers, the graph starts to look like this one: Why does it bend? Remember, polymers in solution are coiled up into balls. A longer chain just corresponds to a bigger coil. At some point, these chains become big enough that they aren't likely to stay coiled independently of each other. They become more and more likely to interact with each other. They become entangled with each other. There are plenty of other factors you would have to take into account in using viscosity to assess molecular weight. One of them is concentration, for example. The more concentrated a polymer solution, the more large molecules you have exerting drag and interacting with each other. Higher concentration leads to a higher viscosity measurement. Consequently, you would have to make sure you accounted for the polymer concentrations if you used this method. Estimate the values of α and for the polymer that provided the plot below. Suppose you have a polymer described by the plot below. If your sample has a molecular weight of 1 million g/mol, what would you expect for intrinsic viscosity under the conditions of this experiment? Suppose you have a calibration plot, below, for a particular polymer. If you measure the intrinsic viscosity of a sample and find that the value is 800 ml / g, what is the molecular weight of the sample? Ethanol (CH CH OH) has an "absolute viscosity" of 1.095 centipoise, whereas ethylene glycol (HOCH CH OH) has an absolute viscosity of 16.2 centipoise. Explain why the two values are so different.

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Imagine adding a small amount of salt to a glass of water, stirring until all the salt has dissolved, and then adding a bit more. You can repeat this process until the salt concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent-solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved salt remains. The concentration of salt in the solution at this point is known as its solubility. The of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium. Referring to the example of salt in water: \[\ce{NaCl}(s)⇌\ce{Na+}(aq)+\ce{Cl-}(aq) \label{11.4.1} \] When a solute’s concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute’s concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated. If we add more salt to a saturated solution of salt, we see it fall to the bottom and no more seems to dissolve. In fact, the added salt does dissolve, as represented by the forward direction of the dissolution equation. Accompanying this process, dissolved salt will precipitate, as depicted by the reverse direction of the equation. The system is said to be at equilibrium when these two reciprocal processes are occurring at equal rates, and so the amount of undissolved and dissolved salt remains constant. Support for the simultaneous occurrence of the dissolution and precipitation processes is provided by noting that the number and sizes of the undissolved salt crystals will change over time, though their combined mass will remain the same. Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be , and they are interesting examples of nonequilibrium states. For example, the carbonated beverage in an open container that has not yet “gone flat” is supersaturated with carbon dioxide gas; given time, the CO concentration will decrease until it reaches its equilibrium value. In an earlier module of this chapter, the effect of intermolecular attractive forces on solution formation was discussed. The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl . Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C H , is approximately 20 times greater than it is in water. Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure \(\Page {1}\)). This is one of the major impacts resulting from the thermal pollution of natural bodies of water. When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure \(\Page {2}\)). The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure \(\Page {3}\)). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.” For many gaseous solutes, the relation between solubility, \(C_g\), and partial pressure, \(P_g\), is a proportional one: \[C_\ce{g}=kP_\ce{g} \nonumber \] where \(k\) is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of : The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas. At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10 mol L . Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere. According to Henry’s law, for an ideal solution the solubility, C , of a gas (1.38 × 10 mol L , in this case) is directly proportional to the pressure, P , of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both C and P , we can rearrange this expression to solve for k. \[\begin{align*} C_\ce{g}&=kP_\ce{g}\\[4pt] k&=\dfrac{C_\ce{g}}{P_\ce{g}}\\[4pt] &=\mathrm{\dfrac{1.38×10^{−3}\:mol\:L^{−1}}{101.3\:kPa}}\\[4pt] &=\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}}\\[4pt] &\hspace{15px}\mathrm{(1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1})} \end{align*} \nonumber \] Now we can use k to find the solubility at the lower pressure. \[C_\ce{g}=kP_\ce{g} \nonumber \] \[\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}×20.7\:kPa\\[4pt] (or\:1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1}×155\:torr)\\[4pt] =2.82×10^{−4}\:mol\:L^{−1}} \nonumber \] Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable. Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr). 7.25 × 10 g in 100.0 mL or 0.0725 g/L Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law. As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid , divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure \(\Page {4}\)). Deviations from are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions. Gases can form . If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure \(\Page {5}\)), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.   We know that some liquids mix with each other in all proportions; in other words, they have infinite mutual solubility and are said to be . Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in Figure \(\Page {6}\)) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline. Liquids that mix with water in all proportions are usually polar substances or substances that form hydrogen bonds. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solvent-solvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom “like dissolves like.” Two liquids that do not mix to an appreciable extent are called . Layers are formed when we pour immiscible liquids into the same container. Gasoline, oil (Figure \(\Page {7}\)), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. The attraction between the molecules of such nonpolar liquids and polar water molecules is ineffectively weak. The only strong attractions in such a mixture are between the water molecules, so they effectively squeeze out the molecules of the nonpolar liquid. The distinction between immiscibility and miscibility is really one of degrees, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility. Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be . Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure \(\Page {8}\)).   The dependence of solubility on temperature for a number of inorganic solids in water is shown by the solubility curves in Figure \(\Page {9}\). Reviewing these data indicate a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate. The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure \(\Page {10}\), take advantage of this behavior. \(\Page {2}\): This video shows the crystallization process occurring in a hand warmer. The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances’ atoms, ions, or molecules. This tendency to dissolve is quantified as substance’s solubility, its maximum concentration in a solution at equilibrium under specified conditions. A saturated solution contains solute at a concentration equal to its solubility. A supersaturated solution is one in which a solute’s concentration exceeds its solubility—a nonequilibrium (unstable) condition that will result in solute precipitation when the solution is appropriately perturbed. Miscible liquids are soluble in all proportions, and immiscible liquids exhibit very low mutual solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature. The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry’s law.

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Most metals have very compact crystal structures involving either the body-centered cubic, face-centered cubic, or hexagonal . Thus every atom in a metal is usually surrounded by 8 or 12 equivalent nearest neighbors. How can each atom be bonded to so many of its fellow atoms? Although there are plenty of electropositive atoms to donate electrons, there are no electronegative atoms to receive them, and so ionic bonding seems unlikely. Ordinary covalent bonding can also be ruled out. Each covalent bond would require one electron from each atom, and no metal has 12 valence electrons. A valuable clue to the nature of bonding in metals is provided by their ability to conduct electricity. Electrons can be fed into one end of a metal wire and removed from the other end without causing any obvious change in the physical and chemical properties of the metal. To account for this freedom of movement modern theories of metallic bonding assume that the valence electrons are completely delocalized; that is, they occupy molecular orbitals belonging to the metallic crystal as a whole. These delocalized electrons are often referred to as an or an . Positive metal ions produced by the loss of these valence electrons can then be thought of as “floating” in this three-dimensional sea. Each ion is held in place by the attraction of the negatively charged electron sea and the repulsion of its fellow positive ions. In order to see how MO theory can be applied to metals, let us first consider the simplest case, lithium. If two lithium atoms are brought together, the 1 core electrons remain essentially unchanged since there is virtually no overlap between them. The 2s orbitals, by contrast, overlap extensively and produce both a bonding and an antibonding orbital. Only the bonding orbital will actually be occupied by the two electrons, as shown in Figure 1. Somewhat higher than these two orbitals are a group of six unoccupied orbitals produced by the overlap of six 2 atomic orbitals (three on each atom). Suppose now we add a third atom to the two already considered so that we form a triangular molecule of formula Li . As shown in the figure, the overlap of three 2 orbitals produces a lower group of three orbitals, while the overlap of three times three 2 orbitals produces a higher group of nine orbitals. Again the total number of molecular orbitals is equal to the number of atomic orbitals from which they are derived. Continuing to add lithium atoms in this fashion, we soon attain a cluster of 25 lithium atoms. The energy-level situation for a cluster this size is a lower group of 25 MO’s, all deriving from 2 atomic orbitals, and a higher group of 75 MO’s, all deriving from 2 atomic orbitals. Note how closely spaced these energy levels have become. This is in line with the tendency for the energy levels to get closer the greater the degree of delocalization. Finally, if we add enough lithium atoms to our cluster to make a visible, weigh-able sample of lithium, say 10 atoms, the energy spacing between the molecular orbitals becomes so small it is impossible to indicate in the figure or even to measure. In effect an electron jumping among these levels can have any energy within a broad band from the lowest to highest. In consequence this view of electronic structure in solids is often referred to as the . It should also be clear from Figure \(\Page {1}\) that all the available molecular orbitals need not be completely filled with electrons. In the case of lithium, for example, a sample containing 10 atoms would have 10 valence electrons. Since each atom would have a single 2 orbital as well as three 2 orbitals, there would be 1 × 10 MO’s in the 2 band and 3 × 10 MO’s in the 2 band. If all electrons were paired, only the 0.5 × 10 MO’s of lowest energy in the 2 band would be required to hold them. Note that there is a nice correspondence between the half-filled 2 band of the macroscopic sample and the half-filled 2 orbital of an individual Li atom. According to the band theory, it is this partial filling which accounts for the high electrical and thermal conductance of metals. If an electric field is applied to a metallic conductor, some electrons can be forced into one end, occupying slightly higher energy levels than those already there. As a consequence of delocalization this increased electronic energy is available throughout the metal. It therefore can result in an almost instantaneous flow of electrons from the other end of the conductor. A similar argument applies to the transfer of thermal energy. Heating a small region in a solid amounts to increasing the energy of motion of atomic nuclei and electrons in that region. Since the nuclei occupy specific lattice positions, conduction of heat requires that energy be transferred among nearest neighbors. Thus when the edge of a solid is heated, atoms in that region vibrate more extensively about their average lattice positions. They also induce their neighbors to vibrate, eventually transferring heat to the interior of the sample. This process can be speeded up enormously if some of the added energy raises electrons to higher energy MO’s within an incompletely filled band. Electron delocalization permits rapid transfer of this energy to other atomic nuclei, some of which may be quite far from the original source. When an energy band is completely filled with electrons, the mechanism just described for electrical and thermal conduction can no longer operate. In such a case we obtain a solid which is a very poor conductor of electricity, or an . At first glance we might expect Be, Mg, and other alkaline earths to be insulators like this. Since atoms of these elements all contain filled 2 subshells, we would anticipate a filled 2 band in the solid for all of them. That this is not the case is due to the relatively small energy difference between the 2 and 2 levels in these atoms. As you can see from Figure \(\Page {1}\), this small separation results in an overlap between the 2 and 2 bands. Thus electrons can move easily from the one band to the other and provide a mechanism for conduction. Figure \(\Page {2}\) shows four different possibilities for band structure in a solid. For a solid to be a conductor, a band must be either partially filled or must overlap a higher unfilled band. When there is a very large energy gap between bands and the lower band is filled, we have an insulator. If the gap is quite small, we get an intermediate situation and the solid is a . All the semimetals found along the stairstep diagonal in the periodic table, notably germanium, have a band structure of this type. In a semiconductor we find that collisions among atoms and electrons in the crystal are occasionally energetic enough to excite an electron into the top band. As a result there are always a small number of electrons in this band and an equal number of (orbitals from which electrons have been removed) in the lower-energy band. The excited electrons can carry electrical current because many different energy levels are available to them. So can the holes—other electrons from the nearly filled band can move up or down into them, a process which decreases or increases the energy of the hole. In a metal, electrical conductivity decreases as the temperature is raised because the nuclei vibrate farther from their rest positions and therefore get in the way of moving valence electrons more often. Exactly the opposite behavior is found for semiconductors. With increasing temperature, more and more electrons are excited to the higher-energy conduction band so that more current can be carried. Excitation of electrons to the conduction band can also be accomplished by a photon, a phenomenon known as photoconduction. Selenium metal is often used in this way as a photocell in light meters and “electric eyes.” The electron-sea model of metals not only explains their electrical properties but their malleability and ductility as well. When one layer of ions in an electron sea moves along one space with respect to the layer below it, a process we can represent pictorially as: The final situation is much the same as the initial. Thus if we hit a metal with a hammer, the crystals do not shatter, but merely change their shape, This is very different from the behavior of ionic crystals. The electron-sea model also enables us to explain, at least partially, why the metallic bond is noticeably stronger for some metals than others. While the alkali metals and some of the alkaline-earth metals can be cut with a knife, metals like tungsten are hard enough to scratch the knife itself. A good indication of how the strength of the metallic bonding varies with position in the periodic table is given by the melting point. As can be seen from Figure \(\Page {3}\), if the melting point of the metals is plotted against the group number for the three long periods, there is a sharp increase from group IA to group VB or VIB, after which there is a leveling off. Finally the melting point again drops to quite low values. A similar behavior is found for other properties such as boiling point, enthalpy of fusion, density, and hardness. The initial increases in the strength of metallic bonding as we move from group IA to VIB can be explained in terms of the number of valence electrons the metal is capable of contributing to the electron sea. The more electrons an atom loses, the larger will be the charge of the positive ion embedded in the electron sea and the greater will be the electron probability density of the electron sea itself. Thus the more electrons which are lost, the more tightly the ions will be held together. Chromium with six valence electrons is thus much harder than sodium with one. This trend cannot continue indefinitely, however. The more electrons that are removed from an atom, the more energy it takes to remove the next electron. Eventually we find that more energy is needed to remove an electron from a metal nucleus than is liberated by placing it in the electron sea. The strength of the bonding thus begins to level off and eventually to drop. It should be pointed out that metallic bonding strength is not solely dependent on the number of valence electrons (or the periodic group number) of an element. Other factors such as atomic radius and type of crystal lattice are also important. Nevertheless it is useful to remember that melting points and other properties related to metallic bond strength reach their maximum at about the middle of each transition series.

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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.

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

One of the most important applications of chemistry is the study of . Fuel basically means a chemical that can provide energy. We use fuel to do nearly everything. Food provides fuel to our bodies. Gasoline provides fuel for our cars. Many different chemicals, from coal and gas to uranium, provide fuel for the power plants that make electric power. When we use a fuel, it becomes a reactant in some type of reaction. Most often, fuels are burned in air to provide heat, and the heat is converted to work. For this reason, we often want to know how much heat we can get out of a chemical reaction. When our systems includes chemicals that participate in a reaction, the system may gain or lose heat because of the reaction, and the system may also do work, or work may be done on the system. How would the system do work? Suppose you burn some propane (C H ). Write a balanced equation for this reaction, and convince yourself that there are more moles of gas present in the products than the reactants. Also, you know that burning propane raises the temperature. If you burn some propane, the system will increase in volume, because there is more gas, and it is hotter. (If you burn the propane in a very strong container, then the volume will stay the same and the pressure will increase.) When the volume of the system increases in an open container, it will push on the surroundings, like the atmosphere, and do pV work. We might be able to use this work (in an internal combustion engine like in a car, for instance) but only if we control the volume. If we do the reaction open to the atmosphere, at constant pressure (atmospheric pressure), then we won't be able to use the work done by the reaction. In chemistry, because reactions are often done at atmospheric pressure, we often want to know how much heat is available from a reaction that occurs at constant pressure. To make this convenient, we can define a new quantity, related to internal energy, which is called , abbreviated H: \[H=E+pV\] Here's why this definition is useful. The work done by a reaction at constant pressure is \[w=-p \Delta V\] where p is the pressure, and ΔV is the change in volume of the reaction system, and the sign of w is negative because the system is doing work. The change in enthalpy for a reaction is \[\Delta H=\Delta (E+pV) = \Delta E + p\Delta V (at\; constant\; p)\] Since ΔE = q + w, \[\Delta H = (q + w) + p\Delta V = (q + w) − w = q\] Thus, the enthalpy change in a reaction tells us exactly how much heat the reaction can provide if it runs at constant pressure. We don't need to worry about calculating the work done by the reaction pushing back the atmosphere because it is already removed from the definition of enthalpy. Enthalpy is a state function because E, p and V are all state functions. Enthalpy doesn't have a molecular meaning like internal energy, but usually pV is small, so enthalpy is similar to internal energy. When ΔH is positive, then heat has entered the system, and the process is called . Endothermic reactions feel cold to the touch because they pull heat from your hand into the reaction system. Evaporation is endothermic, which is why sweat helps us cool off. When ΔH is negative, heat leaves the system, and the process is called . Exothermic processes feel hot. This is why flames will burn you.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/08%3A_Strategies_in_Cedrene_Synthesis

Cedrene represents a very complex tricyclic sesquiterpene. Such complexity called for ingenious approaches for the total synthesis of the ring system. There are several syntheses reported for this ring system. We would discuss a few approaches. Some strategic bond cleavages are shown as broken lines in . G. Stork et.al., (J. Am. Chem. Soc., 77, 1078 (1955); ibid, 83, 3114 (1961)) took advantage of the fact that fused five membered rings would be cis- fused. Such a system would have a crowded ‘concave phase’. These two features formed the basis for Stork’s synthesis of Cedrene. Their synthetic scheme is shown in . The first alkylation set in motion the steric out come of the remaining steps. For his 1969 Cedrene synthesis (J. Am. Chem. Soc., 91, 1557 (1969)), Corey set up the spiro-ring system (cleavage of bond ‘b’). A p- alkylation of phenolate gave the spiro ring (J. Am. Chem. Soc., 84, 788 (1962)). Lewis catalyzed cylisation of the enolate completed the skeleton . Based on biogenetic cyclization concept, Corey achieved a one-step cyclization of the A and B rings onto a preformed C ring (Tet. Lett., 2455 (1972)). The key for success in this scheme is the prior formation of bond ‘d’ followed by the formation of the ‘b’ bond. Note the role of a cyclopropyl ketone that orchestrated the development of a carbonium ion followed by an incipient carbanion to complete the A and B rings in this order. A remarkable biogenetic type cyclization of Nerolidol to Cedrene was reported by Anderson et.al.,(Tet. Lett., 2455 (1972). The cyclizations could be accomplished in two steps. With formic acid a six membered ring is first formed. Further treatments with triflouroacetic acid completed the synthesis. The overall yield was very moderate . In Anderson’s biogenetic-type cyclization, formation of the six membered C ring was first realized. A six membered ring system also served as a key step for free radical cyclization reported by Hee-Yoon Lee. Tandem radical cyclization was their central theme. Such radical cyclizations have been reported for the synthesis of natural products. Hee-Yoon Lee reported application of this strategy to the synthesis of α-Cedrene (Ter. Lett., 7713 (1998)). The success of this scheme hinged on two factors. An approach by Chen et.al., (Tet. Lett., 2961 (1993)) relied on the free radical cyclization reaction on a suitably fuctionalized C ring . The synthesis of C ring is achieved via., a DA reaction. Knoevenagal reaction placed the required chain for completing the A and B rings. Tin hydride reduction generated a free radical at the site of the nitro group, which underwent a tandem cyclization to complete the A and B rings in that order. A Diels-Alder approach to to tricyclic cedrene skeleton was reported by Breitholle et.al., (Can. J. Chem., 54, 1991 (1976) . Alkylation of cyclobutadiene to the requisite chain gave 8.8A after equilibration. The DA reaction proceeded in 36% yield to give a mixture of isomers. The ketone 8.8B did not undergo ring expantion with diazomethane. The ring expansion was finally achieved via the methylene amine 8.8C via diazotisation. The fact that two ring expansion products 8.8D and 8.8E were formed with 8.8D as the major product suggests that the amine 8.8F was the major isomer.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Gay-Lussac's_Law

Dalton's atomic theory led to a new question: if each element has atoms with a characteristic mass, what are those masses? If water is 88.9% O, 11.1% H, what is the atomic mass of O in terms of H (if you assume H is 1.00)? To answer, you need the formula: H O (2 H atoms for 1 O atom) (11.1)/2 = 5.55 → (88.9)/5.55 = 16.0 (this is the atomic weight of O, assuming H is ~1) But early chemists didn't know the formulas, because they didn't know the atomic weights! Dalton assumed that the simplest formula (example: HO) was right, but this was usually wrong! How could they figure out the formulas? In 1808, Gay-Lussac published results that showed what volumes of gases combined with each other in chemical reactions. For instance (O is oxygen, H is hydrogen, N is nitrogen, L is liters): Based on Gay-Lussac's Law, we can guess the following: equal volumes of gas at the same temperature (T) and pressure (P) have the same number of "particles". What are these "particles?" Many of them are , strongly-bonded collection of atoms. Molecules usually remain intact when vaporized into the gas phase. Molecular compounds (compounds containing molecules, as compared to ionic compounds discussed later) are usually made of non-metal elements such as C, O, S, P, H, Cl, etc. How was Gay-Lussac's law received? Dalton didn't believe in it because the densities of the gases seem wrong. Oxygen gas is denser than steam (water gas), even though water is oxygen plus hydrogen. Also, Gay-Lussac didn't push the conclusions of his law as far as he could have because Berthollet (who believed that combining ratios of elements could vary) was his mentor. In 1811, Avogadro explained the problems with Gay-Lussac's law by saying that equal volumes of gases (at the same T and P) have the same number of molecules. The elemental gases were present not as single atoms but as diatomic molecules (such as H , O , N ). Now the equations from Gay-Lussac's law are: This resolves the issue of gas density (O is denser than H O because H weighs less than O). Avogadro's hypothesis could have cleared up all the confusion about formulas, and allowed good atomic weight calculations. But people either ignored him or said it was impossible.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Conjugation/Conjugated_Dienes

A diene is a hydrocarbon chain that has two double bonds that may or may not be adjacent to each other. This section focuses on the delocalization of pi systems by comparing two neighboring double bonds. The arrangements of these double bonds can have varying affects on the compounds reactivity and stability. First identify the longest chain containing both carbons with double bonds in the compound. Then give the lowest possible number for the location of the carbons with double bonds and any other functional groups present (remember when naming alkenes that some groups take priority such as alcohols). Do not forget or any other orientation of the double bond such as (E/Z,cis or trans). Examples: Conjugated dienes are two double bonds separated by a single bond Nonconjugated (Isolated) Dienes are two double bonds are separated by more than one single bond. Cumulated Dienes are two double bond connected to a similar atom. When using electrostatic potential maps, it is observed that the pi electron density overlap is closer together and delocalized in conjugated dienes, while in non conjugated dienes and cummulated dienes the pi electron density is located differently across the molecule. Since having more electron density delocalized makes the molecule more stable conjugated dienes are more stable than non conjugated and cummulated dienes. Conjugated dienes are more stable than non conjugated dienes (both isolated and cumulated) due to factors such as delocalization of charge through resonance and hybridization energy. This can also explain why are much more stable than secondary or even tertiary carbocations. This is all due to the positioning of the pi orbitals and ability for overlap to occur to strengthen the single bond between the two double bonds. The resonance structure shown below gives a good understanding of how the charge is delocalized across the four carbons in this conjugated diene. This delocalization of charges stablizes the conjugated diene: Along with resonance, hybridization energy effect the stability of the compound. For example in 1,3-butadiene the carbons with the single bond are sp2 hybridized unlike in nonconjugated dienes where the carbons with single bonds are sp3 hybridized. This difference in hybridization shows that the conjugated dienes have more 's' character and draw in more of the pi electrons, thus making the single bond stronger and shorter than an ordinary alkane C-C bond (1.54Å). Another useful resource to consider are the heats of hydrogenation of different arrangements of double bonds. Since the higher the heat of hydrogenation the less stable the compound, it is shown below that conjugated dienes (~54 kcal) have a lower heat of hydrogenation than their isolated (~60 kcal) and cumulated diene (~70 kcal) counterparts. Here is an energy diagram comparing differnt types of bonds with their heats of hydrogenation to show relative stability of each molecule: There are two different conformations of conjugated dienes which are s- and s- conformations. s- is when the double bonds are cis in reference to the single bond and s is when the two double bonds are trans in reference to the single bond. The cis conformation is less stable due to the steric interation of hydrogens on carbon. One important use of the cis conformation of a conjugated diene is that it is used . Even though the trans conformation is more stable the cis conformation is used because of the molecule's ability to interconvert and rotate about the single bond. The four pi electrons from the two double bonds are placed in the bonding orbitals with no nodes (2 electrons) and one node (2 electrons). The orbital with the Highest occupied molecular orbital (HOMO) is used in as dienophiles. Here is a clear representation of what is going on in the Highest occupied molecular orbital of the conjugated diene: 1) Draw the following: 1,3-cycloheptadiene 2) Name the following: 3) Draw both s-cis and s-trans conformations of 1,3-pentadiene and state which is more stable and why? 4) What do you think is more stable -1,3-pentadiene or 1,4-pentadiene? and why? 5) Draw the molecular orbital for 1,3-butadiene 1) 2) (Z)-4-Chloro-1,3-hexadiene 3) s-trans is more stable because there is less steric interaction between the methyl and hydrogens unlike in s-cis. 4) 1,3-pentadiene because it is a conjugated diene, which is more stable than an isolated diene such as 1,4-pentadiene. This is due do many reasons such as resonance energy and hybridization. 5)

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

A is a laboratory technique used to precisely measure molar concentration of an unknown solution using a known solution. The basic process involves adding a of one reagent to a known amount of the unknown solution of a different reagent. For instance, you might add a standard base solution to an mystery acid solution. As the addition takes place, the two reagents in the solutions, in this the acid and base, react. You also add an , which is a molecule that changes color when the original reagent (the acid in the mystery solution, say) is completely consumed by reaction with the standard solution reagent. If you know exactly how much standard was added before the color change, you can calculate how many moles of the unknown were present at the beginning, and thus the concentration of the unknown. Many of the standard reagents you might use in the lab, especially HCl and NaOH, which are very common and important, are hard to prepare at precise concentration without titration. The reason is that HCl is purchased as a concentrated solution, which can vary a little in concentration because both the HCl and the water can evaporate. NaOH can be purchased as a solid, but it is which means that it absorbs water from the air. It can absorb so much water that it actually dissolves. For this reason, even if you buy it dry, once you open the bottle, it might start to absorb water, and it would be difficult to know when you measure it what % water it is. Thus, if you work in a biochemistry lab, for instance, you might want to control the pH of your solutions by adding a little bit of dilute HCl or NaOH, because chloride and sodium ions are very common and probably are already included in the solution, but you might want to know how concentrated your solutions are. To determine this, you would use a standard solution made of some easier-to-mass acid or base to titrate the solution you actually want to use. Once titrated, you could dilute it precisely to the concentration you want. Some other reagents you might want standard solutions of react with air; these you might also titrate if they have been waiting a long time so you know what the current concentration is. Titrations might seem a little old-fashioned. Actually, the number of automated titration machines available (try a google search!) suggest that titrations are still very important in industry. One reason might be that titrations can be good for studying newly discovered molecules, for instance to measure the molecular weight and other properties that we will study more later. Traditionally, you take a known mass or volume of the unknown solution and put it in a flask with the indicator. Then you add the standard solution in a buret, which is a special tube for adding solution slowly and measuring the volume added at the end. These days, it might be easier to use a plastic squeeze bottle instead of a buret. You put the standard solution in the squeeze bottle, get the mass of the bottle, do the titration, and then mass the bottle again. Now you know exactly how much standard was added! We have a solution of HCl whose concentration is known imprecisely (~2.5 M). (We made this solution in the previous section on molarity.) We want to determine the concentration more precisely. We have a solution of NaOH that is known to be 5.1079 M. We place 100.00 ml of the HCl solution in a flask with a drop of an indicator that will change color when the solution is no longer acidic. Then we add NaOH slowly until the indicator color changes. At this point, we have added 46.67 ml NaOH. Calculate the precise concentration of the HCl. To answer, we need to know that the reaction is \[HCl + NaOH \rightarrow NaCl + H_{2}O\] So the ratio is 1 HCl:1 NaOH. We calculate the number of moles of NaOH added: \[(5.1079\; mol/L)(46.67\; mL)= 238.4\; mmol\] This is also the number of moles of HCl in the original 100.00 mL of solution, because the reaction ratio is 1:1. To calculate the concentration of the HCl solution, we just divide the number of moles of HCl by the volume. \[(238.4\; mmol)/(100.00\; mL) = 2.384\; M\] We could do this in one step using dimensional analysis: \[(46.67\; \cancel{mL\; NaOH}) \left(\dfrac{5.1079\; \cancel{mol\; NaOH}}{1000\; \cancel{mL}}\right) \left(\dfrac{1\; mol\; HCl}{1\; \cancel{mol\; NaOH}}\right) \left(\dfrac{1}{100.00\; mL}\right)=2.384\; M\] Now, we diluted 250 mL of the original stock solution to 1.00 L to make this solution. What is the concentration of the 10 M HCl, precisely? First we calculate the moles of HCl in the whole "2.5 M" solution, which is equal to the number of moles of HCl in quantity of stock solution ("10 M") that was used to make it (250ml). \[(1.00\; \cancel{L\; of\; diluted\; solution}) \left(\dfrac{2.384\; mol\; HCl}{1\; \cancel{L\; of\; diluted\; solution}}\right) \left(\dfrac{1}{0.250\; L\; of\; conc.\; solution}\right)=9.536\; M\] Why so low? HCl is a gas, and can evaporate out of solution. The stock solution must have been pretty old.

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Benzene, C H , is an organic aromatic compound with many interesting properties. Unlike aliphatic (straight chain carbons) or other cyclic organic compounds, the structure of benzene (3 conjugated π bonds) allows benzene and its derived products to be useful in fields such as health, laboratory, and other applications such as rubber synthesis. Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (Figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections and Hückel's Rule. Many aromatic compounds are however, sweet/pleasant smelling. Eugenol, for example, is extracted from essential oils of cloves and it releases a spicy, clove-like aroma used in perfumes. In addition, it is also used in dentistry as an analgesic. Due to the similarity between benzene and , the two is often confused with each other in beginning organic chemistry students. If you were to count the number of carbons and hydrogens in , you will notice that its molecular formula is . Since the carbons in the ring is with hydrogens (carbon is bound to 2 hydrogens and 2 adjacent carbons), no double bonds are formed in the cyclic ring. In contrast, is only with one hydrogen per carbon, leading to its molecular formula of . In order to stabilize this structure, 3 conjugated π (double) bonds are formed in the benzene ring in order for carbon to have four adjacent bonds. In other words, cyclohexane is not the same as benzene! These two compounds have different molecular formulas and their chemical and physical properties are not the same. The technique can be used by chemists to convert from benzene to cyclohexane by saturating the benzene ring with missing hydrogens. A special catalyst is required to hydrogenate benzene rings due to its unusual stability and configuration. Normal catalytic hydrogenation techniques will not hydrogenate benzene and yield any meaningful products. Benzene can be drawn a number of different ways. This is because benzene's conjugated pi electrons freely resonate within the cyclic ring, thus resulting in its two resonance forms. As the electrons in the benzene ring can resonate within the ring at a fairly high rate, a simplified notation is often used to designate the two different resonance forms. This notation is shown above, with the initial three pi bonds (#1, #2) replaced with an inner ring circle (#3). Alternatively, the circle within the benzene ring can also be dashed to show the same resonance forms (#4). The phenyl group can be formed by taking benzene, and removing a hydrogen from it. The resulting molecular formula for the fragment is C H . NOTE: Although the molecular formula of the phenyl group is C H , the phenyl group would always have something attached to where the hydrogen was removed. Thus, the formula is often written as Ph-R, where Ph refers to the Phenyl group, and R refers to the R group attached to where the hydrogen was removed. Different R groups on the phenyl group allows different benzene derivatives to be formed. Phenol, Ph-OH, or C H OH, for example, is formed when an alcohol (-OH) group displaces a hydrogen atom on the benzene ring. Benzene, for this very same reason, can be formed from the phenyl group by reattaching the hydrogen back its place of removal. Thus benzene, similar to phenol, can be abbreviated Ph-H, or C H . As you can see above, these are only some of the many possibilities of the benzene derived products that have special uses in human health and other industrial fields. Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. In these sections, we will analyze some of the ways these compounds can be named. Some common substituents, like NO , Br, and Cl, can be named this way when it is attached to a phenyl group. Long chain carbons attached can also be named this way. The general format for this kind of naming is: . For example, chlorine (Cl) attached to a phenyl group would be named . Since there is only one substituent on the benzene ring, we do not have to indicate its position on the benzene ring (as it can freely rotate around and you would end up getting the same compound.) Example of simple benzene naming with chlorine and NO as substituents. Instead of using numbers to indicate substituents on a benzene ring, can be used in place of positional markers when there are substituents on the benzene ring (disubstituted benzenes). They are defined as the following: Using the same example above in Figure 9a (1,3-dichlorobenzene), we can use the ortho-, meta-, para- nomenclature to transform the chemical name into m-dichlorobenzene, as shown in the Figure below. Here are some other examples of ortho-, meta-, para- nomenclature used in context: However, the substituents used in ortho-, meta-, para- nomenclature do not have to be the same. For example, we can use chlorine and a nitro group as substituents in the benzene ring. In conclusion, these can be pieced together into a summary diagram, as shown below: In addition to simple benzene naming and OMP nomenclature, benzene derived compounds are also sometimes used as . The concept of a base is similar to the nomenclature of aliphatic and cyclic compounds, where the parent for the organic compound is used as a base (a name for its chemical name. For example, the following compounds have the base names and , respectively. See Nomenclature of Organic Compounds for a review on naming organic compounds. Benzene, similar to these compounds shown above, also has base names from its derived compounds. , as introduced previously in this article, for example, serves as a base when other substituents are attached to it. This is best illustrated in the diagram below. Alternatively, we can use the numbering system to indicate this compound. When the numbering system is used, the carbon where the substituent is attached on the base will be given the first priority and named as carbon #1 (C ). The normal priority rules then apply in the nomenclature process (give the rest of the substituents the lowest numbering as you could). Below is a list of commonly seen benzene-derived compounds. Some of these mono-substituted compounds (labeled in red and green), such as phenol or toluene, can be used in place of benzene for the chemical's base name. Common benzene derived compounds with various substituents. According to the indexing preferences of the , are some of the common names that are retained in the IUPAC (systematic) nomenclature. Other names such as toluene, styrene, naphthalene, or phenanthrene can also be seen in the IUPAC system in the same way. While the use of other common names are usually acceptable in IUPAC, their use are discouraged in the nomenclature of compounds. Nomenclature for compounds which has such discouraged names will be named by the simple benzene naming system. An example of this would include (Note that toluene by itself is retained by the IUPAC nomenclature, but its derivatives, which contains additional substituents on the benzene ring, might be excluded from the convention). For this reason, the 2,4,6-trinitrotoluene, or TNT, as shown in Figure 17, would not be advisable under the IUPAC ( ) nomenclature. In order to correctly name TNT under the IUPAC system, the simple benzene naming system should be used: Since the IUPAC nomenclature primarily rely on the simple benzene naming system for the nomenclature of different benzene derived compounds, the OMP (ortho-, meta-, para-) system is not accepted in the IUPAC nomenclature. For this reason, the OMP system will yield common names that can be converted to systematic names by using the same method as above. For example, o-Xylene from the OMP system can be named 1,2-dimethylbenzene by using simple benzene naming (IUPAC standard). As mentioned previously, the phenyl group (Ph-R, C H -R) can be formed by removing a hydrogen from benzene and attaching a substituent to where the hydrogen was removed. To this phenomenon, we can name compounds formed this way by applying this rule: . For example, a chlorine attached in this manner would be named , and a bromine attached in this manner would be named . (See below diagram) While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below. Although the diagram above might be a little daunting to understand at first, it is not as difficult as it seems after careful analysis of the structure is made. By looking for the longest chain in the compound, it should be clear that the longest chain is eight (8) carbons long (octane, as shown in green) and that a benzene ring is attached to the second position of this longest chain (labeled in red). As this rule suggests that the benzene ring will act as a function group (a substituent) whenever a substituent of more than six (6) carbons is attached to it, the name "benzene" is changed to and is used the same way as any other substituents, such as Putting it all together, the name can be derived as: (phenyl is attached at the second position of the longest carbon chain, octane). The benzyl group (abbv. Bn), similar to the phenyl group, is formed by manipulating the benzene ring. In the case of the benzyl group, it is formed by taking the phenyl group and adding a CH group to where the hydrogen was removed. Its molecular fragment can be written as C H CH -R, PhCH -R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol. Additionally, other substituents can attach on the benzene ring in the presence of the benzyl group. An example of this can be seen in the Figure below: Similar to the base name nomenclature system, the carbon in which the base substituent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible. Under this consideration, the above compound can be named: As benzene derived compounds can be extremely complex, only compounds covered in this article and other commonly named compounds can be named using this flowchart. To demonstrate how this flowchart can be used to name TNT in its common and systematic (IUPAC) name, a replica of the flowchart with the appropriate flow paths are shown below: The compound above contains a benzene ring and thus is aromatic. Benzene unusual stability is caused by how many conjugated pi bonds in its cyclic ring? ____ Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not? At normal conditions, benzene has ___ resonance structures. Which of the following name(s) is/are correct for the following compound? a) nitrohydride benzene b) phenylamine c) phenylamide d) aniline e) nitrogenhydrogen benzene f) All of the above is correct Convert 1,4-dimethylbenzene into its common name. TNT's common name is: ______________________________ Name the following compound using OMP nomenclature: Draw the structure of 2,4-dinitrotoluene. Name the following compound: Which of the following is the correct name for the following compound? a) 3,4-difluorobenzyl bromide b) 1,2-difluorobenzyl bromide c) 4,5-difluorobenzyl bromide d) 1,2-difluoroethyl bromide e) 5,6-difluoroethyl bromide f) 4,5-difluoroethyl bromide Benzyl chloride can be abbreviated Bz-Cl. Benzoic Acid has what R group attached to its phenyl functional group? A single aromatic compound can have multiple names indicating its structure. List the corresponding positions for the OMP system (o-, m-, p-). A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound? a) Cyclohexanol b) Cyclicheptanol c) Phenol d) Methanol e) Bleach f) Cannot determine from the above information Which of the following statements is for the compound, phenol? a) Phenol is a benzene derived compound. b) Phenol can be made by attaching an -OH group to a phenyl group. c) Phenol is highly toxic to the body even in small doses. d) Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane. e) Phenol is used as an antiseptic in minute doses. f) Phenol is amongst one of the three common names retained in the IUPAC nomenclature. False, this compound does not contain a benzene ring in its structure. 3 No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (Figure 1) No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds. 2 b, d p-Xylene 2,4,6-trinitrotoluene p-chloronitrobenzene 4-phenylheptane a False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl. COOH True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene. Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4 The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic. d

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Atoms of an element are not all identical and indivisible as Dalton said. They are made of three mains types of particles. The protons and neutrons together form a very small dense center of the atom, called a . The nucleus contains all the mass of the atom except the small mass of electrons. The electrons move around the nucleus, and occupy a much bigger space than the nucleus, so that most of the atom is empty space. An element is defined by the number of protons, or atomic number, which is equal to the number of electrons in the neutral element. Atoms of an element can have different numbers of neutrons, resulting in different masses. are atoms that have the same number of protons, but different numbers of neutrons. Chemical properties depend mostly on the atomic number, so isotopes are nearly the same chemically. The is the average mass of the atom, including all the different isotopes that are likely to be present. If you want to show what isotope of an element you can use special notation, like this: C. This is read as carbon-12. The 12 is the , or the number of protons + neutrons. Since carbon always has 6 protons, C must have 6 neutrons also. We use some very small units to describe atoms. For instance, the charge of a proton or electron (which we will use as the unit of charge) is 1.602 x 10 coulombs (C). The atomic mass unit, or amu, is 1.661 x 10 g. Sizes of atoms are usually measured in angstroms or Å, which is 1 x 10 m. The diameter of most atoms is 1-5 Å. The diameter of nuclei is roughly 10 Å. Electrons are smaller than nuclei, so most of the atom is empty.

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Why are sharp knives more effective than dull knives (Hint: think about the definition of pressure)? The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively. Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has? Why should you roll or belly-crawl rather than walk across a thinly-frozen pond? Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice. A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa. A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals? 0.809 atm; 82.0 kPa A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals? Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi? 2.2 × 10 kPa During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr and kPa? The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi. Earth: 14.7 lb in ; Venus: 13.1× 10 lb in A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. What is the pressure of this gas in atmospheres and torr? Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in., 1013.9 mbar. (a) 101.5 kPa; (b) 51 torr drop Why is it necessary to use a nonvolatile liquid in a barometer or manometer? The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in: (a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in. Hg, determine the pressure of the gas in: The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg, determine the pressure of the gas in: (a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg, determine the pressure of the gas in: How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers? With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since = + . Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why? Explain how the volume of the bubbles exhausted by a scuba diver ( ) change as they rise to the surface, assuming that they remain intact. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law. One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.” (a) What is the meaning of the term “inversely proportional?” (b) What are the “other things” that must be equal? An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.” (a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure How would the graph in Figure change if the number of moles of gas in the sample used to determine the curve were doubled? How would the graph in Figure change if the number of moles of gas in the sample used to determine the curve were doubled? The curve would be farther to the right and higher up, but the same basic shape. In addition to the data found in , what other information do we need to find the mass of the sample of air used to determine the graph? Determine the volume of 1 mol of CH gas at 150 K and 1 atm, using . 16.3 to 16.5 L Determine the pressure of the gas in the syringe shown in when its volume is 12.5 mL, using: A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can? 3.40 × 10 torr What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr? we must use \(\dfrac{P_1V_1}{T_1} =\dfrac{P_2V_2}{T_2}\) and solve for \(T_1\) \(T_1 = \dfrac{P_1V_1T_2}{P_2V_2}\) Where: \(P_1 = 744\: torr\) \(V_1 = 11.2\: L\) \(P_2 = 744\: torr\) \(V_2 = 13.3\: L\) \(T_2 = 328.15°\: K\) \(\dfrac{(744\: torr)(11.2\: L)(328.15°\: K)}{(744\: torr)(13.3\: L)} = 276°\: K\) 276°K ; 3°C A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure. 12.1 L A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon? A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions? 217 L How many moles of gaseous boron trifluoride, BF , are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF ? 8.190 × 10 mol; 5.553 g Iodine, I , is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I vapor at a pressure of 0.462 atm? 1.) Use the equation \(PV =nRT\) and solve for \(T\) \(T= \dfrac{PV}{nR}\) 2.) convert grams of I to moles of I and convert mL to L \(0.292g\: \ce{I2}\times \dfrac{1\: mole\: \ce{I2}}{253.8g\: \ce{I2}} = 1.15 \times10^{-3}\: moles\: \ce{I2}\) \(73.3\:mL = 0.0733\:L\) 3.) Use these values along with \(R= 0.08206\: \dfrac{atm\:L}{mole\:°K}\) to solve for \(T\) \(T= \dfrac{(0.462\: \cancel{atm})(0.0733\:\cancel{L})}{(1.15\times10^{-3}\: \cancel{moles})(0.08206\: \dfrac{\cancel{atm}\:\cancel{L}}{\cancel{mole}\:°K})} = 359\: °K \) 359°K ; 86°C How many grams of gas are present in each of the following cases? (a) 7.24 × 10 g; (b) 23.1 g; (c) 1.5 × 10 g A high altitude balloon is filled with 1.41 × 10 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr? A cylinder of medical oxygen has a volume of 35.4 L, and contains O at a pressure of 151 atm and a temperature of 25 °C. What volume of O does this correspond to at normal body conditions, that is, 1 atm and 37 °C? 5561 L A large scuba tank ( ) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C? A 20.0-L cylinder containing 11.34 kg of butane, C H , was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C. 46.4 g While resting, the average 70-kg human male consumes 14 L of pure O per hour at 25 °C and 100 kPa. How many moles of O are consumed by a 70 kg man while resting for 1.0 h? For a given amount of gas showing ideal behavior, draw labeled graphs of: For a gas exhibiting ideal behavior: A liter of methane gas, CH , at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H , at STP. Using Avogadro’s law as a starting point, explain why. The effect of chlorofluorocarbons (such as CCl F ) on the depletion of the ozone layer is well known. The use of substitutes, such as CH CH F( ), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP: (a) 1.85 L CCl F ; (b) 4.66 L CH CH F As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 × 10 alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 °C? A balloon that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mount Crumpet? 0.644 atm If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure? If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure? The pressure decreases by a factor of 3. What is the density of laughing gas, dinitrogen monoxide, N O, at a temperature of 325 K and a pressure of 113.0 kPa? 1.) First convert kPa to atm \(113.0\:kPa\times\dfrac{1\:atm}{101.325\:kPa}=1.115\:atm\) 2.) The use the equation \(d=\dfrac{PM}{RT}\) where d = density in g L and M = molar mass in g mol \(d=\dfrac{(1.115\:atm)(44.02\dfrac{g}{\cancel{mol}})}{(0.08206\: \dfrac{\cancel{atm}\:L}{\cancel{mole}\:\cancel{°K}})(325\:\cancel{°K})}=1.84\:\dfrac{g}{L}\) Calculate the density of Freon 12, CF Cl , at 30.0 °C and 0.954 atm. 4.64 g L Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain. A cylinder of O ( ) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder? 38.8 g What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr? 1.) convert torr to atm and °C to °K \(307\:torr=0.404atm\) \(26°C= 300.°K\) 2.) Use the equation \(PV=nRT\) and solve for \(n\) \(n=\dfrac{PV}{RT}\) \(n=\dfrac{(0.404\:\cancel{atm})(0.100\:\cancel{L})}{(0.08206\dfrac{\cancel{atm}\:\cancel{L}}{mol\:\cancel{°K}})(300.\cancel{°K})}=0.00165\:moles\) 3.) Then divide grams by the number of moles to obtain the molar mass: \(\dfrac{0.0494g}{0.00165\:moles}=30.0\dfrac{g}{mole}\) What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr? 72.0 g mol How could you show experimentally that the molecular formula of propene is C H , not CH ? The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula. 88.1 g mol ; PF Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies 125 mL with a pressure of 99.5 kPa at 22 °C? A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO , 805 g O , and 4,880 g N . At 25 degrees C, what is the pressure in the cylinder in atmospheres? 141 atm A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO , 12.0% O , and the remainder N at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.) A sample of gas isolated from unrefined petroleum contains 90.0% CH , 8.9% C H , and 1.1% C H at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.) CH : 276 kPa; C H : 27 kPa; C H : 3.4 kPa A mixture of 0.200 g of H , 1.00 g of N , and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior. Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O is not. If enough O is added to a cylinder of H at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive? Yes A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 × 10 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C? A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 °C. What is the pressure of the carbon monoxide? (See for the vapor pressure of water.) 740 torr In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas? Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO: \(\ce{2HgO}(s)⟶\ce{2Hg}(l)+\ce{O2}(g)\) (a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O produced by decomposition of this amount of HgO; and determine the volume of O from the moles of O , temperature, and pressure. (b) 0.308 L Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: \[\ce{4H2O}(g)+\ce{3Fe}(s)⟶\ce{Fe3O4}(s)+\ce{4H2}(g)\] \[\ce{CCl2F2}(g)+\ce{4H2}(g)⟶\ce{CH2F2}(g)+\ce{2HCl}(g)\] Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN ). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide. Lime, CaO, is produced by heating calcium carbonate, CaCO ; carbon dioxide is the other product. (a) Balance the equation. Determine the grams of CO produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 × 10 L Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C H , and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC , with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons. Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C H , to produce carbon dioxide and water, if the volumes of C H and O are measured under the same conditions of temperature and pressure. 42.00 L What volume of O at STP is required to oxidize 8.0 L of NO at STP to NO ? What volume of NO is produced at STP? Consider the following questions: (a) 18.0 L; (b) 0.533 atm Methanol, CH OH, is produced industrially by the following reaction: \[\ce{CO}(g)+\ce{2H2}(g)\xrightarrow{\textrm{ copper catalyst 300 °C, 300 atm }}\ce{CH3OH}(g)\] Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume. What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO to BaO and O ? 10.57 L O A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N and 1.25 L of O at STP. What is the colorless gas? Ethanol, C H OH, is produced industrially from ethylene, C H , by the following sequence of reactions: \[\ce{3C2H4 + 2H2SO4⟶C2H5HSO4 + (C2H5)2SO4}\] \[\ce{C2H5HSO4 + (C2H5)2SO4 + 3H2O⟶3C2H5OH + 2H2SO4}\] What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%? 5.40 × 10 L One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin? A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.) XeF One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (−NH ) in protein material are allowed to react with nitrous acid, HNO , to form N gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH (NH )COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample? A balloon filled with helium gas is found to take 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%? 4.2 hours Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of . Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses. Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: \(\mathrm{\left(\dfrac{rate\: A}{rate\: B}\right)=\left(\dfrac{molar\: mass\: of\: B}{molar\: mass\: of\: A}\right)^{1/2}}\). Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy: Heavy water, D O (molar mass = 20.03 g mol ), can be separated from ordinary water, H O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H O and D O. Which of the following gases diffuse more slowly than oxygen? F , Ne, N O, C H , NO, Cl , H S F , N O, Cl , H S During the discussion of gaseous diffusion for enriching uranium, it was claimed that UF diffuses 0.4% faster than UF . Show the calculation that supports this value. The molar mass of UF = 235.043930 + 6 × 18.998403 = 349.034348 g/mol, and the molar mass of UF = 238.050788 + 6 × 18.998403 = 352.041206 g/mol. Calculate the relative rate of diffusion of H (molar mass 2.0 g/mol) compared to that of H (molar mass 4.0 g/mol) and the relative rate of diffusion of O (molar mass 32 g/mol) compared to that of O (molar mass 48 g/mol). 1.4; 1.2 A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH Cl forms where gaseous NH and gaseous HCl first come into contact. (Hint: Calculate the rates of diffusion for both NH and HCl, and find out how much faster NH diffuses than HCl.) 51.7 cm Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer. Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows: The distribution of molecular velocities in a sample of helium is shown in . If the sample is cooled, will the distribution of velocities look more like that of H or of H O? Explain your answer. H O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier. What is the ratio of the average kinetic energy of a SO molecule to that of an O molecule in a mixture of two gases? What is the ratio of the root mean square speeds, , of the two gases? A 1-L sample of CO initially at STP is heated to 546 °C, and its volume is increased to 2 L. (a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to \(\sqrt{2}\) times its initial value; is proportional to \(\mathrm{KE_{avg}}\). The root mean square speed of H molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N molecule at 25 °C? Answer the following questions: (a) equal; (b) less than; (c) 29.48 g mol ; (d) 1.0966 g L ; (e) 0.129 g/L; (f) 4.01 × 10 g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases? Gases C, E, and F Explain why the plot of for CO differs from that of an ideal gas. Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain. The gas behavior most like an ideal gas will occur under the conditions in (b). Molecules have high speeds and move through greater distances between collision; they also have shorter contact times and interactions are less likely. Deviations occur with the conditions described in (a) and (c). Under conditions of (a), some gases may liquefy. Under conditions of (c), most gases will liquefy. Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas. For which of the following gases should the correction for the molecular volume be largest: CO, CO , H , He, NH , SF ? SF A 0.245-L flask contains 0.467 mol CO at 159 °C. Calculate the pressure: Answer the following questions: (a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures they behave close enough to ideal gases that they are approximated as such, however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the van der Waals equation (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see ) (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See . (e) low temperatures Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases? Gases C, E, and F Explain why the plot of for CO differs from that of an ideal gas. Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain. (a) high pressure, small volume (b) high temperature, low pressure (c) low temperature, high pressure The gas behavior most like an ideal gas will occur under the conditions in (b). Molecules have high speeds and move through greater distances between collision; they also have shorter contact times and interactions are less likely. Deviations occur with the conditions described in (a) and (c). Under conditions of (a), some gases may liquefy. Under conditions of (c), most gases will liquefy. Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas. For which of the following gases should the correction for the molecular volume be largest: CO, CO , H , He, NH , SF ? SF A 0.245-L flask contains 0.467 mol CO at 159 °C. Calculate the pressure: (a) using the ideal gas law (b) using the van der Waals equation (c) Explain the reason for the difference. (d) Identify which correction (that for P or V) is dominant and why. Answer the following questions: (a) If XX behaved as an ideal gas, what would its graph of Z vs. P look like? (b) For most of this chapter, we performed calculations treating gases as ideal. Was this justified? (c) What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram. (d) What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram. (e) In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas? (a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures they behave close enough to ideal gases that they are approximated as such, however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the van der Waals equation (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see ) (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See . (e) low temperatures

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are long chain, giant organic molecules are assembled from many smaller molecules called . Polymers consist of many repeating monomer units in long chains, sometimes with or between the chains. A polymer is analogous to a necklace made from many small beads (monomers). A chemical reaction forming polymers from monomers is called , of which there are many types. A common name for many synthetic polymer materials is plastic, which comes from the Greek word "plastikos", suitable for molding or shaping. In the following illustrated example, many monomers called styrene are polymerized into a long chain polymer called polystyrene. The squiggly lines indicate that the polymer molecule extends further at both the left and right ends. In fact, polymer molecules are often hundreds or thousands of monomer units long. Many objects in daily use from packing, wrapping, and building materials include half of all polymers synthesized. Other uses include textiles, many electronic appliance casings, CD's, automobile parts, and many others are made from polymers. A quarter of the solid waste from homes is plastic materials - some of which may be recycled as shown in the table below. Some products, such as adhesives, are made to include monomers which can be polymerized by the user in their application. There are many types of polymers including synthetic and natural polymers. Plastics that soften when heated and become firm again when cooled. This is the more popular type of plastic because the heating and cooling may be repeated and the thermoplastic may be reformed. These are plastics that soften when heated and can be molded, but harden permanently. They will decompose when reheated. An example is Bakelite, which is used in toasters, handles for pots and pans, dishes, electrical outlets and billiard balls.

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Esters are known for their distinctive odors and are commonly used for food aroma and fragrances. The general formula of an ester is RCOOR'. Esters are formed through reactions between an acid and an alcohol with the elimination of water. An example of this is the reaction of acetic acid with an alcohol, which yields an acetic ester and water. The part enclosed by the red circle represents the ethyl group from the alcohol and the part enclosed by the green rectangle represents the acetate group from the acid. Esters are named as if the alkyl chain from the alcohol is a substituent. No number is assigned to this alkyl chain. This is followed by the name of the parent chain from the carboxylic acid part of the ester with an remove and replaced with the ending . Esters are formed through reactions between an acid and an alcohol with the elimination of water. An example 1. First, identify the oxygen that is part of the continuous chain and bonded to carbon on both sides. (On one side of this oxygen there will be a carbonyl present but on the other side there won't be.) 2. Second, begin numbering the carbon chains on either side of the oxygen identified in step 1. 3. Next, use this format: [alkyl on side further from the carbonyl] (space) [alkane on the side with the carbonyl] - (In this case: [methyl] [methane]) 4. Finally, change the ending of the alkane on the same side as the carbonyl from -e to -oate. (In this case: methyl methanoate) When an ester group is attached to a ring, the ester is named as a substituent on the ring. Other substituents that exist on either side of the ester are named in the same way as they are on regular alkane chains. The only thing you must make sure of is placing the substituent name on the part of the name that corresponds to the side of the ester that it is on. CH COOCH CH CH COOCH CH CH CH CH CH CH CH Name the following: Answer: propyl ethanoate Answer: 3-bromopentyl 2-chlorobutanoate Answer: ethyl hexanoate Answer: ethyl 3-bromopentanoate Answer: 4-nitrobenzenecarboxylic acid or 4-nitrobenzoic acid )

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Sand, salt, and chalk dust are made up of chunks of solid particles, each containing huge numbers of molecules. You can usually see the individual particles directly, although the smallest ones might require some magnification. At the opposite end of the size scale, we have individual molecules which dissolve in liquids to form homogeneous solutions. There is, however, a vast but largely hidden world in between: particles so tiny that they cannot be resolved by an optical microscope, or molecules so large that they begin to constitute a phase of their own when they are suspended in a liquid. This is the world of which we will survey in this lesson. As you will see, we encounter colloids in the food we eat, the consumer products we buy... and we ourselves are built largely of colloidal materials. Colloids occupy an intermediate place between [particulate] suspensions and solutions, both in terms of their observable properties and particle size. In a sense, they bridge the microscopic and the macroscopic. As such, they possess some of the properties of both, which makes colloidal matter highly adaptable to specific uses and functions. Colloid science is central to biology, food science and numerous consumer products. Colloidal particles need not fall within the indicated size range in all three dimensions; thus fibrous colloids such as many biopolymers may be very extended sizes along one direction. To begin, you need to recall two important definitions: But imagine that you are able to shrink your view of a solution of sugar in water down to the sub-microscopic level at which individual molecules can be resolved: you would see some regions of space occupied by H O molecules, others by sugar molecules, and likely still others in which sugar and H O molecules are momentarily linked together by hydrogen bonding— not to mention the void spaces that are continually appearing and disappearing between molecules as they are jumbled about by thermal motions. As with so many simple definitions, the concept of (and thus of a ) breaks down as we move from the macro-scale into the molecular scale. And it is the region in between these two extremes that constitutes the realm of the colloid. What makes colloidal particles so special is not so much their sizes as it is the manner in which their surface areas increase as their sizes decrease. If we take a sample of matter and cut it up into smaller and smaller chunks, the total surface area will increase very rapidly. Although mass is conserved, surface area is not; as a solid is sliced up into smaller bits, more surfaces are created. These new surfaces are smaller, but there are many more of them; the ratio of surface area to mass can become extremely large. The total surface area increases as the inverse cube of the the face length, so as we make our slices still smaller, the total surface area grows rapidly. In practical situations with real colloids, surface areas can reach hectares (or acres) per mole! Why do we focus so much attention on surface area? The general answer is that surfaces (or more generally, between phases) possess physical and chemical properties of their own. In particular, In normal "bulk" matter, these properties are mostly hidden from us owing to the small amount of surface area in relation to the quantity of matter. But as the particle size diminishes, surface phenomena begin to dominate their properties. The small sizes of colloidal solids allows the properties of their surfaces to dominate their behavior. Colloidal matter commonly exists in the form of colloidal-sized phases of solids, liquids, or gases that are uniformly dispersed in a separate (sometimes called the ) which may itself be a solid, liquid, or gas. Colloids are often classified and given special names according to the particular kinds of phases involved. gas Notes on this table: Very large polymeric molecules such as proteins, starches and other biological polymers, as well as many natural polymers, exhibit colloidal behavior. There is no clear point at which a molecule becomes sufficiently large to behave as a colloidal particle. Colloidal dispersions behave very much like solutions in that they appear to be homogeneous on a macroscopic scale. They are often said to be . The most important feature that distinguishes them from other particulate matter is that: Colloids dispersed in liquids or gases are sufficiently small that they do not settle out under the influence of gravity. This, together with the their small sizes which allows them to pass through most filters, makes it difficult to separate colloidal matter from the phase in which it is dispersed. Colloidal dispersions are distinguished from true solutions by their light-scattering properties. The nature of this scattering depends on the ratio of the particle size in the medium to the wavelength of the light. A collimated beam of light passing through a solution composed of ordinary molecules ( ) tends retain its shape. When such a beam is directed through a colloidal dispersion, it spreads out ( ).→ John Tyndall discovered this effect in 1869. (as it is commonly known) scatters all wavelengths equally. This is in contrast to , which scatters shorter wavelengths more, bringing us blue skies and red sunsets. Tyndall scattering can be seen even in dispersions that are transparent. As the density of particles (or the particle size) increases, the light scattering may become great enough to produce a "cloudy" effect, as in this image of a smoke-filled room. This is the reason that milk, fog, and clouds themselves appear to be white. The individual water droplets in clouds (or the butterfat droplets in milk) are actually transparent, but the intense light scattering disperses the light in all directions, preventing us from seeing through them. Colloidal particles are, like molecules, too small to be visible though an ordinary optical microscope. However, if one looks in a direction perpendicular to the light beam, a colloidal particle will "appear" over a dark background as a tiny speck due to the Tyndall scattering. A microscope specially designed for this application is known as an . Bear in mind that the ultramicroscope (invented in Austria in 1902) does not really allow us to "see" the particle; the scattered light merely indicates where it is at any given instant. If you observe a single colloidal particle through the ultramicroscope, you will notice that it is continually jumping around in an irregular manner. These movements are known as Brownian motion. Scottish botanist Robert Brown discovered this effect in 1827 when observing pollen particles floating in water through a microscope. (Pollen particles are larger than colloids, but they are still small enough to exhibit some Brownian motion.) It is worth noting that Albert Einstein's analysis of Brownian motion in 1901 constituted the first proof of the molecular theory of matter. Brownian motion arises from collisions of the liquid molecules with the solid particle. For large particles, the millions of collisions from different directions cancel out, so they remain stationary. The smaller the particle, the smaller the number of surrounding molecules able to collide with it, and more likely that random fluctuations will occur in the number of collisions from different sides. Simple statistics predicts that every once in a while, the imbalance in collisions from different directions will become great enough to give the particle a real kick! In general, differences in electric potential exist between all phase boundaries. If you have studied electrochemistry, you will know that two dissimilar metals in contact exhibit a "contact potential", and that similar potential differences exist between a metal and a solution in which it is immersed. But this principle extends well beyond ordinary electrochemistry; there are small potential differences even at the water-glass interface in a drinking glass, and the water-air interface above it. Colloids are no exception to this rule; there is always a difference in electric potential between the colloid "phase" and that of the surrounding liquid. Even if the liquid consists of pure water, the polar H O molecules at the colloid's surface are likely to be predominantly oriented with either their oxygen (negative) or hydrogen (positive) ends facing the interface, depending on the electrical properties of the colloid particle itself. Interfacial electrical potential differences can have a variety of origins: Charged colloidal particles will attract an excess of oppositely-charged to their vicinity from the bulk solution, forming a localized "cloud" of compensating charge around each particle. The entire assembly is called an Electric double layers of one kind or another exist at all phase boundaries, but those associated with colloids are especially important. What keeps the colloidal particles suspended in the dispersion medium? How can we force the particles to settle out? These are very important practical matters: You will recall that weak attractive forces act between matter of all kinds. These are known generally as and forces, and they only "take hold" at very close distances. Countering these is the that acts at even shorter distances, but is far stronger; it is the basic reason why two atoms cannot occupy the same space. For very small atomic and molecular sized particles, another thing that keeps them apart is thermal motion. Thus when two molecules in a gas collide, they do so with more than enough kinetic energy to overcome the weak attractive forces between them. As the temperature of the gas is reduced, so is the collisional energy; below its boiling point, the attractive forces dominate and the gas condenses into a liquid. Electrical forces help keep colloids dispersed When particles of colloidal dimension suspended in a liquid collide with each other, they do so with much smaller kinetic energies than is the case for gases, so in the absence of any compensating repulsion forces, we might expect van der Waals or dispersion attractions to win out. This would quickly result in the growth of aggregates sufficiently large to exceed colloidal size and to fall to the bottom of the container. This process is called . So how do stable dispersions such as sols manage to survive? In the preceding section, we saw that each particle with its double layer is more or less electrically neutral. However, when two particles approach each other, each one "sees" mainly the outer part [shown here in blue] of the double layer of the other. These will always have the same charge sign (which depends on the type of colloid and the nature of the medium), so there will be an electrostatic repulsive force that opposes the dispersion force attractions. Electrostatic (coulombic) forces have a strong advantage in this respect because they act over much greater distances do van der Waals forces. But as we will see further on, electrostatic repulsion can lose its effectiveness if the ionic concentration of the medium is too great, or if the medium freezes. Under these conditions, there are other mechanisms that can stabilize colloidal dispersions. Colloids can be divided into two general classes according to how the particles interact with the dispersions medium (often referred to as the "solvent"). In one class of colloids, called ("solvent loving") colloids, the particles contain chemical groups that interact strongly with the solvent, creating a sheath of solvent molecules that physically prevent the particles from coming together. Ordinary is a common example of a lyophilic colloid. It is in fact , since it forms strong hydrogen bonds with water. When you mix Jell-O or tapioca powder to make a gelatine dessert, the material takes up water and forms a stable colloidal gel. Lyophilic (hydrophilic) colloids are very common in biological systems and in foods. Most of the colloids in manufactured products exhibit very little attraction to water: think of oil emulsions or glacially-produced rock dust in river water. These colloids are said to be . Lyophobic colloids are all inherently unstable; they will eventually . However, "eventually" can be a very long time (the settling time for some clay colloids in the ocean is 200-600 years!). For systems in which coagulation proceeds too rapidly, the process can be slowed down by adding a stabilizer. Stabilizers can act by coating the particles with a protective layer such as a polymer as described immediately below, or by providing an ion that is selectively adsorbed by the particle, thereby surrounding it with a charged sheath that will repel similar particles it collides with. Dispersions of these colloids are stabilized by electrostatic repulsion between the electric double layers surrounding the particles which we discussed in the preceding section. has unwittingly been employed since ancient times through the use of natural gums to stabilize pigment particles in inks, paints, and pottery glazes. These gums are also widely used to stabilize foods and personal care products. A lyophobic colloid can be made to masquerade as lyophilic by coating it with something that itself possesses suitable lyophilic properties. Alternatively, attaching a lyophobic material to a colloid of any type can surround the particles with a protective shield that physically prevents the particles from approaching close enough to join together. This method usually employs synthetic polymers and is often referred to as . , which can be tailor-made for specific applications, are now widely employed for both purposes. The polymer can be attached to the central particle either by simple adsorption or by chemical bond formation. and are basically the same thing. Surfactants that serve as cleaning agents are commonly called (from L. "to wipe away, cleanse"). Surfactants are molecules consisting of a hydrophylic "head" connected to a hydrophobic chain. Because such molecules can interact with both "oil" and water phases, they are often said to be . Typical of these is the well known cleaning detergent ("sodium laurel sulfate") CH (CH ) OSO Na . Emulsions are inherently unstable; left alone, they tend to separate into "oil" and "water" phases. Think of a simple salad dressing made by shaking vegetable oil and vinegar. When a detergent-like molecule is employed to stabilize an emulsion, it is often referred to as an . The resulting structure (left) is known as a . Emulsifiers are essential components of many foods. They are widely employed in pharmaceuticals, consumer goods such as lotions and other personal care products, paints and printing inks, and numerous industrial processes. The "dirt" we are trying to remove consists of oily or greasy materials whose hydrophobic nature makes them resistant to the action of pure water. If the water contains amphiphilic molecules such as soaps or cleaning detergents that can embed their hydrophobic ends in the particles, the latter will present a hydrophilic interface to the water and will thus become "solubilized". Soaps and detergents can also disrupt the cell membranes of many types of bacteria, for which they serve as . However, they are generally ineffective against viruses, which do not possess cell membranes. Oils and fats are important components of our diets, but being insoluble in water, they are unable to mix intimately with the aqueous fluid in the digestive tract in which the digestive enzymes are dissolved. In order to enable the lipase enzymes (produced by the pancreas) to break down these lipids into their component fatty acids, our livers produce a mixture of surfactants known as . The great surface area of the micelles in the resulting emulsion enables efficient contact between the lipase enzymes and the lipid materials. The liver of the average adult produces about 500 mL of bile per day. Most of this is stored in the gall bladder, where it is concentrated five-fold by removal of water. As partially-digested material exits the stomach, the gall bladder squeezes bile into the top of the small intestine (the ). In addition to its action as a detergent (which also aids in the destruction of bacteria that may have survived the high acidity of the gastric fluid), the alkaline nature of the bile salts neutralizes the acidity of the stomach exudate. The bile itself consists of of salts of a variety of bile acids, all of which are derived from cholesterol. The cholesterol-like part of the structure is hydrophobic, while the charged end of the salt is hydrophilic. Ordinary emulsions are inherently unstable; they do not form spontaneously, and once formed, the drop sizes are sufficiently large to scatter light, producing a milky appearance. As time passes, the average drop size tends to increase, eventually resulting in gravitational separation of the phases. Microemulsions, in contrast, are thermodynamically stable and can form spontaneously. The drop radii are at the very low end of the colloidal scale, often 100 nm or smaller. This is too small to appreciably scatter visible light, so microemulsions appear visually to be homogenous systems. Microemulsions require the presence of one or more surfactants which increase the flexibility and stability of the boundary regions. This allows them to vary form smaller micelles than surface tension forces would ordinarily allow; in some cases they can form sponge-like bicontinuous mixtures in which "oil" and "water" phases extend throughout the mixture, affording more contact area between the phases. The uses of microemulsions are quite wide-ranging, with drug delivery, polymer synthesis, enzyme-assisted synthesis, coatings, and enhanced oil recovery being especially prominent. Particles of colloidal size can be made in two general ways: processes all require an input of energy as new surfaces are created. For , this is usually accomplished by some kind of grinding process such as in a ball- or roller-mill. Solids and liquids can also be broken into colloidal dimensions by injecting them into the narrow space between a rapidly revolving shaft and its enclosure, thus subjecting them to a strong that tends to pull the two sides of a particle in opposite directions. The application of ultrasound (at about 20 kHz) to a mixture of two immiscible can create liquid-in-liquid dispersions; the process is comparable to what we do when we shake a vinegar-and-oil salad dressing in order to create a more uniform distribution of the two liquids. Numerous methods exist for building colloidal particles from sub-colloidal entities. \[S_2O_3^{2–} + H_2O \rightarrow S + SO_4^{2–} + 2 H^+ + 2 e^– \] \[2 Fe^{3+} + 3 H_2O \rightarrow Fe_2O_3 + 6 H^+\] \[Fe^{3+} + 2 H_2O \rightarrow FeO(OH) + 3 H^+\] That oil-in-vinegar salad dressing you served at dinner the other day has now mostly separated into two layers, with unsightly globs of one phase floating in the other. This is surface chemistry in action! Emulsions are fundamentally unstable because molecules near surfaces (i.e., interfaces between phases) are no longer surrounded by their own kind on all sides. The resulting repulsions between like and unlike exact an energetic cost that must eventually be repaid through processes that reduce the interfacial area. The consequent breakup of the emulsion can proceed through various stages: The time required for these processes to take place is highly variable, and can be extended by the presence of substances. Thus , an emulsion of butterfat in water, is stabilized by some of its natural components. The processes described above that allow colloids to remain suspended sometimes fail when conditions change, or equally troublesome, they work entirely too well and make it impossible to separate the colloidal particles from the medium; this is an especially serious problem in wastewater settling basins associated with sewage treatment and operations such as mining and the pulp-and-paper industries. is the general term that refers to the "breaking" of dispersions so that the colloidal particles can be collected, usually by settling out. The term is often used as a synonym for coagulation, but it is more properly reserved for a special method of effecting coagulation which is described further on. Most coagulation processes act by disrupting the outer (diffuse) part of the electric double layer that gives rise to the electrostatic repulsion between them. Have you ever encountered milk that had previously been frozen? Not likely something you would want to drink! You will see "Do not freeze" labels on many foodstuffs and on colloidal consumer products such as latex house paint. Freezing disrupts the double layer by causing the ions within it to combine into neutral species so that the particles can now approach closely enough for attractive forces to take over, and once they do so, they never let go: coagulation is definitely an irreversible process! Coagulation of water-suspended dispersions can be brought about by raising the ionic concentration of the medium. The added ions will migrate to the oppositely-charged regions of the double layer, thus neutralizing its charges; this effectively reduces the thickness of the double layer, eventually allowing the attractive forces to prevail. Rivers carry millions of tons of colloidal clay into the oceans. If you fly over the mouth of a river such as the Mississippi (shown here in a satellite image), you can sometimes see the difference in color as the clay colloids coagulate due to the action of the salt water. The coagulated clay accumulates as sediments which eventually form a geographical feature called a river delta. A liquid phase dispersed in a solid medium is known as a , but this formal definition does not always convey the full sense of the nature of the "solid". The latter may start out as a powdery or granulated material such as natural gelatin or a hydrophilic polymer, but once the gel has formed, the "solid" part is less a "phase" than a cross-linked network that extends throughout the volume of the liquid, whose quantity largely defines the volume of the entire gel. Hydrogels can contain up to 90% water by weight Most of the gels we commonly encounter have water as the liquid phase, and thus are called ; ordinary gelatin deserts are well known examples. The "solid" components of hydrogels are usually polymeric materials that have an abundance of hydrophilic groups such as hydroxyl (–OH) that readily hydrogen-bond to water and also to each other, creating an irregular, flexible, and greatly-extendable network. These polymers are sometimes synthesized for this purpose, but are more commonly formed by processing natural materials, including natural polymers such as cellulose. Gels are essential components of a huge variety of consumer products ranging from thickening agents in foods and personal care products to cushioning agents in running shoes. You may have noticed that a newly-opened container of yogurt or sour cream appears to be smooth and firm, but once some of the material has been spooned out, little puddles of liquid appear in the hollowed-out depressions. As the spoon is plunged into the material, it pulls the nearby layers of the gel along with it, creating a shearing action that breaks it apart, releasing the liquid. Anyone who has attacked an egg yolk with a cook's whisk, written with a ball-point pen, or spread latex paint on a wall has made use of this phenomenon which is known as . The interior (the cytoplasm) of each cell in the soft tissues of our bodies consists of a variety of inclusions (organelles) suspended in a gel-like liquid phase called the cytosol. Dissolved in the cytosol are a variety of ions and molecules varying from the small to the large; among the latter, proteins and carbohydrates make up the "solid" portion of the gel structure. Embedded within the cytosol is the filament-like which controls the overall shape of the cell and holds the organelles in place. (In free-living cells such as the amoeba, changes in the cytoskeleton enable the organism to alter its shape and move around to engulf food particles.) Be thankful for the gels in your body; without them, you would be little more than a bag of gunge-filled liquid, likely to end up as a puddle on the floor! The individual cells are bound into tissues by the extracellular matrix (ECM) which — on a much larger scale, holds us together and confers an overall structure and shape to the body. The ECM is made of a variety of structural fibers (collagens, elastins) embedded in a gel-like matrix. The usefulness of many industrial and consumer products is strongly dependent on their viscosity and flow properties. Toothpastes, lotions, lubricants, coatings are common examples. Most of the additives that confer desirable flow properties on these products are colloidal in nature; in many cases, they also provide stabilization and prevent phase separation. Since ancient times, various natural gums have been employed for such purposes, and many remain in use today. More recently, manufactured materials whose properties can be tailored for specific applications have become widely available. Examples are colloidal microcrystalline cellulose, carboxymethyl cellulose, and fumed silica. is a fine (5-50 nm), powdery form of SiO of exceptionally low bulk density (as little as 0.002 g cm ); the total surface area of one Kg can be as great as 60 hectares (148 acres). It is made by spraying SiCl (a liquid) into a flame. It is used as a filler, for viscosity and flow control, a gelling agent, and as an additive for strenghthening concrete. Most of the foods we eat are largely colloidal in nature. The function of food colloids generally has less to do with nutritional value than appearance, texture, and "mouth feel". The latter two terms relate to the flow properties of the material, such as spreadability and the ability to "melt" (transform from gel to liquid emulsion) on contact with the warmth of the mouth. is basically an emulsion of lipid oils ("butterfat") dispersed in water and stabilized by phospholipids and proteins. Most of the protein content of milk consists of a group known as which aggregate into a complex micellar structure which is bound together by calcium phosphate units. Homogenizer The stabilizers present in fresh milk will maintain its uniformity for 12-24 hours, but after this time the butterfat globules begin to coalesce and float to the top ("creaming"). In order to retard this process, most milk sold after the early 1940's undergoes in which the oil particles are forced through a narrow space under high pressure. This breaks up the oil droplets into much smaller ones which remain suspended for the useful shelf life of the milk. Before homogenization become common, milk bottles commonly had enlarged tops ↑ to make it easier to skim off the cream that would separate out. The structures of cream, yogurt and ice cream are dominated by the casein aggregates mentioned above. is a complex mixture of several colloid types: Whereas milk is an oil (butterfat)-in-water dispersion, and have a "reversed" (water-in-oil) arrangement. This transformation is accomplished by subjecting the butterfat droplets in cream to violent agitation ( ) which forces the droplets to coalesce into a semisolid mass within which remnants of the water phase are embedded. The greater part of this phase ends up as the by-product . A detailed study of eggs and their many roles in cooking can amount to a mini-course in colloid chemistry in itself. There is something almost magical in the way that the clear, viscous "white" of the egg can be transformed into a white, opaque semi-solid by brief heating, or rendered into more intricate forms by poaching, frying, scrambling, or baking into custards, soufflés, and meringues, not to mention tasty omelettes, quiches, and more exotic delights such as the (Arabic) and (Persian) dishes of the Middle-East. The raw eggwhite is basically a colloidal sol of long-chain protein molecules, all curled up into compact folded forms due to hydrogen bonding between different parts of the same molecule. Upon heating, these bonds are broken, allowing the proteins to unfold. The denuded chains can now tangle and bind to each other, transforming the sol into a cross-linked hydrogel, now so dense that scattered light changes its appearance to opaque white. What happens next depends very much on the skill of the cook. The idea is to drive out enough of the water entrapped within the gel network to achieve the desired density while retaining enough gel structure to prevent it from forming a rubbery mass, as usually happens with hard-boiled eggs. This is especially important when the egg structure is to be incorporated into other food components as in baked dishes. The key to all this is temperature control; the eggwhite proteins begin to coagulate at 65°C and if yolk proteins are present, the mixture is nicely set at about 73°; by 80° the principal (albumin) protein has set, and at much above this the gel network will collapse into an overcooked mass. The temperature limit required to avoid this disaster can be raised by adding milk or sugar; the water part of the milk dilutes the proteins, while sugar molecules hydrogen-bond to them, forming a protective shield that keeps the proton strand separated. This is essential when baking custards, but incorporating a bit of cream into scrambled eggs can similarly help them retain their softness. The other colloidal personalities eggs can display are liquid and solid foams. Instead of applying heat to unfold the proteins, we "beat" them; the shearing force of a whisk or egg-beater helps pull them apart, and the air bubbles that get entrapped in the mixture attract the hydrophobic parts of the unfolded proteins and help hold them in place. Sugar will stabilize the foam by raising its viscosity, but will interfere with protein folding if added before the foam is fully formed. Sugar also binds the residual water during cooking, retarding its evaporation until after the proteins not broken up by beating can be thermally coagulated. have been used since ancient times for both protective and decorative purposes. They consist basically of pigment particles dispersed in vehicle — a liquid capable for forming a stable solid film as the paint "dries". The earliest protective coatings were made by dissolving plant-derived natural polymers (resins) in an oil such as that of linseed. The double-bonds in these oils tends to oxidize when exposed to air, causing it to polymerize into an impervious film. The colloidal pigments were stabilized with naturally-occurring surfactants such as polysaccharide gums. Present-day paints are highly-engineered products specialized for particular industrial or architectural coatings and for marine or domestic use. For environmental reasons, water-based ("latex") vehicles are now preferred. The most critical properties of inks relate to their drying and surface properties; they must be able to flow properly and attach to the surface without penetrating it — the latter is especially critical when printing on a porous material such as paper. Many inks consist of organic dyes dissolved in a water-based solvent, and are not colloidal at all. The ink used in printing newspapers employs colloidal carbon black dispersed in an oil vehicle. The pressure applied by the printing press forces the vehicle into the pores of the paper, leaving most of the pigment particles on the surface. The inks employed in are gels, formulated in such a way that the ink will only flow over the ball and onto the paper when the shearing action of the ball (which rotates as it moves across the paper) "breaks" the gel into a liquid; the resulting liquid coats the ball and is transferred to the paper. As in conventional printing, the pigment particles remain on the paper surface, while the liquid is pressed into the pores and gradually evaporates. Turbidities of 5, 50, and 500 units. Water, whether intended specifically for drinking, or wastewaters such as sewage or from industrial operations such as from pulp-and-paper manufacture (most of which are likely to end up being re-used elsewhere) usually contains colloidal matter that cannot be removed by ordinary sand filters, as evidenced by its turbidity. Even "pristine" surface waters often contain suspended soil sediments that can harbor infectious organisms and may provide them with partial protection from standard disinfection treatments. The sulfates of aluminum (alum) and of iron(III) have long been widely employed for this purpose. Synthetic polymers tailored specifically for these applications have more recently come into use. The usual method of removing turbidity is to add a flocculating agent (flocculant). These are most often metallic salts that can form gel-like hydroxide precipitates, often with the aid of added calcium hydroxide (quicklime) if pH of the water must be raised. The flocculant salts neutralize the surface charges of the colloids, thus enabling them to coagulate; these are engulfed and trapped by fragments of gelatinous precipitate, which are drawn together into larger aggregates by gentle agitation until they become sufficiently large to form flocs which can be separated by settling or filtration. The four major components of soils are mineral sediments, organic matter, water, and air. The water is primarily adsorbed to the mineral and organic materials, but may also share pore spaces with air; pore spaces constitute about half the bulk volume of typical solid. The principal colloidal components of soils are mineral sediments in the form of clays, and the humic materials in the organic matter. In addition to influencing the consistency of soil by binding water molecules, soil colloids play an essential role in storing and exchanging the mineral ions required by plants. These are layered structures based on alumino-silicates or hydrous oxides, mostly of iron or aluminum. Each layer is built of two or three sheets of extended silica or alumina structures linked together by shared oxygen atoms. These layers generally have an overall negative charge owing to the occasional replacement of a Si ion by one of Al . Adjacent layers are separated by a region of adsorbed cations (to neutralize the negative charges) and water molecules, and thus are held together relatively loosely. It is these interlayer regions that enable clays to work their magic by exchanging ions with both the soil water and the roots of plants. The principal organic components of soil are complex substances of indeterminate structure that present –OH and –COOH groups which become increasingly dissociated as the pH increases. This allows them to bind and exchange cations in much the same way as described above.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.03%3A_Rheology

Viscosity measurements are the realm of a field of science called . Rheology is, literally, the study of flow. Another very simple definition, attributed to chemical engineer Chris Macosko at University of Minnesota, is the study of "what happens when you squish stuff". There's an element of force or pressure that comes into play here, too. One of the common ways of assessing properties in rheology is to place a sample between two parallel plates and move one plate with respect to the other. One plate says still and the other one moves. What happens to the liquid between the plates? There ought to be some friction between the stationary plate and the liquid that will keep the liquid still. There also ought to be some friction between the moving plate and the liquid that will make the liquid move along at the same speed as that plate. So at one extreme, the liquid is moving along with the sliding plate and at the other extreme the liquid is perfectly still. If we imagine that the liquid in between these two extremes is divided into very thin layers, then each layer must be moving at a slightly different speed than the next. In the diagram, the symbol, , stands for the speed of the layer of liquid. The arrow beside the layer is meant to convey its relative speed: the top layer is moving the fastest, the next layer is a little slower, and so on; the bottom layer isn't moving at all. There is an important quantity, called the strain rate or shear rate (given as a symbol the Greek letter gamma, γ, with a dot on top) that describes how the speed of the liquid moving along the x-axis (left to right) changes, layer by layer, in the y-direction. What does this picture have to do with viscosity? Well, in order to get that top plate to move, we have to apply some sort of force to it. In rheology the force is given per unit area (like pressure); this force per unit area is described as the stress (given as a symbol the Greek letter sigma, σ). What if we wanted to slide the plate even faster? What would happen to the liquid? Well, the shear rate would go up. The top layer would be moving even faster, and the bottom layer would still be stationary. And what would we have to do to get that faster shear rate? We would have to push the top plate a little harder. The result is a graph that looks like this: The graph says that shear stress and shear rate are directly related. The stress is called "shear stress" because of the direction of the force parallel to the liquid, causing shear strain. When we increase the shear rate, the shear stress also increases proportionally. So, if we did this experiment using water as the liquid between the plates, we would get a linear relationship between shear stress and shear rate. That's how Sir Isaac Newton described the behavior of liquids, so when we see that linear relationship between stress and strain, we describe the liquid as a . What happens if we do the experiment with honey instead of water? What would happen, for example, if we exerted a certain stress to get a certain shear rate with the water, and we wanted to get the same shear rate using honey? Well, the honey is thicker than the water. It's not going to move as easily. We will have to apply a greater stress in order to achieve the same shear rate that we measured with water. That would be true for every possible shear rate. If we were to graph shear stress vs. shear strain with honey, we would get a linear relationship, just like we did with water, but the value of shear stress would always be higher than for water. The slope of the line for honey would be greater than the slope of the line for water. Viscosity is formally defined as the slope of that line. Honey, being thicker than water, requires higher and higher stresses to achieve the same strain rates that you would get using lower stresses with water. In the simple kitchen analog of the experiment, you have to push the spoon harder to stir the honey than you do if you are just stirring water. Honey and water are Newtonian liquids. The relationship is simple: increase the shear stress and you increase the shear rate. Not all liquids behave that way. Polymer solutions don't, for example. They display a non-linear relationship, in which the shear stresses needed to get higher shear rates are just not as high as you would expect. That behavior is described as . What makes polymer solutions do that? Remember that, in solution, an individual polymer molecule tends to coil up into a ball. Normally, those balls are spherical. At high shear rates, they distort into ellipses. That reduces the amount of drag, making the solution a little less viscous than you would expect. The shear stress still increases with higher shear rates, but not as much as if it were a Newtonian fluid. That's not all. Remember, polymers display chain entanglement, especially when they are very large. Chain entanglement increases the viscosity of the solution. These chains are mobile because the polymer is constantly undergoing conformational changes. As the solution gets sheared, some of the polymer chains may wiggle loose from their neighbors, but random conformational changes will always result in new entanglements right away. But what happens at a very high rate of shear strain? The polymer chains still become untangled, but they don't have time to form new entanglements, because everything is whizzing by too quickly. Without those entanglements, viscosity doesn't increase as sharply as you would expect. Some liquids behave just like water, but polymer solutions do not. That's a consequence of the enormously long chain structures of the polymers. It's worth pointing out that there are other things that don't behave in the "normal" way. Colloids often display shear thickening. You can see examples of this in videos of people running across wading pools filled with corn starch and water, a colloidal mixture in which the solid cornstarch particles are just suspended in the water, not dissolved. If you walk across the pool, you sink. If you run, you stay on top. That's shear thickening, and it has to do with how those solid particles move during shearing. When you walk, they just slide past each other, but when you run, they all collide, and come to a stop. A Bingham plastic, on the other hand, actually appears to be a solid until you give it a good shove; then it flows like a liquid. Peanut butter behaves sort of like that. Honey is a Newtonian fluid, but molasses undergoes shear thinning. What might you deduce about the composition of molasses and honey?

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.02%3A_Property_of_Gases

Why does the average person often overlook the presence of gases? Probably because the properties of gases are so unobtrusive. All gases are transparent, and most are colorless. The major exceptions to the second half of this rule are fluorine, F , and chlorine, Cl , which are pale yellow-green; bromine, Br , and nitrogen dioxide, NO , which are reddish brown; and iodine, I , which is violet. Another important property of all gases is their mobility. Every gas will disperse to fill all space, unless prevented from doing so by a solid or liquid barrier or a force. (The force of earth’s gravity, for example, prevents air from escaping our planet.) Moreover, gases are capable of escaping through small holes ( ) in barriers such as plaster of paris or a balloon, even though the human eye sees such materials as continuous and impenetrable. The mobility of gases is also demonstrated by the minimal resistance they present to objects moving through them. You can wave your hand through air much more easily than you can through any liquid. A third general characteristic of gases is their wide variation in density under various conditions. Densities of solids and liquids change by only a few percent when temperature or pressure is doubled or halved. Similar changes in the conditions of a gas can alter its density by a factor of 2. This occurs because the volume of any gas increases greatly with an increase in temperature or with a reduction in pressure. This third characteristic is related to gases abilities to compress and expand. The variable density of gases is made possible by their ability to change volume. This property of gases makes them very versatile, allowing gases to be compressed for storage or heated and expanded to drive a piston (as in an engine).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.06%3A_Batteries_and_Fuel_Cells

Make sure you thoroughly understand the following essential ideas which have been presented below. One of the oldest and most important applications of electrochemistry is to the storage and conversion of energy. You already know that a converts chemical energy to work; similarly, an converts electrical work into . Devices that carry out these conversions are called . In ordinary batteries the chemical components are contained within the device itself. If the reactants are supplied from an external source as they are consumed, the device is called a . The term derives from the older use of this word to describe physical attack or "beating"; Benjamin Franklin first applied the term to the electrical shocks that could be produced by an array of charged glass plates. In common usage, the term "call" is often used in place of battery. For portable and transportation applications especially, a battery or fuel cell should store (and be able to deliver) the maximum amount of energy at the desired rate (power level) from a device that has the smallest possible weight and volume. The following parameters are commonly used to express these attributes: A or is capable of being recharged; its electrode reactions can proceed in either direction. During charging, electrical work is done on the cell to provide the free energy needed to force the reaction in the non-spontaneous direction. A , as expemplified by an ordinary flashlight battery, cannot be recharged with any efficiency, so the amount of energy it can deliver is limited to that obtainable from the reactants that were placed in it at the time of manufacture. The most well-known storage cell is the lead-acid cell, which was invented by Gaston Planté in 1859 and is still the most widely used device of its type. The cell is represented by \[Pb(s) | PbSO_4(s) | H_2SO_4(aq) || PbSO_4(s), PbO_2(s) | Pb(s)\] and the net cell reaction is \[Pb(s) + PbO_2(s) + 2 H_2SO_4(aq) → 2 PbSO_4(s) + 2 H_2O\] The reaction proceeds to the right during discharge and to the left during charging. The state of charge can be estimated by measuring the density of the electrolyte; sulfuric acid is about twice as dense as water, so as the cell is discharged, the density of the electrolyte decreases. The technology of lead-acid storage batteries has undergone remarkably little change since the late 19th century. Their main drawback as power sources for electric vehicles is the weight of the lead; the maximum energy density is only about 35 Ah/kg, and actual values may be only half as much. There are also a few other problems: The most well-known primary battery has long been the common "dry cell" that is widely used to power flashlights and similar devices. The modern dry cell is based on the one invented by Georges Leclanché in 1866. The electrode reactions are \[Zn → Zn^{2+} + 2e^–\] \[2 MnO_2 + 2H^+ + 2e^– → Mn_2O_3 + H_2O\] Despite its name, this cell is not really "dry"; the electrolyte is a wet paste containing NH Cl to supply the hydrogen ions. The chemistry of this cell is more complicated than it would appear from these equations, and there are many side reactions and these cells have limited shelf-lifes due to self discharge. (In some of the older ones, attack by the acidic ammonium ion on the zinc would release hydrogen gas, causing the battery to swell and rupture, often ruining an unused flashlight or other device.) A more modern version, introduced in 1949, is the which employs a KOH electrolyte and a zinc-powder anode which permits the cell to deliver higher currents and avoids the corrosive effects of the acidic ammonium ion on the zinc. What is the cell notation for the LeClanché dry cell? The most important of these are: Clearly, these are all primarily kinetic and mechanistic factors which require a great deal of experimentation to understand and optimize. Conventional batteries supply electrical energy from the chemical reactants stored within them; when these reactants are consumed, the battery is "dead". An alternative approach would be to feed the reactants into the cell as they are required, so as to permit the cell to operate continuously. In this case the reactants can be thought of as "fuel" to drive the cell, hence the term fuel cell. Although fuel cells were not employed for practical purposes until space exploration began in the 1960's, the principle was first demonstrated in 1839 by Sir William Grove, a Welsh lawyer and amateur chemist. At the time, it was already known that water could be decomposed into hydrogen and oxygen by electrolysis; Grove tried recombining the two gases in a simple apparatus, and discovered what he called "reverse electrolysis"— that is, the recombination of H and O into water— causing a potential difference to be generated between the two electrodes: It was not until 1959 that the first working hydrogen-oxygen fuel cell was developed by Francis Thomas Bacon in England. Modern cells employ an alkaline electrolyte, so the electrode reactions differ from the one shown above by the addition of OH to both sides of the equations (note that the net reaction is the same): Although hydrogen has the largest energy-to-mass ratio of any fuel, it cannot be compressed to a liquid at ordinary temperatures. If it is stored as a gas, the very high pressures require heavy storage containers, greatly reducing its effective energy density. Some solid materials capable of absorbing large amount of H can reduce the required pressure. Other fuels such as alcohols, hydrocarbon liquids, and even coal slurries have been used; methanol appears to be an especially promising fuel. One reason for the interest in fuel cells is that they offer a far more efficient way of utilizing chemical energy than does conventional thermal conversion. The work obtainable in the limit of reversible operation of a fuel cell is 229 kJ per mole of H O formed. If the hydrogen were simply burned in oxygen, the heat obtainable would be ΔH = 242 kJ mol , but no more than about half of this heat can be converted into work so the output would not exceed 121 kJ mol . This limit is a consequence of the . The fraction of heat that can be converted into work (\(\eta\)) is a function of how far (in temperature) the heat falls as it flows through the engine and into the surroundings; this fraction is given by \[\eta=\dfrac{1 - T_{high}}{T_{low}}\] At normal environmental temperatures of around 300 K, this would have to be at least 600 K for 50% thermal efficiency. The major limitation of present fuel cells is that the rates of the electrode reactions, especially the one in which oxygen is reduced, tend to be very small, and thus so is the output current per unit of electrode surface. Coating the electrode with a suitable catalytic material is almost always necessary to obtain usable output currents, but good catalysts are mostly very expensive substances such as platinum, so that the resulting cells are too costly for most practical uses. There is no doubt that if an efficient, low-cost catalytic electrode surface is ever developed, the fuel cell would become a mainstay of the energy economy. Certain types of bacteria are able to oxidize organic compounds to carbon dioxide while directly transferring electrons to electrodes. These so-called organisms may make it possible to convert renewable biomass and organic waste directly into electricity without the wasted energy and pollution produced by direct combustion. In one experiment, a graphite electrode immersed in ordinary mud (containing humic materials) was able to produce measurable amounts of electricity. Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.

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Normality expresses concentration in terms of the equivalents of one chemical species that react stoichiometrically with another chemical species. Note that this definition makes an equivalent, and thus normality, a function of the chemical reaction. Although a solution of H SO has a single molarity, its normality depends on its reaction. We define the number of equivalents, , using a reaction unit, which is the part of a chemical species that participates in the chemical reaction. In a precipitation reaction, for example, the reaction unit is the charge of the cation or the anion that participates in the reaction; thus, for the reaction \[\ce{Pb^{2+}}(aq) + 2\ce{I-}(aq) \ce{<=>} \ce{PbI2}(s) \nonumber\] = 2 for Pb and = 1 for I . In an acid–base reaction, the reaction unit is the number of H ions that an acid donates or that a base accepts. For the reaction between sulfuric acid and ammonia \[\ce{H2SO4}(aq) + 2\ce{NH3}(aq) \ce{<=>} 2\ce{NH4+}(aq) + \ce{SO4^{2-}} \nonumber\] = 2 for H SO because sulfuric acid donates two protons, and = 1 for NH because each ammonia accepts one proton. For a complexation reaction, the reaction unit is the number of electron pairs that the metal accepts or that the ligand donates. In the reaction between Ag and NH \[\ce{Ag+}(aq) + 2\ce{NH3}(aq) \ce{<=>} \ce{Ag(NH3)2+}(aq) \nonumber\] = 2 for Ag because the silver ion accepts two pairs of electrons, and = 1 for NH because each ammonia has one pair of electrons to donate. Finally, in an oxidation–reduction reaction the reaction unit is the number of electrons released by the reducing agent or accepted by the oxidizing agent; thus, for the reaction \[2\ce{Fe^{3+}}(aq) + \ce{Sn^{2+}}(aq) \ce{<=>} \ce{Sn^{4+}}(aq) + 2\ce{Fe^{2+}}(aq) \nonumber\] = 1 for Fe and = 2 for Sn . Clearly, determining the number of equivalents for a chemical species requires an understanding of how it reacts. Normality is the number of equivalent weights, , per unit volume. An equivalent weight is the ratio of a chemical species’ formula weight, , to the number of its equivalents, . \[EW = \frac {FW} {n} \nonumber\] The following simple relationship exists between normality, , and molarity, M, \[N = n \times M \nonumber\]

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by Nivaldo Tro. . In addition to these individual basis; please contact What volume does 41.2 g of sodium gas at a pressure of 6.9 atm and a temperature of 514 K occupy? Would the volume be different if the sample were 41.2 g of calcium (under identical conditions)? Based on \(PV=nRT\), and we need to know V, so the equation is rearranged to \(V= nRT/P\). Then, the atomic mass of Sodium =22.99 g/mol \[n = 41.2 \;g = \dfrac{mass}{atomic\; mass} = \dfrac{41.2\;g}{22.99\; g/mol} = 1.79\; mol\] Finally, put every number into the equation \[V=\dfrac{nRT}{P}=\dfrac{(1.79\;mol)(0.0820\;6L (atm)/(K mol))(514\;K}{6.9\; atm} = 10.94\; L\] Because Calcium has different atomic mass than Sodium, so the volume is different. We have a 20.0 L cylinder that is filled with 28.6 g of oxygen gas at the temperature of 401 K. What is the pressure that the oxygen gas is exerting on the cylinder? = A car tire has a maximum rating of 37.0 psi (gauge pressure). While at the temperature of \(11.0^{\circ}C\), the tire is inflated to a volume of 10.4 L and a gauge pressure of 31.0 psi. When driving on a hot day, the tire warms to \(62.0^{\circ}C\) and its volume expands to 10.9 L. Does the pressure in the tire exceed its maximum rating on the hot day? Note: Gauge pressure is defined as the absolute pressure minutes the atmospheric pressure (14.7 psi). Gauge pressure=total pressure-atmospheric pressure 31.0 psi= X -14.7 psi X= 45.7 psi \[P=45.7 psi\cdot \dfrac{1 atm}{14.7 psi}=3.11 atm\] \[T=11^{\circ}C+273= 284 K\] V=10.4 L \[R=0.082057\dfrac{L\cdot atm}{mol\cdot K}\] \[T=62^{\circ}C+273= 335 K\] V=10.9 L \[R=0.082057\dfrac{L\cdot atm}{mol\cdot K}\] B. \[n=\dfrac{3.11 atm\cdot 10.4 L}{0.082057\dfrac{L\cdot atm}{mol\cdot K}\cdot 284 K}\] n= 1.39 mol C. \[P=\dfrac{1.39 mol\cdot 0.082057\dfrac{L\cdot atm}{mol\cdot K}\cdot 335 K}{10.9 L}\] P= 3.51 atm D. \[3.51 atm\cdot \dfrac{14.7 psi}{1 atm}= 51.6 psi\] 51.6 psi-14.7 psi= 36.9 psi The gauge pressure is 36.9 psi and since the maximum rating is 37.0 psi, the pressure does not exceed the maximum rating (although close). A balloon is floating on top of an ocean at a volume of 2.04 L at a pressure of 730 mmHg and a temperature at 20 °C. A sea creature then gets hungry and pulls the balloon down to the bottom of the ocean where the temperature decreases to 5°C while the pressure increases to 1510 mmHg. Suppose the balloon can freely increase, calculate the volume of the balloon. \[PV=nRT\] ---------> \[V_2 =0.950\;L\] Calculate the density of Ne gas at 143 ºC and 4.3 atm. Calculate Temperature in Kelvin: \[T = 143\; ºC + 273 = 416\; K\] Calculate the density of Ne \[\rho = \dfrac{MM \cdot P}{RT}\] \[\rho = \dfrac{\rm 20.2 g/mol \cdot 4.3 \rm atm}{(0.08206 \dfrac{\rm L atm}{\rm mol K}) \rm 416 K}\] \[\rho= \rm 2.54 g/L\] Determine the volume in liters occupied by 64.5 grams of Argon at STP. A. The formula to use to determine the molar volume for this problem is \(PV=nRT\). B. Since we know the numbers to plug in for P and T, we need to find numbers for \(n\) and \(R\). The rate, R, is \(0.08206 \dfrac{L\cdot atm}{mol\cdot K}\). This will always be the R for this equation. To determine n, the number of moles, we must use what is given to us, which is the grams or Ar.mol of \[Ar= 64.5\; g \;of \;Ar \cdot \dfrac{1 \;mol\; of\; Ar}{39.948\; g \;of \;Ar} =1.61 \;mol\; of Ar\] C. Now, we have all the numbers except the one we are looking for, which is V. Plug them in and work out the equation. \[1 atm\cdot V=1.61 mol of Ar \cdot 0.08206 \dfrac{L\cdot atm}{mol\cdot K} \cdot 273 K\] What is the total pressure in (atm) of a mixture of gases in a closed container with the partial pressures as indicated: H, 115 torr; Ar, 105 torr; and N , 204 torr? What is the mass of H, Ar, and N at 28°C and 1.45 L? \[H=(\dfrac{115\;torr}{424\;torr})=0.2712\] \[Ar=(\dfrac{105\;torr}{424\;torr})=0.2476\] \[N_{2}=(\dfrac{204\;torr}{424\;torr})=0.4811\] \[H=(0.2712)(0.558\;atm)=0.151\;atm\] \[Ar=(0.2476)(0.558\;atm)=0.138\;atm\] \[N_{2}=(0.4811)(0.558\;atm)=0.268\;atm\] \[H=(0.2712)(1.45\;L)=0.393L\] \[Ar=(0.2476)(1.45\;L)=0.359L\] \[N_2=(0.4811)(1.45\;L)=0.698L\] \[n(H)=\dfrac{(0.151atm)(0.393L)}{(0.0821\dfrac{L*atm}{mol*K})(301K)}=0.0024mol\] \[n(Ar)=\dfrac{(0.138atm)(0.359L)}{(0.0821\dfrac{L*atm}{mol*K})(301K)}=0.0020mol\] \[n(N_{2})=\dfrac{(0.268atm)(0.698L)}{(0.0821\dfrac{L*atm}{mol*K})(301K)}=0.0076mol\] \[H=(0.0024mol)(1.008\dfrac{g}{mol})=0.0024g\] \[Ar=(0.0020mol)(39.95\dfrac{g}{mol})=0.0799g\] \[N_{2}=(0.0076mol)(28.02\dfrac{g}{mol})=0.2130g\] \[Total Pressure=0.558atm\] \[H=0.0024g\] \[Ar=0.0799g\] \[N_{2}=0.2130g\] Hydrogen gas produced by a chemical reaction is collected in the vapor space of a container holding water. The hydrogen partial pressure is 605 mmHg and the system temperature is 55 °C. What is the total pressure in the container? If 0.1 mole of H was produced in the reaction, what is the total volume of the gas above the water in the container in L? Use the gas constant R of 62.3637 \[\dfrac{L\cdot mmHg}{mol\cdot K}\] Knowing the system temperature is 55 °C, the vapor pressure of the water in the hydrogen/water vapor mixture can be determined. Using Table 11.3 on page 412, the water partial pressure is found to be 118.2 mmHg. From Equation 11.8 we know that \[P_{Total} = P_{H_2O} + P_{H_2}\] P is given as 605 mmHg and P was found to be 118.2 mmHg It is given there is 0.1 mole of H and hydrogen partial pressure 605 mmHg. In the first part of this problem, the total system pressure was found to be 723.2 mmHg. Solving Equation 11.10 for the total number of moles, we find \[n_{total}=\left ( \dfrac{n_{H_{2}}}{P_{H_{2}}} \right )\cdot P_{total}\] Therefore \[n_{total}=\left ( \dfrac{0.1 mol_{H_{2}}}{605mmHg_} \right )\cdot 723.2 mmHg_=0.12 mol\] From the Ideal Gas Law, we know \(PV=n_{total}RT\) Solving for V, we find \[V=n_{total}\left ( \dfrac{RT}{P} \right )\] Convert the system temperature to Kelvin. K = 55 °C + 273.2 = 328.2 K Therefore V is: \[V=0.12mol\left ( \dfrac{62.3637\dfrac{L\cdot mmHg}{mol\cdot K}\cdot 328.2K}{723.2mmHg} \right )=3.4L\] Hydrogen gas is collected over water at 35 °C at a total pressure of 745 mmHg. What is the partial pressure of the hydrogen gas collected? Given that the total volume of the gas collected is 750 ml, calculate the mass of the hydrogen gas collected.

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is the study of heat, energy, and work and how they move. This is important because we have to move or generate heat to stay comfortable in winter and summer; we need to do work by moving things for many different purposes; we need to generate chemical energy to live and grow our bodies, etc. In thermodynamics, we often separate the universe (that is, everything that exists) into 2 parts: the , which is the small part we are interested in, and the which is everything outside the system. This will help us think about how heat, energy and work move between parts of the universe. allow energy and matter (stuff) to enter and leave the system. A pan on the stove is an open system because water can evaporate or be poured in, and heat can enter the pan if the stove is turned on, and leave the pan also. A does not allow matter to enter or leave, but does allow energy to enter or leave. A covered pot on the stove is approximately a closed system. An does not allow either matter or energy to enter or leave. A thermos or cooler is approximately an isolated system. There are no truly isolated systems. are quantities that don't depend on path. Your bank balance is a good example. It doesn't matter how the money entered your bank account, the total amount there at any given time is what it is and you can measure it easily. It doesn't matter if you put it in all at once, or a little bit every month, or put a lot in then spent it slowly... any time you want to know how much is there, you just check. Most of the quantities you know are state functions, like pressure, volume, temperature, location, etc. But some quantities that are important in thermodynamics, like heat and work, are only defined by a process, so they aren't state functions. refers to properties that depend on how much stuff there is. For instance, the volume or pressure created by a sample of gas depend on how much gas is in the sample, so they are extensive. quantities don't depend on how much there is. For instance, temperature, density, etc. If you divide a sample in 2, it does not change temperature. Density is a ratio of 2 extensive properties, so it is intensive.

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A is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time. The is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity. For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. If we measure the concentration of hydrogen peroxide, H O , in an aqueous solution, we find that it changes slowly over time as the H O decomposes, according to the equation: \[\ce{2H2O2}(aq)⟶\ce{2H2O}(l)+\ce{O2}(g) \nonumber \] The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here: \[\begin{align*} \ce{rate\: of\: decomposition\: of\: H_2O_2} &=\mathrm{−\dfrac{change\: in\: concentration\: of\: reactant}{time\: interval}}\\[4pt] &=−\dfrac{[\ce{H2O2}]_{t_2}−[\ce{H2O2}]_{t_1}}{t_2−t_1}\\[4pt] &=−\dfrac{Δ[\ce{H2O2}]}{Δt} \end{align*} \nonumber \] This mathematical representation of the change in species concentration over time is the for the reaction. The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus, \([\ce{H2O2}]_{t_1}\) represents the molar concentration of hydrogen peroxide at some time ; likewise, \([\ce{H2O2}]_{t_2}\) represents the molar concentration of hydrogen peroxide at a later time ; and Δ[H O ] represents the change in molar concentration of hydrogen peroxide during the time interval Δ (that is, − ). Since the reactant concentration decreases as the reaction proceeds, Δ[H O ] is a negative quantity; we place a negative sign in front of the expression because reaction rates are, by convention, positive quantities. Figure \(\Page {1}\) provides an example of data collected during the decomposition of H O . To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period: \[\dfrac{−Δ[\ce{H2O2}]}{Δt}=\mathrm{\dfrac{−(0.500\: mol/L−1.000\: mol/L)}{(6.00\: h−0.00\: h)}=0.0833\: mol\:L^{−1}\:h^{−1}} \nonumber \] Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of: \[\dfrac{−Δ[\ce{H2O2}]}{Δt}=\mathrm{\dfrac{−(0.0625\:mol/L−0.125\:mol/L)}{(24.00\:h−18.00\:h)}=0.0104\:mol\:L^{−1}\:h^{−1}} \nonumber \] This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its . The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its . Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes ( ). A few moments later, the instantaneous rate at a specific moment—call it —would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δ ). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates. The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. If we plot the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H O at any time is given by the slope of a straight line that is tangent to the curve at that time (Figure \(\Page {2}\)). We can use calculus to evaluating the slopes of such tangent lines, but the procedure for doing so is beyond the scope of this chapter. Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure \(\Page {2}\)). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations. The test for urinary glucose relies on a two-step process represented by the chemical equations shown here: \[\ce{C6H12O6 + O2}\underset{\large\textrm{catalyst}}{\xrightarrow{\hspace{45px}}}\ce{C6H10O6 + H2O2} \label{eq1} \] \[\ce{2H2O2 + 2I-}\underset{\large\textrm{catalyst}}{\xrightarrow{\hspace{45px}}}\ce{I2 + 2H2O + O2} \label{eq2} \] Equation \(\ref{eq1}\) depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine (Equation \(\ref{eq2}\)), which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change. The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of , a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a result). Waiting too long to assess the color change can lead to a due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine. The rate of a reaction may be expressed in terms of the change in the amount of any reactant or product, and may be simply derived from the stoichiometry of the reaction. Consider the reaction represented by the following equation: \[\ce{2NH3}(g)⟶\ce{N2}(g)+\ce{3H2}(g) \nonumber \] The stoichiometric factors derived from this equation may be used to relate reaction rates in the same manner that they are used to related reactant and product amounts. The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is: \[\mathrm{−\dfrac{Δmol\: NH_3}{Δ\mathit t}×\dfrac{1\: mol\: N_2}{2\: mol\: NH_3}=\dfrac{Δmol\:N_2}{Δ\mathit t}} \nonumber \] We can express this more simply without showing the stoichiometric factor’s units: \[−\dfrac{1}{2}\dfrac{\mathrm{Δmol\:NH_3}}{Δt}=\dfrac{\mathrm{Δmol\:N_2}}{Δt} \nonumber \] Note that a negative sign has been added to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). If the reactants and products are present in the same solution, the molar amounts may be replaced by concentrations: \[−\dfrac{1}{2}\dfrac{Δ[\ce{NH3}]}{Δt}=\dfrac{Δ[\ce{N2}]}{Δt} \nonumber \] Similarly, the rate of formation of H is three times the rate of formation of N because three moles of H form during the time required for the formation of one mole of N : \[\dfrac{1}{3}\dfrac{Δ[\ce{H2}]}{Δt}=\dfrac{Δ[\ce{N2}]}{Δt} \nonumber \] Figure \(\Page {3}\) illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. We can see from the slopes of the tangents drawn at = 500 seconds that the instantaneous rates of change in the concentrations of the reactants and products are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production: \[\dfrac{2.91×10^{−6}\:M/\ce s}{9.71×10^{−6}\:M/\ce s}≈3 \nonumber \] The first step in the production of nitric acid is the combustion of ammonia: \[\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g) \nonumber \] Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products. Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are: \[−\dfrac{1}{4}\dfrac{Δ[\ce{NH3}]}{Δt}=−\dfrac{1}{5}\dfrac{Δ[\ce{O2}]}{Δt}=\dfrac{1}{4}\dfrac{Δ[\ce{NO}]}{Δt}=\dfrac{1}{6}\dfrac{Δ[\ce{H2O}]}{Δt} \nonumber \] The rate of formation of Br is 6.0 × 10 mol/L/s in a reaction described by the following net ionic equation: \[\ce{5Br- + BrO3- + 6H+ ⟶ 3Br2 + 3H2O} \nonumber \] Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products. \[−\dfrac{1}{5}\dfrac{Δ[\ce{Br-}]}{Δt}=−\dfrac{Δ[\ce{BrO3-}]}{Δt}=−\dfrac{1}{6}\dfrac{Δ[\ce{H+}]}{Δt}=\dfrac{1}{3}\dfrac{Δ[\ce{Br2}]}{Δt}=\dfrac{1}{3}\dfrac{Δ[\ce{H2O}]}{Δt} \nonumber \] The graph in Figure \(\Page {3}\) shows the rate of the decomposition of H O over time: \[\ce{2H2O2 ⟶ 2H2O + O2} \nonumber \] Based on these data, the instantaneous rate of decomposition of H O at = 11.1 h is determined to be 3.20 × 10 mol/L/h, that is: \[−\dfrac{Δ[\ce{H2O2}]}{Δt}=\mathrm{3.20×10^{−2}\:mol\: L^{−1}\:h^{−1}} \nonumber \] What is the instantaneous rate of production of H O and O ? Using the stoichiometry of the reaction, we may determine that: \[−\dfrac{1}{2}\dfrac{Δ[\ce{H2O2}]}{Δt}=\dfrac{1}{2}\dfrac{Δ[\ce{H2O}]}{Δt}=\dfrac{Δ[\ce{O2}]}{Δt} \nonumber \] Therefore: and \[\dfrac{Δ[\ce{O2}]}{Δt}=\mathrm{1.60×10^{−2}\:mol\:L^{−1}\:h^{−1}} \nonumber \] If the rate of decomposition of ammonia, NH , at 1150 K is 2.10 × 10 mol/L/s, what is the rate of production of nitrogen and hydrogen? 1.05 × 10 mol/L/s, N and 3.15 × 10 mol/L/s, H . The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.

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Wöhler synthesis of Urea in 1828 heralded the birth of modern chemistry. The Art of synthesis is as old as Organic chemistry itself. Natural product chemistry is firmly rooted in the science of degrading a molecule to known smaller molecules using known chemical reactions and conforming the assigned structure by chemical synthesis from small, well known molecules using well established synthetic chemistry techniques. Once this art of synthesizing a molecule was mastered, chemists attempted to modify bioactive molecules in an attempt to develop new drugs and also to unravel the mystery of biomolecular interactions. Until the middle of the 20th Century, organic chemists approached the task of synthesis of molecules as independent tailor made projects, guided mainly by chemical intuition and a sound knowledge of chemical reactions. During this period, a strong foundation was laid for the development of mechanistic principles of organic reactions, new reactions and reagents. More than a century of such intensive studies on the chemistry of carbohydrates, alkaloids, terpenes and steroids laid the foundation for the development of logical approaches for the synthesis of molecules. The job of a synthetic chemist is akin to that of an architect (or civil engineer). While the architect could actually see the building he is constructing, a molecular architect called Chemist is handicapped by the fact that the molecule he is synthesizing is too small to be seen even through the most powerful microscope developed to date. With such a limitation, how does he ‘see’ the developing structure? For this purpose, a chemist makes use of spectroscopic tools. How does he cut, tailor and glue the components on a molecule that he cannot see? For this purpose chemists have developed molecular level tools called Reagents and Reactions. How does he clean the debris and produce pure molecules? This feat is achieved by crystallization, distillation and extensive use of Chromatography techniques. A mastery over several such techniques enables the molecular architect (popularly known as organic chemist) to achieve the challenging task of synthesizing the mirade molecular structures encountered in Natural Products Chemistry, Drug Chemistry and modern Molecular Materials. In this task, he is further guided by several ‘thumb rules’ that chemists have evolved over the past two centuries. The discussions on the topics for synthesis, and are beyond the scope of this write-up. Let us begin with a brief look at some of the important We would then discuss Protection and Deprotection of some important functional groups. We could then move on to the Logic of planning Organic Synthesis.

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

Hybridization was introduced to explain molecular structure when the failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109 , 120 , or 180 . According to Valence Shell Electron Pair Repulsion ( ) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles. Carbon is a perfect example showing the value of hybrid orbitals. Carbon's ground state configuration is: According to , carbon should form two covalent bonds, resulting in a CH , because it has two unpaired electrons in its electronic configuration.However, experiments have shown that \(CH_2\) is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH can exist. To form bonds the configuration of carbon must have unpaired electrons. One way CH can be explained is, the 2s and the 3 2p orbitals combine to make four, equal energy . That would give us the following configuration: Now that carbon has four unpaired electrons it can have four equal energy bonds. The hybridization of orbitals is favored because hybridized orbitals are more directional which leads to greater overlap when forming bonds, therefore the bonds formed are stronger. This results in more stable compounds when hybridization occurs. The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules. sp hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized. Hybridization of an s orbital with all three p orbitals (p , p , and p ) results in four sp hybrid orbitals. sp hybrid orbitals are oriented at bond angle of 109.5 from each other. This 109.5 arrangement gives tetrahedral geometry (Figure 4). Because carbon plays such a significant role in organic chemistry, we will be using it as an example here. Carbon's 2s and all three of its 2p orbitals hybridize to form four sp orbitals. These orbitals then bond with four hydrogen atoms through sp -s orbital overlap, creating methane. The resulting shape is tetrahedral, since that minimizes electron repulsion. Remember to take into account lone pairs of electrons. These lone pairs cannot double bond so they are placed in their own hybrid orbital. This is why H O is tetrahedral. sp hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp orbitals. In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp orbitals that align themselves in the trigonal planar structure. The three Al sp orbitals bond with with 1s orbitals from the three hydrogens through sp -s orbital overlap. Similar hybridization occurs in each carbon of ethene. For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap. The remaining sp orbitals on each carbon are bonded with each other, forming a bond between each carbon through sp -sp orbital overlap. This leaves us with the two p orbitals on each carbon that have a single carbon in them. These orbitals form a ? bonds through p-p orbital overlap, creating a double bond between the two carbons. Because a double bond was created, the overall structure of the ethene compound is linear. However, the structure of each molecule in ethene, the two carbons, is still trigonal planar. sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals. Figure 1: Notice how the energy of the electrons lowers when hybridized. These p orbitals come into play in compounds such as ethyne where they form two addition? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180 . In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. The two frontal lobes of the sp orbitals face away from each other forming a straight line leading to a linear structure. These two sp orbitals bond with the two 1s orbitals of the two hydrogen atoms through sp-s orbital overlap. The hybridization in ethyne is similar to the hybridization in magnesium hydride. For each carbon, the 2s orbital hybridizes with one of the 2p orbitals to form two sp hybridized orbitals. The frontal lobes of these orbitals face away from each other forming a straight line. The first bond consists of sp-sp orbital overlap between the two carbons. Another two bonds consist of s-sp orbital overlap between the sp hybridized orbitals of the carbons and the 1s orbitals of the hydrogens. This leaves us with two p orbitals on each carbon that have a single carbon in them. This allows for the formation of two ? bonds through p-p orbital overlap. The linear shape, or 180° angle, is formed because electron repulsion is minimized the greatest in this position. Using the , try to figure out the hybridization (sp, sp , sp ) of the indicated atom and indicate the atom's shape. 1. The carbon. 2. The oxygen. 3. The carbon on the right. 1. sp - Trigonal Planar The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp . 2. sp - Tetrahedral Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp orbital. That makes 4 orbitals, aka sp . 3. sp - Linear The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp. An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/20%3A_Carboxylic_Acids_and_Nitriles

This chapter presents a straightforward discussion of the chemistry of carboxylic acids (formerly called “fatty acids”) and nitriles. As usual, we begin with a description of how the compounds are named. We then consider the subtleties of their structure, and how these structural features influence physical properties, such as boiling point. We place considerable emphasis on the dissociation of carboxylic acids and the effect of substituents on acid strength. We have already encountered a number of methods for preparing carboxylic acids. We shall review these methods, and learn about two additional procedures. The only reactions of carboxylic acids that we study in detail in this chapter are reduction and decarboxylation, although two other common reactions of carboxylic acids, alpha substitution and nucleophilic acyl substitution, will be described in later chapters. We will then look at the formation of nitriles and their chemical reactivity; and our discussion of carboxylic acid and nitrile chemistry concludes with a look at the infrared and NMR spectra of these compounds, with emphasis on the characterization of carboxylic acids.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.07%3A_Carbohydrates

Carbohydrates are the most abundant biomolecules on earth and are widely used by organisms for structural and energy-storage purposes. In particular, glucose is the most commonly used monosaccharide, thus, this is why all of the pathways that we have covered start with glucose. However, many microorganisms are able to utilize more complex carbohydrates for energy. Let’s look at the structures of different carbohydrates and their use in microbial metabolism. Monosaccarides are the building blocks (monomers) for the synthesis of polymers. These sugars are classified by the length of the chain and the position of the carbonyl. Glucose and Ribose are shown below. Glyderaldehyde and dihydroxyacetone are shown below. Monosaccharides of four or more carbon atoms are typically more stable when they adopt cyclic, or ring, structures. This is a nucleophilic addition the results in a hemiacetal. There hemiacetal formed when the sugar cyclizes is a new chiral center. Two possible orientations can be formed. Disaccharides are carbohydrates composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a linkage. Three common disaccharides are the grain sugar , made of two glucose molecules; the milk sugar , made of a and a molecule; and the table sugar , made of a glucose and a fructose molecule. There are different types of glycosidic linkages. They are characterized by the numbering of the alcohols that are linked in the ether. And the anomer of the sugar. Thus, this is alpha-1,4-maltose. The human body is unable to metabolize maltose or any other disaccharide directly from the diet because the molecules are too large to pass through the cell membranes of the intestinal wall. Therefore, an ingested disaccharide must first be broken down by hydrolysis into its two constituent monosaccharide units. In the body, such hydrolysis reactions are catalyzed by enzymes such as or . Polysaccharides are very large polymers composed of hundreds to thousands of monosaccharides. These structures are used for energy storage or, in the case of cellulose, structural components. Starch is a mixture of two polysaccharides and is an important component of grains (wheat, rice, barley, etc.). This will again be important in bread and beer fermentations. These two polymers are and . is a straight chain polysaccharide (shown below). is a branched-chain polysaccharide. (cartoon shown below)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.05%3A_Acid_Base_Titration

Make sure you thoroughly understand the following essential concepts: The objective of an acid-base titration is to determine \(C_a\), the nominal concentration of acid in the solution. In its simplest form, titration is carried out by measuring the volume of the solution of strong base required to complete the reaction \[\ce{H_nA + n OH- → n A- + n H_2O} \label{0-1}\] in which \(n\) is the number of in the acid. The point at which this reaction is just complete is known as the . This is to be distinguished from the , which is the value we observe experimentally. A hydrogen atom (sometimes called an "acidic" hydrogen) is one that can be donated to a strong base — that is, to an OH ion. Thus in acetic acid HCOO , only the hydrogen in the carboxyl group is considered "replaceable". What we actually measure, of course, is the volume of titrant delivered by the burette. Learning to properly control the stopcock at the bottom of the burette usually requires some instruction and practice, as does the reading of the volume. For highly precise work, the concentration of the titrant itself must be determined in a separate experiment known as "standardization". A plot showing the pH of the solution as a function of the quantity of base added is known as a . These plots can be constructed by plotting the pH as a function of either the volume of base added, or the \(ƒ\) which is simply the number of moles of base added per mole of acid present in the solution. In most of the titration curves illustrated in this section, we plot pH as a function of \(ƒ\). It's worth taking some time to thoroughly familiarize yourself with the general form of a titration curve such as the one shown below, in which a weak acid HA is titrated with a strong base, typically sodium hydroxide. The equivalence point occurs at the pH at which the equivalent fraction ƒ of base added is unity. At this point, the reaction \[HA + OH^– → AB^– + H_2O \label{1-1}\] is stoichiometrically complete; a solution initially containing moles of a monoprotic acid HA will now be identical to one containing the same number of moles of the conjugate base A . At the ƒ = 0.5, the concentrations of the conjugate species are identical: [HA] = [A ]. This, of course corresponds to a (hence the relatively flat part of the curve) whose pH is the same as . As base is added beyond ƒ = 1, the pH begins to level off, suggesting that another buffered system has come into play. In this case it involves the solvent (water) and hydroxide ion: {H O} ≈ {OH-}. A similar effect is seen at the low-pH side of the curve when a strong acid is titrated, as in the plot for the titration of HCl below. In this case, the buffering is due to {H O ) ≈ {H2O}. How can this be? Surely, the concentration of OH , even when the pH approaches 14, cannot be anything like that of [H O] which will be about 55.5 in most solutions! This fine point (along with the mention of H O/OH buffering) is rarely mentioned in elementary courses because the theory behind it involves some rather esoteric elements of solution thermodynamics. However, in case you are curious, note that the curly brackets in {H O} ≈ {OH } denote not concentrations. And by convention the activity of a pure liquid (H O in this case) is unity. At a pH of around 12, pOH = 2, [OH ] = .01. At this rather high ion concentration, {OH } will be somewhat smaller than this, but the two activities will be similar enough to produce the buffering effect we observe. The pH of the solution at its equivalence point will be 7 if we are titrating a strong acid with strong base, as in HCl + NaOH → H O + NaCl. However, if the acid is weak, as in the above plot, the solution will be alkaline. This pH can be calculated from and in a manner exactly analogous to that used for calculating the pH of a solution of a weak acid in water. It is important to understand that the equivalent fraction ƒ of base that must be added to reach the equivalence point is independent of the of the acid and of its in the solution. The whole utility of titration as a means of quantitative analysis rests on this independence; we are in all cases measuring only the in the sample undergoing titration. Although the strength of an acid has no effect on the location of the equivalence point, it does affect the of the titration curve and can be estimated on a plot of the curve. The weaker the acid being titrated, the higher the initial pH (at ƒ=0), and the smaller will be the vertical height of the plot near the equivalence point. As we shall see later, this can make it difficult to locate the equivalence point if the acid is extremely weak. As shown in plot above, the of a weak acid can be estimated by noting the pH that corresponds to the half-titration point ƒ = 0.5. Recalling that the pH is controlled by the ratio of conjugate species concentrations \[pH = pK_a + \log \dfrac{[A^]}{[HA]} \label{1-2}\] it will be apparent that this equation reduces to pH = when the titration is half complete (that is, when [HA] = [A ]), the pH of the solution will be identical to the pK of the acid. This equation does not work for strong acids owing to the strong buffering that occurs at the very low pH at which ƒ = 0.5. As indicated here, the buffering has nothing to do with the acid HCl itself (which does not exist as such in water), but rather with its dissociation products H O and OH , "the strongest acid and base that can exist in water." The following two principles govern the detailed shape of a titration curve: It is important to understand the reasons for these two relations. The second is the simplest to explain. Titration of an acid HA with a base such as NaOH results in a solution of NaA; that is, a solution of the conjugate base A . Being a base, it will react with water to yield an excess of hydroxide ions, leaving a slightly alkaline solution. Titration of a weak base with an acid will have the opposite effect. The extent of the jump in the pH at the equivalence point is determined by a combination of factors. In the case of a weak acid, for example, the initial pH is likely to be higher, so the titration curve starts higher. Further, the weaker the acid, the stronger will be its conjugate base, so the higher will be the pH at the equivalence point. These two factors raise the bottom part of the titration curve. The upper extent of the curve is of course limited by the concentration and strength of the titrant. These principles are clearly evident in the above plots for the titrations of acids and bases having various strengths. Notice the blue curves that represent the titration of pure water (a very weak acid) with strong acid or base. When both the titrant and sample are "strong", we get long vertical plots at ƒ = 1. Adding even half a drop of titrant can take us across the equivalence point! When one of the reactants is weak, the pH changes rapidly at first until buffering sets in. In (C), the onset of H O/OH- buffering near ƒ=1 makes the equivalence point more difficult to locate. "Weak/weak" titrations tend to be problematic as the buffered regions move closer to ƒ=1. The equivalence point pH of 7 in these examples reflects the near-equality of and of the reactants. It can be difficult to reliably detect the equivalence point in the titration of boric acid ( = 9.3) or of other similarly weak acids from the shape of the titration curve*. *An interesting student laboratory experiment that employs an auxiliary reagent (mannitol) to make boric acid stronger and thus more readily titratable was described in . 2012, 89, 767-770. The problem here is that An extreme example occurs in the with a strong acid or base. At these extremes of pH the concentrations of H O and of OH are sufficiently great that a competing buffer system (either H O /H O or H O/OH , depending on whether the solution is highly acidic or highly alkaline) comes into play. The above plots clearly show that the most easily-detectable equivalence points occur when an acid with is titrated with a strong base such as sodium hydroxide (or a base is titrated with a strong acid.) In practice, many of the titrations carried out in research, industry, and clinical practice involve of more than one acid. Examples include natural waters, physiological fluids, fruit juices, wine making, brewing, and industrial effluents. For titrating these kinds of samples, the use of anything other than a strong titrant presents the possibility that the titrant may be weaker than one or more of the "stronger" components in the sample, in which case it would be incapable of titrating these components to completion. In terms of proton-free energies, the proton source (the acidic titrant) would be unable to deliver an equivalent quantity of protons to the (stronger) component of the mixture. There will be as many equivalence points as there are replaceable hydrogens in an acid. Thus in the extremely important , equivalence points are seen at both ƒ=1 and ƒ=2: In general, there are two requirements for a clearly discernible jump in the pH to occur in a polyprotic titration: The effect of the first point is seen by comparing the titration curves of two diprotic acids, sulfurous and succinic. The appearance of only one equivalence point in the latter is a consequence of the closeness of the first and second acid dissociation constants. The 's of (below, left) are sufficiently far apart that its titration curve can be regarded as the superposition of those for two independent monoprotic acids having the corresponding 's. This reflects the fact that the two acidic –OH groups are connected to the same central atom, so that the local negative charge that remains when HSO is formed acts to suppress the second dissociation step. *It can be shown that in the limit of large , the ratio of K /K for a symmetrical dicarboxylic acid HOOC-(CH ) - COOH converges to a value of 4. In , the two –COOH groups are physically more separated and thus tend to dissociate independently*. Inspection of the species distribution curves for succinic acid (above, right) reveals that the fraction of the ampholyte HA can never exceed 75 percent. That is, there is no pH at which the reaction H A → HA + H can be said to be “complete” while at the same time the second step HA → A + H has occurred to only a negligible extent. Thus the rise in the pH that would normally be expected as HA is produced will be prevented by consumption of OH in the second step which will be well underway at that point; only when all steps are completed and hydroxide ion is no longer being consumed will the pH rise. Two other examples of polyprotic acids whose titration curves do not reveal all of the equivalence points are and phosphoric acids. Owing to the leveling effect, the apparent of H SO is so close to = 0.01 that the effect is the same as in succinic acid, so only the second equivalence point is detected. In , the third equivalence point (for HPO ) is obscured by H O-OH buffering as explained previously. Whether or not the equivalence point is revealed by a distinct "break" in the titration curve, it will correspond to a unique hydrogen ion concentration which can be calculated in advance. There are many ways of determining the equivalence point of an acid-base titration. The traditional method of detecting the equivalence point has been to employ an dye, which is a second acid-base system in which the protonated and deprotonated forms differ in color, and whose is close to the pH expected at the equivalence point. If the acid being titrated is not a strong one, it is important to keep the indicator concentration as low as possible in order to prevent its own consumption of OH from distorting the titration curve. The observed color change of an indicator does not take place sharply, but occurs over a range of about 1.5 to 2 pH units. Indicators are therefore only useful in the titration of acids and bases that are sufficiently strong to show a definite break in the titration curve. Some plants contain coloring agents that can act as natural pH indicators. These include cabbage (shown), beets, and hydrangea flowers. For a strong acid - strong base titration, almost any indicator can be used, although phenolphthalein is most commonly employed. For titrations involving weak acids or bases, as in the acid titration of sodium carbonate solution shown here, the indicator should have a pK close to that of the substance being titrated. When titrating a polyprotic acid or base, multiple indicators are required if more than one equivalence point is to be seen. The s of phenolphthalein and methyl orange are 9.3 and 3.7, respectively. The pH meter detects the voltage produced when the H ions in the solution displace Na ions from a thin glass membrane that is dipped into the solution. A more modern way of finding an equivalence point is to follow the titration by means of a pH meter. Because it involves measuring the electrical potential difference between two electrodes, this method is known as . Until around 1980, pH meters were too expensive for regular use in student laboratories, but this has changed; potentiometry is now the standard tool for determining equivalence points. Plotting the pH after each volume increment of titrant has been added can yield a titration curve as detailed as desired, but there are better ways of locating the equivalence point. The most common of these is to take the first or second derivatives of the plot: (pH)/d or (pH)/d (of course, for finite increments of pH and volume, these terms would be expressed as Δ(pH)/Δ and Δ (pH)/Δ .) A curve locates the inflection point by finding where the at which the pH changes is zero. The , showing rate-of-change of pH against titrant volume, locates the inflection point which is also the equivalence point In a of pH-vs-volume of titrant added, the inflection point is located visually as half-way along the steepest part of the curve. The idealized plots shown above are unlikely to be seen in practice. When the titration is carried out manually, the titrant is added in increments, so even the simple titration curve must be constructed from points subject to uncertainties in volume measurement and pH (especially if the latter is visually estimated by color change of an indicator.) If this data is then converted to differential form, these uncertainties add a certain amount of "noise" to the data. A second-derivative plot uses pH readings on both sides of the equivalence point, making it easier to locate in the presence of noise. Locating the equivalence point depends very strongly on correct reading of only one or two pH readings near the top of the plot. A simple curve, plotted from a small number of pH readings, will not always unambiguously locate the equivalence point. The "noise" in differential plots can usually be minimized by keeping the titrant and analyte concentrations above 10 . Monitoring the pH by means of an indicator or by potentiometry as described above are the standard ways of detecting the equivalence point of a titration. However, we have already seen that in certain cases involving polyprotic acids or bases, some of the equivalence points are obscured by their close proximity to others, or by the buffering that occurs near the extremes of the pH range. Similar problems can arise when the solution to be titrated contains several different acids, as often happens when fluids connected with industrial processes must be monitored. Acid-base neutralization reactions HA + B → A + BH are always exothermic; when protons fall from their level in the acid to that in the base, most of the free energy drop gets released as heat. If the acid and base are both strong (i.e., totally dissociated), the for the reaction \[\ce{H3O+ + OH- → 2 H2O}\] is -68 kJ mol . See this page for more on thermometric titrations, including many examples. Note also the on this topic in the " s" section near the end of this page. A typical thermometric titration curve consists of two branches, beginning with a steep rise in temperature as the titrant being added reacts with the analyte, liberating heat. Once the equivalence point is reached, the rise quickly diminishes as heat production stops. Then, as the mixture begins to cool, the plot assumes a negative slope. Although a rough indication of the equivalence point can be estimated by extrapolating the linear parts of the curve (blue dashed lines), the differential methods described above are generally preferred. Acids and bases are , meaning that their solutions conduct electric current. The conductivity of such solutions depends on the concentrations of the ions, and to a lesser extent, on the nature of the particular ions. Any chemical reaction in which there is a change in the total quantity of ions in the solution can usually be followed by monitoring the conductance. Acid-base titrations fall into this category. Consider, for example, the titration of hydrochloric acid with sodium hydroxide. This can be described by the equation \[\ce{H^{+} + Cl^{–} + Na^{+} + OH^{–} → H2O + Na^{+} + Cl^{–}} \label{2-1}\] which shows that two of the four species of ions being combined disappear at the equivalence point. During the course of the titration, the conductance of the solution falls as H and ions are consumed. At the equivalence point the conductance passes through a minimum, and then rises as continued addition of titrant adds more and OH ions to the solution. Each kind of ion makes its own contribution to the solution conductivity. If we could observe the contribution of each ion separately, see that the slopes for H and OH are much greater. This reflects the much greater conductivities of these ions owing to their uniquely rapid movement through the solution by hopping across water molecules. However, because the conductances of individual ions cannot be observed directly, conductance measurements always register the total conductances of ions in the solution. The change in conductance that is actually observed during the titration of HCl by sodium hydroxide is the sum of the ionic conductances shown above. For most ordinary acid-base titrations, conductimetry rarely offers any special advantage over regular volumetric analysis using indicators or potentiometry. This is especially true if the acid being titrated is weak; if the is much below 2, the rising salt line (Na when titrating with NaOH) will overwhelm the fall in the contribution the small amount of H makes to the conductance, thus preventing any minimum in the total conductance curve from being seen. However, in some special cases such as those illustrated below, conductimetry is the only method capable of yielding useful results. Because pure H SO is a neutral molecule, it makes no contribution to the conductance, which rises to a maximum at the equivalence point. In four years of college lab sessions, many Chemistry majors will likely carry out fewer than a dozen titrations. However, in the real world, time is money, and long gone are the days when technicians were employed full time just to titrate multiple samples in such enterprises as breweries, food processing (such as blending of canned orange juice), clinical labs, and biochemical research. A titration is carried out by adding a sufficient volume \Vo of the titrant solution to a known volume \Vt of the solution being titrated. This addition continues until the is reached. The end point is our experimental approximation of the at which the acid-base reaction is stoichiometrically complete (ƒ = 1). The quantity we actually measure at the end point is the volume V_ep of titrant delivered to the solution undergoing titration. The solution being titrated is often referred to as the (the substance being "analyzed") or, less commonly, as the . We shall employ the latter term in what follows. In a simple titration of a monoprotic acid HA by a base B, the equivalence point corresponds to completion of the reaction \[\ce{HA + B → A^{–} + BH^{+} }\label{3-1}\] where equimolar quantities of HA and B have been combined; that is, = (where represents the number of moles.) Recall that the number of moles is given by the product of the volume and concentration: L × mol L = mol, mL x mMol ml = mmol. Because we are measuring the of titrant rather than the number of moles, we need to use its to link the two quantities. So if we are titrating the base B with acid HA, the end point is reached when a volume V of HA has been added. The number of moles of HA we have added at the end point is given by the product of its volume and concentration \[M_{HA} = V_{HA} \times C_{HA} \label{3-2}\] And because the reaction \ref{3-1} is now complete, = ; thus, \[V_{HA} C_{HA} = V_B C_B \label{3-3}\] Equation \ref{3-3} is i In any titration, both the volume and the concentration of the titrant are known, so the unknown concentration is easily calculated. In titrations carried out in the laboratory, the titrant is delivered by a burette that is usually calibrated in milliliters, so it is more convenient to express in millimoles and in millimoles/mL (mMol ml ); . 50.0 mL of 0.100 M hydrochloric acid is titrated with 0.200 M sodium hydroxide. What volume of NaOH solution will have been added at the equivalence point? First, find the number of moles of HCl in the titrand: \[\begin{align*} M_{\ce{HCl}} &= C_{\ce{HCl}} \times V_{\ce{HCl}} \\[4pt] &= (0.100\, mMol\, mL^{–1}) \times (50.0\, mL) \\[4pt] &= 5.0 \,mMol\,\ce{HCl} \end{align*}\] This same number of moles of NaOH must be delivered by the burette in order to reach the equivalence point ( , = 5.0 mMol.) The volume of NaOH solution required is = ( / ) = (5.0 mMol) / (0.200 mMol/mL) =

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/05%3A_Atoms_and_the_Periodic_Table/5.04%3A_The_Bohr_Atom

Make sure you thoroughly understand the following essential ideas: are widely employed in science to help understand things that cannot be viewed directly. The idea is to imagine a simplified system or process that might be expected to exhibit the basic properties or behavior of the real thing, and then to test this model against more complicated examples and modify it as necessary. Although one is always on shaky philosophical ground in trying to a model with reality, there comes a point when the difference between them becomes insignificant for most practical purposes. The demonstration by Thompson in 1867 that all atoms contain units of negative electric charge led to the first science-based model of the atom which envisaged the electrons being spread out uniformly throughout the spherical volume of the atom. Ernest Rutherford, a New Zealander who started out as Thompson's student at Cambridge, distrusted this "plum pudding" model (as he called it) and soon put it to rest; Rutherford's famous alpha-ray bombardment experiment (carried out, in 1909, by his students Hans Geiger and Ernest Marsden) showed that nearly all the mass of the atom is concentrated in an extremely small (and thus extremely dense) body called the nucleus. This led him to suggest the of the atom, in which the electrons revolve in orbits around the nuclear "sun". Even though the planetary model has long since been discredited, it seems to have found a permanent place in popular depictions of the atom, and certain aspects of it remain useful in describing and classifying atomic structure and behavior. The planetary model of the atom assumed that the electrostatic attraction between the central nucleus and the electron is exactly balanced by the centrifugal force created by the revolution of the electron in its orbit. If this balance were not present, the electron would either fall into the nucleus, or it would be flung out of the atom. The difficulty with this picture is that it is inconsistent with a well established fact of classical electrodynamics which says that whenever an electric charge undergoes a change in velocity or direction (that is, , which must happen if the electron circles around the nucleus), it must continually radiate energy. If electrons actually followed such a trajectory, all atoms would act is miniature broadcasting stations. Moreover, the radiated energy would come from the kinetic energy of the orbiting electron; as this energy gets radiated away, there is less centrifugal force to oppose the attractive force due to the nucleus. The electron would quickly fall into the nucleus, following a trajectory that became known as the "death spiral of the electron". According to classical physics, no atom based on this model could exist for more than a brief fraction of a second. Niels Bohr was a brilliant Danish physicist who came to dominate the world of atomic and nuclear physics during the first half of the twentieth century. Bohr suggested that the planetary model could be saved if one new assumption were made: certain "special states of motion" of the electron, corresponding to different orbital radii, would not result in radiation, and could therefore persist indefinitely without the electron falling into the nucleus. Specifically, Bohr postulated that the angular momentum of the electron, (the mass and angular velocity of the electron and in an orbit of radius \(r\)) is restricted to values that are integral multiples of \(h/2\pi\). The radius of one of these allowed Bohr orbits is given by \[r=\dfrac{nh}{2\pi m u}\] in which is Planck's constant, is the mass of the electron, is the orbital velocity, and can have only the integer values 1, 2, 3, etc. The most revolutionary aspect of this assumption was its use of the variable integer ; this was the first application of the concept of the to matter. The larger the value of , the larger the radius of the electron orbit, and the greater the potential energy of the electron. As the electron moves to orbits of increasing radius, it does so in opposition to the restoring force due to the positive nucleus, and its potential energy is thereby raised. This is entirely analogous to the increase in potential energy that occurs when any mechanical system moves against a restoring force— as, for example, when a rubber band is stretched or a weight is lifted. Thus what Bohr was saying, in effect, is that the atom can exist only in certain discrete energy states: the energy of the atom is . Bohr noted that this quantization nicely explained the observed emission spectrum of the hydrogen atom. The electron is normally in its smallest allowed orbit, corresponding to = 1; upon excitation in an electrical discharge or by ultraviolet light, the atom absorbs energy and the electron gets promoted to higher quantum levels. These higher excited states of the atom are unstable, so after a very short time (around 10 sec) the electron falls into lower orbits and finally into the innermost one, which corresponds to the atom's . The energy lost on each jump is given off as a photon, and the frequency of this light provides a direct experimental measurement of the difference in the energies of the two states, according to the Planck-Einstein relationship ν. Bohr's theory worked; it completely explained the observed spectrum of the hydrogen atom, and this triumph would later win him a Nobel prize. The main weakness of the theory, as Bohr himself was the first to admit, is that it could offer no good explanation of these special orbits immunized the electron from radiating its energy away. The only justification for the proposal, other than that it seems to work, comes from its analogy to certain aspects of the behavior of vibrating mechanical systems. In order to produce a tone when plucked, a guitar string must be fixed at each end (that is, it must be a ) and must be under some tension. Only under these conditions will a transverse disturbance be countered by a restoring force (the string's tension) so as to set up a sustained vibration. Having the string tied down at both ends places a very important on the motion: the only allowed modes of vibration are those whose wavelengths produce zero displacements at the bound ends of the string; if the string breaks or becomes unattached at one end, it becomes silent. In its lowest-energy mode of vibration there is a single wave whose point of maximum displacement is placed at the center of the string. In musical terms, this corresponds to the fundamental note to which the string is tuned; in terms of the theory of vibrations, it corresponds to a "quantum number" of 1. Higher modes, known as (and in music, as ), contain 2, 3, 4 and more points of maximum displacement ( ) spaced evenly along the string, separated by points of zero displacement ( ). These correspond to successively higher quantum numbers and higher energies. The vibrational states of the string are quantized in the sense that an integral number of antinodes must be present. Note again that this condition is imposed by the boundary condition that the ends of the string, being fixed in place, must be nodes. Because the locations of the nodes and antinodes do not change as the string vibrates, the vibrational patterns are known as . A similar kind of quantization occurs in other musical instruments; in each case the vibrations, whether of a stretched string, a column of air, or of a stretched membrane. The analogy with the atom can be seen by imagining a guitar string that has been closed into a circle. The circle is the electron orbit, and the boundary condition is that the waves must not interfere with themselves along the circle. This condition can only be met if the circumference of an orbit can exactly accommodate an integral number of wavelengths. Thus only certain discrete orbital radii and energies are allowed, as depicted in the two diagrams below. If a guitar string is plucked so harshly that it breaks, the restoring force and boundary conditions that restricted its motions to a few discrete harmonically related frequencies are suddenly absent; with no constraint on its movement, the string's mechanical energy is dissipated in a random way without musical effect. In the same way, if an atom absorbs so much energy that the electron is no longer bound to the nucleus, then the energy states of the atom are no longer quantized; instead of the line spectrum associated with discrete energy jumps, the spectrum degenerates into a continuum in which all possible electron energies are allowed. The energy at which the of an atom begins is easily observed spectroscopically, and serves as a simple method of experimentally measuring the energy with which the electron is bound to the atom. Hydrogen, the simplest atom, also has the simplest line spectrum (line spectra were briefly introduced in the previous chapter.) The hydrogen spectrum was the first to be observed (by Ånders Ångström in the 1860's). Johannn Balmer, a German high school teacher, discovered a simple mathematical formula that related the wavelengths of the various lines that are observable in the visible and near-uv parts of the spectrum. This set of lines is now known as the . The four lines in the visible spectrum (designated by α through δ) were the first observed by Balmer. Notice how the lines crowd together as they approach the in the near-ultraviolet part of the spectrum. Once the electron has left the atom, it is in an unbound state and its energy is no longer quantized. When such electrons return to the atom, they possess random amounts of kinetic energies over and above the binding energy. This reveals itself as the radiation at the short-wavelength end of the spectrum known as the radiation. Other named sets of lines in the hydrogen spectrum are the (in the ultraviolet) and the and series in the infrared. Each spectral line represents an energy difference between two possible states of the atom. Each of these states corresponds to the electron in the hydrogen atom being in an "orbit" whose radius increases with the quantum number . The lowest allowed value of is 1; because the electron is as close to the nucleus as it can get, the energy of the system has its minimum (most negative) value. This is the "normal" (most stable) state of the hydrogen atom, and is called the . If a hydrogen atom absorbs radiation whose energy corresponds to the difference between that of =1 and some higher value of , the atom is said to be in an excited state. Excited states are unstable and quickly decay to the ground state, but not always in a single step. For example, if the electron is initially promoted to the =3 state, it can decay either to the ground state or to the =2 state, which then decays to =1. Thus this single =1→3 excitation can result in the three emission lines depicted in the diagram above, corresponding to =3→1, =3→2, and =2→1. If, instead, enough energy is supplied to the atom to completely remove the electron, we end up with a hydrogen ion and an electron. When these two particles recombine (H + e → H), the electron can initially find itself in a state corresponding to any value of , leading to the emission of many lines. The lines of the hydrogen spectrum can be organized into different according to the value of at which the emission terminates (or at which absorption originates.) The first few series are named after their discoverers. The most well-known (and first-observed) of these is the Balmer series, which lies mostly in the visible region of the spectrum. The Lyman lines are in the ultraviolet, while the other series lie in the infrared. The lines in each series crowd together as they converge toward the which corresponds to ionization of the atom and is observed as the beginning of the emission. Note that the of hydrogen (from its ground state) is 1312 kJ mol . Although an infinite number of -values are possible, the number of observable lines is limited by our ability to resolve them as they converge into the continuum; this number is around a thousand. The line spectra we have been discussing are produced when electrons which had previously been excited to values of greater than 1 fall back to the =1 ground state, either directly, or by way of intermediate- states. But if light from a continuous source (a hot body such as a star) passes through an atmosphere of hydrogen (such as the star's outer atmosphere), those wavelengths that correspond to the allowed transitions are absorbed, and appear as dark lines superimposed on the continuous spectrum. These dark absorption lines were first observed by William Wollaston in his study of the solar spectrum. In 1814, Joseph von Fraunhofner (1787-1826) re-discovered them and made accurate measurements of 814 lines, including the four most prominent of the Balmer lines.

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Humanity's first chemical knowledge was mostly , like metal working, ceramics, cooking, etc. Early civilizations learned to control fire, to cast metals and make alloys, to make glass and ceramics, and so forth. The first chemical thinking, as opposed to chemical applications, asked: What is ? Matter is stuff. It's what we are made of, what the earth and the air are made of. Matter is anything that has mass... what is ? It's the amount of stuff. Not how much space it occupies (that's ) but how much stuff is there. We measure mass using , which is how strongly the stuff in question is attracted to the earth by gravity. What is matter made of? Well, one philosopher of ancient Greece proposed that all matter is made of . He observed that water can "become air" by , or become solid by freezing into ice. He reasoned therefore that water can convert into everything, and matter is made of water. Now, we call those changes . The water is still water when it boils and turns into steam. The water is still water when it freezes into ice. We changed its , not its nature. Another Greek philosopher said that everything was made of : when air becomes less compressed, it becomes fire, and when more compressed, it turns into water, stones, and so forth. He offered the proof that when you breathe through open lips, the air is warm, and when you compress it by breathing through puckered lips, it's cold, and condenses into liquid or solid. Air turning into stone would be a , in modern terms. In terms of Dalton's Atomic Theory, a chemical change means that the atoms form new combinations, like one atom of A combining with an atom of B. Before A and B were separate, but now they are attached. That's a chemical change. Others proposed 5 elements, with distinct shapes: , , , etc. For a long while, the four element model (earth, air, fire, water) was popular. (In Greece it was proposed by a man who was asked to become king of his city, but created a democracy instead. Then he declared himself a god and jumped into a volcano. It's said that the volcano tossed back his sandals to prove he wasn't a god.) This model, which Plato and Aristotle also used, suggested that all matter was composed of these four elements in different ratios. For instance, when wood is burned, you get smoke (air), ash (earth), pitch (a viscous liquid, here identified with water), and fire, so wood is made of all these things. Wood burning is a great example of a chemical change. From a modern perspective, in all these theories an important element was lacking: experimentation. The Greeks preferred thinking to trying things, and you might say that it shows in their theories. For them, the fundamental difference was what we now call the or the . , and are the three main states of matter. The three states are fundamentally different in nature. Gases take up as much space as they can, in whatever shape they can: the molecules are far apart, and try to spread out. Liquids change shape but have a constant volume. Solids have a constant shape and volume. Now we know that you can change the state of matter just by changing the temperature and pressure, and the molecules stay the same. So the wrong theories of the Greeks were based on not recognizing the difference between physical changes, which are changes in the state of matter, and chemical changes, changes in the combination of atoms. is anything that has volume and mass. is the amount of matter in a substance. involve processes that change the form of a substance but not its chemical composition. involves processes in which a new substances are formed with new chemical compositions. is the force resulting from a mass being pulled towards a particular location as a result of gravity. The three are: solid, liquid, and gas. A is an object with a fixed shape and volume. are fluids have a constant volume but are capable of changing their shape. are fluids that do not have constant shape or volume so they take up as much space as they can within a given container.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Complex_Molecular_Synthesis_(Salomon)/06%3A_Amino_Acids_and_Alkaloids

The alkaloids are a diverse family of nitrogen-containing natural products that generally are produced from amino acids in plants. Phenyl rings derived from aromatic amino acids may often be discerned embedded in the skeletons of some alkaloids. For example, the A ring of colchicine is derived from L-phenylalanine and the A rings of cephalotaxine and morphine are derived from L-tyrosine. Interestingly, the remaining carbons of the above mentioned alkaloids are also derived exclusively from L- phenylalanine or L-tyrosine. The loss of aromaticity that is common during such biosyntheses is an example of the unusual synthetic strategies that must be adopted in Nature owing to a limited selection of available starting materials. The aromatic rings of polyketides (see chapter 5) arise from acetyl-CoA by a culminating in dehydrocyclization of intermediate poly-β-ketoalkanoic acids. In the biosynthetsis of the aromatic amino acids L-phenyl-alanine, L-tyrosine, and L-tryptophan, the aromatic rings are assembled by a more starting with an aldol condensation of phosphoenol pyruvate (PEP) and erythrose 4-phosphate (E4P). These starting materials are available from glucose metabolism (see chapter 2). Cyclization of the enol tautomer of the resulting 3-deoxy-D-arabinoheptulosonic acid 7-phosphate ( ) is reminiscent of the polyene cyclizations that are initiated by allylic pyrophosphates which are encountered in the biosynthesis of terpenes (see chapter 4). Dehydration and reduction then provide shikimic acid ( ), the intermediate for which this biosynthetyic pathway is named. Phosphorylation of and transetherification with a second molecule of PEP leads to a pivotal intermediate, chorismic acid ( ). Appendage of the final three carbons of tyrosine and phenylalanine is achieved by a Claisen, i.e. [3.3] sigmatropic, rearrangement of that produces prephenic acid ( ). Decarboxylative elimination generates phenylpyruvic acid from while oxidation and decarboxylation of the resulting vinylogous b-keto acid affords p- hydroxyphenylpyruvic acid. Transamination of the arylpyruvic acids (see → on section 5.3) delivers the corresponding α-amino acids L-phenylalanine ( ) and L-tyrosine ( ). Amino acids are not the only building blocks incorporated into alkaloids. Thus, for example, some alkaloids incorporate starting materials of terpenoid origin. Quinine is assembled in Nature by the union of L-tryptophan with secologanin, a monoterpene. Interestingly, neither the tryptophane origin of the aromatic portion of quinine nor the terpenoid biogenisis of secologanin are at all obvious. Much more obvious is the presence of L-tryptophan and an isopentenyl group embedded in the skeleton of lysergic acid. Polyketide fragments and nonaromatic amino acids may also serve as building blocks for alkaloids. For example, lycopodine is derived in nature from two molecules of L-lysine and one of acetoacetyl CoA.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids

When you have completed Chapter 28, you should be able to Two types of nucleic acids are found in cells—deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). These highly complex substances are built up from a number of simpler units, called nucleotides. Each nucleotide consists of three parts: a phosphoric acid residue, a sugar and a nitrogen‑containing heterocyclic base. Thus, in order to understand the biochemistry of the nucleic acids, you must first study the chemistry of the sugars (see Chapter 25) and simple heterocyclic systems. We have already discussed certain aspects of the structure of heterocyclic ring systems during our study of aromaticity (Sections 15.5–15.6). You may find it helpful to review this chapter. Chapter 28 examines the structure and replication of DNA and then describes the structure and synthesis of RNA. The chapter closes with a brief study of the role played by RNA in the biosynthesis of proteins.

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

A resonance form is another way of drawing a Lewis dot structure for a given compound. Equivalent Lewis structures are called resonance forms. They are used when there is more than one way to place double bonds and lone pairs on atoms. Resonance structures arise when there are more than one way to draw a Lewis dot diagram that satisfies the . Remember the octet rule is where the atom gains, loses, or shares electrons so that the outer electron shell has eight electrons. We draw them when one structure does not accurately show the real structure. There are some basic principle on the resonance theory. First resonance structures are not real, they just show possible structures for a compound. Resonance structures are not in equilibrium with each other. Resonance structures are not isomers. Isomers have different arrangement of both atoms and electrons. Resonance forms differ in arrangement of electrons. Resonance structures are a better depiction of a Lewis dot structure because they clearly show bonding in molecules. Not all resonance structures are equal there are some that are better than others. The better ones have minimal formal charges, negative formal charges are the most electronegative atoms, and bond is maximized in the structure. The more resonance forms a molecule has makes the molecule more stable. They are drawn with a double-headed arrow between them to show the actual structure is somewhere between the resonance structures. These structures used curved arrow notation to show the movement of the electrons in one resonance form to the next. Formal charges are used in Chemistry to determine the location of a charge in a molecule and determine how good of a Lewis structure it will be. Remember, the best resonance structure is the one with the least formal charge. This is why formal charges are very important. Atoms that are missing one or more electrons will have a positive charge. An atom with many electrons will have a negative charge. Assigning to an atom is very useful in resonance forms. Formal charge is calculated using this format: # of valence electrons- (#non bonding electrons + 1/2 #bonding electrons) Curved arrow notation is used in showing the placement of electrons between atoms. The tail of the arrow begins at the electron source and the head points to where the electron will be. Make sure the arrows are clear including the single and half headed arrow. The reader must know the flow of the electrons. Benzene is commonly seen in Organic Chemistry and it has a resonance form. Benzene has two resonance structures, showing the placements of the bonds. Another example of resonance is ozone. Ozone is represented by two different Lewis structures. The difference between the two structures is the location of double bond. There are several things that should be checked before and after drawing the resonance forms. First know where the nonbonding electrons are, keep track of formal charges on atoms, and do not break sigma bonds. Finally, after drawing the resonance form make sure all the atoms have eight electrons in the outer shell. Checking these will make drawing resonance forms easier. When drawing a resonance structure there are three rules that need to be followed for the structures to be correct: Approaches for moving electrons are move pi electrons toward a positive charge or toward an another pi bond. Move a single nonbonding electron towards a pi bond. Move lone pair electrons toward a pi bond and when electrons can be moved in more than one direction, move them to the more electronegative atom. After drawing resonance structures check the net charge of all the structures. For example, if a structure has a net charge of +1 then all other structures must also have a net charge of +1. If not, the structure is not correct. Always check the net charge after each structure. These important details can ensure success in drawing any Resonance structure. 1) 2) Fill in any lone pair electrons and identify any pi bond electrons. 3) Fill in any lone pair electrons and identify any pi bond electrons. 4) 5)

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A third gas law may be derived as a corollary to and laws. Suppose we double the thermodynamic temperature of a sample of gas. According to Charles’s law, the volume should double. Now, how much pressure would be required at the higher temperature to return the gas to its original volume? According to Boyle’s law, we would have to double the pressure to halve the volume. Thus, if the volume of gas is to remain the same, doubling the temperature will require doubling the pressure. This law was first stated by the Frenchman Joseph Gay-Lussac (1778 to 1850). According to , Mathematically, \[P\propto T\text{ or }P=k_{\text{G}}T\text{ or }\frac{P}{T}=k_{\text{G}} \nonumber \] where is the appropriate proportionality constant. Gay-Lussac’s law tells us that it may be dangerous to heat a gas in a closed container. The increased pressure might cause the container to explode, as you can see in the video below. The video shows very, very cold nitrogen gas in a bottle being warmed by the air. Since the bottle's volume is relatively constant, as the temperature of the nitrogen gas (formed when the liquid nitrogen boils) increases, so does the pressure inside the bottle until, finally, BOOM! A container is designed to hold a pressure of 2.5 atm. The volume of the container is 20.0 cm , and it is filled with air at room temperature (20°C) and normal atmospheric pressure. Would it be safe to throw the container into a fire where temperatures of 600°C would be reached? Using the common-sense method, we realize that the pressure will increase at the higher temperature, and so \[P_{\text{2}}=\text{1}\text{.0 atm }\times \frac{\text{(273}\text{.15 + 600) K}}{\text{(273}\text{.15 + 20) K}}=\text{3}\text{.0 atm} \nonumber \] This would exceed the safe strength of the container. Note that the volume of the container was not needed to solve the problem. This concept works in reverse, as well. For instance, if we subject a gas to lower temperatures than their initial state, the external atmosphere can actually force the container to shrink. The following video demonstrates how a sample of hot gas, when cooled will collapse a container. A syringe barrel is filled with hot steam (vaporized water) and a plunger placed to cap off the end. The syringe is then placed in a beaker of ice water to cool the internal gas. When the temperature of the water vapor decreases, the pressure exerted by the vapor decreases as well. This leads to a difference in pressure between the vapor inside the barrel and the atmosphere. Atmospheric pressure then pushes the plunger into the barrel.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.5%3A_Colloids

As a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, when we make a solution, we prepare a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. A group of mixtures called (or ) exhibit properties intermediate between those of suspensions and solutions (Figure \(\Page {1}\)). The particles in a colloid are larger than most simple molecules; however, colloidal particles are small enough that they do not settle out upon standing. The particles in a colloid are large enough to scatter light, a phenomenon called the . This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure \(\Page {2}\). Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out. The term “colloid”—from the Greek words , meaning “glue,” and , meaning “like”—was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units. Analogous to the identification of solution components as “solute” and “solvent,” the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the and the substance or solution throughout which the particulate is dispersed is called the . Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table \(\Page {1}\). We can prepare a colloidal system by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods: A few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air. We can prepare an by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an , a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents. Condensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. We can prepare a red colloidal suspension of iron(III) hydroxide by mixing a concentrated solution of iron(III) chloride with hot water: \[\mathrm{Fe^{3+}}_{(aq)}+\mathrm{3Cl^-}_{(aq)}+\mathrm{6H_2O}_{(l)}⟶\mathrm{Fe(OH)}_{3(s)}+\mathrm{H_3O^+}_{(aq)}+\mathrm{3Cl^-}_{(aq)} \label{11.6.1} \] A colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate: \[\ce{Au}^{3+}+ \ce{3e}^− \rightarrow \ce{Au} \label{11.6.2} \] Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids. Pioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, \(\ce{K_2CO_3}\), from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula \(\ce{C_{17}H_{35}CO_2Na}\) and contains an uncharged nonpolar hydrocarbon chain, the \(\mathrm{C_{17}H_{35}-}\) unit, and an ionic carboxylate group, the \(-\mathrm{\sideset{ }{_{2}^{-}}{CO}}\) unit (Figure \(\Page {3}\)). ents (soap substitutes) also contain nonpolar hydrocarbon chains, such as \(\mathrm{C_{12}H_{25}—}\), and an ionic group, such as a sulfate \(—\mathrm{\sideset{ }{_{3}^{-}}{OSO}}\), or a sulfonate \(—\mathrm{\sideset{ }{_{3}^{-}}{SO}}\) (Figure \(\Page {4}\)). Soaps form insoluble calcium and magnesium compounds in hard water; detergents form water-soluble products—a definite advantage for detergents. The cleaning action of soaps and detergents can be explained in terms of the structures of the molecules involved. The hydrocarbon (nonpolar) end of a soap or detergent molecule dissolves in, or is attracted to, nonpolar substances such as oil, grease, or dirt particles. The ionic end is attracted by water (polar), illustrated in Figure \(\Page {5}\). As a result, the soap or detergent molecules become oriented at the interface between the dirt particles and the water so they act as a kind of bridge between two different kinds of matter, nonpolar and polar. Molecules such as this are termed since they have both a hydrophobic (“water-fearing”) part and a hydrophilic (“water-loving”) part. As a consequence, dirt particles become suspended as colloidal particles and are readily washed away. The blowout of the Deepwater Horizon oil drilling rig on April 20, 2010, in the Gulf of Mexico near Mississippi began the largest marine oil spill in the history of the petroleum. In the 87 days following the blowout, an estimated 4.9 million barrels (210 million gallons) of oil flowed from the ruptured well 5000 feet below the water’s surface. The well was finally declared sealed on September 19, 2010. Crude oil is immiscible with and less dense than water, so the spilled oil rose to the surface of the water. Floating booms, skimmer ships, and controlled burns were used to remove oil from the water’s surface in an attempt to protect beaches and wetlands along the Gulf coast. In addition to removal of the oil, attempts were also made to lessen its environmental impact by rendering it “soluble” (in the loose sense of the term) and thus allowing it to be diluted to hopefully less harmful levels by the vast volume of ocean water. This approach used 1.84 million gallons of the oil dispersant Corexit 9527, most of which was injected underwater at the site of the leak, with small amounts being sprayed on top of the spill. Corexit 9527 contains 2-butoxyethanol (C H O ), an amphiphilic molecule whose polar and nonpolar ends are useful for emulsifying oil into small droplets, increasing the surface area of the oil and making it more available to marine bacteria for digestion (Figure \(\Page {6}\)). While this approach avoids many of the immediate hazards that bulk oil poses to marine and coastal ecosystems, it introduces the possibility of long-term effects resulting from the introduction of the complex and potential toxic components of petroleum into the ocean’s food chain. A number of organizations are involved in monitoring the extended impact of this oil spill, including the National Oceanic and Atmospheric Administration. Dispersed colloidal particles are often electrically charged. A colloidal particle of iron(III) hydroxide, for example, does not contain enough hydroxide ions to compensate exactly for the positive charges on the iron(III) ions. Thus, each individual colloidal particle bears a positive charge, and the colloidal dispersion consists of charged colloidal particles and some free hydroxide ions, which keep the dispersion electrically neutral. Most metal hydroxide colloids have positive charges, whereas most metals and metal sulfides form negatively charged dispersions. All colloidal particles in any one system have charges of the same sign. This helps keep them dispersed because particles containing like charges repel each other. We can take advantage of the charge on colloidal particles to remove them from a variety of mixtures. If we place a colloidal dispersion in a container with charged electrodes, positively charged particles, such as iron(III) hydroxide particles, would move to the negative electrode. There, the colloidal particles lose their charge and coagulate as a precipitate. The carbon and dust particles in smoke are often colloidally dispersed and electrically charged. The charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust (Figure \(\Page {7}\). This is one of the important methods used to clean up the smoke from a variety of industrial processes. The process is also important in the recovery of valuable products from the smoke and flue dust of smelters, furnaces, and kilns. There are also ionic air filters designed for home use to improve indoor air quality. When we make gelatin, such as Jell-O, we are making a type of colloid (Figure \(\Page {8}\)). Gelatin sets on cooling because the hot aqueous mixture of gelatin coagulates as it cools and the whole mass, including the liquid, sets to an extremely viscous body known as a , a colloid in which the dispersing medium is a solid and the dispersed phase is a liquid. It appears that the fibers of the dispersing medium form a complex three-dimensional network, the interstices being filled with the liquid medium or a dilute solution of the dispersing medium. Because the formation of a gel is accompanied by the taking up of water or some other solvent, the gel is said to be hydrated or solvated. Pectin, a carbohydrate from fruit juices, is a gel-forming substance important in jelly making. Silica gel, a colloidal dispersion of hydrated silicon dioxide, is formed when dilute hydrochloric acid is added to a dilute solution of sodium silicate. Canned Heat is a gel made by mixing alcohol and a saturated aqueous solution of calcium acetate. Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity, and they are often electrically charged. Colloids are widespread in nature and are involved in many technological applications.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/07%3A_Solids_and_Liquids/7.09%3A_Polymers_and_Plastics

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 bold terms in the context of this topic. Plastics and natural materials such as rubber or cellulose are composed of very large molecules called polymers. Polymers are constructed from relatively small molecular fragments known as monomers that are joined together. Wool, cotton, silk, wood and leather are examples of natural polymers that have been known and used since ancient times. This group includes biopolymers such as proteins and carbohydrates that are constituents of all living organisms. Synthetic polymers, which includes the large group known as plastics, came into prominence in the early twentieth century. Chemists' ability to engineer them to yield a desired set of properties (strength, stiffness, density, heat resistance, electrical conductivity) has greatly expanded the many roles they play in the modern industrial economy. This Module deals mostly with synthetic polymers, but will include a synopsis of some of the more important natural polymers. It will close with a summary of some of the very significant environmental problems created by the wide use of plastics. Let's begin by looking at an artificial polymer that is known to everyone in the form of flexible, transparent : . It is also the simplest polymer, consisting of random-length (but generally very long) chains made up of two-carbon units. You will notice some "fuzziness" in the way that the polyethylene structures are represented above. The squiggly lines at the ends of the long structure indicate that the same pattern extends indefinitely. The more compact notation on the right shows the minimal repeating unit enclosed in brackets overprinted with a dash; this means the same thing and is the preferred way of depicting polymer structures. In most areas of chemistry, a "pure substance" has a definite structure, molar mass, and properties. It turns out, however, that few polymeric substances are uniform in this way. This is especially the case with synthetic polymers, whose molecular weights cover a range of values, as may the sequence, orientation, and connectivity of the individual monomers. So most synthetic polymers are really rather than pure substances in the ordinary chemical sense of the term. Their molecular weights are typically distributed over a wide range. Figure \(\Page {1}\): Polymers Don't be misled by chemical formulas that depict polymers such as polyethylene as reasonably straight chains of substituted carbon atoms. Free rotation around C—C bonds allows long polymer molecules to curl up and and tangle very much like spaghetti (Figure \(\Page {2}\)). Thus polymers generally form solids. There are, however, ways in which certain polymers can be partially oriented. Polymers can be classified in ways that reflect their chemical makeup, or perhaps more importantly, their properties and applications. Many of these factors are strongly interdependent, and most are discussed in much more detail in subsequent sections of this page. The physical properties of a polymer such as its strength and flexibility depend on: The spaghetti-like entanglements of polymer molecules tend to produce amorphous solids, but it often happens that some parts can become sufficiently aligned to produce a region exhibiting crystal-like order, so it is not uncommon for some polymeric solids to consist of a random mixture of amorphous and crystalline regions. As might be expected, shorter and less-branched polymer chains can more easily organize themselves into ordered layers than have can long chains. Hydrogen-bonding between adjacent chains also helps, and is very important in fiber-forming polymers both synthetic (Nylon 6.6) and natural (cotton cellulose). Pure crystalline solids have definite melting points, but polymers, if they melt at all, exhibit a more complex behavior. At low temperatures, the tangled polymer chains tend to behave as rigid glasses. For example, the natural polymer that we call becomes hard and brittle when cooled to liquid nitrogen temperature. Many synthetic polymers remain in this state to well above room temperature. The melting of a crystalline compound corresponds to a sudden loss of long-range order; this is the fundamental reason that such solids exhibit definite melting points, and it is why there is no intermediate form between the liquid and the solid states. In amorphous solids there is no long-range order, so there is no melting point in the usual sense. Such solids simply become less and less viscous as the temperature is raised. In some polymers (known as ) there is a fairly definite softening point that is observed when the thermal kinetic energy becomes high enough to allow internal rotation to occur within the bonds and to allow the individual molecules to slide independently of their neighbors, thus rendering them more flexible and deformable. This defines the . Depending on the degree of crystallinity, there will be a higher temperature, the , at which the crystalline regions come apart and the material becomes a viscous liquid. Such liquids can easily be injected into molds to manufacture objects of various shapes, or extruded into sheets or fibers. Other polymers (generally those that are highly cross-linked) do not melt at all; these are known as . If they are to be made into molded objects, the polymerization reaction must take place within the molds — a far more complicated process. About 20% of the commercially-produced polymers are thermosets; the remainder are thermoplastics. Copolymerization is an invaluable tool for "tuning" polymers so that they have the right combination of properties for an application. For example, homopolymeric polystyrene is a rigid and very brittle transparent thermoplastic with a glass transition temperature of 97°C. Copolymerizing it with acrylonitrile yields an alternating "SAN" copolymer in which t is raised to 107°, making it useable for transparent drink containers. A polymer that is composed of identical monomeric units (as is polyethylene) is called a . are built up from more than one type of monomer. Artificial heteropolymers are more commonly known as copolymers. Polymers may also be classified as straight-chained or branched, leading to forms such as these: The monomers can be joined end-to-end, and they can also be to provide a harder material: If the cross-links are fairly long and flexible, adjacent chains can move with respect to each other, producing an polymer or In a linear polymer such as polyethylene, rotations around carbon-carbon single bonds can allow the chains to bend or curl up in various ways, resulting in the spaghetti-like mixture of these different we alluded to above. But if one of the hydrogen atoms is replaced by some other entity such as a methyl group, the relative orientations of the individual monomer units that make up a linear section of any carbon chain becomes an important characteristic of the polymer. occurs because rotation around carbon-carbon double bonds is not possible — unlike the case for single bonds. Any pair of unlike substituents attached to the two carbons is permanently locked into being on the same side ( ) or opposite sides ( ) of the double bond. If the carbon chain contains double bonds, then becomes possible, giving rise to two different possible configurations (known as diastereomers) at each unit of the chain. This seemingly small variable can profoundly affect the nature of the polymer. For example, the latex in natural rubber is made mostly of -polyisoprene, whereas the isomer (known as gutta percha latex) has very different (and generally inferior) properties. The tetrahedral nature of carbon bonding has an important consequence that is not revealed by simple two-dimensional structural formulas: atoms attached to the carbon can be on one side or on the other, and these will not be geometrically equivalent if all four of the groups attached to a single carbon atom are different. Such carbons (and the groups attached to them) are said to be and can exist in two different three-dimensional forms known as For an individual carbon atom in a polymer chain, two of its attached groups will ordinarily be the chain segments on either side of the carbon. If the two remaining groups are different (say one hydrogen and the other methyl), then the above conditions are satisfied and this part of the chain can give rise to two enantiomeric forms. A chain that can be represented as (in which the orange and green circles represent different groups) will have multiple chiral centers, giving rise to a huge number of possible enantiomers. In practice, it is usually sufficient to classify chiral polymers into the following three classes of , usually referred to as . The tacticity of a polymer chain can have a major influence on its properties. Atactic polymers, for example, being more disordered, cannot crystallize. One of the major breakthroughs in polymer chemistry occurred in the early 1950s when the German chemist Karl Ziegler discovered a group of catalysts that could efficiently polymerize ethylene. At about the same time, Giulio Natta (Italian) made the first isotactic (and crystalline) polyethylene. The Ziegler-Natta catalysts revolutionized polymer chemistry by making it possible to control the stereoregularity of these giant molecules. The two shared the 1963 Nobel Prize in Chemistry Polymers are made by joining small molecules into large ones. But most of these monomeric molecules are perfectly stable as they are, so chemists have devised two general methods to make them react with each other, building up the backbone chain as the reaction proceeds. This method (also known as ) requires that the monomers possess two or more kinds of functional groups that are able to react with each other in such a way that parts of these groups combine to form a small molecule (often H O) which is eliminated from the two pieces. The now-empty bonding positions on the two monomers can then join together . This occurs, for example, in the synthesis of the Nylon family of polymers in which the eliminated H O molecule comes from the hydroxyl group of the acid and one of the amino hydrogens: Note that the monomeric units that make up the polymer are not identical with the starting components. or polymerization involves the rearrangement of bonds within the monomer in such a way that the monomers link up directly with each other: In order to make this happen, a chemically active molecule (called an ) is needed to start what is known as a . The manufacture of polyethylene is a very common example of such a process. It employs a initiator that donates its unpaired electron to the monomer, making the latter highly reactive and able to form a bond with another monomer at this site. In theory, only a single chain-initiation process needs to take place, and the chain-propagation step then repeats itself indefinitely, but in practice multiple initiation steps are required, and eventually two radicals react ( ) to bring the polymerization to a halt. As with all polymerizations, chains having a range of molecular weights are produced, and this range can be altered by controlling the pressure and temperature of the process. Note: the left panels below show the polymer name and synonyms, structural formula, glass transition temperature, melting point/decomposition temperature, and (where applicable) the used to facilitate recycling. (Lexan®) = 145°C, = 225°C. (PET, ) = 76°C, = 250°C. Thin and very strong films of this material are made by drawing out the molten polymer in both directions, thus orienting the molecules into a highly crystalline state that becomes "locked-in" on cooling. Its many applications include food packaging (in foil-laminated drink containers and microwaveable frozen-food containers), overhead-projector film, weather balloons, and as aluminum-coated reflective material in spacecraft and other applications. (a ) = 50°C, = 255°C. Nylon has a fascinating history, both scientific and cultural. It was invented by DuPont chemist Wallace Carothers (1896-1937). The common form has six carbon atoms in both parts of its chain; there are several other kinds. Notice that the two copolymer sub-units are held together by bonds, the same kinds that join amino acids into proteins. Nylon 6.6 has good abrasion resistance and is self-lubricating, which makes it a good engineering material. It is also widely used as a fiber in carpeting, clothing, and tire cord. For an interesting account of the development of Nylon, see by Ann Gaines (1971) (Orlon, Acrilan, "acrylic" fiber) = 85°C, = 318°C. = –78°C, = 100°C. LDPE HDPE Control of polymerization by means of catalysts and additives has led to a large variety of materials based on polyethylene that exhibit differences in densities, degrees of chain branching and crystallinity, and cross-linking. Some major types are low-density (LDPE), linear low density (LLDPE), high-density (HDPE). LDPE was the first commercial form (1933) and is used mostly for ordinary "plastic bags", but also for food containers and in six-pack soda can rings. Its low density is due to long-chain branching that inhibits close packing. LLDPE has less branching; its greater toughness allows its use in those annoyingly-thin plastic bags often found in food markets. A "very low density" form (VLDPE) with extensive short-chain branching is now used for plastic stretch wrap (replacing the original component of Saran Wrap) and in flexible tubing. HDPE has mostly straight chains and is therefore stronger. It is widely used in milk jugs and similar containers, garbage containers, and as an "engineering plastic" for machine parts. (Plexiglass, Lucite, Perspex) = 114°C, = 130-140°C. = –10°C, = 173°C. PP Polypropylene is used alone or as a copolymer, usually with with ethylene. These polymers have an exceptionally wide range of uses — rope, binder covers, plastic bottles, staple yarns, non-woven fabrics, electric kettles. When uncolored, it is translucent but not transparent. Its resistance to fatigue makes it useful for food containers and their lids, and flip-top lids on bottled products such as ketchup. = 95°C, = 240°C. PS Polystyrene is transparent but rather brittle, and yellows under uv light. Widely used for inexpensive packaging materials and "take-out trays", foam "packaging peanuts", CD cases, foam-walled drink cups, and other thin-walled and moldable parts. = 30°C PVA is too soft and low-melting to be used by itself; it is commonly employed as a water-based emulsion in paints, wood glue and other adhesives. ("vinyl", "PVC") = 85°C, = 240°C. PVC (polychloroprene) = –70°C < –90°C Neoprene, invented in 1930, was the first mass-produced synthetic rubber. It is used for such things as roofing membranes and wet suits. Polybutadiene substitutes a hydrogen for the chlorine; it is the major component (usually admixed with other rubbers) of tires. Synthetic rubbers played a crucial role in World War II SBS (styrene-butadiene-styrene) rubber is a block copolymer whose special durability makes it valued for tire treads. ( , PTFE) Decomposes above 350°C. This highly-crystalline fluorocarbon is exceptionally inert to chemicals and solvents. Water and oils do not wet it, which accounts for its use in cooking ware and other anti-stick applications, including personal care products. These properties — non-adhesion to other materials, non-wetability, and very low coefficient of friction ("slipperyness") — have their origin in the highly electronegative nature of fluorine whose atoms partly shield the carbon chain. Fluorine's outer electrons are so strongly attracted to its nucleus that they are less available to participate in London (dispersion force) interactions. ) Sublimation temperature 450°C. Kevlar is known for its ability to be spun into fibers that have five times the tensile strength of steel. It was first used in the 1970s to replace steel tire cords. Bullet-proof vests are one of it more colorful uses, but other applications include boat hulls, drum heads, sports equipment, and as a replacement for asbestos in brake pads. It is often combined with carbon or glass fibers in composite materials. The high tensile strength is due in part to the extensive hydrogen bonding between adjacent chains. Kevlar also has the distinction of having been invented by a woman chemist, Stephanie Kwolek. The thermoplastic materials described above are chains based on relatively simple monomeric units having varying degrees of polymerization, branching, bending, cross-linking and crystallinity, but with each molecular chain being a discrete unit. In thermosets, the concept of an individual molecular unit is largely lost; the material becomes more like a gigantic extended molecule of its own — hence the lack of anything like a glass transition temperature or a melting point. These properties have their origins in the nature of the monomers used to produce them. The most important feature is the presence of multiple reactive sites that are able to form what amount to cross-links at every center. The phenolic resins, typified by the reaction of phenol with formaldehyde, illustrate the multiplicity of linkages that can be built. Polymers derived from plants have been essential components of human existence for thousands of years. In this survey we will look at only those that have major industrial uses, so we will not be discussing the very important biopolymers and . Polysaccharides are polymers of ; they play essential roles in energy storage, signaling, and as structural components in all living organisms. The only ones we will be concerned with here are those composed of , the most important of the six-carbon . Glucose serves as the primary fuel of most organisms. Glucose, however, is highly soluble and cannot be easily stored, so organisms make polymeric forms of glucose to set aside as , from which glucose molecules can be withdrawn as needed. In humans and higher animals, the reserve storage polymer is . It consists of roughly 60,000 glucose units in a highly branched configuration. Glycogen is made mostly in the liver under the influence of the hormone which triggers a process in which digested glucose is polymerized and stored mostly in that organ. A few hours after a meal, the glucose content of the blood begins to fall, and glycogen begins to be broken down in order to maintain the body's required glucose level. In plants, these glucose-polymer reserves are known as . Starch granules are stored in seeds or tubers to provide glucose for the energy needs of newly-germinated plants, and in the twigs of deciduous plants to tide them over during the winter when (the process in which glucose is synthesized from CO and H O) does not take place. The starches in food grains such as rice and wheat, and in tubers such as potatoes, are a major nutritional source for humans. Plant starches are mixtures of two principal forms, and . Amylose is a largely-unbranched polymer of 500 to 20,000 glucose molecules that curls up into a helical form that is stabilized by internal hydrogen bonding. Amylopectin is a much larger polymer having up to two million glucose residues arranged into branches of 20 to 30 units. Extensive hydrogen bonding between the chains causes native cellulose to be about 70% crystalline. It also raises the melting point (>280°C) to above its combustion temperature. The structures of starch and cellulose appear to be very similar; in the latter, every other glucose molecule is "upside-down". But the consequences of this are far-reaching; starch can dissolve in water and can be digested by higher animals including humans, whereas cellulose is insoluble and undigestible. Cellulose serves as the principal structural component of green plants and (along with lignin) in wood. is one of the purest forms of cellulose and has been cultivated since ancient times. Its ability to absorb water (which increases its strength) makes cotton fabrics especially useful for clothing in very warm climates. Cotton also serves (along with treated wood pulp) as the source the industrial production of cellulose-derived materials which were the first "plastic" materials of commercial importance. A variety of plants produce a sap consisting of a colloidal dispersion of -polyisoprene. This milky fluid is especially abundant in the rubber tree ( ), from which it drips when the bark is wounded. After collection, the latex is coagulated to obtain the solid rubber. Natural rubber is thermoplastic, with a glass transition temperature of –70°C. -polyisoprene Raw natural rubber tends to be sticky when warm and brittle when cold, so it was little more than a novelty material when first introduced to Europe around 1770. It did not become generally useful until the mid-nineteenth century when Charles Goodyear found that heating it with sulfur — a process he called — could greatly improve its properties. Vulcanization creates disulfide cross-links that prevent the polyisoprene chains from sliding over each other. The degree of cross-linking can be controlled to produce a rubber having the desired elasticity and hardness. More recently, other kinds of chemical treatment (such as epoxidation) have been developed to produce rubbers for special purposes. " is a famous commercial slogan that captured the attitude of the public around 1940 when synthetic polymers were beginning to make a major impact in people's lives. What was not realized at the time, however, were some of the problems these materials would create as their uses multiplied and the world became more wary of "chemicals". (DuPont dropped the "through chemistry" part in 1982.) Many kinds of polymers contain small molecules — either unreacted monomers, or substances specifically added (plasticizers, uv absorbers, flame retardants, etc.) to modify their properties. Many of these smaller molecules are able to diffuse through the material and be released into any liquid or air in contact with the plastic — and eventually into the aquatic environment. Those that are used for building materials (in mobile homes, for example) can build up in closed environments and contribute to indoor air pollution. Formation of long polymer chains is a complicated and somewhat random process that is never perfectly stoichiometric. It is therefore not uncommon for some unreacted monomer to remain in the finished product. Some of these monomers, such as formaldehyde, styrene (from polystyrene, including polystyrene foam food take-out containers), vinyl chloride, and bisphenol-A (from polycarbonates) are known carcinogens. Although there is little evidence that the small quantities that diffuse into the air or leach out into fluids pose a quantifiable health risk, people are understandably reluctant to tolerate these exposures, and public policy is gradually beginning to regulate them. (PFOA), the monomer from which Teflon is made, has been the subject of a 2004 lawsuit against a DuPont factory that contaminated groundwater. Small amounts of PFOA have been detected in gaseous emissions from hot fluorocarbon products. These substances are compounded into certain types of plastics to render them more flexible by lowering the glass transition temperature. They accomplish this by taking up space between the polymer chains and acting as lubricants to enable the chains to more readily slip over each other. Many (but not all) are small enough to be diffusible and a potential source of health problems. Polyvinyl chloride polymers are one of the most widely-plasticized types, and the odors often associated with flexible vinyl materials such as garden hoses, waterbeds, cheap shower curtains, raincoats and upholstery are testament to their ability to migrate into the environment. The well-known "new car smell" is largely due to plasticizer release from upholstery and internal trim. There is now an active movement to develop non-diffusible and "green" plasticizers that do not present these dangers. To complicate matters even further, many of these small molecules have been found to be physiologically active owing to their ability to mimic the action of hormones or other signaling molecules, probably by fitting into and binding with the specialized receptor sites present in many tissues. The evidence that many of these chemicals are able to act in this way at the cellular level is fairly clear, but there is still some dispute whether many of these pose actual health risks to adult humans at the relatively low concentrations in which they commonly occur in the environment. There is, however, some concern about the effects of these substances on non-adults and especially on fetuses, given that endocrines are intimately connected with sexual differentiation and neurological development which continues up through the late teens. Most commonly-used polymers are not readily biodegradable, particularly under the anaerobic conditions of most landfills. And what decomposition does occur will combine with rainwater to form that can contaminate nearby streams and groundwater supplies. Partial photodecomposition, initiated by exposure to sunlight, is a more likely long-term fate for exposed plastics, resulting in tiny broken-up fragments. Many of these materials are less dense than seawater, and once they enter the oceans through coastal sewage outfalls or from marine vessel wastes, they tend to remain there indefinitely. of polymeric materials containing chlorine (polyvinyl chloride, for example) is known to release compounds such as dioxins that persist in the environment. Incineration under the right conditions can effectively eliminate this hazard. Disposed products containing (Teflon-coated ware, some personal-care, waterproofing and anti-stick materials) break down into perfluorooctane sulfonate which has been shown to damage aquatic animals. There are two general types of hazards that polymers can introduce into the aquatic environment. One of these relates to the release of small molecules that act as hormone disrupters as described above. It is well established that small aquatic animals such as fish are being seriously affected by such substances in many rivers and estuarine systems, but details of the sources and identities of these molecules have not been identified. One confounding factor is the release of sewage water containing human birth-control drugs (which have a feminizing effect on sexual development) into many waterways. The other hazard relates to pieces of plastic waste that aquatic animals mistake for food or become entangled in. These dangers occur throughout the ocean, but are greatly accentuated in regions known as gyres. These are regions of the ocean in which a combination of ocean currents drives permanent vortices that tend to collect and concentrate floating materials. The most notorious of these are the Great Pacific Gyres that have accumulated astounding quantities of plastic waste. The huge quantity (one estimate is 10 metric tons per year) of plastic materials produced for consumer and industrial use has created a gigantic problem of what to do with plastic waste which is difficult to incinerate safely and which, being largely non-biodegradable, threatens to overwhelm the capacity of landfills. An additional consideration is that production most of the major polymers consumes non-renewable hydrocarbon resources. Plastic water bottles (left) present a special recycling problem because of their widespread use in away-from-home locations. Plastics recycling has become a major industry, greatly aided by enlightened trash management policies in the major developed nations. However, it is plagued with some special problems of its own: Some of the major recycling processes include In order to facilitate efficient recycling, a set of seven has been established (the seventh, not shown below, is "other"). These codes are stamped on the bottoms of many containers of widely-distributed products. Not all categories are accepted by all local recycling authorities, so residents need to be informed about which kinds should be placed in recycling containers and which should be combined with ordinary trash. The large number of rubber tires that are disposed of, together with the increasing reluctance of landfills to accept them, has stimulated considerable innovation in the re-use of this material, especially in the construction industry.

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Esters are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an ester the hydrogen in this group is replaced by a hydrocarbon group. This could be an alkyl group like methyl or ethyl, or one containing a benzene ring such as a phenyl or benzyl group. The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is: Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group replacing the hydrogen. In each case, be sure that you can see how the names and formulae relate to each other. Notice that the acid is named by counting up the total number of carbon atoms in the chain - including the carbonyl carbon. So, for example, \(CH_3CH_2COOH\) is propanoic acid, and \(CH_3CH_2COO\) is the propanoate group. Animal and vegetable fats and oils are composed of long-chain, complicated esters. The physical differences observed between a fat (like butter) and an oil (like sunflower oil) are due to differences in melting points of the mixture of esters they contain. If the melting point of the substance is below room temperature, it will be a liquid - an oil. If the melting point is above room temperature, it will be a solid - a fat. The causes of the differences in melting points are discussed below. Esters can be made from and . This is discussed in detail on another page; in general terms, the two combine together, losing a molecule of water in the process. Consider a very simple ester such as ethyl ethanoate. The figure below shows its formation from ethanoic acid and ethanol. The same process can be carried out for more complicated alcohols. The diagram below shows the structure of propane-1,2,3-triol (commonly known as glycerol). Just as with the ethanol in the previous equation, I've drawn this back-to-front to make the next diagrams clearer. Normally, it is drawn with the -OH groups on the right-hand side. By the esterification process shown above, three ethanoate groups can be formed. Lengthening each carbon chain creates a triglyceride, otherwise known as a fat. The acid CH (CH ) COOH is called octadecanoic acid, frequently referred to by its common name, stearic acid. The full name for the ester of this with propane-1,2,3-triol is propane-1,2,3-triyl trioctadecanoate, unsurprisingly known by by its common name of glyceryl tristearate. If the fat or oil is saturated, the acid from which it is derived has no carbon-carbon double bonds in its chain. Stearic acid is a saturated acid; therefore glyceryl tristearate is a saturated fat. If the acid has just one carbon-carbon double bond somewhere in the chain, it is called mono-unsaturated. If it has more than one carbon-carbon double bond, it is polyunsaturated. The acids below are saturated acids, and will therefore form saturated fats and oils: Oleic acid is a typical mono-unsaturated acid: Linoleic and linolenic acids are typical polyunsaturated acids. The terms "omega-6" and "omega-3" have become popular in the context of fats and oils. Linoleic acid is an omega-6 acid. This indicates that the first carbon-carbon double bond starts on the sixth carbon from the CH end. Linolenic acid is an omega-3 acid for the same reason. Because of their relationship with fats and oils, all of the acids above are sometimes described as fatty acids. Small esters have boiling points which are similar to those of aldehydes and ketones with the same number of carbon atoms. Esters, like aldehydes and ketones, are polar molecules and so have as well as van der Waals dispersion forces. However, they do not form ester-ester hydrogen bonds, so their boiling points are significantly lower than those of an acid with the same number of carbon atoms. Small esters are fairly soluble in water but solubility decreases with increasing chain length, as shown below: The reason for this trend in solubility is that although esters cannot hydrogen bond with each other, they can hydrogen bond with water molecules. One of the partially-positive hydrogen atoms in a water molecule can be sufficiently attracted to one of the lone pairs on one of the oxygen atoms in an ester, forming a hydrogen bond. Dispersion forces and dipole-dipole attractions are also present. Forming these intermolecular attractions releases some of the energy needed to solvate the ester. As chain length increases, the hydrocarbon portion forces itself between water molecules, breaking the relatively strong hydrogen bonds between water molecules without offering an energetic compensation; furthermore, the water molecules are forced into an ordered alignment along the chain, decreasing the entropy in the system. This makes the process thermodynamically less favorable, and so solubility decreases. Fats and oils are not water soluble. The chain lengths are so great that far too many hydrogen bonds between water molecules must be broken; this is not an energetically profitable arrangement. Melting points determine whether the substance is a fat (a solid at room temperature) or an oil (a liquid at room temperature). Fats normally contain saturated chains. These allow more effective van der Waals dispersion forces between the molecules: more energy is required to separate the chains, increasing the melting point. A greater number of double bonds, or , in the molecules results in a lower melting point, because the van der Waals forces are less effective. The efficacy of derWaals The hydrocarbon chains are, of course, in constant motion in the liquid, but it is possible for them to lie tidily when the substance solidifies. If the chains in one molecule can lie tidily, that means that neighboring molecules can get close. That increases the attractions between one molecule and its neighbors and so increases the melting point. Unsaturated fats and oils have at least one carbon-carbon double bond in at least one chain. Rotation about a carbon-carbon double bond is constrained, locking a permanent kink into the chain. This makes packing molecules close together more difficult. If the chains cannot pack well, the van der Waals forces will be less effective. This effect is much stronger for molecules in which the hydrocarbon chains at either end of the double bond are arranged - to each other, as shown in the figure below: In the - conformation, the effect is not as marked. It is, however, rather more than the diagram below suggests because of the changes in bond angles around the double bond compared with the rest of the chain. -fats and oils have higher melting points than their corresponding - conformations because the packing is not interrupted to the same degree. Naturally occurring unsaturated fats and oils tend to adopt the - conformations. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/03%3A_Covalent_Bonding/3.07%3A_Polarity

As we already learned, the atoms engaged in covalent bonding share electrons in order to fulfill the octet rule. However, this electron sharing can take place on an equal or unequal basis. If the atoms involved in covalent bonding are of equal electronegativities (which occurs only if they are the same atoms), then sharing takes place on an equal basis and there is no bias in the amount of time the bonding electrons spend around each atom. The hydrogen molecule (H ) shown below is an example of this. The electronic cloud surrounding the two atoms is highly symmetrical, and the H-H bond is said to be Now consider the case of hydrogen chloride, H-Cl. Hydrogen and chlorine are engaged in covalent bonding, but the electronegativity of chlorine is higher than that of hydrogen. The greater tendency of chlorine to attract electrons results in unequal sharing between the two atoms. The bonding electrons spend more time around chlorine than around hydrogen. They are still being shared, but chlorine behaves as if it carried a negative charge, and hydrogen behaves as if it carried a positive charge. as is the case in ionic molecules. In covalent molecules they are referred to as , or poles, because they are analogous of the poles of a magnet. The positive pole is indicated by ∂ , and the negative pole by ∂ . The two together constitute a , and the bond in question is said to be   A polar bond is sometimes represented as a vector, with an arrow pointing in the direction of the more electronegative atom. The following are valid representations for polar bonds. Since the difference in electronegativity between two bonding atoms can be zero or very large, there is a , ranging from nonpolar to highly polar bonds. In an extreme case where the difference in electronegativity is vary large, the bond ceases to be covalent and becomes ionic. Bond polarity is measured by the . This parameter is reported in units (D). General Chemistry textbooks typically contain tables of dipole moments for different types of bonds. For example, the dipole moment for the C-H bond is 0.3 D, whereas that for the H-Cl bond is 1.09 D. Every covalent bond is either polar or nonpolar. When all the dipoles for all the covalent bonds that make up a molecule are added together as vectors, the result is the of the entire molecule. . Obviously, it is possible to have nonpolar molecules made up of polar bonds, as long as the corresponding dipoles add up to zero. Some examples are shown below. Refer to chapter 2 in your textbook for a more comprehensive discussion of polarity and dipoles. One must be careful in deciding whether a molecule is polar or nonpolar based purely on a two-dimensional representation. Molecules are three-dimensional, and direction is as important as magnitude when it comes to adding vectors. For example, a two-dimensional representation of the methylene chloride molecule (CH Cl ) shown below might lead to the erroneous conclusion that it is nonpolar when in fact it is polar. Many organic molecules are made up of long hydrocarbon chains with many C-H bonds. Since the difference in electronegativity between carbon and hydrogen is very small, the C-H bond has a very small dipole moment, and hydrocarbons are for the most part considered nonpolar molecules. However, the introduction of a relatively polar bond in such structures dominates the entire molecule, rendering it polar. The polarity of molecules affects their physical properties. As a rule of thumb and other factors being similar, the higher the polarity of the molecule, the higher the value of properties such as melting and boiling point. The solubility of molecules in solvents is also largely determined by polarity. The rule “ ” makes reference to the fact that polar molecules dissolve better in polar solvents, and nonpolar molecules dissolve better in nonpolar solvents. Water and oil don’t mix because water is highly polar and oil is largely made up of hydrocarbon chains, which are nonpolar. Conversely, water and alcohol do mix because they are both of very similar polarities. For a more comprehensive discussion refer to chapter 2 of your textbook.

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

means how much of something there is in a given volume, kind of like density, except that it describes . A solution is some compound, called the that is dissolved in another, more abundant compound, called the . To be a solution, the molecules or ions of the solute must be separated from each other and surrounded by solvent molecules or ions. If very small bits of one compound are mixed into another compound, but not actually dissolved into molecules or ions, that is called a (a solid in a liquid) or (two liquids). In a solution, both the solute and solvent can be any phase, solid, liquid or gas. The concentration of a solution is the amount of solute divided by the total amount of solution, usually. However, there are many different units used for concentration, and some of them assume that there is so much more solvent than solute that you can use amount of solvent instead of amount of solution. There are many, many units used for concentration. Some are mostly used to describe concentrated solutions (that have a lot of solute) and others are mostly used to describe dilute solutions (that have very little solute). The most common unit in chemistry is (abbreviated M), which is moles of solute divided by liters of solution. \[Molarity=\frac{(moles\; of\; solute)}{(liters\; of\; solution)}\] For example, if you dissolve 1 mol of NaCl in 1L of water, that is a 1 M (read "1 molar") NaCl solution. Another common unit is weight %, which means \[Weight\; (Mass)\; \%=\frac{(mass\; of\; solute)}{(mass\; of\; solution)}\; \times 100\%\] This is often used for very concentrated solutions. For very dilute solutions, you'll see ppm (parts per million) or ppb (parts per billion): \[ppm=\frac{(mass\; of\; solute)}{(mass\; of\; solution)}\; \times 10^{6}\] \[ppb=\frac{(mass\; of\; solute)}{(mass\; of\; solution)}\; \times 10^{9}\] Many compounds are important even at these very low concentrations. Some chemicals used in agriculture or industry are called "endocrine disrupters" and studies suggest that they can be dangerous for living things at the ppb level. (Example abstract of a scientific paper: note that ppm is described as "high concentration.") In semiconductors, which are what computer chips and LEDs are made of, ppm-level solid solutions enable the essential properties. HCl (hydrogen chloride) is a really nasty, dangerous gas, but dissolved in water it makes a convenient acid (hydrochloric acid) for many applications in the lab or in industry. When working with an HCl solutions in the lab, we often want to measure the mass of volume of solution used and know how many moles of HCl we added. Unfortunately, when you buy HCl, it usually comes as concentrated HCl, and the bottle will say something like "32% by weight, density 1.1593." Convert this to molarity. To solve this, we need to think of it like a unit conversion. A good trick for dealing with % quantities is just to translate that into g/g, like this: 32 weight % = (32 g HCl)/(100 g solution). Now we do our usual unit conversion: \[\left(\dfrac{32\; \cancel{g\; HCl}}{100\; \cancel{g\; solution}}\right) \left(\dfrac{1\; mol\; HCl}{36.46\; \cancel{g\; HCl}}\right) \left(\dfrac{1.1593\; \cancel{g}}{1\; \cancel{mL\; solution}}\right) \left(\dfrac{\cancel{1000\; mL}}{1\; L}\right)=10\; M\; HCl\] Here we started with units of g HCl/g solution, and converted the g HCl to moles, and the g solution to liters using the density, which is (g/ml). Notice that the number of significant figures is fairly low. This is because the gas HCl can evaporate out of the solution, just like the water, so the concentration might change a little over time. If you want to know the concentration more precisely, see the next section on titrations to find out how to measure it. Now suppose you want to make 2.5 M HCl using the 10 M HCl. You will need to dilute it, which means adding solvent to decrease the concentration. If you want to make 1 L of 2.5 M HCl, how much 10 M HCl do you dilute? To solve this, we can still think about it like a unit conversion. We want 1 L of 2.5 M HCl solution. In the first step, we will convert to the number of moles of HCl needed to make that solution. In the second step, we will find how many ml of 10 M solution have this number of moles HCl. \[(1\; \cancel{L\; of\; 2.5\; M\; HCl}) \left(\dfrac{2.5\; \cancel{mol\; HCl}}{1\; \cancel{L\; of\; 2.5\; M\; HCl}}\right) \left(\dfrac{1000\; mL\; of\; 10\; M\; HCl}{10\; \cancel{mol\; HCl}}\right)=250\; mL\; of\; 10\; M\; HCl\] To make the solution, we will take 250 mL of 10 M HCl, and we will put it in a 1 L volumetric flask, and then add water slowly, with mixing, until the volume reaches the 1 L mark. The reason we do it this way, instead of just adding 750 mL of water, is that the density can change because of the solute. If we use the volumetric flask, we can be sure that we have exactly 2.5 mol HCl in exactly 1 L of solution.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.04%3A_Beneficiation

Beneficiation is any process which removes the gangue minerals from ore to produce a higher grade product, and a waste stream. Beneficiation may involve physical or chemical processes. Often, as in the case of panning for gold, the desired ore or metal is denser than the gangue. The latter can be suspended in a stream of water and flushed away. The iron ore magnetite, Fe O , is ferrimagnetic. It can be separated from abundant deposits of taconite by grinding to a fine slurry in water. Passing this suspension over powerful electromagnets removes the Fe O , more than doubling the concentration of Fe. Both the physical beneficiation processes just described produce large quantities of tailings-water suspensions of the gangue. Usually the silicate and other particles are trapped in a settling basin, but in one case large quantities of taconite tailings have been dumped into Lake Superior for several decades. This is a partly physical and partly chemical beneficiation used to concentrate ores of copper, lead, and zinc. These metals have considerable affinity for sulfur. (the lists sulfide ores for each of them.) The ore is ground and suspended with water in large tanks. Flotation agents such as the xanthates are added and a stream of air bubbles passed up through the tank. The sulfur-containing end of the xanthate molecule is attracted to the desired metal, while the hydrocarbon end tends to avoid water (like the ). The metal-containing particles are swept to the top of the tank in a froth much akin to soap bubbles. The concentrated ore may then be skimmed from the water surface. By means of flotation, ores containing as little as 0.3% copper can be concentrated to 20 to 30% copper, making recovery of the metal economically feasible. Many ores can be chemically separated from their gangue by means of acid or base. Thus copper ore is often leached with acid which dissolves copper oxides and carbonates and leaves behind most silicates: \[\text{CuCo}_3 + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4 (aq) + \text{H}_2\text{O} + \text{CO}_2 \label{1} \] In the bauxite is dissolved in 30% NaOH: \[\text{Al}_2\text{O}_3(s) + \text{OH}^-(aq) + \text{3H}_2\text{O}(l) \rightarrow \text{2Al(OH}^-_4(aq) \label{2} \] and then reprecipitated by adding acid. Impurities such as Fe O and SiO are eliminated in this way. Many sulfide and carbonate ores are roasted in air in order to convert them into oxides, a form more suitable for further processing: \[\text{2ZnS}(s) + \text{3O}_2(g) \rightarrow \text{2ZnO}(s) + \text{SO}_2(g)\label{3} \] \[\text{PbCO}_3 (s) \rightarrow \text{PbO}(s) + \text{Co}_2(g) \nonumber \] The production of SO in Equation \(\ref{3}\) is a notorious source of air pollution, including .

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Qualitative_Analysis

Qualitative Analysis is the determination of non-numerical information about a chemical species, a reaction, etc. Examples would be observing that a reaction is creating gas that is bubbling out of solution or observing that a reaction results in a color change. Qualitative analysis is not as reliable as quantitative analysis but is often far easier, faster and cheaper to perform. This chapter discusses how to perform a systematic analysis on inorganic material to ascertain its composition. Thumbnail: Lead(II) iodide precipitates when solutions of potassium iodide and lead(II) nitrate are combined. (CC BY- ; ).

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates

This chapter is designed to provide you with an overview of the biologically important group of compounds known as carbohydrates. Many of the compounds you will encounter while studying this chapter may appear to have very complex structures, but much of their chemistry can be readily understood in terms of the concepts and reactions discussed in earlier chapters of the course. The chapter begins with an explanation of the classification schemes used to simplify the study of carbohydrates. We make extensive use of Fischer projection formulas throughout the chapter. We place considerable emphasis on gaining an appreciation of the configurations of carbohydrates, particularly of the aldoses. We describe the disadvantages of representing monosaccharides by open-chain structures, and at this point, introduce you to cyclic representations—called Haworth projections—of these substances. We describe the mutarotation of glucose, explaining it in terms of the existence of anomers. We then examine some reactions of monosaccharides, including the formation of ethers and esters, the formation of glycosides, and reduction and oxidation. We discuss the structures of some common disaccharides and polysaccharides, and conclude the chapter with a brief explanation of the role played by carbohydrates in cell recognition.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/20%3A_Organic_Chemistry/20.1%3A_Hydrocarbons

The largest database of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated at 10 —an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities. The simplest contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. This leads to differences in geometries and in the hybridization of the carbon orbitals. , or , contain only single covalent bonds between carbon atoms. Each of the carbon atoms in an alkane has hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. The Lewis structures and models of methane, ethane, and pentane are illustrated in Figure \(\Page {1}\). Carbon chains are usually drawn as straight lines in Lewis structures, but one has to remember that Lewis structures are not intended to indicate the geometry of molecules. Notice that the carbon atoms in the structural models (the ball-and-stick and space-filling models) of the pentane molecule do not lie in a straight line. Because of the hybridization, the bond angles in carbon chains are close to 109.5°, giving such chains in an alkane a zigzag shape. The structures of alkanes and other organic molecules may also be represented in a less detailed manner by condensed structural formulas (or simply, ). Instead of the usual format for chemical formulas in which each element symbol appears just once, a condensed formula is written to suggest the bonding in the molecule. These formulas have the appearance of a Lewis structure from which most or all of the bond symbols have been removed. Condensed structural formulas for ethane and pentane are shown at the bottom of Figure \(\Page {1}\), and several additional examples are provided in the exercises at the end of this chapter. A common method used by organic chemists to simplify the drawings of larger molecules is to use a (also called a line-angle structure). In this type of structure, carbon atoms are not symbolized with a C, but represented by each end of a line or bend in a line. Hydrogen atoms are not drawn if they are attached to a carbon. Other atoms besides carbon and hydrogen are represented by their elemental symbols. Figure \(\Page {2}\) shows three different ways to draw the same structure. Draw the skeletal structures for these two molecules: Each carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):   Draw the skeletal structures for these two molecules:     Identify the chemical formula of the molecule represented here:   There are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of C H . Location of the hydrogen atoms:   Identify the chemical formula of the molecule represented here:   C H All alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of C H . The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table \(\Page {1}\)) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change. Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula C H : They are called -butane and 2-methylpropane (or isobutane), and have the following Lewis structures: The compounds -butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The -butane molecule contains an , meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term , or the prefix , to refer to a chain of carbon atoms without branching. The compound 2–methylpropane has a branched chain (the carbon atom in the center of the Lewis structure is bonded to three other carbon atoms) Identifying isomers from Lewis structures is not as easy as it looks. Lewis structures that look different may actually represent the same isomers. For example, the three structures in Figure \(\Page {3}\) all represent the same molecule, -butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms. The International Union of Pure and Applied Chemistry ( ) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. The nomenclature for alkanes is based on two rules: When more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending replaces at the end of the name of an electronegative substituent (in ionic compounds, the negatively charged ion ends with like chloride; in organic compounds, such atoms are treated as substituents and the ending is used). The number of substituents of the same type is indicated by the prefixes (two), (three), (four), and so on (for example, indicates two fluoride substituents). Name the molecule whose structure is shown here:     The four-carbon chain is numbered from the end with the chlorine atom. This puts the substituents on positions 1 and 2 (numbering from the other end would put the substituents on positions 3 and 4). Four carbon atoms means that the base name of this compound will be butane. The bromine at position 2 will be described by adding 2-bromo-; this will come at the beginning of the name, since bromo- comes before chloro- alphabetically. The chlorine at position 1 will be described by adding 1-chloro-, resulting in the name of the molecule being 2-bromo-1-chlorobutane. Name the following molecule:   3,3-dibromo-2-iodopentane We call a substituent that contains one less hydrogen than the corresponding alkane an alkyl group. The name of an is obtained by dropping the suffix of the alkane name and adding :   The open bonds in the methyl and ethyl groups indicate that these alkyl groups are bonded to another atom. Name the molecule whose structure is shown here:   The longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right—this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)—so we take our name for two carbons and attach at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane. Name the following molecule:   4-propyloctane Some hydrocarbons can form more than one type of alkyl group when the hydrogen atoms that would be removed have different “environments” in the molecule. This diversity of possible alkyl groups can be identified in the following way: The four hydrogen atoms in a methane molecule are equivalent; they all have the same environment. They are equivalent because each is bonded to a carbon atom (the same carbon atom) that is bonded to three hydrogen atoms. (It may be easier to see the equivalency in the ball and stick models in Figure \(\Page {3}\). Removal of any one of the four hydrogen atoms from methane forms a methyl group. Likewise, the six hydrogen atoms in ethane are equivalent and removing any one of these hydrogen atoms produces an ethyl group. Each of the six hydrogen atoms is bonded to a carbon atom that is bonded to two other hydrogen atoms and a carbon atom. However, in both propane and 2–methylpropane, there are hydrogen atoms in two different environments, distinguished by the adjacent atoms or groups of atoms: Note that alkyl groups do not exist as stable independent entities. They are always a part of some larger molecule. The location of an alkyl group on a hydrocarbon chain is indicated in the same way as any other substituent: Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction: \[\ce{CH4}(g)+\ce{2O2}(g)⟶\ce{CO2}(g)+\ce{2H2O}(g) \nonumber \] Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH , is the principal component of natural gas. Butane, C H , used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (Figure \(\Page {5}\)). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids. In a , another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction: The C–Cl portion of the chloroethane molecule is an example of a , the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter. Organic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. You have likely heard of unsaturated fats. These are complex organic molecules with long chains of carbon atoms, which contain at least one double bond between carbon atoms. Unsaturated hydrocarbon molecules that contain one or more double bonds are called . Carbon atoms linked by a double bond are bound together by two bonds, one σ bond and one π bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats. Ethene, C H , is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure \(\Page {6}\)); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism. Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules. Polymers (from Greek words meaning “many” and meaning “parts”) are large molecules made up of repeating units, referred to as monomers. Polymers can be natural (starch is a polymer of sugar residues and proteins are polymers of amino acids) or synthetic [like polyethylene, polyvinyl chloride (PVC), and polystyrene]. The variety of structures of polymers translates into a broad range of properties and uses that make them integral parts of our everyday lives. Adding functional groups to the structure of a polymer can result in significantly different properties (see the discussion about Kevlar later in this chapter). An example of a polymerization reaction is shown in Figure \(\Page {7}\). The monomer ethylene (C H ) is a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of –CH – units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films). Polyethylene is a member of one subset of synthetic polymers classified as plastics. Plastics are synthetic organic solids that can be molded; they are typically organic polymers with high molecular masses. Most of the monomers that go into common plastics (ethylene, propylene, vinyl chloride, styrene, and ethylene terephthalate) are derived from petrochemicals and are not very biodegradable, making them candidate materials for recycling. Recycling plastics helps minimize the need for using more of the petrochemical supplies and also minimizes the environmental damage caused by throwing away these nonbiodegradable materials. Plastic recycling is the process of recovering waste, scrap, or used plastics, and reprocessing the material into useful products. For example, polyethylene terephthalate (soft drink bottles) can be melted down and used for plastic furniture, in carpets, or for other applications. Other plastics, like polyethylene (bags) and polypropylene (cups, plastic food containers), can be recycled or reprocessed to be used again. Many areas of the country have recycling programs that focus on one or more of the commodity plastics that have been assigned a recycling code (Figure \(\Page {8}\)). These operations have been in effect since the 1970s and have made the production of some plastics among the most efficient industrial operations today. The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix with the suffix . The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:   Molecules of 1-butene and 2-butene are structural isomers; the arrangement of the atoms in these two molecules differs. As an example of arrangement differences, the first carbon atom in 1-butene is bonded to two hydrogen atoms; the first carbon atom in 2-butene is bonded to three hydrogen atoms. The compound 2-butene and some other alkenes also form a second type of isomer called a geometric isomer. In a set of geometric isomers, the same types of atoms are attached to each other in the same order, but the geometries of the two molecules differ. Geometric isomers of alkenes differ in the orientation of the groups on either side of a \(\mathrm{C=C}\) bond. Carbon atoms are free to rotate around a single bond but not around a double bond; a double bond is rigid. This makes it possible to have two isomers of 2-butene, one with both methyl groups on the same side of the double bond and one with the methyl groups on opposite sides. When structures of butene are drawn with 120° bond angles around the -hybridized carbon atoms participating in the double bond, the isomers are apparent. The 2-butene isomer in which the two methyl groups are on the same side is called a -isomer; the one in which the two methyl groups are on opposite sides is called a -isomer (Figure \(\Page {9}\)). The different geometries produce different physical properties, such as boiling point, that may make separation of the isomers possible: Alkenes are much more reactive than alkanes because the \(\mathrm{C=C}\) moiety is a reactive functional group. A π bond, being a weaker bond, is disrupted much more easily than a σ bond. Thus, alkenes undergo a characteristic reaction in which the π bond is broken and replaced by two σ bonds. This reaction is called an addition reaction. The hybridization of the carbon atoms in the double bond in an alkene changes from to during an addition reaction. For example, halogens add to the double bond in an alkene instead of replacing hydrogen, as occurs in an alkane:   Provide the IUPAC names for the reactant and product of the halogenation reaction shown here:   The reactant is a five-carbon chain that contains a carbon-carbon double bond, so the base name will be pentene. We begin counting at the end of the chain closest to the double bond—in this case, from the left—the double bond spans carbons 2 and 3, so the name becomes 2-pentene. Since there are two carbon-containing groups attached to the two carbon atoms in the double bond—and they are on the same side of the double bond—this molecule is the isomer, making the name of the starting alkene -2-pentene. The product of the halogenation reaction will have two chlorine atoms attached to the carbon atoms that were a part of the carbon-carbon double bond:   This molecule is now a substituted alkane and will be named as such. The base of the name will be pentane. We will count from the end that numbers the carbon atoms where the chlorine atoms are attached as 2 and 3, making the name of the product 2,3-dichloropentane. Provide names for the reactant and product of the reaction shown:   reactant: cis-3-hexene, product: 3,4-dichlorohexane Hydrocarbon molecules with one or more triple bonds are called ; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The -hybridized carbons involved in the triple bond have bond angles of 180°, giving these types of bonds a linear, rod-like shape. The simplest member of the alkyne series is ethyne, C H , commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:   The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix is used to indicate a triple bond in the chain. For example, \(\mathrm{CH_3CH_2C≡CH}\) is called 1-butyne. Describe the geometry and hybridization of the carbon atoms in the following molecule:   Carbon atoms 1 and 4 have four single bonds and are thus tetrahedral with hybridization. Carbon atoms 2 and 3 are involved in the triple bond, so they have linear geometries and would be classified as hybrids. Identify the hybridization and bond angles at the carbon atoms in the molecule shown:   carbon 1: , 180°; carbon 2: , 180°; carbon 3: , 120°; carbon 4: , 120°; carbon 5: , 109.5° Chemically, the alkynes are similar to the alkenes. Since the \(\mathrm{C≡C}\) functional group has two π bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example:   Acetylene and the other alkynes also burn readily. An acetylene torch takes advantage of the high heat of combustion for acetylene. Benzene, C H , is the simplest member of a large family of hydrocarbons, called . These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, C H , are: There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:   Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene. One possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:   Since the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:   Draw three isomers of a six-membered aromatic ring compound substituted with two bromines.   Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. The chemistry of these compounds is called organic chemistry. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring structures with delocalized π electron systems.

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The use of food by organisms is termed nutrition. Vitamins and minerals necessary for biochemical processes. There are three general categories of food: (1) Essential fiber which are non-digestible polysaccharide material, essential for normal functioning of animal digestive systems (i.e. colon), (2) Energy-yielding nutrients which are protein, carbohydrate and lipid and (3) Micronutrients. Animals are unable to synthesize certain amino acids (humans can only make 10 of the 20 common amino acids). The amino acids that an animal is unable to synthesize must be obtained from the diet (i.e. by consuming plants or microorganisms), and these amino acids are termed "essential amino acids". Excess dietary protein becomes a source of metabolic energy Protein is an important source of nitrogen in the diet. Protein within the body is constantly turning over (i.e. being degraded and resynthesized). Furthermore, there is a general demand for protein synthesis when an organism is growing. The Nitrogen Balance refers to the relationship between the supply and demand for nitrogen (i.e. protein) within an organism. Carbohydrates are also an essential structural component of nucleic acids, nucleotides, glycoproteins and glycolipids. However, the principle role of carbohydrate in the diet is production of metabolic energy. Fatty acids and triacylglycerols can be used as fuel by many tissues in the human body. Phospholipids are essential components of all biological membranes "Dietary fiber" refers to molecules that cannot be broken down by enzymes in the human body. Vitamins are essential nutrients that are required in the diet because they cannot be synthesized by human metabolic enzymes. Often, only trace levels are required, but a shortage can result in disease or death. Coenzymes are low molecular weight molecules that provide unique chemical functionalities for certain enzyme/coenzyme complexes. Summary of water soluble and fat soluble vitamins: Common name Chemical name Related cofactor(s) Water Soluble Vitamins Vitamin B1 Thiamine Thiamine pyrophosphate Vitamin B2 Riboflavin Flavin adenine dinucleotide (FAD) Flavin mononucleotide (FMN) Vitamin B6 Pyridoxal, pyridoxine, pyridoxamine Pyridoxal phosphate Vitamin B12 Cobalamin 5'-deoxyadenosylcobalamin Methylcobalamin Niacin Nicotinic acid Nicotinamide adenine dinucleotide (NAD+) Nicotinamide adenine dinucleotide phosphate (NADP+) Vitamin B3 Pantothenic acid Coenzyme A Biotin Biotin-lysine conjugates (biocytin) Lipoic acid Lipoyl-lysine conjugates (lipoamide) Folic acid Tetrahydrofolate Vitamin C L-ascorbate Fat Soluble Vitamins Vitamin A Retinol Vitamin D2 Ergocalciferol Vitamin D3 Cholecalciferol Vitamin E a-Tocopherol Vitamin K Thiamine is the precursor of thiamine pyrophosphate (TPP): Nicotinamide is an essential part of two important coenzymes: nicotinamide adenine dinucleotide (NAD+) and nicotinamide adenine dinucleotide phosphate (NADP+). Riboflavin is a constituent of riboflavin 5'-phosphate (flavin mononucleotide, or FMN) and flavin adenine dinucleotide (FAD). The nucleotide part of the molecule does not enter into any chemistry, but is important for recognition and binding to enzymes that will use FMN or FAD as a cofactor. The isoalloxazine ring is the core structure of the different flavin molecules. It is yellow in color and the word "flavin" is derived from the latin word for yellow, flavus. Pantothenic acid is a component of coenzyme A (CoA). The two main functions of CoA are: a-hydrogen of the acyl group for removal as a proton Both of these functions involve the reactive sulfhydryl group through the formation of thioester linkages with acyl groups The biologically active form of vitamin B6 is pyridoxal-5-phosphate (PLP), however, the nutritional requirements can be met by either pyridoxine, pyridoxal or pyridoxol. PLP participates in a wide variety of reactions involving amino acids, including: These involve bonds to the amino acid Ca as well as side chain carbons. The wide variety of reactions is due to the ability of PLP to form stable Schiff base adducts with a-amino groups of amino acids: Vitamin B is not made by any animal or plant, it is produced by only a few species of bacteria. Once in the food chain, vitamin B12 is obtained by animals by eating other animals, but plants are sadly deficient. Therefore, herbivorous animals (and vegetarians) can suffer a deficiency. The structure contains a cobalt ion, coordinated within a corrin ring structure: Vitamin B (cyanocobalamin) is converted in the body into two coenzymes: Vitamin B coenzymes participate in three types of reactions: L-Ascorbate is a reducing sugar (has a reactive ene-diol structure) that is involved in the following biochemical processes: Almost all animals can synthesize vitamin C (its in the pathway of carbohydrate synthesis). Humans and great apes have suffered a mutation in the last enzyme in the pathway of synthesis for L-ascorbate (mutation occurred about 10-40 million years ago). Since that time, all great apes (of which humans are a member) must get L-ascorbate from their diet (fresh fruits and vegetable contain an abundance). Thus, for the great apes L-ascorbate is a "vitamin" (another way of looking at it is that all great apes suffer an in-born error in metabolism). Humans still have the gene for the enzyme to make vitamin C. However, it has suffered a couple of deletions that introduce a frame shift mutation, in addition to numerous point mutations. Biotin acts as a mobile carboxyl group carrier in a variety of enzymatic carboxylation reactions. Lipoic acid contains two sulfur atoms that can exist as a disulfide bonded pair, or as two free sulfhydrils. Conversion between the two forms involves a redox reaction (the two free sulfhydrils represent the reduced form). Lipoic acid is typically found covalently attached to a lysine side chain in enzymes that use it as a cofactor, as a lipoamide complex. Folic acid derivatives (i.e. "folates") are acceptors and donors of one-carbon units for all oxidation levels of carbon (except for the most oxidized form - CO2. See biotin above). Vitamin A occurs as an ester (Retinyl ester), aldehyde (Retinal) or acidic form (Retinoic acid). Cholecalciferol is produced in the skin of animals by the action of U.V. light on the precursor molecule 7-dehydrocholesterol. Vitamin E: Tocopherol a-Tocopherol is a potent antioxidant, however, molecular details of its function are not clearly understood. Vitamin K is essential to the blood-clotting process. Vitamin K is required for the post-translational modification to produce g-carboxy glutamic acid from glutamic acid. Such modified residues can bind Ca2+, which is an essential part of the process in the clotting cascade. g-carboxy glutamic acids in their structure

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The molecular formula of a hydrocarbon provides information about the possible structural types it may represent. For example, consider compounds having the formula \(\ce{C5H8}\). The formula of the five-carbon alkane pentane is \(\ce{C5H12}\) so the difference in hydrogen content is 4. This difference suggests such compounds may have a triple bond, two double bonds, a ring plus a double bond, or two rings. Some examples are shown here, and there are at least fourteen others! Hence, as with alkanes, a consistent nomenclature system needs to be adopted that can separate the nature of these unsaturated chemicals. The simplest are the alkenes, which are hydrocarbons which have functional groups and are unsaturated hydrocarbons with the molecular formula is \(\ce{CnH2n}\), which is also the same molecular formula as . Alkenes are named using the same general naming rules for alkanes, except that the suffix is now -ene. Here is a chart containing the systemic name for the first twenty straight chain alkenes. Did you notice how there is no methene? Because it is impossible for a carbon to have a double bond with nothing. The parent structure is the longest chain containing both carbon atoms of the double bond. If the alkene contains only one double bond and that double bond is terminal (the double bond is at one end of the molecule or another) then it is not necessary to place any number in front of the name. If the double bond is not terminal (if it is on a carbon somewhere in the center of the chain) then the carbons should be numbered in such a way as to give the first of the two double-bonded carbons the lowest possible number, and that number should precede the "ene" suffix with a dash, as shown below. Name the following compounds: a: b. 1-pentene or pent-1-ene 2-ethyl-1-hexene or 2-ethylhex-1-ene a: Name the following compound (hint: give the double bond the lowest possible numbers regardless of substituent placement). b. Draw a structure for 4-methyl-2-pentene. 4-methylpent-1-ene If there is more than one double bond in an alkene, all of the bonds should be numbered in the name of the molecule - even terminal double bonds. The numbers should go from lowest to highest, and be separated from one another by a comma. The IUPAC numerical prefixes are used to indicate the number of double bonds. Note that the numbering of "2-4" above yields a molecule with two double bonds separated by just one single bond. Double bonds in such a condition are called "conjugated", and they represent an enhanced stability of conformation, so they are energetically favored as reactants in many situations and combinations. Double bonds can exist as geometric isomers and these isomers are designated by using either the cis / trans designation or the more flexible E / Z designation. Isomers have the two largest groups are on the same side of the double bond (left structure above) and trans Isomers have the two largest groups are on opposite sides of the double bond (right structure above).E/Z nomenclature If there are 3 or 4 non-hydrogen different atoms attached to the alkene then use the E, Z system. E = entgegan ("trans") Z = zusamen ("cis") Priority of groups is based on the atomic mass of attached atoms (not the size of the group). An atom attached by a multiple bond is counted once for each bond. fluorine atom > isopropyl group > n-hexyl group deuterium atom > hydrogen atom -CH -CH=CH > -CH CH CH Try to name the following compounds using both cis-trans and E/Z conventions: a: b: 4-methylpent-1-ene What is the name of this molecule? In this diagram this is a cis conformation. It has both the substituents going upward. This molecule would be called (cis) 5-chloro-3-heptene.) Trans would look like this What is the name of this molecule? In this example it is E-4-chloro-3-heptene. It is E because the Chlorine and the CH CH are the two higher priorities and they are on opposite sides. vi. A hydroxyl group gets precedence over the double bond. Therefore alkenes containing alchol groups are called alkenols. And the prefix becomes --enol. And this means that now the alcohol gets lowest priority over the alkene. vii. Lastly remember that alkene substituents are called alkenyl. Suffix --enyl. Remove the -ane suffix and add -ylene. There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl. That you should know are... Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring. Cyclopentene is an example of an endocyclic double bond. Methylenecylopentane is an example of an exocyclic double bond. Name the following compounds: a: b: 1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene. 1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene. Name the following compounds: a: b: 1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene. 1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene. Draw structures for the following 2-vinyl-1,3-cyclohexadiene Try to name the following compounds... 1-pentene or pent-1-ene 2-ethyl-1-hexene or 2-ethylhex-1-ene Try to draw structures for the following compounds... CH –CH=CH–CH –CH CH –CH –CH=CH–CH –CH –CH b. Give the double bond the lowest possible numbers regardless of substituent placement. • Try to name the following compound... • Try to draw a structure for the following compound... 4-methyl-2-pentene Name the following structures: v. Draw (Z)-5-Chloro-3-ethly-4-hexen-2-ol. I. trans-8-ethyl-3-undecene II. E-5-bromo-4-chloro-7,7-dimethyl-4-undecene III. Z-1,2-difluoro-cyclohexene IV. 4-ethenylcyclohexanol. V. ( ) ),

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Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule. If a molecule contains both a double and a triple bond, the carbon chain is numbered so that the first multiple bond gets a lower number. If both bonds can be assigned the same number, the double bond takes precedence. The molecule is then named "n-en-n-yne", with the double bond root name preceding the triple bond root name (e.g. 2-hepten-4-yne). Here are the molecular formulas and names of the first ten carbon straight chain alkynes. The more commonly used name for ethyne is acetylene, which used industrially. Like previously mentioned, the IUPAC rules are used for the naming of alkynes. Find the longest carbon chain that includes both carbons of the triple bond. Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes. For example: 4-chloro-6-diiodo-7-methyl-2-nonyne After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order. For example: 2,2,10-triiodo-5-methyl-3-decyne If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond. 5- methyl-7-octyn-3-ol When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne. For example: 4-methyl-1,5-octadiyne Substituents containing a triple bond are called alkynyl. For example: 4-bromo-1-chloro-1-ethynylcyclohexane Here is a table with a few of the alkynyl substituents: A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example: Name or draw out the following molecules: 1. 4,4-dimethyl-2-pentyne 2. 4-Penten-1-yne 3. 1-ethyl-3-dimethylnonyne 4.

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Optical activity is an effect of an optical isomer's interaction with plane-polarized light. , or enantiomers, have the same sequence of atoms and bonds but are different in their 3D shape. Two enantiomers are nonsuperimposible mirror images of one another (i.e., ), with the most common cited example being our hands. Our left hand is a mirror image of our right, yet there is no way our left thumb can be over our right thumb if our palms are facing the same way and placed over one another. Optical isomers also have no axis of symmetry, which means that there is no line that bisects the compound such that the left half is a mirror image of the right half. Optical isomers have basically the same properties (melting points, boiling points, etc.) but there are a few exceptions (uses in biological mechanisms and optical activity). There are drugs, called enantiopure drugs, that have different effects based on whether the drug is a racemic mixture or purely one enantiomer. For example, d-ethambutol treats tuberculosis, while l-ethambutol causes blindness. Optical activity is the interaction of these enantiomers with plane-polarized light. Optical activity was first observed by the French physicist Jean-Baptiste Biot. He concluded that the change in direction of plane-polarized light when it passed through certain substances was actually a rotation of light, and that it had a molecular basis. His work was supported by the experimentation of Louis Pasteur. Pasteur observed the existence of two crystals that were mirror images in tartaric acid, an acid found in wine. Through meticulous experimentation, he found that one set of molecules rotated polarized light clockwise while the other rotated light counterclockwise to the same extent. He also observed that a mixture of both, a (or ), did not rotate light because the optical activity of one molecule canceled the effects of the other molecule. Pasteur was the first to show the existence of chiral molecules. An enantiomer that rotates plane-polarized light in the positive direction, or clockwise, is called dextrorotary [(+), or d-], while the enantiomer that rotates the light in the negative direction, or counterclockwise, is called levorotary [(-), or l-]. When both d- and l- isomers are present in equal amounts, the mixture is called a racemic mixture. In the picture above, you can see that unpolarized light passes through a filter so that only waves that oscillate in a certain direction can pass through. When these waves interact with an optically active material, they are rotated either clockwise or counterclockwise, depending on the enantiomer. In the case of the image above, the light is rotated clockwise so the substance is the dextrorotary . Optical activity is measured by a polarimeter, and is dependent on several factors: concentration of the sample, temperature, length of the sample tube or cell, and wavelength of the light passing through the sample. Rotation is given in +/- degrees, depending on whether the sample has d- (positive) or l- (negative) enantiomers. The standard measurement for rotation for a specific chemical compound is called the specific rotation, defined as an angle measured at a path length of 1 decimeter and a concentration of 1g/ml. The specific rotation of a pure substance is an intrinsic property. In solution, the formula for specific rotation is: \[ [\alpha]^T_\lambda = \dfrac{\alpha}{I\cdot c}\] where

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Silicone polymers, more properly called do not have carbon as part of the backbone structure. Although silicon is in the same group as carbon in the periodic table, it has quite different chemistry. Many silanes are known which are analogous to the hydrocarbons with Si-Si bonds. These compounds are not very stable and hence not very useful. Silicones on the other hand have an alternating -Si-O- type structure. This basic structural unit is found in many rocks and minerals in nature including common sand. Various organic groups such as methyl or the benzene ring may be bonded to the silicon as shown below. Silicones are water repellent, heat stable, and very resistant to chemical attack. They find many uses in oils, greases, and rubberlike materials. Silicone oils are very desirable since they do not decompose at high temperature and do not become viscous. Other silicones are used in hydraulic fluids, electrical insulators and moisture proofing agent in fabrics. The preparation of dimethyl silicon dichloride, or dimethyldichlorosilane, is the first step in the production of modern dimethylsilicone (polydimethylsiloxane) products. As explained in US Patent 2,380,995, Eugene G. Rochow describes the reaction of elemental silicon with gaseous methyl chloride within a tube furnace at 300˚C. 2 CH Cl + Si → (CH ) SiCl Silicones have a number of medical applications because they are chemically inert. Medical devices composed of silicone may be approved by the FDA for permanent or temporary implantation. Catheters, tubing, gastric bags, drains, and endoscopic windows are examples of consumable medical devices that are often molded from silicone. Breast implants, stents, and prostheses are examples of permanent implants often molded from silicone. Silicone rubber approved for use in FDA devices approved for permanent implantation differs from that used in medical consumables is several important ways. Implant grade silicone is of long linear chain length often exceeding one million molecular weight. Implant grade silicone rubber has had low molecular weight silicone oils, added for improved dispersion of silica fume fillers, removed thru high temperature vacuum mixing. Implant grade silicone rubber is normally cross-linked using a platinum catalyst. Silicone rubber is often used in medical devices because it can be heat sterilized. Most silicone consumables are removed from hot press molds while hot, saving expensive chilling cycles and simplifying mold design. Silicone Rubber tubing used in medical practice can cause some problems. Gas permeability of dimethylsilicone is high enough to cause bubbles to form in silicone tubing often used in pumps that deliver medication to patients. Some medications are sensitive to oxygen permeate. A simple way to demonstrate this effect is to fill a section of silicone tubing with water then tie off both ends excluding any air pockets. The water filled tubing may be draped over any suitable ledge for several hours. Notice air cavities form in tube lumen. A good deal of controversy has involved the the use of silicone in polyurethane bags as breast implants. Again they were used because they were thought to be very inert and resistant to dissolving or other reactions. Reports have cited increased cancer risk and severe immune responses from possible leakage of the silicone from the implants. Some scientists dispute these findings. The liquids solution of sodium silicate is already in the form polymer. The silicate is alternating atoms of silicon and oxygen in long chains. When the ethanol is added, it bridges and connects the chains by cross-linking them. The analogy of a chain-link fence is a good picture of the idea of chains that are cross linked. That is what the ethanol and the silicate are doing to form this super ball.

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The was established by Lavoisier, although others had used it before. It said that in any chemical reaction, the total mass of products is the same as the total mass of reactants. No matter is created or destroyed during the reaction. There was a debate over whether elements always combine in exactly the same ratio, which is called . It was well known that the elemental ratios in many materials are approximately constant. Water is made by burning hydrogen in oxygen. The composition is about 11.1% hydrogen and 88.9% oxygen, by mass. The oxides of metals also generally have consistent compositions. For instance, mercuric oxide is 92.6% mercury, and mercurous oxide is 96.2% mercury; no bigger or smaller ratios were known. Berthollet, a respected scientist, argued that the consistent ratios observed arise from the conditions of the experiment; for instance, the most insoluble or volatile composition will be preferentially produced because it removes itself from the reaction. In the absence of such influences, he believed the ratios could vary continuously. Another scientist, Proust, said that the proportions were always the same, and eventually persuaded Berthollet of the . Proust was more careful to study only pure compounds, and knew how precise his measurements were, so that even though the exact numbers didn't always come out the same, once he rounded to the correct number of significant figures, the ratios were the same, no matter how the material was prepared. Review the difference between accuracy and precision on the previous page ( ), and how to use the correct number of . (Some types of materials, like minerals or alloys, can have variable proportions; Berthollet was also right that in solution compositions might be possible that were not observed in the isolated substances. Berthollet's arguments on composition were inspired by his correct understanding of chemical equilibrium, but because he was wrong about definite proportions for pure compounds, this contribution to scientific knowledge was not recognized for a long time.) Studying the data of people like Proust and Lavoisier, Dalton noticed a remarkable pattern. Carbon and nitrogen both combine with oxygen in several different definite ratios to form several different products. Likewise sulfur and phosphorus, which is why there are sulfates and sulfites. For example, carbon combines with oxygen in the ratios of 3:4 and 3:8 (3g carbon for each 4g of oxygen). Of course, 2 x 4 = 8. One compound had exactly twice as much oxygen as the other. Nitrogen combines with oxygen in ratios of 7:4, 7:8, and 7:16. In each case, the amount of oxygen doubles. This is called the . These laws all made good sense if each element has atoms that combine as whole atoms. An is a very small, distinct thing, like a ball. Now we know that atoms can be divided into smaller parts, but an atom is the smallest amount of an element you can have, because if you divide it into smaller parts, the properties will change. John Dalton's atomic theory: In more modern terminology, we say that atoms of different elements are combined to make molecules. Chemical reactions change how the atoms are combined, but the number of each type of atom doesn't change.

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Make sure you thoroughly understand the following essential ideas: Consider a simple reaction such as the gas-phase synthesis of hydrogen iodide from its elements: \[H_2 + I_2 \rightarrow 2 HI\] Suppose you combine arbitrary quantities of \(H_2\), \(I_2\) and \(HI\). Will the reaction create more HI, or will some of the HI be consumed as the system moves toward its equilibrium state? The concept of the , which is the focus of this short lesson, makes it easy to predict what will happen. In the previous section we defined the for the reaction In the general case in which the concentrations can have any arbitrary values (including zero), this expression is called the (the term is also commonly used.) and its value is denoted by \(Q\) (or \(Q_c\) or \(Q_p\) if we wish to emphasize that the terms represent molar concentrations or partial pressures.) If the terms correspond to concentrations, then the above expression is called the and its value is denoted by \(K\) (or \(K_c\) or \(K_p\)). is thus the special value that \(Q\) has when the reaction is at equilibrium The value of in relation to serves as an index how the composition of the reaction system compares to that of the equilibrium state, and thus it indicates the direction in which any net reaction must proceed. For example, if we combine the two reactants and at concentrations of 1 mol L each, the value of will be 0÷1=0. The only possible change is the conversion of some of these reactants into products. If instead our mixture consists only of the two products and , will be indeterminately large (1÷0) and the only possible change will be in the reverse direction. For example, if we combine the two reactants and at concentrations of 1 mol L each, the value of will be 0÷1=0. The only possible change is the conversion of some of these reactants into products. If instead our mixture consists only of the two products and , will be indeterminately large (1÷0) and the only possible change will be in the reverse direction. It is easy to see (by simple application of the Le Chatelier principle) that the ratio of immediately tells us whether, and in which direction, a net reaction will occur as the system moves toward its equilibrium state. A schematic view of this relationship is shown below: It is very important that you be able to work out these relations for yourself, not by memorizing them, but from the definitions of \(Q\) and \(K\). It is very important that you be able to work out these relations for yourself, not by memorizing them, but from the definitions of \(Q\) and \(K\). The equilibrium constant for the oxidation of sulfur dioxide is = 0.14 at 900 K. \[\ce{2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)} \nonumber\] If a reaction vessel is filled with SO at a partial pressure of 0.10 atm and with O and SO each at a partial pressure of 0.20 atm, what can you conclude about whether, and in which direction, any net change in composition will take place? The value of the equilibrium quotient for the initial conditions is \[ Q= \dfrac{p_{SO_3}^2}{p_{O_2}p_{SO_2}^2} = \dfrac{(0.10\; atm)^2}{(0.20 \;atm) (0.20 \; atm)^2} = 1.25\; atm^{-1} \nonumber\] Since > , the reaction is not at equilibrium, so a net change will occur in a direction that decreases . This can only occur if some of the SO is converted back into products. In other words, the reaction will "shift to the left". The formal definitions of and are quite simple, but they are of limited usefulness unless you are able to relate them to real chemical situations. The following diagrams illustrate the relation between and from various standpoints. Take some time to study each one carefully, making sure that you are able to relate the description to the illustration. For the reaction \[N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)} \nonumber\] This equilibrium condition is represented by the red curve that passes through all points on the graph that satisfy the requirement that \[Q = \dfrac{[NO_2]^2}{ [N_2O_4]} = 0.0059 \nonumber\] There are of course an infinite number of possible 's of this system within the concentration boundaries shown on the plot. Only those points that fall on the red line correspond to of this system (those for which \(Q = K_c\)). The line itself is a plot of [NO ] that we obtain by rearranging the equilibrium expression \[[NO_2] = \sqrt{[N_2O_4]K_c} \nonumber\] If the system is initially in a non-equilibrium state, its composition will tend to change in a direction that moves it to one that is on the line. Two such non-equilibrium states are shown. The state indicated by has \(Q > K\), so we would expect a net reaction that reduces by converting some of the NO into N O ; in other words, the equilibrium "shifts to the left". Similarly, in state < , indicating that the forward reaction will occur. The blue arrows in the above diagram indicate the successive values that Q assumes as the reaction moves closer to equilibrium. The slope of the line reflects the stoichiometry of the equation. In this case, one mole of reactant yields two moles of products, so the slopes have an absolute value of 2:1. One of the simplest equilibria we can write is that between a solid and its vapor. In this case, the equilibrium constant is just the vapor pressure of the solid. Thus for the process \[I_{2(s)} \rightleftharpoons I_{2(g)} \nonumber\] all possible equilibrium states of the system lie on the horizontal red line and is independent of the quantity of solid present (as long as there is at least enough to supply the relative tiny quantity of vapor.) So adding various amounts of the solid to an empty closed vessel (states and ) causes a gradual buildup of iodine vapor. Because the equilibrium pressure of the vapor is so small, the amount of solid consumed in the process is negligible, so the arrows go straight up and all lead to the same equilibrium vapor pressure. The decomposition of ammonium chloride is a common example of a heterogeneous (two-phase) equilibrium. Solid ammonium chloride has a substantial vapor pressure even at room temperature: \[NH_4Cl_{(s)} \rightleftharpoons NH_{3(g)} + HCl_{(g)}\] Arrow traces the states the system passes through when solid NH Cl is placed in a closed container. Arrow represents the addition of ammonia to the equilibrium mixture; the system responds by following the path back to a new equilibrium state which, as the Le Chatelier principle predicts, contains a smaller quantity of ammonia than was added. The unit slopes of the paths and reflect the 1:1 stoichiometry of the gaseous products of the reaction.

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

are long chain giant organic molecules are assembled from many smaller molecules called . consist of many repeating monomer units in long chains. A polymer is analogous to a necklace made from many small beads (monomers). Many monomers are alkenes or other molecules with double bonds which react by to their unsaturated double bonds. The electrons in the double bond are used to bond two monomer molecules together. This is represented by the red arrows moving from one molecule to the space between two molecules where a new bond is to form. The formation of polyethylene from ethylene (ethene) may be illustrated in the graphic on the left as follows. In the complete polymer, all of the double bonds have been turned into single bonds. No atoms have been lost and you can see that the monomers have just been joined in the process of addition. A simple representation is -[A-A-A-A-A]-. Polyethylene is used in plastic bags, bottles, toys, and electrical insulation. Non-stick coating for cooking utensils, chemically-resistant specialty plastic parts, Gore-Tex usually give elastomer . In this kind of polymerization, molecular rings are opened in the formation of a polymer. Here epsilon-caprolactam, a 6-carbon cyclic monomer, undergoes ring opening to form a Nylon 6 homopolymer, which is somewhat similar to but not the same as Nylon 6,6 alternating copolymer.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/Magnetism

In 1923, Lewis wrote that the study of spectra and magnetism are the 2 best ways to learn about chemical bonding. Some types of spectra are discussed . In this section, we will describe the magnetic evidence Lewis used. You are familiar with normal magnets, like the ones used to stick restaurant menus to fridges. In this section, however, we are more interested in the magnetic properties of molecular materials, rather than metallic or semi-ionic solids. Most materials are non-magnetic. The molecular materials that are magnetic still wouldn't stick to your fridge, because they only really act magnetic in a magnetic field. Normally, each molecular magnetic has a random direction because the interactions between the molecules are weak. (Just like how molecular materials melt at lower temperature than metals or rocks). If you put them in a magnetic field, if they are cold enough, they will start to line up with the field. If they are too hot, they will continue to move randomly. This is called . Let's start by reviewing electric dipoles. An is something that had separated electrical charges. For instance, an HF molecule or a water molecule have an electric defined as \[\mu = q \times d\] where q is the partial charge on each end of the molecule and d is the distance between the charges. Generally, solvents with large dipoles have high . In Coulomb's law, the dielectric constant D reduces the force between charges. \[F= \frac{kQq}{Dr^{2}}\] Bigger D means less force. This explains how ionic substances dissolve in solvents with large dipole moments and large D. When the ions separate and have water between them, they no longer attract each other strongly. Although you can have a point charge, magnets are always dipoles (have a North and South pole). Just like solvents and other materials have dielectric constants, they also have magnetic permeabilities. How do you measure or magnetic dipoles in molecules? The old way is to put the molecular material between the poles of a big magnet and see what forces are present. For instance, if you put a paramagnetic material (that is cold enough) between the N and S magnetic poles, it will line up with the field and be attracted by the field. Thus if you weigh a paramagnetic material in a magnetic field, it will be heavier than without the field. Most materials are , and get lighter in magnetic fields, because they are repelled by the field. So the easiest way to see if a material is magnetic is to weigh it with and without a magnetic field. (Now there is a fancier method also, called SQUID, which is too complicated to explain here.) Lewis correctly realized that molecules with an odd number of electrons (such as NO, with 7+8 =15, but not CO, with 6+8 =14) have some unusual properties. They are paramagnetic, usually strongly colored, and quite reactive. The huge majority of molecules have even numbers of electrons, and are diamagnetic. Thus, paramagnetism seems to be a result of unpaired electrons, while diamagnetism is a result of paired electrons. Transition metal compounds are often colored and magnetic; most compounds of C,H,O,N etc (such as in your body) have little color, are diamagnetic, and not so reactive. Basically, almost all molecules have all their electrons in pairs.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.06%3A_Avogadro's_Law

For most solids and liquids it is convenient to obtain the amount of substance (and the number of particles, if we want it) from the mass. In the section on numerous such calculations using molar mass were done. In the case of gases, however, accurate measurement of mass is not so simple. Think about how you would weigh a balloon filled with helium, for example. Because it is buoyed up by the air it displaces, such a balloon would force a balance pan instead of down, and a negative weight would be obtained. The mass of a gas can be obtained by weighing a truly empty container (with a perfect vacuum), and then filling and re-weighing the container. But this is a time-consuming, inconvenient, and sometimes dangerous procedure. (Such a container might —explode inward—due to the difference between atmospheric pressure outside and zero pressure within.) A more convenient way of obtaining the amount of substance in a gaseous sample is suggested by the data on molar volumes in Table \(\Page {1}\). Remember that a molar quantity (a quantity divided by the amount of substance) refers to the same number of particles. The data in Table \(\Page {1}\), then, indicate that for a variety of gases, 6.022 × 10 molecules occupy almost exactly the same volume (the ) if the temperature and pressure are held constant. We define for gases as 0°C and 1.00 atm (101.3 kPa) to establish convenient conditions for comparing the molar volumes of gases. The molar volume is close to 22.4 liters (22.4 dm ) for virtually all gases. That was first suggested in 1811 by the Italian chemist (1776 to 1856). Consequently it is called or . Avogadro’s law has two important messages. First, it says that molar volumes of gases are the same at a given temperature and pressure. Therefore, even if we do not know what gas we are dealing with, we can still find the amount of substance. The image below demonstrates this concept. All 3 balloons are full of different gases, yet have the same number of moles and therefore the same volume (22.4 Liters).     Second, we expect that if a particular volume corresponds to a certain number of molecules, twice that volume would contain twice as many molecules. In other words, the volume corresponds to the amount of substance, the volume corresponds to the amount, and so on. In general, if we the volume by some factor, say , then we also the amount of substance by that same factor . Such a relationship is called direct proportionality and may be expressed mathematically as \[\text{V \(\propto\) n}\label{1} \] where the symbol \(\propto\) means “is proportional to.” For a simple demonstration of this concept, play with Concord Consortium's tool shown below, which allows you to manipulate the number of gas molecules in an a certain area and observe the effects on the volume. Try beginning with the default 120 molecules and observing the volume. Then cut the number of molecules in half to 60 and see what affect that has on the volume... Any proportion, such as Equation \(\ref{1}\) can be changed to an equivalent equation if one side is multiplied by a proportionality constant, such a in Equation\(\ref{2}\): \[\text{V} = \text{k}_A\text{n}\label{2} \] If we know for a gas, we can determine the amount of substance from Equation \(\ref{2}\). The situation is complicated by the fact that the volume of a gas depends on pressure and temperature, as well as on the amount of substance. That is, will vary as temperature and pressure change. Therefore we need quantitative information about the effects of pressure and temperature on the volume of a gas before we can explore the relationship between amount of substance and volume.

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

\(\ce{HCl}\) Hint: 0.4 mol/L Calculate using ideal gas law. Apply ideal gas law to solve stoichiometry problems. \(\ce{HCl}\) \(\ce{NaOH}\) \(\ce{NaOH}\) Hint: 0.64 mol/L Solve stoichiometric problem. \(\ce{H2S + SO2 \rightarrow H2O + S_{(solid)}}\), unbalanced \(\ce{H2S}\) Hint: 93 degrees C Apply ideal gas law to solve stoichiometry problems. \(\ce{AgNO3}\) \(\ce{HI}\) \(\ce{AgI}\) \(\ce{AgNO3}\) Hint: 0.10 M \(\ce{AgNO3}\) \(\ce{HI}\) \(\ce{AgI}\) \(\ce{HI}\) Hint: 0.020 mol/L \(\ce{AgNO3}\) \(\ce{HI}\) \(\ce{AgI}\) \(\ce{HI}\) Hint: 0.49 atm Depending on the numerical values you use, you may get the pressure in other units. \(\ce{AgNO3}\) \(\ce{HI}\) \(\ce{AgI}\) \(\ce{HI}\) Hint: 300 K Note the relationship of this problem with the previous one.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.4%3A_Integrated_Rate_Laws

The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called . We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level. Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions. An equation relating the rate constant \(k\) to the initial concentration \([A]_0\) and the concentration \([A]_t\) present after any given time \(t\) can be derived for a first-order reaction and shown to be: \[\ln\left(\dfrac{[A]_t}{[A]_0}\right)=−kt \nonumber \] or alternatively \[\ln\left(\dfrac{[A]_0}{[A]_t}\right)=kt \nonumber \] or \[[A]=[A]_0e^{−kt} \nonumber \] The rate constant for the first-order decomposition of cyclobutane, \(\ce{C4H8}\) at 500 °C is 9.2 × 10 s : \[\ce{C4H8⟶2C2H4} \nonumber \] How long will it take for 80.0% of a sample of C H to decompose? We use the integrated form of the rate law to answer questions regarding time: \[\ln\left(\dfrac{[A]_0}{[A]}\right)=kt \nonumber \] There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [ ] , [ ], and , and need to find . The initial concentration of C H , [ ] , is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of or 0.200 Rearranging the rate law to isolate and substituting the provided quantities yields: \[\begin{align*} t&=\ln\dfrac{[x]}{[0.200x]}×\dfrac{1}{k}\\[4pt] &=\mathrm{\ln\dfrac{0.100\:mol\: L^{−1}}{0.020\:mol\: L^{−1}}×\dfrac{1}{9.2×10^{−3}\:s^{−1}}}\\[4pt] &=\mathrm{1.609×\dfrac{1}{9.2×10^{−3}\:s^{−1}}}\\[4pt] &=\mathrm{1.7×10^2\:s} \end{align*} \nonumber \] Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation: \[\textrm{I-131 ⟶ Xe-131 + electron} \nonumber \] The decay is first-order with a rate constant of 0.138 d . All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 solution of this substance to decay to Xe-131? 16.7 days We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format: \[\begin{align} \ln[A]&=(−k)(t)+\ln[A]_0 \label{in1st}\\[4pt] y&=mx+b \end{align} \nonumber \] A plot of \(\ln[A]\) versus \(t\) for a first-order reaction is a straight line with a slope of \(−k\) and an intercept of \(\ln[A]_0\). If a set of rate data are plotted in this fashion but do result in a straight line, the reaction is not first order in \(A\). Show that the data in this Figure can be represented by a first-order rate law by graphing ln[H O ] versus time. Determine the rate constant for the rate of decomposition of H O from this data. The data from this Figure with the addition of values of ln[H O ] are given in Figure \(\Page {1}\). The plot of ln[H O ] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law. The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H O ] versus time where: \[\ce{slope}=\dfrac{\textrm{change in }y}{\textrm{change in }x}=\dfrac{Δy}{Δx}=\dfrac{Δ\ln[\ce{H2O2}]}{Δt} \nonumber \] In order to determine the slope of the line, we need two values of ln[H O ] at different values of (one near each end of the line is preferable). For example, the value of ln[H O ] when is 6.00 h is −0.693; the value when = 12.00 h is −1.386: \[\begin{align*} \ce{slope}&=\mathrm{\dfrac{−1.386−(−0.693)}{12.00\: h−6.00\: h}}\\[4pt] &=\mathrm{\dfrac{−0.693}{6.00\: h}}\\[4pt] &=\mathrm{−1.155×10^{−2}\:h^{−1}}\\[4pt] k&=\mathrm{−slope=−(−1.155×10^{−1}\:h^{−1})=1.155×10^{−1}\:h^{−1}} \end{align*} \nonumber \] Graph the following data to determine whether the reaction \(A⟶B+C\) is first order. The plot of ln[A] vs. t is not a straight line. The equation is not first order:     The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law: \[\ce{Rate}=k[A]^2 \nonumber \] For these second-order reactions, the integrated rate law is: \[\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \label{int2nd} \] where the terms in the equation have their usual meanings as defined earlier. The reaction of butadiene gas (C H ) with itself produces C H gas as follows: \[\ce{2C4H6}(g)⟶\ce{C8H12(g)} \nonumber \] The reaction is second order with a rate constant equal to 5.76 × 10 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 , what is the concentration remaining after 10.0 min? We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have: \[\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \nonumber \] We know three variables in this equation: [ ] = 0.200 mol/L, = 5.76 × 10 L/mol/min, and = 10.0 min. Therefore, we can solve for [ ], the fourth variable: \[\begin{align*} \dfrac{1}{[A]}&=\mathrm{(5.76×10^{−2}\:L\: mol^{−1}\:min^{−1})(10\:min)+\dfrac{1}{0.200\:mol^{−1}}}\\[4pt] \dfrac{1}{[A]}&=\mathrm{(5.76×10^{−1}\:L\: mol^{−1})+5.00\:L\: mol^{−1}}\\[4pt] \dfrac{1}{[A]}&=\mathrm{5.58\:L\: mol^{−1}}\\[4pt] [A]&=\mathrm{1.79×10^{−1}\:mol\: L^{−1}} \end{align*} \nonumber \] Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present. If the initial concentration of butadiene is 0.0200 , what is the concentration remaining after 20.0 min? 0.0196 mol/L The integrated rate law for our second-order reactions has the form of the equation of a straight line: \[\begin{align*} \dfrac{1}{[A]}&=kt+\dfrac{1}{[A]_0}\\[4pt] y&=mx+b \end{align*} \nonumber \] A plot of \(\dfrac{1}{[A]}\) versus for a second-order reaction is a straight line with a slope of and an intercept of \(\dfrac{1}{[A]_0}\). If the plot is not a straight line, then the reaction is not second order. Test the data given to show whether the dimerization of C H is a first- or a second-order reaction. In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C H ] versus and compare it with a plot of \(\mathrm{\dfrac{1}{[C_4H_6]}}\) versus . The values needed for these plots follow. The plots are shown in Figure \(\Page {2}\). As you can see, the plot of ln[C H ] versus is not linear, therefore the reaction is not first order. The plot of \(\dfrac{1}{[\ce{C4H6}]}\) versus is linear, indicating that the reaction is second order. Does the following data fit a second-order rate law? Yes. The plot of \(\dfrac{1}{[A]}\) vs. is linear:     For zero-order reactions, the differential rate law is: \[\ce{Rate}=k[A]^0=k \nonumber \] A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants. The integrated rate law for a zero-order reaction also has the form of the equation of a straight line: \[\begin{align*} [A]&=−kt+[A]_0 \label{intzero}\\[4pt] y&=mx+b \end{align*} \nonumber \] A plot of \([A]\) versus \(t\) for a zero-order reaction is a straight line with a slope of and an intercept of [ ] . Figure \(\Page {3}\) shows a plot of [NH ] versus for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO ). The decomposition of NH on hot tungsten is zero order; the plot is a straight line. The decomposition of NH on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant: \[\ce{slope}=−k=\mathrm{1.3110^{−6}\:mol/L/s} \nonumber \]   The is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H O decreases from 1.000 to 0.500 . During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 to 0.250 ; during the third half-life, it decreases from 0.250 to 0.125 . The concentration of H O decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants. We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows: \[\begin{align*} \ln\dfrac{[A]_0}{[A]}&=kt\\[4pt] t&=\ln\dfrac{[A]_0}{[A]}×\dfrac{1}{k} \end{align*} \nonumber \] If we set the time equal to the half-life, \(t_{1/2}\), the corresponding concentration of at this time is equal to one-half of its initial concentration. Hence, when \(t=t_{1/2}\), \([A]=\dfrac{1}{2}[A]_0\). Therefore: \[\begin{align*} t_{1/2}&=\ln\dfrac{[A]_0}{\dfrac{1}{2}[A]_0}×\dfrac{1}{k}\\[4pt] &=\ln 2×\dfrac{1}{k}=0.693×\dfrac{1}{k} \end{align*} \nonumber \] Thus: \[t_{1/2}=\dfrac{0.693}{k} \nonumber \] We can see that the half-life of a first-order reaction is inversely proportional to the rate constant . A fast reaction (shorter half-life) will have a larger ; a slow reaction (longer half-life) will have a smaller . Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in Figure \(\Page {4}\). The half-life for the decomposition of H O is 2.16 × 10 s: \[\begin{align*} t_{1/2}&=\dfrac{0.693}{k}\\[4pt] k&=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{2.16×10^4\:\ce s}=3.21×10^{−5}\:\ce s^{−1} \end{align*} \nonumber \] The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d . What is the half-life for this decay? 5.02 d.   We can derive the equation for calculating the half-life of a second order as follows: \[\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \nonumber \] or \[\dfrac{1}{[A]}−\dfrac{1}{[A]_0}=kt \nonumber \] If \[t=t_{1/2} \nonumber \] then \[[A]=\dfrac{1}{2}[A]_0 \nonumber \] and we can write: Thus: \[t_{1/2}=\dfrac{1}{k[A]_0} \nonumber \] For a second-order reaction, \(t_{1/2}\) is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.   We can derive an equation for calculating the half-life of a zero order reaction as follows: \[[A]=−kt+[A]_0 \nonumber \] When half of the initial amount of reactant has been consumed \(t=t_{1/2}\) and \([A]=\dfrac{[A]_0}{2}\). Thus: \[\begin{align*} \dfrac{[A]_0}{2}&=−kt_{1/2}+[A]_0\\[4pt] kt_{1/2}&=\dfrac{[A]_0}{2} \end{align*} \nonumber \] and \[t_{1/2}=\dfrac{[A]_0}{2k} \nonumber \] The half-life of a zero-order reaction increases as the initial concentration increases. Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table \(\Page {1}\).   Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction. The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases. \([A]=−kt+[A]_0\) with \(t_{1/2}=\dfrac{[A]_0}{2k}\) \(\ln[A]=−kt+ \ln[A]_0\) with \(t_{1/2}=\dfrac{0.693}{k}\) \(\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0}\) with \(t_{1/2}=\dfrac{1}{[A]_0k}\)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.E%3A_Kinetics_(Exercises)

What is the difference between average rate, initial rate, and instantaneous rate? First, a general reaction rate must be defined to know what any variation of a rate is. The reaction rate is defined as the measure of the change in concentration of the reactants or products per unit time. The rate of a chemical reaction is not a constant and rather changes continuously, and can be influenced by temperature. Rate of a reaction can be defined as the disappearance of any reactant or appearance of any product. Thus, an average rate is the average reaction rate over a given period of time in the reaction, the instantaneous rate is the reaction rate at a specific given moment during the reaction, and the initial rate is the instantaneous rate at the very start of the reaction (when the product begins to form). The instantaneous rate of a reaction can be denoted as \[ \lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t} \nonumber \] Ozone decomposes to oxygen according to the equation \(\ce{2O3}(g)⟶\ce{3O2}(g)\). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O and the formation of oxygen. For the general reaction, aA ---> bB, the rate of the reaction can be expressed in terms of the disappearance of A or the appearance of B over a certain time period as follows. \[- \dfrac{1}{a}\dfrac{\Delta [A]}{\Delta t} = - \dfrac{1}{b}\dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{\Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{\Delta [D]}{\Delta t}\] We want the rate of a reaction to be positive, but the change in the concentration of a reactant, A, will be negative because it is being used up to be transformed into product, B. Therefore, when expressing the rate of the reaction in terms of the change in the concentration of A, it is important to add a negative sign in front to ensure the overall rate positive. Lastly, the rate must be normalized according to the stoichiometry of the reaction. In the decomposition of ozone to oxygen, two moles of ozone form three moles of oxygen gas. This means that the increase in oxygen gas will be 1.5 times as great as the decrease in ozone. Because the rate of the reaction should be able to describe both species, we divide the change in concentration by its stoichiometric coefficient in the balanced reaction equation to deal with this issue. Therefore, the rate of the reaction of the decomposition of ozone into oxygen gas can be described as follows: \[Rate=-\frac{Δ[O3]}{2ΔT}=\frac{Δ[O2]}{3ΔT}\] $$Rate=-\frac{Δ[O3]}{2ΔT}=\frac{Δ[O2]}{3ΔT}\] In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction \(\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)\). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl and F and the formation of ClF . In this problem we are asked to write the equation that relates rate expressions in terms of disappearance of the reactants of the equation and in terms of the formation of the product. A reaction rate gives insight to how rate is affected as a function of concentration of the substances in the equation. Rates can often be expressed on graphs of concentration vs time expressed in change (\({\Delta}\)) of concentration and time and in a short enough time interval, the instantaneous rate can be approximated. If we were to analyze the reaction given, the graph would demonstrate that Cl decreases, that F decreases 3 times as quickly, and then ClF increases at a rate doubles. The reactants are being used and converted to product so they decrease while products increase. For this problem, we can apply the general formula of a rate to the specific aspects of a problem where the general form follows: \[aA+bB⟶cC+dD\nonumber \]. And the rate can then be written as \(rate=-\frac {1}{a}\frac{{\Delta}[A]}{{\Delta}t}\) \(=-\frac {1}{b}\frac{{\Delta}[B]}{{\Delta}t}\) \(=\frac {1}{c}\frac{{\Delta}[C]}{{\Delta}t}\) \(=\frac {1}{d}\frac{{\Delta}[D]}{{\Delta}t}.\) Here the negative signs are used to keep the convention of expressing rates as positive numbers. In this specific case we use the stoichiometry to get the specific rates of disappearance and formation (back to what was said in the first paragraph). So, the problem just involves referring the to the equation and its balanced coefficients. Based upon the equation we see that Cl is a reactant and has no coefficient, F has a coefficient of 3 and is also used up, and then ClF is a product that increases two-fold with a coefficient of 2. So, the rate here can be written as: \[rate=-\frac{{\Delta}[Cl_2]}{{\Delta}t}=-\frac {1}{3}\frac{{\Delta}[F_2]}{{\Delta}t}=\frac {1}{2}\frac{{\Delta}[ClF_3]}{{\Delta}t}\nonumber \] \[\ce{rate}=+\dfrac{1}{2}\dfrac{Δ[\ce{CIF3}]}{Δt}=−\dfrac{Δ[\ce{Cl2}]}{Δt}=−\dfrac{1}{3}\dfrac{Δ[\ce{F2}]}{Δt}\nonumber \] A study of the rate of dimerization of C H gave the data shown in the table: \[\ce{2C4H6⟶C8H12}\nonumber \] 1.) The average rate of dimerization is the change in concentration of a reactant per unit time. In this case it would be: \(rate\) \(of\) \(dimerization=-\frac{\Delta [C_4H_6]}{\Delta t}\) Rate of dimerization between 0 s and 1600 s: \(rate\) \(of\) \(dimerization=-\frac{5.04×10^{-3}M-1.00×10^{-2}M}{1600 s-0 s}\) Rate of dimerization between 1600 s and 3200 s: \(rate\) \(of\) \(dimerization=-\frac{3.37×10^{-3}M-5.04×10^{-3}M}{3200 s-1600 s}\) 2.) The instantaneous rate of dimerization at 3200 s can be found by graphing time versus [C H ]. Because you want to find the rate of dimerization at 3200 s, you need to find the slope between 1600 s and 3200 s and also 3200 s and 4800 s. For the slope between 1600 s and 3200 s use the points (1600 s, 5.04 x 10 M) and (3200 s, 3.37 x 10 M) \(\frac{3.37×10^{-3}M-5.04×10^{-3}M}{3200 s-1600 s}\) \(\frac{-0.00167 M}{1600 s}\) \(-1.04×10^{-6}\frac{M}{s}\) For the slope between 3200 s and 4800 s use the points (3200s, 3.37 x 10 M) and (4800s, 2.53 x 10 M) \(\frac{2.53×10^{-3}M-3.37×10^{-3}M}{4800 s-3200 s}\) \(\frac{-8.4×10^{-4} M}{1600 s}\) \(-5.25×10^{-7}\frac{M}{s}\) Take the two slopes you just found and find the average of them to get the instantaneous rate of dimerization. \(\frac{-1.04×10^{-6}\frac{M}{s}+-5.25×x10^{-7}\frac{M}{s}}{2}\) \(\frac{-1.565×10^{-6}\frac{M}{s}}{2}\) \(-7.83×10^-7\frac{M}{s}\) The instantaneous rate of dimerization is and the units of this rate is . 3.) The average rate of formation of C H at 1600 s and the instantaneous rate of formation at 3200 s can be found by using our answers from part a and b. If you look back up at the original equation, you could see that C H and C H are related in a two to one ratio. For every two moles of C H used, there is one mole of C H produced. For this reaction, the average rate of dimerization and the average rate of formation can be linked through this equation: \(\frac{-1}{2}\frac{\Delta [C_4H_6]}{\Delta t}=\frac{\Delta [C_8H_{12}]}{\Delta t}\) Notice that reactant side is negative because the reactants are being used up in the reaction. So, for the average rate of formation of C H at 1600 s, use the rate of dimerization between 0 s and 1600 s we found earlier and plug into the equation: The average rate of formation for C H at 1600 s is \(1.55×10^{-6}\frac{M}{s}\). The rate of formation will be positive because products are being formed. The instantaneous rate of formation for C H can be linked to the instantaneous rate of dimerization by this equation: \(\frac{-1}{2}\frac{d[C_4H_6]}{dt}=\frac{d[C_8H_{12}]}{dt}\) So, for the instantaneous rate of formation for C H at 3200 s, use the value of instantaneous rate of dimerization at 3200 s found earlier and plug into the equation: The instantaneous rate of formation for C H at 3200 s is \(-3.92×10^-7\frac{M}{s}\) A study of the rate of the reaction represented as \(2A⟶B\) gave the following data: Equations: \(\frac{-\bigtriangleup A}{\bigtriangleup time}\) and Rate=\(\frac{-\bigtriangleup A}{2\bigtriangleup time}=\frac{\bigtriangleup B}{time}\) Solve: 1.)The change in A from 0s to 10s is .625-1=-.375 so \(\frac{-\bigtriangleup A}{\bigtriangleup time}\)=.375/10= Similarly, the change in A from 10 to 20 seconds is .370-.625=-.255 so \(\frac{-\bigtriangleup A}{\bigtriangleup time}\)=.255/20-10= 2.) We can estimate the rate law graphing the points against different order equations to determine the right order. Zero Order: \[\frac{d[A]}{dt}=-k\nonumber \] \[\int_{A_{\circ}}^{A}d[A]=-k\int_{0}^{t}dt\nonumber \] \[[A]=-kt+[A_{\circ}]\nonumber \] First Order: \[\frac{d[A]}{dt}=-k[A]\nonumber \] \[\int_{A_{\circ}}^{A}\frac{d[A]}{[A]}=-kdt\nonumber \] \[Ln(A)=-kt+Ln(A_{\circ})\nonumber \] Second Order: \[\frac{d[A]}{dt}=-k[A]^{2}\nonumber \] \[\int_{A\circ}^{A}\frac{d[A]}{[A]^{2}}=-k\int_{0}^{t}dt\nonumber \] \[\frac{1}{[A]}=kt+\frac{1}{[A_{\circ}]}\nonumber \] Now that we have found the linear from of each order we will plot the points vs an [A] y-axis, a Ln(A) y-axis, and a 1/[A] y-axis. whichever of the plots has the most linear points will give us a good idea of the order and the slope will be the k value. Here we notice that the second order is most linear so we conclude the Rate to be.. \[\frac{-d[A]}{2dt}=k[A]^{2}\nonumber \] At 15 seconds [A]=.465 and from the slope of the graph we find k=.116.so if we plug this data in and multiply both sides by 2 to get rid of the 2 in the denominator on the left side of the equation we find that the rate of disappearance of A is .05 M/s where the units are equivalent to [mol*L *s ] 3.) Using the equation \(\frac{-\bigtriangleup A}{2\bigtriangleup time}=\frac{\bigtriangleup B}{time}\) we divide the rates in part a and b in half to get .0188 M/s from 0 to 10 seconds and .025 M/s for the estimated instantaneous rate at 15s. (a) average rate, 0 − 10 s = 0.0375 mol L s ; average rate, 12 − 18 s = 0.0225 mol L s ; (b) instantaneous rate, 15 s = 0.0500 mol L s ; (c) average rate for B formation = 0.0188 mol L s ; instantaneous rate for B formation = 0.0250 mol L s Consider the following reaction in aqueous solution: \[\ce{5Br-}(aq)+\ce{BrO3-}(aq)+\ce{6H+}(aq)⟶\ce{3Br2}(aq)+\ce{3H2O}(l)\nonumber \] If the rate of disappearance of Br ( ) at a particular moment during the reaction is 3.5 × 10 , what is the rate of appearance of Br ( ) at that moment? Define the rate of the reaction. Recall: For the general reaction: aA + bB → cC+ dD \(rate =- \frac{\Delta[A]}{a\Delta{t}}=- \frac{\Delta[B]}{b\Delta{t}}= \frac{\Delta[C]}{c\Delta{t}}=\frac{\Delta[D]}{d\Delta{t}}\) So, for the reaction: \(5Br^−(aq)+BrO^−_3(aq)+6H^+→3Br_2(aq)+3H_2O(l)\) The rate would be: \(rate =- \frac{\Delta[Br^-]}{5\Delta{t}}=- \frac{\Delta[BrO^-_3]}{\Delta{t}}= -\frac{\Delta[H^+]}{6\Delta{t}}=\frac{\Delta[Br_2]}{3\Delta{t}}=\frac{H_2O}{3\Delta{t}}\) Since we are given the rate for the disappearance of \(Br^-\)(aq) is \(3.5x10^-4 Ms^{-1}\), and we want to find the rate of appearance of \(Br_2\)(aq). Therefore we set the two rates equal to each other. \(rate =- \frac{\Delta[Br^-]}{5\Delta{t}}= \frac{\Delta[Br_2]}{3\Delta{t}}\) And,\(-\frac{\Delta[Br^-]}{\Delta{t}}= -3.5x10^{-4} Ms^{-1}\) So, \(3.5x10^{-4} Ms^{-1}\) = \(\frac{5}{3}\frac{\Delta[Br_2]}{\Delta{t}}\) Now solve the equation. \(\frac{(3.5x10^{-4})(3)}{5} = \frac{\Delta[Br_2]}{\Delta{t}}\) \(\frac{\Delta[Br_2]}{\Delta{t}} = 2.1 x 10^{-4} Ms^{-1}\) \(\frac{\Delta[Br_2]}{\Delta{t}} = 2.1 x 10^{-4} Ms^{-1}\) Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium. Molarity of Hydrochloric Acid Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference. According to the collision theory, there are many factors that cause a reaction to happen, with three of the factors being how often the molecules or atoms collide, the molecules' or atoms' orientations, and if there is sufficient energy for the reaction to happen. So, if the angle of the plunger is changed, the atom that is shot (a lone Oxygen atom in this case) will hit the other molecule (CO in this case) at a different spot and at a different angle, therefore changing the orientation and the number of proper collisions will most likely not cause for a reaction to happen. Thanks to the simulation, we can see that this is true: depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other (no reaction happens). In this particular case, the rate of the reaction will decrease because, by changing the angle, the molecules or atoms won't collide with the correct orientation or as often with the correct orientation. In the PhET Reactions & Rates interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature? Based on the Collision Theory, a reaction will only occur if the molecules collide with proper orientation and with sufficient energy required for the reaction to occur. The minimum energy the molecules must collide with is called the activation energy (energy of transition state). Increasing the concentration of reactants increases the probability that reactants will collide in the correct orientation since there are more reactants in the same volume of space. Therefore, increasing the concentration of reactants would increase the rate of the reaction. Decreasing the concentration of reactants would decrease the rate of reaction because the overall number of possible collisions would decrease. Temperature is directly related the the kinetic energy of molecules and activation energy \(E_a\) is the minimum energy required for a reaction to occur and doesn't change for a reaction. Increasing the temperature increases the kinetic energy of the reactants meaning the reactants will move faster and collide with each other more frequently. Therefore, increasing the temperature increase the rate of the reaction. Decreasing the temperature decreases the rate of reaction since the molecules will have less kinetic energy, move slower, and therefore collide with each other less frequently. In the PhET Reactions & Rates interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options. a. On the simulation, we select the default setting and the reaction A+BC. In the default setting, we see frequent collisions, a low initial temperature, and a total average energy lower than the energy of activation. The collision theory states that the rate of a reaction is directly proportional to (the fraction of molecules with required orientation), (fractions of collisions with required energy), and (collision frequency). Although we see moving and frequently colliding reactants, the rate of the forward reaction is actually slow because it takes a long time for the products, AB and C, to start appearing. This is mainly because the fractions of collisions with required energy is low, coming from the average energy of the molecules being lower than the energy of activation. b. The reaction proceeds at an even faster rate. Again, the collision theory states that the rate of a reaction is directly proportional to (the fraction of molecules with required orientation), (fractions of collisions with required energy), and (collision frequency). Because molecules have a higher amount of energy, they have more kinetic energy. With an increased kinetic energy, the molecules not only collide more but also increase in the fraction of collision. However, the forward reaction and the backward reaction both proceed at a fast rate, so both happen almost simultaneously. It takes a shorter time for both reactions to happen. With both of the reactions adding up together overall, there is eventually a state of equilibrium. The process at which equilibrium is reached, however, is faster. Therefore, the amount of products of A+BC stays the same after a while. How do the rate of a reaction and its rate constant differ? The rate of a reaction or reaction rate is the change in the concentration of either the reactant or the product over a period of time. If the concentrations change, the rate also changes. Rate for A → B: The rate constant (k) is a proportionality constant that relates the reaction rates to reactants. If the concentrations change, the rate constant does not change. For a reaction with the general equation: \(aA+bB→cC+dD \) the experimentally determined rate law usually has the following form: Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions: (a) 2; (b) 1 Tripling the concentration of a reactant increases the rate of a reaction nine times. With this knowledge, answer the following questions: How much and in what direction will each of the following affect the rate of the reaction: \(\ce{CO}(g)+\ce{NO2}(g)⟶\ce{CO2}(g)+\ce{NO}(g)\) if the rate law for the reaction is \(\ce{rate}=k[\ce{NO2}]^2\)? (a) The process reduces the rate by a factor of 4. (b) Since CO does not appear in the rate law, the rate is not affected. How will each of the following affect the rate of the reaction: \(\ce{CO}(g)+\ce{NO2}(g)⟶\ce{CO2}(g)+\ce{NO}(g)\) if the rate law for the reaction is \(\ce{rate}=k[\ce{NO2},\ce{CO}]\) ? Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction \(\ce{NO + O3⟶NO2 + O2}\) is first order with respect to both NO and O with a rate constant of 2.20 × 10 L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 × 10 and [O ] = 5.9 × 10 ? 4.3 × 10 mol/L/s Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: \[\ce{^{32}_{15}P⟶^{32}_{16}S + e-}\nonumber \] Rate = 4.85 × 10 \(\mathrm{day^{-1}\:[^{32}P]}\) What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 ? The rate constant for the radioactive decay of C is 1.21 × 10 year . The products of the decay are nitrogen atoms and electrons (beta particles): \[\ce{^6_{14}C⟶^{6}_{14}N + e-}\nonumber \] \[\ce{rate}=k[\ce{^6_{14}C}]\nonumber \] What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 × 10 ? 7.9 × 10 mol/L/year What is the instantaneous rate of production of N atoms Q12.3.8 in a sample with a carbon-14 content of 1.5 × 10 ? The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 × 10 L/mol/s. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 × 10 ? Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men: Determine the rate equation, the rate constant, and the overall order for this reaction. rate = ; = 2.0 × 10 mol/L/h (about 0.9 g/L/h for the average male); The reaction is zero order. Under certain conditions the decomposition of ammonia on a metal surface gives the following data: Determine the rate equation, the rate constant, and the overall order for this reaction. Nitrosyl chloride, NOCl, decomposes to NO and Cl . \[\ce{2NOCl}(g)⟶\ce{2NO}(g)+\ce{Cl2}(g)\nonumber \] Determine the rate equation, the rate constant, and the overall order for this reaction from the following data: Before we can figure out the rate constant first we must first determine the basic rate equation and rate order. The basic rate equation for this reaction, where n is the rate order of NOCl and k is the rate constant, is \[rate = k[NOCl]^n\nonumber \] since NOCl is the reactant in the reaction. In order to figure out the order of the reaction we must find the order of [NOCl] as it is the only reactant in the reaction. To do this we must examine how the rate of the reaction changes as the concentration of NOCl changes. As [NOCl] doubles in concentration from 0.10 M to 0.20 M the rate goes from 8.0 x 10 to 3.2 x 10 (3.2 x 10 (mol/L/h))/(8.0 x 10 (mol/L/h)) = 4 so we conclude that as [NOCl] doubles, the rate goes up by 4. Since 2 = 4 we can say that the order of [NOCl] is 2 so our updated rate law is \[rate = k[NOCl]^2\nonumber \] Now that we have the order, we can substitute the first experimental values from the given table to find the rate constant, k (8.0 x 10 (mol/L/h)) = k(0.10 M) so \[k= \dfrac{8.0 \times 10^{-10}}{ (0.10\, M)^2} = 8 \times 10^{-8} M^{-1} sec^{-1}\nonumber \] We were able to find the units of k using rate order, when the rate order is 2 units of k are M x sec So the rate equation is: rate = k[NOCl] , it is second order, and k = 8 x 10 M x sec Overall rate law : \[rate = \underbrace{(8 \times 10^{-8})}_{\text{1/(M x sec)}} [NOCl]^2\nonumber \] rate = [NOCl] ; = 8.0 × 10 L/mol/s; second order From the following data, determine the rate equation, the rate constant, and the order with respect to for the reaction \(A⟶2C\). A. Using the experimental data, we can compare the effects of changing [A] on the rate of reaction by relating ratios of [A] to ratios of rates \[ \frac{2.66 \times 10^{-2}}{1.33 \times 10^{-2}} = 2\nonumber \] and \[ \frac{1.52 \times 10^{-6}}{3.8 \times 10^{-7}} = 4\nonumber \] B. From this we know that doubling the concentration of A will result in quadrupling the rate of reaction. The order of this reaction is 2. C. We can now write the rate equation since we know the order: \[rate=k[A]^2\nonumber \] D. By plugging in one set of experimental data into our rate equation we can solve for the rate constant, k: \[3.8 \times 10^{-7} = k \times (1.33 \times 10^{-2})^{2}\nonumber \] \[k = \frac{3.8 \times 10^{-7}}{1.769 \times 10^{-4}}\nonumber \] \[k= .00215 M^{-1}s^{-1}\nonumber \] \(k= .00215 M^{-1}s^{-1}\) 2nd Order Nitrogen(II) oxide reacts with chlorine according to the equation: \[\ce{2NO}(g)+\ce{Cl2}(g)⟶\ce{2NOCl}(g)\nonumber \] The following initial rates of reaction have been observed for certain reactant concentrations: What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl ? What is the rate constant? What are the orders with respect to each reactant? For the general equation, \(aA + bB \rightarrow cC + dD\) The rate can be written as \(rate = k[A]^{m}[B]^{n}\) where k is the rate constant, and m and n are the reaction orders. For our equation \(2NO(g) + Cl_{2}(g) \rightarrow 2NOCl(g)\) the \(rate = k[NO]^{m}[Cl_{2}]^{n}\) Now, we need to find the reaction orders. Reaction orders can only be found through experimental values. We can compare two reactions where one of the reactants has the same concentration for both trials, and solve for the reaction order. \(\frac{rate_{1}}{rate_{2}}=\frac{[NO]_{1}^{m}[Cl_{2}]_{1}^{n}}{[NO]_{2}^{m}[Cl_{2}]_{2}^{n}}\) We can use the data in the table provided. If we plug in the values for rows 1 and 2, we see that the values for the concentration of Cl will cancel, leaving just the rates and the concentrations of NO. \(\frac{1.14}{4.56}=\frac{[0.5]^{m}}{[1.0]^{m}}\) We can now solve for m, and we find that m =2. This means that the reaction order for [NO] is 2. Now we must find the value of n. To do so, we can use the same equation but with the values from rows 2 and 3. This time, the concentration of NO will cancel out. \(\frac{4.56}{9.12}=\frac{[0.5]^{n}}{[1.0]^{n}}\) When we solve for n, we find that n = 1. This means that the reaction order for [Cl ] is 1. We are one step closer to finishing our rate equation. \(rate = k[NO]^{2}[Cl_{2}]\) Finally, we can solve for the rate constant. To do this, we can use one of the trials of the experiment, and plug in the values for the rate, and concentrations of reactants, then solve for k. \(1.14 mol/L/h = k[0.5 mol/L]^{2}[0.5mol/L]\) \(k=9.12L^{2}mol^{-2}h^{-1}\) So, our final rate equation is: \(rate = (9.12 L^{2} mol^{-2}h^{-1})[NO]^{2}[Cl_{2}]\) *A common mistake is forgetting units. Make sure to track your units throughout the process of determining your rate constant. Be careful because the units will change relative to the reaction order. rate = [NO] [Cl] ; = 9.12 L mol h ; second order in NO; first order in Cl Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: \[\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g)\nonumber \] Determine the rate equation, the rate constant, and the orders with respect to each reactant from the following data: Determine the rate equation, the rate constant, and the orders with respect to each reactant. The rate constant and the orders can be determined through the differential rate law. The general form of the differential rate law is given below: aA + bB + cC ==> products where A, B, and C are the concentrations of the reactants, k is the rate constant, and n,m, and p refer to the order of each reactant. To find the orders of each reactant, we see that when [NO] doubles but [H ] doesn't change, the rate quadruples, meaning that [NO] is a second order reaction ([NO] ). When [H ] doubles but [NO] doesn't change, the rate doubles, meaning that [H ] is a first order reaction. So the rate law would look something like this: Rate = k[NO] [H ] We can use this rate law to determine the value of the rate constant. Plug in the data for reactant concentration and rate from one of the trials to solve for k the rate constant. In this case, we chose to use the data from trial 1 from the second column of the data table. 2.835x10 = k[0.3] [0.35] k = .09 M /s For the reaction \(A⟶B+C\), the following data were obtained at 30 °C: 1. The rate equation for an \(n\) order reaction is given as \(\frac{dr}{dt}={k}{[A]^n}\). Where \([A]\) is the concentration in M, and \(\frac{dr}{dt}\) is the rate in M/s. We can then use each set of data points, plug its values into the rate equation and solve for \(n\). Note you can use any of the data points as long as the concentration corresponds to its rate. Rate equation 1: \(4.17 \times {10}^{-4}={k}{[0.230]^n}\) Rate equation 2: \(9.99 \times {10}^{-4}={k}{[0.356]^n}\) We divide Rate equation 1 by Rate equation 2 in order to cancel out k, the rate constant. \({\frac{4.17 \times {10}^{-4}}{9.99 \times {10}^{-4}}} = {\frac{k[0.230]^n}{k[0.356]^n}} \) \({0.417}={0.646^n}\) Now the only unknown we have is \(n\). Using logarithm rules one can solve for it. \(ln{\: 0.417}={n \cdot ln{\: 0.646}}\) \(\frac{ln{\: 0.417}}{ln{\:0.646}}=n=2\) The rate equation is second order with respect to A and is written as \(\frac{dr}{dt}={k}{[A]^2}\). 2. We can solve for \(k\) by plugging in any data point into our rate equation \(\frac{dr}{dt}={k}{[A]^2}\). Using the first data points for instance \( [A]=0.230 \:\frac{mol}{L}\) and \( \frac{dr}{dt} = 4.17 \times {10}^{-4} \:\frac{mol}{L \cdot s}\)] we get the equation \(4.17 \times {10}^{-4} \:\frac{mol}{L \cdot s}={k}{[0.230 \:\frac{mol}{L}]^2}\) Which solves for \(k=7.88 \times {10}^{-3} \frac{L}{mol \cdot s}\) Since we know this is a second order reaction the appropriate units for \(k\) can also be written as \( \frac{1}{M \cdot s}\) (a) The rate equation is second order in A and is written as rate = [ ] . (b) = 7.88 × 10 L mol s For the reaction \(Q⟶W+X\), the following data were obtained at 30 °C: What is the order of the reaction with respect to [ ], and what is the rate equation? Order: 2 k=0.231 \(M^{-1}s^{-1}\) The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N O , dissolved in chloroform, CHCl , is 6.2 × 10 min . \[\ce{2N2O5⟶4NO2 + O2}\nonumber \] What is the rate of the reaction when [N O ] = 0.40 ? The first step is to write the rate law. We know the general formula for for a first-order rate law. It is as follows: Rate=k[A] We now plug in [N O ] in for [A] in our general rate law. We also plug in our rate constant (k), which was given to us. Now our equation looks as follows: Rate=(6.2x10 min )[N O ] We now plug in our given molarity. [N O ]=0.4 M. Now our equation looks as follows: Rate=(6.2x10 min )(0.4 M) We now solve our equation. Rate=(6.2x10 min )(0.4 M)= 2.48x10 M/min. Use significant figures and unit conversion to round 2.48x10 M/min to 2.5 × 10 (moles)L min (a) 2.5 × 10 mol/L/min The annual production of HNO in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel. The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O , what is the rate of formation of NO when the oxygen concentration is 0.50 and the nitric oxide concentration is 0.75 ? The rate constant for the reaction is 5.8 × 10 L /mol /s. To determine the rate law for an equation we need to look at its slow step. Since both equation a and c are fast, equation b can be considered the slow step of the reaction. The slow step is also considered the rate determining step of the system. Hence, The rate determining step is the second step because it's the slow step. rate of production of \(NO_2 = k [A]^m [B]^n \) \(rate = k [NO]^2 [O_2]^1~M/s\) \(rate = (5.8*10^{-6}) [0.75]^2 [0.5]^1 ~M/s\) \(rate = 1.6*10^{-6}~M/s\) \(rate = 1.6*10^{-6}~M/s\) The following data have been determined for the reaction: \[\ce{I- + OCl- ⟶ IO- + Cl-}\nonumber \] Determine the rate equation and the rate constant for this reaction. Using the reactants, we can form the rate law of the reaction: $$ r=k[OCl^-]^n[I^-]^m \] From there, we need to use the data to determine the order of both \([OCl^-]\) and \([I^-]\). In doing so, we need to compare \(r_1\) to \(r_2\) such that: \[ \frac {r_1}{r_2} = \frac {(0.10^m)(0.050^n)}{(0.20^m)(0.050^n)} = \frac {3.05 \times 10^{-4}}{6.20 \times 10^{-4}} \] \[ 0.5^m = 0.5 \] \[ m = 1 \] We can "cross out" the concentration of \([OCl^-]\) because it has the same concentration in both of the trials used. Now that we know m (\([I^-]\)) has a first order of 1. We cannot "cross out" \([I^-]\) to find \([OCl^-]\) because no two trials have the same concentration. In order to solve for n we will plug in 1 for m. \[ \frac {r_1}{r_3} = \frac {(0.10^{1})(0.050^n)}{(0.30^{1})(0.010^n)} = \frac {3.05 \times 10^{-4}}{1.83 \times 10^{-4}} \] \[ \frac {1}{3} (5^{n}) = 1.6666667 \] \[ 5^{n} = 5 \] \[ n = 1 \] Since we know that orders of both n and m are equal to one, we can not substitute them into the rate law equation along with the respective concentrations (from either the first, second, or third reaction) and solve for the rate constant, k. \[ r=k[OCl^-]^n[I^-]^m \] \[ 3.05 * 10^{-4}= k[0.05]^1[0.10]^1 \] \[ k = 6.1 * 10^{-2} \frac {L}{mol \times s} \] Thus the overall rate law is: $$ r = (6.1 * 10^{-2} \frac {L}{mol \times s})[OCl^-,I^-] \] The units for K depend on the overall order of the reaction. To find the overall order we add m and n together. By doing this we find an overall order of 2. This is why the units for K are $$ \frac {L}{mol \times s} \] rate = [I ,OCl ]; = 6.1 × 10 L mol s In the reaction \[2NO + Cl_2 → 2NOCl\nonumber \] the reactants and products are gases at the temperature of the reaction. The following rate data were measured for three experiments: The rate equation can be determined by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction \(A+B\rightarrow products\), for example, we need to determine and the exponents and in the following equation: \[rate=k[A]^m[B]^n\nonumber \] To do this, the initial concentration of B can be kept constant while varying the initial concentration of A and calculating the initial reaction rate. This information would deduce the reaction order with respect to A. The same process can be done to find the reaction order with respect to B. In this particular example, \[\frac{rate_2}{rate_3}=\frac{k[A_2]^m[B_2]^n}{k[A_3]^m[B_3]^n}\nonumber \] So taking the values from the table, \[\frac{4.0*10^{-2}}{1.0*10^{-2}}=\frac{k[1.0]^m[1.0]^n}{k[0.5]^m[1.0]^n}\nonumber \] and by canceling like terms, you are left with \[\frac{4.0*10^{-2}}{1.0*10^{-2}}=\frac{[1.0]^m}{[0.5]^m}\nonumber \] Now, solve for m \(4=2^m\Longrightarrow m=2\) Because m=2, the reaction with respect to \(NO\) is 2. You can repeat the same process to find n. \[\frac{rate_3}{rate_1}=\frac{k[A_3]^m[B_3]^n}{k[A_1]^m[B_1]^n}\nonumber \] Taking the values from the table, \[\frac{1.0*10^{-2}}{5.1*10^{-3}}=\frac{k[0.5]^m[1.0]^n}{k[0.5]^m[0.5]^n}\nonumber \] and by canceling like terms, you are left with \[\frac{1.0*10^{-2}}{5.1*10^{-3}}=\frac{[1.0]^n}{[0.5]^n}\nonumber \] Now this time, solve for n \(2=2^n\Longrightarrow n=1\) Because n=1, the reaction with respect to \(Cl_2\) is 1. So the rate equation is\[rate=k[NO]^2[Cl_2]^1\nonumber \] To find the overall rate order, you simply add the orders together. Second order + first order makes the The rate constant is calculated by inserting the data from any row of the table into the experimentally determined rate law and solving for k. For a third order reaction, the units of k are \(frac{1}{atm^2*sec}\). Using Experiment 1, \[rate=k[NO]^2[Cl_2]^1\Longrightarrow 5.1*10^{-3} \frac{atm}{sec}=k[0.5m atm]^2[0.5 atm]^1\nonumber \] \[k=0.0408 \frac{1}{atm^2*sec}\nonumber \] \(NO\) is second order. \(Cl_2\) is first order. Overall reaction order is three. b) \(k=0.0408\; atm^{-2}*sec^{-1}\) Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of at varying times. To determine the order of a reaction when given the data series, one must graph the data how it is, graph it as natural log of [A], and graph it as 1/[A]. Whichever method yields a straight line will determine the order. Respective of the methods of graphing above, if a straight line is yielded by the first graphing method its a 0 order, if by the second method it's a 1st order, and the third graphing method its a 2nd order. When the order of the graph is known, a series of equations, given in the above image, can be used with the various points on the graph to determine the value of k. We can see that we need an initial value of A and a final value of A, and both of these would be given by the data. Use the data provided to graphically determine the order and rate constant of the following reaction: \(\ce{SO2Cl2 ⟶ SO2 + Cl2}\) Use the data to graphically determine the order and rate constant of the following reaction. slope= -2.0 x 10 The In this graph, ln(concentration) vs time is linear, indicating that the . k=-slope of line Plotting a graph of ln[SO Cl ] versus reveals a linear trend; therefore we know this is a first-order reaction: = −2.20 × 10 s Use the data provided in a graphical method to determine the order and rate constant of the following reaction: \[2P⟶Q+W\nonumber \] Pure ozone decomposes slowly to oxygen, \(\ce{2O3}(g)⟶\ce{3O2}(g)\). Use the data provided in a graphical method and determine the order and rate constant of the reaction. To identify how the concentrations changes a function of time, requires solving the appropriate differential equation (i.e., the differential rate law). The zero-order rate law predicts in a linear decay of concentration with time The 1st-order rate law predicts in an exponential decay of concentration with time The 2nd-order rate law predicts in an reciprocal decay of concentration with time The plot is not linear, so the reaction is not zero order. The plot is not linear, so the reaction is not first order. The plot is nicely linear, so the reaction is second order. To a second order equation, \( 1/[A] \ = k*t + 1/[A_0] \) Thus, the value of K is the slope of the graph Time vs \( \frac{1}{\ce{O3}}\), = 50.3*10^6 L mol h The plot is nicely linear, so the reaction is second order. = 50.1 L mol h From the given data, use a graphical method to determine the order and rate constant of the following reaction: \[2X⟶Y+Z\] In order to determine the order of the reaction we need to plot the data using three different graphs. All three graphs will have time in seconds as the x-axis, but the y-axis is what will differ. One graph will plot concentration versus time, the second will plot natural log of concentration versus time, and the other will plot 1/concentration versus times. Whichever graph results in a line, we know that must be the order of the reaction. If we get a line using the first graph, it will be zero order, if it is a line for the second graph it will be first order, and if it is a line for the third graph it will be a second order reaction. Now lets plot the data to determine the order. We can clearly see that the third graph, which plots 1/M versus time, is a straight line while the other two are slightly curved. Therefore, we can determine that the rate of this reaction is second order. This also tells us that the units of the rate constant which should be M s for a second order reaction. To determine the rate constant, called k, we simple need to figure out the slope of the third graph since that is the order of this reaction. To find the slope of the line, we take two points and subtract the y values and then divide them by the difference of the x values. This is how to do it: Use the points (5, 10.101) and (40, 80). Now use these to get the slop, aka the rate constant: (80-10.101)/(40-5) = 1.997 = k So the rate constant for this second order reaction is 1.997 M s . What is the half-life for the first-order decay of phosphorus-32? \(\ce{(^{32}_{15}P⟶^{32}_{16}S + e- )}\) The rate constant for the decay is 4.85 × 10 day . This is a first order reaction, so we can use our half life equation below: \[t_{1/2}=\frac{0.693}{k}\nonumber \] The rate constant is given to us in units per day. All we have to do, is to plug it into the equation. \[t_{1/2}=\frac{0.693}{4.85*10^{-2}}\nonumber \] \[=14.3\; days\nonumber \] 14.3 d What is the half-life for the first-order decay of carbon-14? \(\ce{(^6_{14}C⟶^7_{14}N + e- )}\) The rate constant for the decay is 1.21 × 10 year . To find the half life, we need to use the first-order half-life equation. All half life reactions undergo first order reactions. The half-life equation for first order is \[t_{1/2}=ln2/k \nonumber \]with k being the rate constant. The rate constant for carbon-14 was given as \(1.21 × 10^{-4} year^{−1}\). Plug it in the equation. \[t_{1/2}=ln2/(1.21 × 10^{−4} year^{−1})\nonumber \] and solve for \( t_{1/2}\). When you calculate it, the half life for carbon-14 is 5.73*10 The half-life for carbon-14 is calculated to be 5.73*10 What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 ? The rate constant for this second-order reaction is 8.0 × 10 L/mol/s. The half-life of a reaction, t , is the amount of time that is required for a reactant concentration to decrease by half compared to its initial concentration. When solving for the half-life of a reaction, we should first consider the order of reaction to determine it's rate law. In this case, we are told that this reaction is second-order, so we know that the integrated rate law is given as: \[\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0­}\nonumber \] Isolating for time, we find that: \[t_{1/2} = \dfrac{1}{k[A]_0­}\nonumber \] Now it is just a matter of substituting the information we have been given to calculate \(t_{1/2}\), where the rate constant, \({k}\), is equal to 8.0 × 10 L/mol/s and initial concentration, \({[A]_0}\), is equal to 0.15 : \[t_{1/2} = \dfrac{1}{(8.0×10^{-8})(0.15)} = {8.33×10^7 seconds}\nonumber \] 8.33 × 10 s What is the half-life for the decomposition of O when the concentration of O is 2.35 × 10 ? The rate constant for this second-order reaction is 50.4 L/mol/h. Since the reaction is second order, its half-life is \[t_{1/2}=\dfrac{1}{(50.4M^{-1}/h)[2.35×10^{-6}M]}\nonumber \] So, half-life is 8443 hours. The reaction of compound to give compounds and was found to be second-order in . The rate constant for the reaction was determined to be 2.42 L/mol/s. If the initial concentration is 0.500 mol/L, what is the value of t ? As mentioned in the question the reaction of compound will result in the formation of compounds C and D. This reaction was found to be second-order in . Therefore, we should use the second order equation for half-life which relates the rate constant and initial concentrations to the half-life: \[t_{\frac{1}{2}}=\frac{1}{k[A]_{0}}\nonumber \] Since we were given (rate constant) and Initial concentration of A, we have everything needed to calculate the half life of A. \[k=0.5\frac{\frac{L}{mol}}{s}\nonumber \] \[[A]_{0}=2.42\frac{mol}{L}\nonumber \] When we plug in the given information notice that the units cancel out to seconds. \[t_{\frac{1}{2}}=\frac{1}{\frac{2.42Lmol^{-}}{s}[0.500\frac{mol}{L}]}=0.826 s\nonumber \] 0.826 s The half-life of a reaction of compound to give compounds and is 8.50 minutes when the initial concentration of is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to or (b) second order with respect to ? Organize the given variables: (half-life of ) \(t_{1/2}=8.50min\) (initial concentration of ) \([A]_{0}=0.150mol/L\) (target concentration of ) \([A]=0.0300mol/L\) Find the the rate constant k, using the half-life formulas for each respective order. After finding k, use the integrated rate law respective to each order and the initial and target concentrations of to find the time it took for the concentration to drop. (half-life) \(t_{1/2}=\frac{ln(2)}{k}=\frac{0.693}{k}\) (rearranged for k) \(k=\frac{0.693}{t_{1/2}}\) (plug in t = 8.50 min) \(k=\frac{0.693}{8.50min}=0.0815min^{-1}\) (integrated rate law) \(ln[A]=-kt+ln[A]_{0}\) (rearranged for t) \(ln(\frac{[A]}{[A]_{0}})=-kt\) \(-ln(\frac{[A]}{[A]_{0}})=kt\) \(ln(\frac{[A]}{[A]_{0}})^{-1}=kt\) \(ln(\frac{[A]_{0}}{[A]})=kt\) \(t=\frac{ln(\frac{[A]_{0}}{[A]})}{k}\) (plug in variables) \(t=\frac{ln(\frac{0.150mol/L}{0.0300mol/L})}{0.0815min^{-1}}=\frac{ln(5.00)}{0.0815min^{-1}}=19.7min\) (half-life) \(t_{1/2}=\frac{1}{k[A]_{0}}\) (rearranged for k) \(k=\frac{1}{t_{1/2}[A]_{0}}\) (plug in variables) \(k=\frac{1}{(8.50min)(0.150mol/L)}=\frac{1}{1.275min\cdot mol/L}=0.784L/mol\cdot min\) (integrated rate law) \(\frac{1}{[A]}=kt+\frac{1}{[A]_{0}}\) (rearranged for t) \(\frac{1}{[A]}-\frac{1}{[A]_{0}}=kt\) \(t=\frac{1}{k}(\frac{1}{[A]}-\frac{1}{[A]_{0}})\) (plug in variables) \(t=\frac{1}{0.784L/mol\cdot min}(\frac{1}{0.0300mol/L}-\frac{1}{0.150mol/L})=\frac{1}{0.784L/mol\cdot min}(\frac{80}{3}L/mol)=34.0min\) a) 19.7 min b) 34.0 min Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 × 10 g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 × 10 g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant. The first step is to solve for the order or the reaction. This can be done by setting up two expressions which equate the rate to the rate constant times the molar concentration of penicillin raised to the power of it's order. Once we have both expressions set up, we can divide them to cancel out k (rate constant) and use a basic logarithm to solve for the exponent, which is the order. It will look like this. rate(mol/L/min)=k[M] (1.0 x 10 )=k[2.0 x 10 ] (1.5 x 10 )=k[3.0 x 10 ] (2/3)=(2/3) *A single ratio equation can also be set up to solve for the reaction order: *\[\frac{rate_{1}}{rate_{2}}=\frac{k[Penicillin]_{1}^{x}}{k[Penicillin]_{2}^{x}}\nonumber \] *We then solve for x in a similar fashion. *\[\frac{1.0x10^{-10}}{1.5x10^{-10}}=\frac{[2.0x10^{-6}]^{x}}{[3.0x10^{-6}]^{x}}\nonumber \] Now that we have the order of the reaction, we can proceed to solve for the value of the rate constant. Substituting x=1 into our first equation yields the expression: (1 x 10 )=k[2.0 x 10 ] k=(1 x 10 )/(2 x 10 ) k= (5 x 10 ) min We have a unit of min because we divided (mol/L/min) by molarity, which is in (mol/L), yielding a unit of min . We were given two important pieces of information to finish the problem. It is stated that the enzyme has a molecular weight of 3 × 10 g/mol, and that we have a one liter solution that contains (0.15 x 10 g) of penicillinase. Dividing the amount of grams by the molecular weight yields . (0.15 x 10 ) g / (3 x 10 ) g/mol = (5 x 10 ) mol Now that we have the amount of moles, we can divide our rate constant by this value. (5 x 10 ) min / (5 x 10 ) mol = The reaction is first order with = 1.0 × 10 mol min Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)? This problem is asking us for the percentage of radioactivity remaining after a certain time for both isotopes after 48 hours. We must identify an equation that will help us solve this and we can determine that we can determine this information using the first order equation. This equation Ln(N/N )= -kt tells that the Natural log of the fraction remaining is equal to the rate constant times time. To determine the rate constant, we can also compute .693 over the half-life given in the information. For Technetium-99 we can determine the rate constant by plugging into the second equation: .693/6 hrs= .1155 h Now that we have the rate constant we can plug in : Ln(N/N )=-(.1155h )(48h) so Ln(N/N )=-5.544 and if we take the inverse of the natural log, we get (N/N )=3.9x10 and if we multiply this by 100, we get .39% remaining. We can do this same process for Thallium-201 and plugin: .693/73 hrs= .009493151 h and when we plug this into the first order equation we get: Ln(N/N )=-(.009493h )(48h) so Ln(N/N )=-.45567248 and when we take the inverse of the natural log, we get (N/N )=.6340 and when multiplied by 100, we get 63.40% remaining which makes sense since its half-life is 73 hours and only 48 hours have passed, half of the amount has yet to be consumed. Technetium-99: 0.39% Thallium-201: 63.40% There are two molecules with the formula C H . Propene, \(\ce{CH_3CH=CH_2}\), is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic: When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of 5.95 × 10 s . What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C? Use the equation \[ t{_1}{_/}{_2} = \frac{ln2} k\nonumber \] since this is a first-order reaction. You can tell that this is a due to the units of measurement of the rate constant, which is s . Different orders of reactions lead to different rate constants, and a rate constant of s will always be first order. Plug into the equation, and you get half life = 1164.95 seconds. To convert this to hours, we would divide this number by 3600 seconds/hour, to get . Use the integrated first order rate law \[ln\frac{[A]}{[A]_0} = -kt\nonumber \]. In this equation, [A] represents the initial amount of compound present at time 0, while [A] represents the amount of compound that is left after the reaction has occurred. Therefore, the fraction \[\frac{[A]}{[A]_0}\nonumber \] is equal to the fraction of cyclopropane that remains after a certain amount of time, in this case, 0.75 hours. Substitute x for the fraction of \[\frac{[A]}{[A]_0}\nonumber \] into the integrated rate law: \[ln\frac{[A]}{[A]_0} = -kt\nonumber \] \[ln(x) = -5.95x10^{-4}(0.75)\nonumber \] \[x=e^{(-0.000595)(0.75)}\nonumber \] = 0.20058 = 20%. So, the half life is 0.324 hours, and 20% of the cyclopropane will remain as 80% will have formed propene. 0.324 hours. ; 20% remains Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the nuclear equation is \(\ce{^{18}_9F ⟶ _8^{18}O + ^0_{1}e^+}\).) Physicians use F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment. a) The nuclear decay of an isotope of an element is represented by the first order equation: ln(N/N0) = −kt Where t is time, N0 is the initial amount of the substance, N is the amount of the substance after time t, and k is the rate constant. We can rearrange the equation and isolate k so that we could solve for the rate constant: k = [-ln(N/N0)] / t We are given that fluorine-18 has a half-life of 109.7 minutes. Since we have the half-life, we can choose an arbitrary value for N and use half of that value for N. In this case, we choose 100 for N and 50 for N. Now we can plug in those values into the equation above and solve for k. k = [-ln(50/100)] / 109.7 k = 0.6931 / 109.7 = 0.006319 min b) For this problem, we are able to use the same equation from part a: ln(N/N0) = −kt However, this time we are given the amount of time elapsed instead of the half-life, and we are asked to determine the percent of fluorine-18 radioactivity remaining after that time. In this problem, we must plug in values for N0, k (determined from part a), and t. But first, since we are given the elapsed time in hours, we must convert it into minutes: 5.59 hours x (60 minutes / 1 hours) = 335.4 minutes This gives us the value for t. We also have values for k (0.006319 min ) and N (again an arbitrary number.) Now we can plug values into the original equation, giving us: ln(N/100) = −(0.006319)(335.4) We solve this equation by taking the exponential of both sides: e = e where e equals 1 and now we can just solve for N: N/100 = e N = [e ] x 100 = 12.0 Since 100 was used as the initial amount and 12.0 was determined as the remaining amount, 12.0 can be used as the percentage of remaining amount of radioactivity of fluorine-18. c) This part of the question is much like the previous two parts, but this time we are given the initial amount of radioactivity, the final amount of radioactivity and we are asked do determine how long it took for that amount of radioactivity to decay. We are able to use the same equation: ln(N/N0) = −kt However, now we are given N and N and we have already determined k from before. We are told that 99.99% of the radioactivity has decayed, so we can use 100 and 0.01 for N and N respectively. We plug these values in to the equation, solve for t, and get ln(0.01/1000) = −0.006319t -9.21 = −0.006319t t = 1458 minutes a) 0.006319 min b) 12.0% c) 1458 minutes Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for \(\dfrac{1}{64}\) of the initial dose to remain in the athlete’s body? 252 days for first order reaction: t = 0.693 / k k = 0.693 / 42 k = 0.0165 for first order reaction: [A] = [A] e 1/64 initial means that: [A] = 1/64 [A] therefore: 1/64 [A] = [A] e t = 252 days Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years. In order to find out what year King Richard III died, set [A]/[A ] (the percent of carbon-14 still contained) equal to 0.5 or use the equation N(t) = N e Using the first equation: \(A/A_{0}\) = \(0.5^{t/t_{1/2}}\) plug in the given numbers \(.9379 = 0.5^{t/5730}\) and solve for t. \(ln.9379\) = \((t/5730)(ln0.5)\) (using the rule of logs) \(-.0641\) = \((t/5730)(-.693)\) \(-367.36\) = \(-.693t\) \(t = 530.1 years\) Using \(N(t) = N_{0}e^{-rt}\) this problem is solved by the following: \(1/2 = e^{-5730r}\) \(r = 0.000121\) Now that we know what r is, we can use this value in our original formula and solve for t, the amount of years that have passed. This time, we use 93.78, the percent of the carbon-14 remaining as N(t) and 100 as the original, N . \(93.78 = 100e^{-0.000121t}\) \(t = 530.7\) years Another way of doing this is by using these two equations: λ = \(\dfrac{0.693}{t_{1/2}}\) and \(\dfrac{n_{t}}{n_{0}}\) = -λt \(n_{t}\) = concentration at time t (93.79) \(n_{0}\) = initial concentration (100) First solve for lambda or the decay constant by plugging in the half life. Then plug in lambda and the other numbers into the second equation, and solve for t- which should equal to 530.1 years as well. If we want to find out what year King Richard III died, we take the current year, 2017, and subtract 530 years. Doing this, we find that King Richard III died in the year 1487. King Richard III died in the year 1487 Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data: First we need to understand what the question is asking for: the average rate constant. The average rate constant is the variable "k" when discussing kinetics and it can be defined as the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances. Knowing that we need to find K in this first order reaction, we can look to formulas that include "k," initial and final concentrations \([A]_o and [A]_t\), and half life time "t." Since this is a first order reaction, we can look to the first order equations, and doing that we find one that includes the variables given in the question: \[\ln[A]_t=-kt+\ln[A]_o\nonumber \] For the first reaction, we have an initial concentration of 4.88 M, and a percentage decomposed. To find the final concentration, we must multiply the initial concentration by the percentage decomposed to know how much decomposed, and subtract that from the original to find out how much is left: 4.88M x 0.52= 2.54 M and 4.88M-2.54M=2.34M Now, we have the variables we need, and we plug it into the equation above: \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[2.34M]=-k(300s)+\ln[4.88M]\) k=\({-(\ln[2.34M]-\ln[4.88M])}\over 300\) \(k=2.45x10^{-3}\) Since it asks for the rate constant of each experiment, we now must do the same procedure for each data set to find the rate constant: \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[1.66M]=-k(300s)+\ln[3.52M]\) k=\({-(\ln[1.66M]-\ln[3.52M])}\over 300\) \(k=2.51x10^{-3}\) \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[1.07M]=-k(300s)+\ln[2.29M]\) k=\({-(\ln[1.07M]-\ln[2.29M])}\over 300\) \(k=2.54x10^{-3}\) \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[0.834M]=-k(300s)+\ln[1.81M]\) k=\({-(\ln[0.834M]-\ln[1.81M])}\over 300\) \(k=2.58x10^{-3}\) \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[3.49M]=-k(180s)+\ln[5.33M]\) k=\({-(\ln[3.49M]-\ln[5.33M])}\over 180\) \(k=2.35x10^{-3}\) \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[2.60M]=-k(180s)+\ln[4.05M]\) k=\({-(\ln[2.60M]-\ln[4.05M])}\over 180\) \(k=2.46x10^{-3}\) \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[1.89M]=-k(180s)+\ln[2.95M]\) k=\({-(\ln[1.89M]-\ln[2.95M])}\over 180\) \(k=2.47x10^{-3}\) \(\ln[A]_t=-kt+\ln[A]_o\) \(\ln[1.11M]=-k(180s)+\ln[1.72M]\) k=\({-(\ln[1.11M]-\ln[1.72M])}\over 180\) \(k=2.43x10^{-3}\) For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene \(\ce{(CH2=CH–CH=CH2)}\) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene: The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 × 10 s at 150 °C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 °C with an initial pressure of 55 torr. Since this is a first order reaction, the integrated rate law is: \([A_{t}]=[A_{0}]e^{-kt}\) Use the integrated rate law to find the partial pressure at 30 minutes: Use \(A_0\) = 55 torr, t = 30 min, and k = \(2.0 * 10^{-4}s^{-1}\) to solve the integrated rate law equation: \([A_{30}]=(55 torr)*e^{-(2.0x10^{-4}\frac{1}{sec})(30min\cdot\frac{60sec}{1 min})}\) Solve this equation to get: \([A_{30}]=(55 torr)*e^{-0.36}\) \(A_{30}]\) = 38.37 torr. Find the initial concentration using the ideal gas law. The ideal gas law is given by \(PV = nRT → n = \frac{PV}{RT}\). Use this form of the gas law to solve for the initial concentration n. Use V = 0.53L, R = 0.08206 \(\frac{L*atm}{mol*L}\), T = 423.15 K, and P = \(\frac{1 atm}{760}\) = 0.07237 atm . Solve the ideal gas equation using these values: \(n=\frac{(55torr)(0.53L)}{(0.08206\frac{L*atm}{mol*K})(423.15K)} = 0.00110\) moles cyclobutene. Now find the initial concentration of cyclobutene \(A_0\) using the equation \([A_0] = \frac{n}{V}\): \(A_0 = \frac{n}{V} = \frac{0.00110 moles}{0.53 L} = 0.00208 M\) Find the concentration of cyclobutene at 30 minutes by using the integrated rate law given above, using time t = 30 minutes, or 1800 seconds. \([A_{30}]=(0.00208M)e^{-0.36}= 0.00145M\) So at 30 minutes, the cyclobutene concentration is 0.00145 M, and the partial pressure is 38.37 torr. Partial Pressure: 38.37 torr. Concentration: 0.00145 M Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction? The two factors that may prevent a collision from producing a chemical reaction are: 1. In order for chemical reactions to occur, molecules require enough velocity to overcome the minimum activation energy needed to break the old bonds and form new bonds with other molecules. At higher temperatures, the molecules possess the minimum amount of kinetic energy needed which ensures the collisions will be energetic enough to lead to a reaction. 2. Two molecules have to collide in the right orientation in order for the reaction to occur. Molecules have to orient properly for another molecule to collide at the right activation state. When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs? There has to be contact between reactants for a reaction to occur. The more the reactants collide, the more often reactions can occur. Factors that determine reaction rates include concentration of reactants, temperature, physical states of reactants, surface area, and the use of a catalyst. The reaction rate usually increases as the concentration of a reactant increases. Increasing the temperature increases the average kinetic energy of molecules, causing them to collide more frequently, which increases the reaction rate. When two reactants are in the same fluid phase, their particles collide more frequently, which increases the reaction rate. If the surface area of a reactant is increased, more particles are exposed to the other reactant therefore more collisions occur and the rate of reaction increases. A catalyst participates in a chemical reaction and increases the reaction rate without changing itself. What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction? Activation energy is the energy barrier that must be overcome in order for a reaction to occur. To get the molecules into a state that allows them to break and form bonds, the molecules must be contorted (deformed, or bent) into an unstable state called the transition state. The transition state is a high-energy state, and some amount of energy – the activation energy – must be added in order for the molecule reach it. Because the transition state is unstable, reactant molecules don’t stay there long, but quickly proceed to the next step of the chemical reaction.The activated complex is the highest energy of the transition state of the reaction. Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures. How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate. Collision theory states that the rates of chemical reactions depend on the fraction of molecules with the correct orientation, fraction of collisions with required energy, and the collision frequency. Because the fraction of collisions with required energy is a function of temperature, as temperature increases, the fraction of collisions with required energy also increases. The kinetic energy of reactants also increases with temperature which means molecules will collide more often increasing collisions frequency. With increased fraction of collisions with required energy and collisions frequency, the rate of chemical reaction increases. Arrhenius equation, that temperature and the rate constant are related. \[k=Ae^{\frac {E_a}{RT}}\] where k is the rate constant, A is a specific constant, R is 8.3145 J/K, Ea is the reaction-specific activation energy in J, and T is temperature in K. We see from the equation that k is very sensitive to changes in the temperature. The rate of a certain reaction doubles for every 10 °C rise in temperature. By finding the difference in temperature, 45 °C - 25 °C, we get 20 °C. Since the rate of the reaction doubles every 10 °C increase in temperature and the rate of the reaction experienced a 20 °C increase in temperature, we see that the reaction rate doubled twice (2 = 4). As a result, the . Following the same process as in part a, we get the difference in temperature to be 70 °C. Since the rate of the reaction doubles every 10 °C increase in temperature and the system experienced a 70 °C change, we see that the reaction doubled seven times (2 = 128). We can see the . (a) 4-times faster (b) 128-times faster In an experiment, a sample of NaClO was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher? First off, it is important to recognize that this decomposition reaction is a , which can be written as follows: \(\mathrm2NaClO_3\to2NaCl + 3O_2\) Understanding this, it is important to be able to then be able to recognize which equation would be most useful given the initial conditions presented by the question. Since we are dealing with time, percentage of material left, and temperature, the only viable equation that could relate all of this would be the Arrhenius Equation, which is written as follows: \(\mathrm \ln(\frac{k_2}{k_1}) = \frac {Ea}{R}({\frac1{t_1}}-{\frac{1}{t_{2}}})\) , this problem does not give us enough information such as what the activation energy is or the initial temperature in order to mathematically solve this problem. Additionally, the problem tells us to approximate how long the decomposition would take, which means we are asked to answer this question conceptually based on our knowledge of thermodynamics and reaction rates. As a general rule of thumb, we know that for every 10˚C rise in temperature the rate of reaction doubles. Since the question tells us that there is a 20˚C rise in temperature we can deduce that the reaction rate doubles twice, as per the general rule mentioned before. This means , or would be 4 times faster than the reaction rate at the initial temperature. We can gut check this answer by recalling how an increase in the average kinetic energy (temperature) decreases the time it takes for the reaction to take place and increase the reaction rate. Thus, if we increase the temperature we should have a faster reaction rate. The rate constant at 325 °C for the decomposition reaction \(\ce{C4H8⟶2C2H4}\) is 6.1 × 10 s , and the activation energy is 261 kJ per mole of C H . Determine the frequency factor for the reaction. Using the Arrhenius equation allows me to find the frequency factor, A. k=Ae ​ k, Ea, R, and T are all known values. k, Ea, and T are given in the problem as 6.1x10 , 261 kJ, and 598 K, respectively. So, plugging them into the equation gives: 6.1x10 s =Ae Take e and get 1.59 x 10 . Divide k, 6.1 x 10 , by 1.59 x 10 and get A=3.9 x 10 s \(\mathrm{3.9×10^{15}\:s^{−1}}\) The rate constant for the decomposition of acetaldehyde (CH CHO), to methane (CH ), and carbon monoxide (CO), in the gas phase is 1.1 × 10 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition. The equation for relating the rate constant and activation energy of a reaction is the Arrhenius equation: \[ln (\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})\] In this problem, all the variables are given except for the E (activation energy). k = 1.1 × 10 L/mol/s T = 703 K k = 4.95 L/mol/s T = 865 K R = 8.314 J/(mol K) (Ideal Gas Constant) Now plug in all these values into the equation, and solve for E . \[ln (\frac{4.95\frac{L}{mol×s}}{1.1 × 10^{-2}\frac{L}{mol×s}}) = \frac{E_a}{8.314 × 10^{-3}\frac{kJ}{mol×K}} (\frac{1}{703} - \frac{1}{865})\] E = 190 kJ (2 sig figs) An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate? 43.0 kJ/mol In terms of collision theory, to which of the following is the rate of a chemical reaction proportional? Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H , and iodine, I . The value of the rate constant, , for the reaction was measured at several different temperatures and the data are shown here: What is the value of the activation energy (in kJ/mol) for this reaction? 177 kJ/mol The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data: The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \[\ce{C12H22O11 + H2O ⟶ C6H12O6 + C6H12O6}\nonumber \] follows a first-order rate equation for the disappearance of sucrose: Rate = [C H O ] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) = 108 kJ = 2.0 × 10 s = 3.2 × 10 s (b) 1.81 × 10 h or 7.6 × 10 day. (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state. Use the to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first \(A+BC⟶AB+C\) reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why? Use the to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first \(A+BC⟶AB+C\) reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the atom with varying angles, but with more Total energy than the transition state. What happens when the atom hits the molecule from different directions? Why? The atom has enough energy to react with ; however, the different angles at which it bounces off of without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur. Why are elementary reactions involving three or more reactants very uncommon? In general, can we predict the effect of doubling the concentration of on the rate of the overall reaction \(A+B⟶C\) ? Can we predict the effect if the reaction is known to be an elementary reaction? No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate equation. Yes. If the reaction is an elementary reaction, then doubling the concentration of doubles the rate. Phosgene, COCl , one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by: Define these terms: What is the rate equation for the elementary termolecular reaction \(A+2B⟶\ce{products}\)? For \(3A⟶\ce{products}\)? We are given that both of these reactions are elementary termolecular. The molecularity of a reaction refers to the number of reactant particles that react together with the proper and energy and orientation. Termolecular reactions have three atoms to collide simultaneously. As it is termolecular, and there are no additional reactants aside from the three given in each reaction, there are no intermediate reactions. The rate law for elementary reactions is determined by the stoichiometry of the reaction without needed experimental data. The basic rate form for the elementary step is what follows: \(rate= {k} \cdot {reactant \ 1}^{i} \cdot {reactant \ 2}^{ii} \cdot ... \) Where i and ii are the stochiometric coefficient from reactant 1 and 2 respectively. For: \(3A \rightarrow products \) \({k} \cdot {A}^3 = rate\) For: \(A + 2B \rightarrow products \) \({k} \cdot {[A]} \cdot {[B]}^2 = rate\) Note that the order of these reactions are both three. Rate = [ , ] ; Rate = [ ] Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same? (a) \(\ce{Cl2 + CO ⟶ Cl2CO}\) \(\ce{rate}=k\ce{[Cl2]^{3/2}[CO]}\) (b) \(\ce{PCl3 + Cl2 ⟶ PCl5}\) \(\ce{rate}=k\ce{[PCl3,Cl2]}\) (c) \(\ce{2NO + H2 ⟶ N2 + H2O}\) \(\ce{rate}=k\ce{[NO,H2]}\) (d) \(\ce{2NO + O2 ⟶ 2NO2}\) \(\ce{rate}=k\ce{[NO]^2[O2]}\) (e) \(\ce{NO + O3 ⟶ NO2 + O2}\) \(\ce{rate}=k\ce{[NO,O3]}\) An elementary reaction is a chemical reaction in which the reactants directly form products in a single step. In another words, the rate law for the overall reaction is same as experimentally found rate law. Out of 5 options, option (b),(d), and (e) are such reactions Write the rate equation for each of the following elementary reactions: Rate equations are The rate law of a reaction can be found using a rate constant (which is found experimentally), and the initial concentrations of reactants. A general solution for the equation \(aA + bB \rightarrow cC + dD\) is \(rate = k[A]^{m}[B]^{n}\) where m and n are reaction orders. However, reaction orders are found experimentally, and since we do not have experimental data for these reactions, we can disregard that part of the equation. To find the rate laws, all we have to do is plug the reactants into the rate formula. This is only due to the case that these are elementary reactions. Further reading on elementary reactions can be found on Libre Texts. a. O ⟶ O + O To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k". Rate = [O ] b. O + Cl ⟶ O + ClO To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k". c. ClO + O ⟶ Cl + O To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k". Rate = [ClO,O] d. O + NO ⟶ NO + O To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k". Rate = [O ,NO] e. NO + O ⟶ NO + O To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k". Rate = [NO ,O] (a) Rate = [O ]; (b) Rate = [O ,Cl]; (c) Rate = [ClO,O]; (d) Rate = [O ,NO]; (e) Rate = [NO ,O] Nitrogen(II) oxide, NO, reacts with hydrogen, H , according to the following equation: \[\ce{2NO + 2H2 ⟶ N2 + 2H2O}\nonumber \] What would the rate law be if the mechanism for this reaction were: \[\ce{2NO + H2 ⟶ N2 + H2O2\:(slow)}\nonumber \] \[\ce{H2O2 + H2 ⟶ 2H2O\:(fast)}\nonumber \] The rate law of the mechanism is determined by the slow step of the reaction. Since the slow step is an elementary step, the rate law can be drawn from the coefficients of the chemical equation. So therefore, the rate law is as follows: rate=k[NO] [H ]. Since both NO and H are reactants in the overall reaction (therefore are not intermediates in the reaction), no further steps have to be done to determine the rate law. Consider the reaction CH + Cl → CH Cl + HCl (occurs under light) The mechanism is a chain reaction involving Cl atoms and CH radicals. Which of the following steps does not terminate this chain reaction? Chain reactions involve reactions that create products necessary for more reactions to occur. In this case, a reaction step will continue the chain reaction if a radical is generated. Radicals are highly reactive particles, so more reactions in the chain will take place as long as they are present. The chlorine is considered a free radical as it has an unpaired electron; for this reason it is very reactive and propagates a chain reaction. It does so by taking an electron from a stable molecule and making that molecule reactive, and that molecule goes on to react with stable species, and in that manner a long series of "chain" reactions are initiated. A chlorine radical will continue the chain by completing the following reaction: \({Cl \cdot}+{CH_4} \rightarrow {CH_3 \cdot}+{HCl} \) The \({CH_3}\) generated by this reaction can then react with other species, continuing to propagate the chain reaction. Option 1 is incorrect because the only species it produces is \({CH_3Cl}\), a product in the overall reaction that is unreactive. This terminates the chain reaction because it fails to produce any \(Cl\) or \(CH_3\) radicals that are necessary for further propagating the overall reaction. Option 2 is the correct answer because it produces a \(Cl\) radical. This \(Cl\) radical can continue the chain by colliding with \(CH_4\) molecules. Option 3 is incorrect because it fails to produce a radical capable of continuing the chain. Option 4 is incorrect because it produces \(Cl_2\), a molecule that does not react unless additional light is supplied. Therefore, this step breaks the chain. Answer: Option 2: \({CH_3}+{HCl} \rightarrow {CH_4}+{Cl}\) Experiments were conducted to study the rate of the reaction represented by this equation. \[\ce{2NO}(g)+\ce{2H2}(g)⟶\ce{N2}(g)+\ce{2H2O}(g)\nonumber \] Initial concentrations and rates of reaction are given here. Consider the following questions: Step 1: \(\ce{NO + NO ⇌ N2O2}\) Step 2: \(\ce{N2O2 + H2 ⇌ H2O + N2O}\) Step 3: \(\ce{N2O + H2 ⇌ N2 + H2O}\) Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction. S12.6.10 1. i) Find the order for [NO] by using experiment 3 and 4 where [H ] is constant Notice that [NO] doubles from experiment 3 to 4 and the rate quadruples. So the order for [NO] is ii) Find the order for [H ] by using experiment 1 and 2 where [NO] is constant Notice that [H ] doubles from experiment 1 to 2 and the rate doubles as well. So the order for [H ] is 2. Put in the order for each product as the exponents for the corresponding reactant. \(rate = k [NO]^2 [H_2]\) 3. Put in the concentrations and the rate from one of the experiments into the rate law and solve for k. (Here, experiement 1 is used but any of them will work) \(rate = k [NO]^2 [H_2]\) \(.00018 = k [.006]^2 [.001]\) \(k = 5000 M^{-2}s^{-1}\) 4. Plug in values for experiment 2 into the rate law equation and solve for the concentration of NO \(.00036=5000[NO]^2[.001]\) \([NO]^2= 7.2 x 10^{-5}\) \([NO] = .0085 M\) 5. Write the rate laws for each step and then see which matches the rate law we found in question 2. The rate determining step (the slow step) is the one that gives the rate for the overall reaction. Because of this, only those concentrations will influence the overall reaction, contrary to what we would believe if we just looked at the overall reaction. Step 1: \(NO + NO \rightleftharpoons N_2O_2\) \(rate =k_1[NO]^2\) This rate law is not the same as the one we calculate in question 2 so this be the rate determining step Step 2: \(N_2O_2+H_2 \rightleftharpoons N_2O + N_2O\) \(rate = k_2[N_2O_2,H_2]\) Since \(N_2O_2\) is an intermediate you must replace it in the rate law equation. Intermediates can not be in the rate law because they do not appear in the overall reaction. Here you can take the reverse of equation 1 (k ) and substitute the other side (the reactants of equation 1) for the intermediate in the rate law equation. \[rate_1 = rate_{-1}\nonumber \] \[k_1[NO]^2 = k_{-1}[N_2O_2]\nonumber \] \[[N_2O_2] = \frac{k_1[NO]^2}{k_{-1}}\nonumber \] \(rate= \frac{k_2k_{1}[NO]^2[H_2]}{k_{-1}}\) Overall: \(rate={k[NO]^2[H_2]}\) This so it is the rate determining step. So \(N_2O_2+H_2 \rightleftharpoons N_2O + N_2O\) is the rate determining step step. (a) NO: 2 \(\ce {H2}\) : 1 (b) Rate = [NO] [H ]; (c) = 5.0 × 10 mol L min ; (d) 0.0050 mol/L; (e) Step II is the rate-determining step. The reaction of CO with Cl gives phosgene (COCl ), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises: 1. To write the overall reaction you have to identify the intermediates and leave them out. The easiest way to do this is to write out all the products and reactants and cross out anything that is on both sides. Cl (g) + CO(g) + + ) + + COCl (g) In this you will cross out the 2Cl(g) molecules and the COCl(g). What is left after that is the overall reaction. Cl (g) + CO(g) + COCl (g) 2. For part two you will just list the intermediates that you crossed out. Cl and COCl are intermediates 3. Each rate law will be the rate equal to the rate constant times the concentrations of the reactants reaction 1 (forward) rate=k [Cl ] ( reverse) rate=k [Cl] reaction 2 rate=k [CO,Cl] Reaction 3 rate=k [COCl,Cl] 4. The overall rate law is based off the slowest step (step #2), since it is the rate determining step, but Cl is present in that rate law so we have to replace it with an equivalent that does not contain an intermediate. To do this you use the equilibrium since the rates are the same you can set up the rate laws of the forward and reverse equal to each other. k [Cl ] = k [Cl] [Cl]= k [Cl ]/k rate=k [CO]k [Cl ]/k Steps to replacing and intermediate Account for the increase in reaction rate brought about by a catalyst.   Compare the functions of homogeneous and heterogeneous catalysts.   Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl F , catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is: \[\ce{O3 \xrightarrow{sunlight} O2 + O}\\ \ce{O3 + Cl ⟶ O2 + ClO}\\ \ce{ClO + O ⟶ Cl + O2}\nonumber \] Is NO a catalyst for the decomposition? Explain your answer. For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed: (a) For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed: (a) (b) For each of the following reaction diagrams, estimate the activation energy ( ) of the reaction: (a) (b) For each of the following reaction diagrams, estimate the activation energy ( ) of the reaction: (a) (b)

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Nomenclature

Chemical is the names we use for chemicals. For instance, H O is called "water", and CH (the gas you burn in a stove) is called "methane." You should learn the chemical nomenclature here on this page now, so that you will be able to understand when it is used. Here is some important info about how we write chemicals. There are lots of elements and you don't need to memorize them all. Here are a few that you should learn right now, though, because they are common or important, so that you won't be confused when they are mentioned later. They are organized by their type. is another word for positive ion. The common positive ions are the ions of the alkali and alkaline earth metals and ammonium, NH . The alkali metals form +1 cations, such as Na and K . The alkaline earth metals form +2 cations, such as Ca and Mg . The hydrogen ion, H is a very common cation. For these cations, you can call them "[element name] ion", such as sodium ion or calcium ion. You'll also see transition metal cations or main group metal cations, but it is harder to predict what charge they will have, especially because some of them can have different charges, like iron, which is commonly Fe or Fe . The charge on a transition metal cation can also be indicated using Roman numerals in parentheses, which looks like Fe(II) or Fe(III). The Roman numerals you will need to know for chemistry are: For cations that have uncertain charge, you should call them "[element name](charge in Roman numerals) ion." For instance, iron(II) ion or sometimes just Fe(II). Sometimes people use special names for these ions, in which the higher charge ion is called "[name]-ic ion" and the lower charge ion is called "[name]-ous ion," such as ferrous for Fe(II) and ferric for Fe(III), or cuprous ion for Cu(I) and cupric ion for Cu(II). I think this is most common for Fe, and I've never heard anyone call nickel(II) nickelous ion because that sounds ridiculous. Here's a list of common cations with less predictable charges: Elements not on the list above, that you may see soon anyway: zinc(II): Zn , cadmium(II): Cd , cobalt(II): Co , manganese(II): Mn , nickel(II): Ni , chromium(III): Cr . is another word for negative ion. Common negative ions are the halide ions, formed from the halogen elements: fluoride, F ; chloride, Cl ; bromide, Br ; and iodide, I . As you may have noticed, the names of anions have "-ide" at the end when they are formed from elements. Other examples include oxide, O , sulfide, S , and nitride, N . There are also many important polyatomic anions, which means anions that include more than one atom. These include toxic cyanide ion, CN , common hydroxide ion, OH , and peroxide ion, O . Other important anions include acetate ion (C H O ), which is in vinegar, the chlorate ion (ClO ), the perchlorate ion (ClO ) which is often explosive, the nitrate ion (NO ), the carbonate ion (CO ) found in shells, the sulfate ion (SO ), and the phosphate ion (PO ). All of these end in "-ate", which means that they have more oxygen. Also, notice that "per-___-ate" means more oxygen than just "-ate", as in perchlorate. Less common but still important are some "-ite" anions, which have less oxygen, such as nitrite (NO ), sulfite (SO ), chlorite (ClO ) and hypochlorite (ClO ). Notice that "hypo-___-ite" means less oxygen than just "-ite" as in hypochlorite. Sulfite and nitrite are used to preserve foods. Sulfite salts are used in wine, dried fruit and preserved radish (mu). Nitrite salts are used in preserved meats. One more rule says that if you take an anion like carbonate or sulfate and add one hydrogen ion, then you call that "bicarbonate" (HCO ) or "bisulfate" (HSO ). Or you might see it called "hydrogen carbonate" or "hydrogen sulfate." Note that because we added a hydrogen ion, the charge on the bicarbonate ion is one less than the charge on the carbonate ion. Also, note that "disulfate" is S O , quite different from bisulfate. Ionic compounds are compounds that include at least two components, a positive ion and a negative ion. Often the positive ion is a metal element ion and the negative ion is a non-metal ion. To name an ionic compound, you usually just give the cation followed by the anion, such as "sodium chloride" or "ammonium nitrate." If the cation is the type that could have different charges, than you should say what the charge is, such as "mercury(I) iodide" or "cupric sulfate." Acid usually means an anion combined with the hydrogen ion as the cation. For instance, HCl is a common acid, which is the hydrogen ion and the chloride anion. If the anion ends in "-ide" then usually the acid is called "hydro-___-ic acid" such as "hydrochloric acid" for HCl. You'll see this for all the "hydrohalic acids" which are H + a halogen, such as "hydrofluoric acid" or "hydroiodic acid." You might also see "hydrocyanic acid" for HCN. If the anion ends in "-ate" than you call the acid "___-ic acid," such as "sulfuric acid," which is H SO , or "nitric acid." HNO . If the anion ends in "-ite" than the acid name is "___-ous acid." such as "hypochlorous acid" for HClO. Notice that earlier "-ic" and "-ous" meant more and less charge for cations, such as ferric and ferrous ions of iron. Now it also means more and less oxygen in acids. Non-metal compounds are often called covalent compounds. They are named following a different rule from ionic compounds. You will need these "prefixes" which indicate how many of each type of atom are present: The prefixes come from Greek. You will put the element that is more left on the periodic table first, unless it is oxygen, which is always last unless it is in a compound with fluorine. This follows the same pattern as ionic compounds. In ionic compounds, the cation is written first, and you will notice that it is usually more to the left in the periodic table than the anion, which is written last. When you name covalent compounds, the atom that's more like an anion is written last. Fluorine is always most "anionic," and oxygen is next most "anionic," so they will always be last. (Fluorine is actually most electronegative, but we will study this concept much later, which is why right now I'm calling it "anion-like.") If both atoms are in the same group (same column of the periodic table) then the lower one is named first. Notice that the two most "anion-like," F and O, are in the upper right of the periodic table. The atom written second, that's more "anion-like" is named like an anion, with the "-ide" ending. For example, CO: carbon is on the left, so we can write "monocarbon monoxide." Actually people usually just call it "carbon monoxide." You can skip "mono" for the first element. For instance, SO is called "sulfur trioxide" and N O is called "dinitrogen tetroxide." XeO is xenon dioxide, even though xenon is more to the right than oxygen, because oxygen is more like an anion than anything except fluorine. If the compound involves hydrogen, then you can leave out the prefixes, such as "hydrogen chloride" for HCl or "hydrogen sulfide" for H S, because the numbers of each atom can be predicted as if it were an ionic substance. But actually many compounds of hydrogen have special names, such as "ammonia" for NH , "methane" for CH , "borane" for BH , "silane" for SiH and "phosphine" for PH . You should learn the first two of these now.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Introduction

is different from physics. The chemical theories that we use to understand and predict aren't written in math. They also aren't based on anything we can see directly, like a ball falling. Instead, they are based in images, patterns, symmetry and imagination. How do we know what metaphors we can use to imagine, and predict accurately, things that we can't see? When you think about it, it's amazing how much chemists were able to figure out. John Dalton proposed his around 1805. He said: Scientists scoffed, laughed at him, or at least remained skeptical, for at least 50 years after that, because how could he know? But he was almost completely right. A little later, in 1830, Jons Jacob Berzelius proposed that even if you have the same (the same number of atoms of each element) you can have two different (a combination of atoms), because the atoms are arranged differently. The first example discovered was silver fulminate (AgCNO, very explosive) and silver cyanate (AgOCN, a non-explosive grayish powder). And then in 1874 Jacobus Henricus van't Hoff proposed that when carbon atoms form bonds to four other atoms, those other atoms around them have a particular arrangement in space: a . A critic said: "A Dr. J. H. van't Hoff who is employed at the Veterinary School in Utrecht appears to find exact chemical research unsuited to his tastes. He finds it more suitable to mount Pegasus (obviously loaned from the Veterinary School) and to proclaim ... how, during his flight to the top of the chemical Parnassus, the atoms appeared to be arranged in the universe." (In ancient Greek mythology, Pegasus is a horse with wings, and Parnassus is a mountain associated with art and knowledge because the Muses were said to live there.) You can tell from that how weird it seemed, at the time, to claim a particular spatial arrangement of tiny particles that nobody had ever seen. But van't Hoff, like Dalton and Berzelius, was right. How did they manage this? That will be one of the questions we answer in this class. In fact, before the physicists had admitted that atoms exist, before they had been proven directly, chemists had already published many showing how atoms were arranged in molecules. is the study of the composition and structure of chemical substances and the changes that they undergo. An is one type of atom. An is the smallest particle of an element that retains all of the element's properties. A is a combination of atoms of the same element. A substance's chemical involves having a piece of the substance being made of the same number of atoms as well as the the same type(s) of atoms.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Introduction_to_Organometallic_Chemistry_(Ghosh_and_Balakrishna)

Organometallic chemistry is the study of organometallic compounds, chemical compounds containing at least one chemical bond between a carbon atom of an organic molecule and a metal, including alkaline, alkaline earth, and transition metals, and sometimes broadened to include metalloids like boron, silicon, and tin. This text explores the organometallic compounds of s- and p-block elements; structure and bonding aspects of main group elements: Lewis structure,VSEPR theory,Bent’s rule, steric numbers, molecular shapes; methods of preparation of organometallic compounds; reactivity of organometallic compounds; Zeigler-Natta polymerization catalysts; classification of ligands; phosphines; organometallic compounds of zinc, cadmium and mercury. Organometallic transition metal complexes; the 18 Valence Electron Rule; synthesis and stability, transition metal alkyls, hydrides, carbonyls, phosphines, alkene, allyl, diene and the cyclopentadienyl complexes; oxidative additions and reductive eliminations; insertion and elimination reactions; nucleophilic and electrophilic addition and abstraction reactions; application in homogeneous catalysis Thumbnail: The ball-and-stick model of diisobutylaluminium hydride showing aluminium as pink, carbon as black, and hydrogen as white. (Public Domain; via )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantitative_NMR_(Larive_and_Korir)

Quantitative nuclear magnetic resonance (Q-NMR) measurements are especially useful for analysis of complex samples where high specificity is needed or when a pure standard of the analyte is not available. Often the sample can be analyzed with a minimum of preparation. Q-NMR has utility in various areas including: This text provides a brief overview of basic NMR theory, a tutorial discussing the practical aspects of Q-NMR, a virtual experiment utilizing an NMR simulator, an exercise that makes use of real NMR data you can download and analyze, and a Q-NMR lab experiment on the determination of malic acid in fruit juice. Several Q-NMR applications are highlighted and links to additional resources are provided along with a guide for instructors.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.0%3A_Prelude_to_Kinetics

The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun’s rays is critical to the lizard’s survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. In the absence of warmth, the lizard is an easy meal for predators. From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. When planning to run a chemical reaction, we should ask at least two questions. The first is: “Will the reaction produce the desired products in useful quantities?” The second question is: “How rapidly will the reaction occur?” A reaction that takes 50 years to produce a product is about as useful as one that never gives a product at all. A third question is often asked when investigating reactions in greater detail: “What specific molecular-level processes take place as the reaction occurs?” Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled. The study of chemical kinetics concerns the second and third questions—that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. In this chapter, we will examine the factors that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to determine and describe the rate at which reactions occur.   ).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/00%3A_Front_Matter/02%3A_InfoPage

Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new   access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ).   and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. .     .

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

Rubber is an example of an elastomer type polymer, where the polymer has the ability to return to its original shape after being stretched or deformed. The rubber polymer is coiled when in the resting state. The elastic properties arise from the its ability to stretch the chains apart, but when the tension is released the chains snap back to the original position. The majority of rubber polymer molecules contain at least some units derived from conjugated diene monomers (see ). Such conjugated diene monomers have a constructive backbone of at least four carbon atoms with a double-single-double bond reactive core (C=C-C=C ). Most if not practically all such dienes undergo 1,4-addition to the polymer chain, where 1 and 4 refer to the 1st and 4th carbons of the backbone unit, which become single-bonded to the rest of the polymer chain. The diene's double bonds turn into single bonds, and the single bond between them turns into a Z or E configured double bond, depending on the polymerization conditions. The unit's backbone thus becomes like this (-C-C=C-C-). Rubber gets its elasticity when the formed double bond gets the Z configuration. For 1,3-butadiene, Z is equivalent to a and E is equivalent to a configuration. Natural rubber is an addition polymer that is obtained as a milky white fluid known as latex from a tropical rubber tree. Natural rubber is from the monomer isoprene (2-methyl-1,3-butadiene), which is a conjugated diene hydrocarbon as mentioned above. In natural rubber, most of the double fonds formed in the polymer chain have the Z configuration, resulting in natural rubber's elastomer qualities. Charles Goodyear accidentally discovered that by mixing sulfur and rubber, the properties of the rubber improved in being tougher, resistant to heat and cold, and increased in elasticity. This process was later called vulcanization after the Roman god of fire. Vulcanization causes shorter chains to cross link through the sulfur to longer chains. The development of vulcanized rubber for automobile tires greatly aided this industry. Important conjugated dienes used in synthetic rubbers include isoprene (2-methyl-1,3-butadiene), 1,3-butadiene, and chloroprene (2-chloro-1,3-butadiene). Polymerized 1,3-butadiene is mostly referred to simply as polybutadiene. Polymerized chloroprene was developed by DuPont and given the trade name . In a number of cases, monomers which are not dienes are also used for certain types of synthetic rubber, often copolymerized with dienes. Some of the most commercially important addition polymers are the copolymers. These are polymers made by polymerizing a mixture of two or more monomers. An example is (SBR) - which is a copolymer of 1,3-butadiene and styrene which is mixed in a 3 to 1 ratio, respectively. SBR rubber was developed during World War II when important supplies of natural rubber were cut off. SBR is more resistant to abrasion and oxidation than natural rubber and can also be vulcanized. More than 40% of the synthetic rubber production is SBR and is used in tire production. A tiny amount is used for bubble-gum in the unvulcanized form. is copolymerized from butadiene and acrylonitrile (H C=CH-CN). is copolymerized from isobutylene [which is methylpropene H C=C(CH ) ] and a small percentage of isoprene. and other compounds, chemically called , are not from conjugated dienes but have repeating units like -O-SiR - where R is some organic radical group like methyl. The polymer rubber chains exist in random loose clumps in the unstretched state. At the nipple end of the balloon, there is lots of rubber and therefore many, many polymer chains - still loosely coiled. These chains can be pierced without popping the balloon because the the chains can still be stretched. This is because they allow the skewer in between the chains without breaking the chains or the bonds that connect them. But on the sides of the balloon, these chains are stretched almost to their limit and very far apart. The piercing is too much for the stretched chains and they break apart, and the balloon pops.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.04%3A_The_Nernst_Equation

Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. The we discussed in a refer to cells in which all dissolved substances are at unit , which essentially means an "effective concentration" of 1 M. Similarly, any gases that take part in an electrode reaction are at an effective pressure (known as the ) of 1 atm. If these concentrations or pressures have other values, the cell potential will change in a manner that can be predicted from the principles you already know. Suppose, for example, that we reduce the concentration of \(Zn^{2+}\) in the \(Zn/Cu\) cell from its standard effective value of 1 M to an to a much smaller value: \[Zn(s) | Zn^{2+}(aq, 0.001\,M) || Cu^{2+}(aq) | Cu(s)\] This will reduce the value of \(Q\) for the cell reaction \[Zn(s) + Cu^{2+} → Zn^{2+} + Cu(s)\] thus making it more spontaneous, or "driving it to the right" as the would predict, and making its free energy change \(\Delta G\) more negative than \(\Delta G°\), so that \(E\) would be more than \(E^°\). The relation between the actual cell potential \(E\) and the standard potential \(E^°\) is developed in the following way. We begin with the which relates the standard free energy change (for the complete conversion of products into reactants) to the standard potential \[\Delta G° = –nFE° \] By analogy we can write the more general equation \[\Delta G = –nFE\] which expresses the change in free energy for any extent of reaction— that is, for any value of the \(Q\). We now substitute these into the expression that relates \(\Delta G\) and \(\Delta G°\) which you will recall from the chapter on chemical equilibrium: \[\Delta G = \Delta G° + RT \ln Q\] which gives \[–nFE = –nFE° + RT \ln Q \] which can be rearranged to \[ \color{red} {\underbrace{E=E° -\dfrac{RT}{nF} \ln Q}_{\text{applicable at all temperatures}}} \label{Nernst Long}\] This is the that relates the cell potential to the standard potential and to the activities of the electroactive species. Notice that the cell potential will be the same as \(E°\) only if \(Q\) is unity. The Nernst equation is more commonly written in base-10 log form and for 25 °C: \[ \color{red} {\underbrace{E=E° -\dfrac{0.059}{n} \log_{10} Q}_{\text{Applicable at only 298K}}} \label{Nernst Short}\] The Nernst equation tells us that a half-cell potential will change by 59 millivolts per 10-fold change in the concentration of a substance involved in a one-electron oxidation or reduction; for two-electron processes, the variation will be 28 millivolts per decade concentration change. Thus for the dissolution of metallic copper \[Cu_{(s)} \rightarrow Cu^{2+} + 2e^–\] the potential \[E = (– 0.337) – 0.0295 \log_{10} [Cu^{2+}]\] becomes more positive (the reaction has a greater tendency to take place) as the cupric ion concentration decreases. This, of course, is exactly what the predicts; the more dilute the product, the greater the extent of the reaction. The equation just above for the Cu/Cu half-cell raises an interesting question: suppose you immerse a piece of copper in a solution of pure water. With = [Cu ] = 0, the potential difference between the electrode and the solution should be infinite! Are you in danger of being electrocuted? You need not worry; without any electron transfer, there is no charge to zap you with. Of course it won't be very long before some Cu ions appear in the solution, and if there are only a few such ions per liter, the potential reduces to only about 20 volts. More to the point, however, the system is so far from equilibrium (for example, there are not enough ions to populate the electric double layer) that the Nernst equation doesn't really give meaningful results. Such an electrode is said to be un . What ionic concentration is needed to poise an electrode? I don't really know, but I would be suspicious of anything much below 10 M. Ions of opposite charge tend to associate into loosely-bound ion pairs in more concentrated solutions, thus reducing the number of ions that are free to donate or accept electrons at an electrode. For this reason, the Nernst equation cannot accurately predict half-cell potentials for solutions in which the total ionic concentration exceeds about 10 M. How the cell potential really depends on concentration! The Nernst equation accurately predicts cell potentials only when the equilibrium quotient term Q is expressed in . Ionic activities depart increasingly from concentrations when the latter exceed 10 to 10 M, depending on the sizes and charges of the ions. If the Nernst equation is applied to more concentrated solutions, the terms in the reaction quotient Q must be expressed in "effective concentrations" or of the electroactive ionic species. The activity coefficient \(\gamma\)) relates the concentration of an ion to its activity a in a given solution through the relation \[a = \gamma c\] Since electrode potentials measure activities directly, activity coefficients can be determined by carrying out appropriate EMF measurements on cells in which the concentration of the ion of interest is known. The resulting Es can then be used to convert concentrations into activities for use in other calculations involving equilibrium constants. As most of us recall from our struggles with balancing redox equations in beginning chemistry courses, many electron-transfer reactions involve hydrogen ions and hydroxide ions. The standard potentials for these reactions therefore refer to the pH, either 0 or 14, at which the appropriate ion has unit activity. Because multiple numbers of H or OH ions are often involved, the potentials given by the Nernst equation can vary greatly with the pH. It is frequently useful to look at the situation in another way by considering what combinations of potential and pH allow the stable existence of a particular species. This information is most usefully expressed by means of a -vs.-pH diagram, also known as a . As was noted in connection with the shaded region in the figure below, water is subject to decomposition by strong oxidizing agents such as Cl and by reducing agents stronger than H . The reduction reaction can be written either as \[2H^+ + 2e^– \rightarrow H_{2}(g) \] or, in neutral or alkaline solutions as \[2H_2O + 2 e^– \rightarrow H_{2}(g) + 2 OH^–\] These two reactions are equivalent and follow the same Nernst equation \[E_{H^+/H_2} = E_{H^+/H_2}^o + \dfrac{RT}{nF} \ln \left( \dfrac{[H^+]^2} {P_{H_2}} \right)\] which, at 25°C and unit H partial pressure reduces to \[E = E° - \dfrac{0.059}{2} × 2 pH = –0.059\; pH\] Similarly, the oxidation of water \[2H_2O \rightarrow O_{2}(g) + 4 H^+ + 2 e^–\] is governed by the Nernst equation \[ E_{O_2/H_2O} = E_{O_2/H_2O}^o + \dfrac{RT}{nF} \ln \left( P_{O_2}[H^+]^4 \right)\] which similarly becomes = 1.23 – 0.059 pH, so the -pH plots for both processes have identical slopes and yield the stability diagram for water in Figure \(\Page {2}\). This Pourbaix diagram has special relevance to electrochemical corrosion of metals. Thus metals above hydrogen in the activity series will tend to undergo oxidation (corrosion) by reducing H ions or water. Because chlorine is widely used as a disinfectant for drinking water, swimming pools, and sewage treatment, it is worth looking at its stability diagram. Note that the effective bactericidal agent is not Cl itself, but its oxidation product hypochlorous acid HOCl which predominates at pH values below its pK of 7.3. Note also that Each solid line represents a combination of E and pH at which the two species on either side of it can coexist; at all other points, only a single species is stable. Note that equilibria between species separated by diagonal lines are dependent on both E and pH, while those separated by horizontal or vertical lines are affected by pH only or E only, respectively. Stability diagrams are able to condense a great amount of information into a compact representation, and are widely employed in geochemistry and corrosion engineering. The Pourbaix diagram for iron is one of the more commonly seen examples. The +3 oxidation state is the only stable one in environments in which the oxidation level is controlled by atmospheric O . This is the reason the Earth’s crust contains iron oxides, which developed only after the appearance of green plants which are the source of O . Iron is attacked by H to form H and Fe(II); the latter then reacts with O to form the various colored Fe(III) oxides that constitute “rust”. Numerous other species such as oxides and hydrous oxides are not shown. A really “complete” diagram for iron would need to have at least two additional dimensions showing the partial pressures of O and CO . From your study of thermodynamics you may recall that the process solute → solute is accompanied by a fall in free energy, and therefore is capable of doing work on the surroundings; all that is required is some practical way of capturing this work. One way of doing this is by means of a concentration cell such as Cu | CuNO || CuNO | Cu cathode: Cu + 2 → Cu anode: Cu → Cu + 2 net: Cu → Cu which represents the transport of cupric ion from a region of higher concentration to one of lower concentration. The driving force for this process is the free energy change ΔG associated with the concentration gradient (C – C ), sometimes known as the free energy of dilution: \[ΔG_{dilution} = RT \ln(C_2 – C_1)\] Note, however, that Cu ions need not physically move between the two compartments; electron flow through the external circuit creates a "virtual" flow as copper ions are created in the low-concentration side and discharged at the opposite electrode. Nitrate ions must also pass between the cells to maintain . The Nernst equation for this cell is \[E = E^° - \left(\dfrac{0.059}{n}\right) \log_{10} Q = 0 - 0.29 \log_{10} 0.1 = +0.285 \,V\] Note that \(E^°\) for a concentration cell is always zero, since this would be the potential of a cell in which the electroactive species are at unit activity in both compartments. \(E^°\) for a concentration cell is zero,

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.6%3A_Reaction_Mechanisms

A balanced equation for a chemical reaction indicates what is reacting and what is produced, but it reveals nothing about how the reaction actually takes place. The (or reaction path) is the process, or pathway, by which a reaction occurs. A chemical reaction often occurs in steps, although it may not always be obvious to an observer. The decomposition of ozone, for example, appears to follow a mechanism with two steps: \[\ce{O3}(g)⟶\ce{O2}(g)+\ce{O}\\ \ce{O}+\ce{O3}(g)⟶\ce{2O2}(g) \label{12.7.1} \] We call each step in a reaction mechanism an . Elementary reactions occur exactly as they are written and cannot be broken down into simpler steps. Elementary reactions add up to the overall reaction, which, for the decomposition, is: \[\ce{2O3}(g)⟶\ce{3O2}(g) \label{12.7.2} \] Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called . While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction does not involve the collision and reaction of two ozone molecules. Rather, it involves a molecule of ozone decomposing to an oxygen molecule and an intermediate oxygen atom; the oxygen atom then reacts with a second ozone molecule to give two oxygen molecules. These two elementary reactions occur exactly as they are shown in the reaction mechanism. The of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a involves the rearrangement of a reactant species to produce one or more molecules of product: \[A⟶\ce{products} \label{12.7.2b} \] The rate equation for a unimolecular reaction is: \[\ce{rate}=k[A] \label{12.7.3} \] A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction: \[\ce{O3 ⟶ O2 + O} \label{12.7.4} \] illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism. However, some unimolecular reactions may have only a single reaction in the reaction mechanism. (In other words, an elementary reaction can also be an overall reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C H , to ethylene, C H , occurs via a unimolecular, single-step mechanism:   For these unimolecular reactions to occur, all that is required is the separation of parts of single reactant molecules into products. Chemical bonds do not simply fall apart during chemical reactions. Energy is required to break chemical bonds. The activation energy for the decomposition of C H , for example, is 261 kJ per mole. This means that it requires 261 kilojoules to distort one mole of these molecules into activated complexes that decompose into products:   In a sample of C H , a few of the rapidly moving C H molecules collide with other rapidly moving molecules and pick up additional energy. When the C H molecules gain enough energy, they can transform into an activated complex, and the formation of ethylene molecules can occur. In effect, a particularly energetic collision knocks a C H molecule into the geometry of the activated complex. However, only a small fraction of gas molecules travel at sufficiently high speeds with large enough kinetic energies to accomplish this. Hence, at any given moment, only a few molecules pick up enough energy from collisions to react. The rate of decomposition of C H is directly proportional to its concentration. Doubling the concentration of C H in a sample gives twice as many molecules per liter. Although the fraction of molecules with enough energy to react remains the same, the total number of such molecules is twice as great. Consequently, there is twice as much C H per liter, and the reaction rate is twice as fast: \[\ce{rate}=−\dfrac{Δ[\ce{C4H8}]}{Δt}=k[\ce{C4H8}] \label{12.7.5} \] A similar relationship applies to any unimolecular elementary reaction; the reaction rate is directly proportional to the concentration of the reactant, and the reaction exhibits first-order behavior. The proportionality constant is the rate constant for the particular unimolecular reaction. The collision and combination of two molecules or atoms to form an activated complex in an elementary reaction is called a . There are two types of bimolecular elementary reactions: \[A+B⟶\ce{products} \label{12.7.6} \] and \[2A⟶\ce{products} \label{12.7.7} \] For the first type, in which the two reactant molecules are different, the rate law is first-order in and first order in : \[\ce{rate}=k[A,B] \label{12.7.8} \] For the second type, in which two identical molecules collide and react, the rate law is second order in : \[\ce{rate}=k[A,A]=k[A]^2 \label{12.7.9} \] Some chemical reactions have mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide: \[\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \label{12.7.10} \] (see Figure \(\Page {1}\)) Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is one example: \[\ce{O}(g)+\ce{O3}(g)⟶\ce{2O2}(g) \label{12.7.12} \] An elementary involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps: \[\ce{2NO + O2 ⟶ 2NO2}\\ \ce{rate}=k[\ce{NO}]^2[\ce{O2}] \label{12.7.13} \] Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps: \[\ce{2NO + Cl2 ⟶ 2NOCl}\\ \ce{rate}=k[\ce{NO}]^2[\ce{Cl2}] \label{12.7.14} \]   It's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the (or rate-determining step) of the reaction Figure \(\Page {2}\). As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, we must determine the overall rate law from experimental data and deduce the mechanism from the rate law (and sometimes from other data). The reaction of NO and provides an illustrative example: \[\ce{NO2}(g)+\ce{CO}(g)⟶\ce{CO2}(g)+\ce{NO}(g) \nonumber \] For temperatures above 225 °C, the rate law has been found to be: \[\ce{rate}=k[\ce{NO2},\ce{CO}] \nonumber \] The reaction is first order with respect to NO and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is that this is the mechanism for this reaction at high temperatures. At temperatures below 225 °C the reaction is described by a rate law that is second order with respect to NO : \[\ce{rate}=k[\ce{NO2}]^2 \nonumber \] This is consistent with a mechanism that involves the following two elementary reactions, the first of which is slower and is therefore the rate-determining step: \[\ce{NO2}(g)+\ce{NO2}(g)⟶\ce{NO3}(g)+\ce{NO}(g)\:\ce{(slow)}\\ \ce{NO3}(g)+\ce{CO}(g)⟶\ce{NO2}(g)+\ce{CO2}(g)\:\ce{(fast)} \nonumber \] The rate-determining step gives a rate law showing second-order dependence on the NO concentration, and the sum of the two equations gives the net overall reaction. In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving an reaction, the rate law for the overall reaction may be more difficult to derive. An elementary reaction is at equilibrium when it proceeds in both the forward and reverse directions at equal rates. Consider the dimerization of NO to N O , with used to represent the rate constant of the forward reaction and used to represent the rate constant of the reverse reaction: \[\ce{NO + NO ⇌ N2O2} \nonumber \] \[\ce{rate_{forward}=rate_{reverse}} \nonumber \] \[k_1[\ce{NO}]^2=k_{−1}[\ce{N2O2}] \nonumber \] If N O was an intermediate in a mechanism, this expression could be rearranged to represent the concentration of N O in the overall rate law expression using algebraic manipulation: \[\mathrm{\left(\dfrac{k_1[NO]^2}{k_{−1}}\right)=[N_2O_2]} \nonumber \] However, once again, intermediates cannot be listed as part of the overall rate law expression, though they can be included in an individual elementary reaction of a mechanism. Example \(\Page {1}\) will illustrate how to derive overall rate laws from mechanisms involving equilibrium steps preceding the rate-determining step. Nitryl chloride (NO Cl) decomposes to nitrogen dioxide (NO ) and chlorine gas (Cl ) according to the following mechanism: Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression. For the overall reaction, simply sum the three steps, cancel intermediates, and combine like formulas: \[\ce{2NO2Cl}(g)⟶\ce{2NO2}(g)+\ce{Cl2}(g) \nonumber \] Next, write the rate law expression for each elementary reaction. Remember that for elementary reactions that are part of a mechanism, the rate law expression can be derived directly from the stoichiometry: \[\begin{align*} k_1\ce{[NO2Cl]2}&=k_{−1}\ce{[ClO2,N2O,ClO]}\\ k_2\ce{[N2O,ClO2]}&=k_{−2}\ce{[NO2,NOCl]}\\ \ce{Rate}&=k_3\ce{[NOCl,ClO]} \end{align*} \nonumber \] The third step, which is the slow step, is the rate-determining step. Therefore, the overall rate law expression could be written as Rate = [NOCl,ClO]. However, both NOCl and ClO are intermediates. Algebraic expressions must be used to represent [NOCl] and [ClO] such that no intermediates remain in the overall rate law expression. Now substitute these algebraic expressions into the overall rate law expression and simplify: \[\begin{align*} \ce{rate}&=k_3\left(\dfrac{k_2\ce{[N2O,ClO2]}}{k_{−2}\ce{[NO2]}}\right)\left(\dfrac{k_1\ce{[NO2Cl]^2}}{k_{−1}\ce{[ClO2,N2O]}}\right)\\ \ce{rate}&=\dfrac{k_3k_2k_1\ce{[NO2Cl]^2}}{k_{−2}k_{−1}\ce{[NO2]}} \end{align*} \nonumber \] Notice that this rate law shows an dependence on the concentration of one of the product species, consistent with the presence of an equilibrium step in the reaction mechanism. Atomic chlorine in the atmosphere reacts with ozone in the following pair of elementary reactions: \(\ce{Cl}+\ce{O3}(g)⟶\ce{ClO}(g)+\ce{O2}(g)\hspace{20px}(\textrm{rate constant }k_1)\) Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression. The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The overall rate of a reaction is determined by the rate of the slowest step, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.

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

Moles are a convenient unit used in chemistry to convert between amounts of a substance in grams and numbers of atoms or molecules. This is useful because we usually measure how much of a molecule is used or produced in a reaction by massing it, but as a chemical equation shows, the reaction will happen between atoms or molecules. For example, suppose we combine 1.0 g of calcium oxide (CaO) with 1.0 g of water (H O). The product we get is Ca(OH) . Here's the equation: \[CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(s\; or\; aq)\] This is balanced. Thus every molecule of water reacts with one CaO formula unit (it's not called a molecule because it's an ionic solid, and each Ca ion is surrounded with oxide ions that it interacts with equally). How much calcium hydroxide is produced by this reaction? Once all of one reactant has been used, whatever is left of the other will stop reacting, because of the law of definite proportions: we won't change the ratio of O:H:Ca in the product. So will we get solid calcium hydroxide with calcium oxide left over, or will we have water left over, and thus get Ca(OH) (aq)? To answer this question, we can convert both masses (1 g of each) to the number of molecules or formula weights, but this would be inconvenient because the number would be very very big! Instead, we use . A mole (abbreviation: mol) is like a pair, which means 2 of something. You can have a pair of people, a pair of apples, whatever. A mole is 6.022 x 10 of something. This is a convenient quantity because it converts amu (atomic mass units) to grams. The atomic weight of carbon is (on average) 12.011 amu/atom. It is also 12.011g/mol. In other words, 1g = 6.02 x 10 amu. Usually, a mol of a substance is a useful, practical amount, somewhere between a few grams and a few kg. The number of things in a mole, 6.022 x 10 , is called Avogadro's number, and abbreviated as N . It is named after Avogadro, the scientist who proposed that a liter of any gas at the same temperature and pressure has the same number of molecules in it. To summarize: So the way to answer the question above is to convert both quantities to moles. The maximum amount of product that can be formed is the smaller number of moles. The formula weight is just the sum of the atomic weights. \[(1.0\; \cancel{ g\; CaO}) \left(\dfrac{1\; \cancel{mol\; CaO}}{56.08\; \cancel{ g\; CaO}}\right) \left(\dfrac{10^{3}\; mmol\; CaO}{1\; \cancel{mol\; CaO}}\right) = 17.8\; mmol\; CaO\] \[(1.0\; \cancel{ g\; H_{2}O}) \left(\dfrac{1\; \cancel{mol\; H_{2}O}}{18.01\; \cancel{ g\; H_{2}O}}\right) \left(\dfrac{10^{3}\; mmol\; H_{2}O}{1\; \cancel{mol\; H_{2}O}}\right)= 55.5\; mmol\; H_{2}O\] After we make 17.8 mmol (milimoles) of Ca(OH) , we will use up all the CaO, so the reaction won't continue. The maximum amount of Ca(OH) possible to make is 17.8 mmol. If we wanted to know the (maximum mass of product) of Ca(OH) , we could do it in a one-step calculation like this: \[(1.0\; \cancel{ g\; CaO}) \left(\dfrac{1\; \cancel{mol\; CaO}}{56.08\; \cancel{ g\; CaO}}\right)\left(\dfrac{1 \; \cancel{mol\; Ca(OH)_{2}}}{1 \; \cancel{mol\; CaO}}\right) \left(\dfrac{74.09\; g\; Ca(OH)_{2}}{1\; \cancel{ mol\; Ca(OH)_{2}}}\right) = 1.3\; g\; Ca(OH)_{2}\] Here, we knew that the (or limiting reagent), which is the reactant that will run out first, is CaO because the masses are the same, the coefficients in the equation are the same, and the formula weight of CaO is bigger than the molecular weight of water. So we start with the limiting reactant mass, convert it to moles (using 1 mol = 56.08 g), then "convert" between moles of CaO and moles of Ca(OH) using the coefficients from the balanced equation (1 mol of CaO produces 1 mol of Ca(OH) ), then we convert to g of Ca(OH) (using 1 mol = 74.09 g). This is just an example of using dimensional analysis to convert units. We check to make sure we have always multiplied by 1 (because 1 mol CaO = 56g CaO, so (1 mol CaO/56 g CaO)=1), and that the units cancel out to leave the correct final units (g Ca(OH) ), and we can be pretty sure that we got it right. Here's a slightly more complicated example. This time, we add 2.0 g of water to 2.5 g of Li O. This will produce LiOH as the major product. What is the most LiOH (in g) that could be produced, also called the theoretical yield? To answer, first we need to write and balance the chemical equation. It's going to look pretty similar to the previous one, because this is a similar reaction. \[Li_{2}O(s) + H_{2}O(l) \rightarrow 2LiOH(s\; or\; aq)\] Which is the limiting reactant? The formula weights are 18.01 g and 29.88 g. Water is still in , which means it will be left over. Here's the unit conversion: \[(2.5\; \cancel{ g\; Li_{2}O}) \left(\dfrac{1\; \cancel{mol\; Li_{2}O}}{29.88\; \cancel{ g\; Li_{2}O}}\right)\left(\dfrac{2 \; \cancel{mol\; LiOH}}{1 \; \cancel{mol\; Li_{2}O}}\right) \left(\dfrac{23.95\; g\; LiOH}{1\; \cancel{ mol\; LiOH}}\right) = 4.0\; g\; LiOH\] This time we used 1 mol Li O producing 2 mol LiOH, using the coefficients from the balanced equation.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/15%3A_Water/15.11%3A_Colloids

Imagine that you are sailing on a yacht. The engine suddenly breaks down and you are stranded in the middle of the ocean. You call the Coast Guard on your radio, but cannot give them an exact location, because your isn't working. Fortunately, you have a smoke flare, which you fire off. The dense colored smoke shows the Coast Guard where you are so that they can rescue you. In using the flare, you are taking advantage of a mixture called a colloid. A is a heterogeneous mixture whose particle size is intermediate between those of a solution and a suspension. The dispersed particles are spread evenly throughout the , which can be a solid, liquid, or gas. Because the dispersed particles of a colloid are not as large as those of a suspension, they do not settle out upon standing. The table below summarizes the properties and distinctions between solutions, colloids, and suspensions. Colloids are unlike solutions because their dispersed particles are much larger than those of a solution. The dispersed particles of a colloid cannot be separated by filtration, but they scatter light—a phenomenon called the . When light is passed through a true solution, the dissolved particles are too small to deflect the light. However, the dispersed particles of a colloid, being larger, do deflect light. The Tyndall effect is the scattering of visible light by colloidal particles. You have undoubtedly "seen" a light beam as it passes through fog, smoke, or a scattering of dust particles suspended in air. All three are examples of colloids. Suspensions may scatter light, but if the number of suspended particles is sufficiently large, the suspension may simply be opaque, and the light scattering will not occur. The table below lists examples of colloidal systems, most of which are very familiar. The dispersed phase describes the particles, while the dispersion medium is the material in which the particles are distributed. Another property of a colloidal system is observed when the colloids are studied under a light microscope. The colloids scintillate, reflecting brief flashes of light because the colloidal particles move in a rapid and random fashion. This phenomenon, called Brownian motion, is caused by collisions between the small colloidal particles and the molecules of the dispersion medium. Butter and mayonnaise are examples of a class of colloids called . An emulsion is a colloidal dispersion of a liquid in either a liquid or a solid. A stable emulsion requires an emulsifying agent to be present. Mayonnaise is made in part of oil and vinegar. Since oil is nonpolar and vinegar is an aqueous solution and polar, the two do not mix, and quickly separate into layers. However, the addition of egg yolk causes the mixture to become stable and not separate. Egg yolk is capable of interacting with both the polar vinegar and the nonpolar oil. The egg yolk is called the emulsifying agent. Soap acts as an emulsifying agent between grease and water. Grease cannot be simply rinsed off your hands, or another surface, because it is insoluble. However, the soap stabilizes a grease-water mixture because one end of a soap molecule is polar, and the other end is nonpolar. This allows the grease to be removed from your hands or your clothing by washing with soapy water.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/09%3A_Gases/9.6%3A_Non-Ideal_Gas_Behavior

Thus far, the ideal gas law, , has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered. One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, Vm) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with: \[\mathrm{Z=\dfrac{molar\: volume\: of\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}{molar\: volume\: of\: ideal\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}}=\left(\dfrac{PV_m}{RT}\right)_\ce{measured} \nonumber \] Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. Figure \(\Page {1}\) shows plots of Z over a large pressure range for several common gases. As is apparent from Figure \(\Page {1}\), the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas. Particles of a hypothetical ideal gas have and . In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas (Figure \(\Page {2}\)). The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not as predicted by Boyle’s law. At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure \(\Page {3}\)). This change is more pronounced at low temperatures because the molecules have lower relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another. There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.   The constant corresponds to the strength of the attraction between molecules of a particular gas, and the constant corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in the ideal gas law is \(\dfrac{n^2a}{V^2}\), and the “correction” to the volume is . Note that when is relatively large and is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, . Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table \(\Page {1}\). At low pressures, the correction for intermolecular attraction, , is more important than the one for molecular volume, . At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in Figure \(\Page {1}\). The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing ). At very high pressures, the gas becomes less compressible (Z increases with ), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume. Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of . Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case. A 4.25-L flask contains 3.46 mol CO at 229 °C. Calculate the pressure of this sample of CO : (a) From the ideal gas law: (b) From the van der Waals equation: \(\left(P+\dfrac{n^2a}{V^2}\right)×(V−nb)=nRT⟶P=\dfrac{nRT}{(V−nb)}−\dfrac{n^2a}{V^2}\) \(P=\mathrm{\dfrac{3.46\:mol×0.08206\:L\:atm\:mol^{−1}\:K^{−1}×502\: K}{(4.25\:L−3.46\:mol×0.0427\:L\:mol^{−1})}−\dfrac{(3.46\:mol)^2×3.59\:L^2\:atm\:mol^2}{(4.25\:L)^2}}\) This finally yields = 32.4 atm. (c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions. A 560-mL flask contains 21.3 g N at 145 °C. Calculate the pressure of N : 46.562 atm 46.594 atm The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions. Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Gases/Diffusion_and_Effusion

In this section we discuss movements of gases. This is actually a very complicated field (called fluid dynamics) and we will not go very deep. is the movement of a gas through a tiny hole into a vacuum. We want to know the rate of effusion, which is how much gas moves through the hole per unit time. We assume that the gas particles don't bump into each other while they move through the hole (this means it's a narrow hole in a thin wall). So the rate of effusion just depends on how often the particles bump the hole. This depends on their density and speed. Because at a given temperature, all gases have the same kinetic energy, their speed is inversely proportional to m , the square root of the mass. Thus, the relative effusion rates for different gases at the same temperature is \[\frac{Effusion\; rate\; for\; gas\; 1}{Effusion\; rate\; for\; gas\; 2} = \frac{M_{2}^{\frac{1}{2}}}{M_{1}^{\frac{1}{2}}}\] is a more complicated process. It means the movement of gases through each other or the spreading of one gas through another. Because there are many collisions, the gases move much slower than we might expect from the average speeds near 400-700 m/s. (This is why it will take a moment to smell perfume when someone walks into a room.) Technically, many processes that sound like this are not exactly simple diffusion. We have to be careful about whether there are pressure differences or flows of gases (like wind). If there are then the process isn't simple diffusion and it won't follow the equations for simple diffusion perfectly. (The equation for simple diffusion is the same as for effusion, but for different reasons, see below.) For our purposes, when you want to predict relative rates of movement of gases, you can start with the effusion/diffusion equation. It will be exactly right in a few situations, and close enough in some others. The other situations you can study in a more advanced class if you are interested. Why is the diffusion equation the same as the effusion equation, but for a different reason? In simple diffusion, 2 gases move in opposite directions through a medium with the same pressure everywhere. If the pressure in the medium is constant, then the collisions of one gas with the medium are balanced by the collisions of the other gas. The momentum given to the medium by one gas in an average collision is mV, where m is the mass and V is the diffusion velocity (which is different from the average speed of the particles: it's the overall rate of movement of the gas). The number of collisions is proportional to nv, where n is the number of particles and v is their average speed. Since there is no pressure difference, \[\left(m_{1}V_{1}\right) \left(n_{1}v_{1}\right) = \left(m_{2}V_{2}\right) \left(n_{2}v_{2}\right)\] When we rearrange, the relative diffusion flux (nV, amount of particles moving times speed of diffusion) of the gases 1 and 2 is \[\frac{n_{1}V_{1}}{n_{2}V_{2}} = \frac{m_{2}v_{2}}{m_{1}v_{1}}\]

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

Meso compounds are achiral compounds that has multiple chiral centers. It is superimposed on its mirror image and is optically despite its stereocenters. In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, result in . Cyclic compounds may also be meso. If A is a meso compound, it should have two or more stereocenters, an internal plane, and the stereochemistry should be . An interesting thing about single bonds or sp -orbitals is that we can rotate the substituted groups that attached to a stereocenter around to recognize the internal plane. As the molecule is rotated, its stereochemistry does not change. For example: Another case is when we rotate the whole molecule by 180 degree. Both molecules below are still meso. Remember the internal plane here is depicted on two dimensions. However, in reality, it is three dimensions, so be aware of it when we identify the internal mirror. This molecule has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; However, it has two chiral carbons and is consequentially a meso compound. This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound. Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cyclobutane. They do not necessarily have to be two stereocenters, but can have more. When the of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+). Beside meso, there are also other types of molecules: enantiomer, , and identical. Determine if the following molecules are meso. Answer key: A C, D, E are meso compounds.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.01%3A_Energy_Heat_and_Work

Make sure you thoroughly understand the following essential ideas: Energy is one of the most fundamental and universal concepts of physical science, but one that is remarkably difficult to define in way that is meaningful to most people. This perhaps reflects the fact that energy is not a “thing” that exists by itself, but is rather an attribute of matter (and also of electromagnetic radiation) that can manifest itself in different ways. It can be observed and measured only indirectly through its effects on matter that acquires, loses, or possesses it. The concept that we call energy was very slow to develop; it took more than a hundred years just to get people to agree on the definitions of many of the terms we use to describe energy and the interconversion between its various forms. But even now, most people have some difficulty in explaining what it is; somehow, the definition we all learned in elementary science ("the capacity to do work") seems less than adequate to convey its meaning. Although the term "energy" was not used in science prior to 1802, it had long been suggested that certain properties related to the motions of objects exhibit an endurance which is incorporated into the modern concept of "conservation of energy". René Descartes (1596-1650) stated it explicitly: When God created the world, He "caused some of its parts to push others and to transfer their motions to others..." and thus "He motion". In the 17th Century, the great mathematician Gottfried Leibniz (1646-1716) suggested the distinction between ("live force") and ("dead force"), which later became known as kinetic energy (1829) and potential energy (1853). Whatever energy may be, there are basically two kinds. is associated with the of an object, and its direct consequences are part of everyone's daily experience; the faster the ball you catch in your hand, and the heavier it is, the more you feel it. Quantitatively, a body with a mass and moving at a velocity possesses the kinetic energy /2. A rifle shoots a 4.25 g bullet at a velocity of 965 m s . What is its kinetic energy? The only additional information you need here is that 1 J = 1 kg m s : KE = ½ × (.00425 kg) (965 m s ) = 1980 J is energy a body has by virtue of its . But there is more: of some kind that tends to move it to a location of lower potential energy. Think of an arrow that is subjected to the force from a stretched bowstring; the more tightly the arrow is pulled back against the string, the more potential energy it has. More generally, the restoring force comes from what we call a — a gravitational, electrostatic, or magnetic field. We observe the consequences of l potential energy all the time, such as when we walk, but seldom give it any thought. If an object of mass is raised off the floor to a height , its potential energy increases by , where is a proportionality constant known as the ; its value at the earth's surface is 9.8 m s . Find the change in potential energy of a 2.6 kg textbook that falls from the 66-cm height of a table top onto the floor. PE = = (2.6 kg)(9.8 m s )(0.66 m) = 16.8 kg m s = 16.8 J Similarly, the potential energy of a particle having an electric charge depends on its location in an electrostatic field. Electrostatic potential energy plays a major role in chemistry; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of atoms and molecules. "Chemical energy" usually refers to the energy that is stored in the chemical bonds of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy. In an exothermic chemical reaction, the electrons and nuclei within the reactants undergo rearrangment into products possessing lower energies, and the difference is released to the environment in the form of heat. Transitions between potential and kinetic energy are such an intimate part of our daily lives that we hardly give them a thought. It happens in walking as the body moves up and down. Our bodies utilize the chemical energy in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of chemical energy to other forms. it can neither be created nor destroyed. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal. Kinetic energy is associated with motion, but in two different ways. For a macroscopic object such as a book or a ball, or a parcel of flowing water, it is simply given by ½  . But as we mentioned above, when an object is dropped onto the floor, or when an exothermic chemical reaction heats surrounding matter, the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as . We observe the effects of this as a rise in the temperature of the surroundings. The temperature of a body is direct measure of the quantity of thermal energy is contains. Once kinetic energy is thermalized, only a portion of it can be converted back into potential energy. The remainder simply gets dispersed and diluted into the environment, and is effectively lost. To summarize, then: You might at first think that a book sitting on the table has zero kinetic energy since it is not moving. But if you think about it, the earth itself is moving; it is spinning on its axis, it is orbiting the sun, and the sun itself is moving away from the other stars in the general expansion of the universe. Since these motions are normally of no interest to us, we are free to adopt an arbitrary scale in which the velocity of the book is measured with respect to the table; on this so-called , the kinetic energy of the book can be considered zero. We do the same thing with potential energy. If the book is on the table, its potential energy with respect to the surface of the table will be zero. If we adopt this as our zero of potential energy, and then push the book off the table, its potential energy will be negative after it reaches the floor. Energy is measured in terms of its ability to perform work or to transfer heat. Mechanical work is done when a force f displaces an object by a distance : = × . The basic unit of energy is the . One joule is the amount of work done when a force of 1 newton acts over a distance of 1 m; thus 1 J = 1 N-m. The newton is the amount of force required to accelerate a 1-kg mass by 1 m/sec , so the basic dimensions of the joule are kg m s . The other two units in wide use. the and the (British thermal unit) are defined in terms of the heating effect on water. Because of the many forms that energy can take, there are a correspondingly large number of units in which it can be expressed, a few of which are summarized below. 1 calorie will raise the temperature of 1 g of water by 1 C°. The “dietary” calorie is actually 1 kcal. An average young adult expends about 1800 kcal per day just to stay alive. (you should know this definition) 1 J = 10 ergs 1 erg = 1 d-cm = 1 g cm s 1 J = 2.78 × 10 watt-hr 1 w-h = 3.6 kJ 1 bboe = 6.1 GJ 1 cmge = 37-39 mJ 1 toce = 29 GJ Heat and work are both measured in energy units, so they must both represent energy. How do they differ from each other, and from just plain “energy” itself? In our daily language, we often say that "this object contains a lot of heat", but this kind of talk is a no-no in thermodynamics! It's ok to say that the object is "hot", meaning that its temperature is high. The term "heat" has a special meaning in thermodynamics: it is a in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a than its surroundings (the rest of the world). Heat and work are and cannot be stored Thermal energy can only flow from a higher temperature to a lower temperature. It is this flow that constitutes "heat". Use of the term "flow" of heat recalls the 18th-century notion that heat is an actual substance called “caloric” that could flow like a liquid. Transfer of thermal energy can be accomplished by bringing two bodies into physical contact (the kettle on top of the stove, or through an electric heating element inside the kettle). Another mechanism of thermal energy transfer is by radiation; a hot object will convey energy to any body in sight of it via electromagnetic radiation in the infrared part of the spectrum. In many cases, a combination of modes will be active. Thus when you place a can of beer in the refrigerator, both processes are operative: the can radiates heat to the cold surfaces around it, and absorbs it by direct conduction from the ambient air. Work refers to the transfer of energy some means that does not depend on temperature difference. Work, like energy, can take various forms, the most familiar being mechanical and electrical. arises when an object moves a distance Δ against an opposing force : =  Δ N^{-m}\] with \(1\, N^{-m}\) = 1\, J.\) is done when a body having a charge moves through a potential difference Δ . Work, like heat, exists only when energy is being transferred.When two bodies are placed in thermal contact and energy flows from the warmer body to the cooler one,we call the process “heat”. A transfer of energy to or from a system by any means other than heat is called “work”. Work can be completely converted into heat (by friction, for example), but heat can only be partially converted to work. is accomplished by means of a , the most common example of which is an ordinary gasoline engine. The science of thermodynamics developed out of the need to understand the limitations of steam-driven heat engines at the beginning of the Industrial Age. A fundamental law of Nature, the Second Law of Thermodynamics, states that the complete conversion of heat into work is impossible. Something to think about when you purchase fuel for your car!

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Most metals are chiefly useful in elemental form, but they usually occur in compounds on the earth’s surface. An is a naturally occurring material from which one or more useful elements or compounds may be obtained at a cost that is economically feasible. As can be seen from the table, most metal ores are oxides, carbonates, or sulfides. A few of the least electropositive metals occur as the element. Whether a mineral can usefully be regarded as an ore or not depends on such factors as how it is, its exact , and whether there is a suitable for extracting the metal. As the more accessible higher-grade ores become exhausted, less accessible and lower-grade ores are becoming increasingly utilized with a consequent shift in the meaning of the word ore. It is conceivable that many silicates could become sources of metals, notably aluminum. Currently, though, silicates are expensive decompose chemically, and silicates form ores only for relatively expensive metals like zirconium and beryllium. The processing of ores may be divided into three steps. Often concentration or other is required to remove worthless material ( ) or to convert the mineral into an appropriate form for subsequent processing. The second and most-important step is of the metal from a positive oxidation state. This may involve elevated temperatures, chemical reducing agents, electrolysis, or some combination of these treatments. Usually a third step, , is required to achieve the purity (or precise mixed composition in the case of an alloy) desired in the final product.

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One of the challenges of using the second law of thermodynamics to determine if a chemical reaction is spontaneous is that we must determine the entropy change for the system the entropy change for the surroundings. A second challenge when working with a chemical reaction is that we need to take into account the mixing of the substances, an issue that does not occur when observing the phase change if a pure substance. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard . This new property is called the  (\(G\)) (or simply the ), and it is defined in terms of a system’s enthalpy and entropy as the following: \[G=H−TS \nonumber \] Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following: \[ΔG^º_\ce{sys}=ΔH^º_\ce{sys}−TΔS^º_\ce{sys} \nonumber \] (For simplicity’s sake, the subscript “sys” will be omitted henceforth.) We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression: \[ΔS_\ce{univ}=ΔS+\dfrac{q_\ce{surr}}{T} \nonumber \] The first law requires that \(q_{surr} = −q_{sys}\), and at constant pressure \(q_{sys} = ΔH\), and so this expression may be rewritten as the following: \[ΔS_\ce{univ}=ΔS−\dfrac{ΔH}{T} \nonumber \] ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following: \[−TΔS_\ce{univ}=ΔH−TΔS \nonumber \] Comparing this equation to the previous one for free energy change shows the following relation: \[ΔG=−TΔS_\ce{univ} \label{6} \] The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, \(ΔS_{univ}\). Table \(\Page {1}\) summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators. nonspontaneous in the forward direction, as written, but moves spontaneously in the reverse direction, as written, to reach equilibrium Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system that have undergone some change. A convenient and common approach to the calculation of free energy changes for physical changes and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in Example \(\Page {1}\). \[ ΔG°=ΔH°−TΔS° \label{7} \] It is important to understand that for phase changes, \(\Delta G^º\) tells you if the phase change is spontaneous or not; will it happen, or not happen. For chemical reactions, \(\Delta G^º\) tells you the of a reaction. In other words, \(\Delta G^º\) for a reaction tells you how much product will be present at equilibrium. A reaction with \(\Delta G^º\) < 0 is considered ; there will be more products than reactants when the reaction reaches equilibrium. A reaction with \(\Delta G^º\) > 0 is considered  ; there will be more reactants than products when the reaction reaches equilibrium. Use standard enthalpy and entropy data from Tables T1 or T2 to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this physical change for a pure substance? The process of interest is the following: \[\ce{H2O}(l)⟶\ce{H2O}(g) \label{\(\Page {8}\)} \] The standard change in free energy may be calculated using the following equation: \[ΔG^\circ_{298}=ΔH°−TΔS° \label{\(\Page {9}\)} \] From Tables T1 or T2, here are the data: Combining at 298 K: \[\begin{align*} ΔH°&=ΔH^\circ_{298}=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \\[4pt] &=[−241.82\: kJ−(−285.83)]\:kJ/mol \\[4pt] &=44.01\: kJ/mol \\[4pt] ΔS° &=ΔS^\circ_{298}=S^\circ_{298}(\ce{H2O}(g))−S^\circ_{298}(\ce{H2O}(l)) \\[4pt] &=188.8\:J/mol⋅K−70.0\:J/K \\[4pt] &=118.8\:J/mol⋅K \end{align*} \nonumber \] then use Equation \ref{7}: \[ΔG°=ΔH°−TΔS° \nonumber \] Converting everything into kJ and combining at 298 K: \[\begin{align*}ΔG^\circ_{298} &=ΔH°−TΔS° \\[4pt] &=44.01\: kJ/mol−(298\: K×118.8\:J/mol⋅K)×\dfrac{1\: kJ}{1000\: J} \end{align*} \nonumber \] \[\mathrm{44.01\: kJ/mol−35.4\: kJ/mol=8.6\: kJ/mol} \nonumber \] At 298 K (25 °C) \(ΔG^\circ_{298}>0\), and so boiling is nonspontaneous (not spontaneous) at 298 K. Use standard enthalpy and entropy data from Tables T1 or T2 to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the extent of this reaction at 298 K? \[\ce{C2H6}(g)⟶\ce{H2}(g)+\ce{C2H4}(g) \nonumber \] \(ΔG^\circ_{298}=\mathrm{102.0\: kJ/mol}\); the reaction is reactant-favored at equilibrium at 25 °C. There will be more \(\ce{C2H6}(g)\) than \(\ce{H2}(g)\) and \(\ce{C2H4}(g) \) at equilibrium Free energy changes may also use the standard free energy of formation \( (ΔG^\circ_\ce{f})\), for each of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of formation, \( (ΔG^\circ_\ce{f})\) is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction \[m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D}, \nonumber \] the standard free energy change at room temperature may be calculated as \[ \begin{align} ΔG^\circ_{298}&=ΔG° \\[4pt] &=∑νΔG^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants})\\[4pt] &=[xΔG^\circ_\ce{f}(\ce{C})+yΔG^\circ_\ce{f}(\ce{D})]−[mΔG^\circ_\ce{f}(\ce{A})+nΔG^\circ_\ce{f}(\ce{B})]. \end{align} \nonumber \] Consider the decomposition of yellow mercury(II) oxide. \[\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+ \ce{ 1/2 O2(g)} \nonumber \] Calculate the standard free energy change at room temperature, \(ΔG^\circ_{298}\), using: Do the results indicate the reaction to be product-favored or reactant-favored at equilibrium? The required data are available in Tables T1 or T2 and are shown here. (a) Using free energies of formation: \[ \begin{align*} ΔG^\circ_{298}&=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) \\[4pt] &=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \\[4pt] & \mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} \end{align*} \nonumber \] (b) Using enthalpies and entropies of formation: \[\begin{align*}ΔH^\circ_{298}&=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) \\[4pt] &=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \\[4pt] &\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} \\[4pt] ΔS^\circ_{298} &=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) \\[4pt] &=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \\[4pt] & \mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} \end{align*} \nonumber \] then we can use Equation \ref7} directly: \[\begin{align*}ΔG°&=ΔH°−TΔS°\\[4pt] &=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} \\[4pt] &=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} \end{align*} \nonumber \] Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is reactant-favored at equilibrium at room temperature. Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Tables T1 or T2). Do the results indicate the reaction to be product-favored or reactant-favored at equilibrium at 25 °C? \[\ce{C2H4}(g)⟶\ce{H2}(g)+\ce{C2H2}(g) \nonumber \] 141.5 kJ/mol, reactant-favored at equilibrium As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. In a similar, but not identical fashion, some chemical reactions can switch from being product-favored at equilibrium, to being reactant-favored at equilibrium, depending on the temperature. The numerical value of \(\Delta G^º\) is always dependent on the temperature. In this section we are determining whether or not the sign of \(\Delta G^º\) is dependent on the temperature. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered: \[ ΔG^º=ΔH^º−TΔS^º \nonumber \] The extent of a process, as reflected in the arithmetic sign of its standard free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (Kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes: These four scenarios are summarized in Table \(\Page {1}\) Sign of \(\Delta H^o\) Sign of \(\Delta S^o\) Sign of \(\Delta G^o\) The sign of \(\Delta G^o\)  depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.   The incomplete combustion of carbon is described by the following equation: \[\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber \] Does the sign of \(\Delta G^º\) of this process depend upon temperature? Combustion processes are exothermic (\(ΔH^º < 0\)). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, \(ΔS^º > 0\)). The reaction is therefore product-favored at equilibrium (\(ΔG^º < 0\)) at all temperatures. Popular chemical hand warmers generate heat by the air-oxidation of iron: \[\ce{4Fe}(s)+\ce{3O2}(g)⟶\ce{2Fe2O3}(s) \nonumber \] Does the sign of \(\Delta G^o\) of this process depend upon temperature? Δ and Δ are both negative; the reaction is product-favored at equilibrium at low temperatures. When considering the conclusions drawn regarding the temperature dependence of the sign of Δ , it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is reactant-favored at equilibrium at one temperature but product-favored at equilibrium at another temperature will necessarily undergo a change in “extent” (as reflected by its Δ ) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which Δ is plotted on the axis versus on the axis: \[ΔG^º=ΔH^º−TΔS^º \nonumber \] \[y=b+mx \nonumber \] Such a plot is shown in Figure \(\Page {2}\). A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependence for the sign of Δ as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative Δ ) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the -intercept of the line, that is, the value of for which Δ is zero: \[ΔG^º=0=ΔH^º−TΔS^º \nonumber \] \[T=\dfrac{ΔH^º}{ΔS^º} \nonumber \] Thus, saying a process is product-favored at equilibrium at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which Δ for the process is zero.  In this discussion, we have used two different descriptions for the meaning of the sign of Δ . You should be aware of the meaning of each description.  a) This description is used to predict the ratio of the product and reactant concentrations at equilibrium. In this description, we use the thermodynamic term Δ  to tell us the same information as the equilibrium constant, K. When Δ  < 0, K > 1, and the reaction will be product-favored at equilibrium. When Δ  > 0, K< 1, and the reaction is reactant-favored at equilibrium. When Δ  = 0, K =1, and the reaction will have roughly equal amounts of products and reactants at equilibrium. In all cases, the reaction will form a mixture of products and reactants at equilibrium. We use the sign and magnitude of Δ  to tell us how much product will be made if the reaction is allowed to reach equilibrium. b) This description is much more complicated because it involves two different interpretations of how a reaction at standard state occurs. One interpretation involves the hypothetical process in which the reaction proceeds from a starting point of pure reactants to a finishing point of pure products, with all substances isolated in their own containers under standard state conditions. In the second, more realistic interpretation, the reaction starts with all reactants and all products in their standard state in one container. We then allow this specific mixture to react an infinitesimally small amount so that we can obtain a rate of change in free energy with respect to the extent of reaction when all reactants and products are mixed and (essentially) in their standard states. Although each interpretation describes a different reaction scenario, the value of the difference in free energy and the value of the rate of change in free energy are the same number. If Δ  < 0, we say that the reaction is spontaneous, meaning that the reaction would proceed in the forward direction, as written, to form pure products in their standard state. If Δ  > 0, we say that the reaction is nonspontaneous, meaning that the reaction would proceed in the reverse direction, as written, to form pure reactants in their standard state. If Δ  = 0, we say that the neither the reactants nor the products are favored to be formed. A detailed treatment of the meaning of ΔGº can be found in the paper, "Free Energy versus Extent of Reaction" by Richard S. Treptow, Journal of Chemical Education, , Volume 73 (1), 51-54. As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Tables T1 or T2 to estimate the boiling point of water. The process of interest is the following phase change: When this process is at equilibrium, Δ = 0, so the following is true: \[0=ΔH°−TΔS°\hspace{40px}\ce{or}\hspace{40px}T=\dfrac{ΔH°}{ΔS°} \nonumber \] Using the standard thermodynamic data from Tables T1 or T2, \[\begin{align*} ΔH°&=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\\ &=\mathrm{−241.82\: kJ/mol−(−285.83\: kJ/mol)=44.01\: kJ/mol} \nonumber \end{align*} \nonumber \] \[\begin{align*} ΔS°&=ΔS^\circ_{298}(\ce{H2O}(g))−ΔS^\circ_{298}(\ce{H2O}(l)) \nonumber\\ &=\mathrm{188.8\: J/K⋅mol−70.0\: J/K⋅mol=118.8\: J/K⋅mol} \nonumber \end{align*} \nonumber \] \[T=\dfrac{ΔH°}{ΔS°}=\mathrm{\dfrac{44.01×10^3\:J/mol}{118.8\:J/K⋅mol}=370.5\:K=97.3\:°C} \nonumber \] The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Tables T1 or T2.). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point. Use the information in Tables T1 or T2 to estimate the boiling point of CS . 313 K (accepted value 319 K).  The free energy change for a process may be viewed as a measure of its driving force. A negative value for Δ represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When Δ is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium). In the chapter on equilibrium the , , was introduced as a convenient measure of the status of an equilibrium system. Recall that is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When is lesser than the equilibrium constant, , the reaction will proceed in the forward direction until equilibrium is reached and = . Conversely, if > , the process will proceed in the reverse direction until equilibrium is achieved. The free energy change for a process taking place with reactants and products present under nonstandard conditions, Δ , is related to the standard free energy change, Δ °, according to this equation: \[ΔG=ΔG°+RT\ln Q \label{eq10A} \] is the gas constant (8.314 J/K mol), is the kelvin or absolute temperature, and is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in Example \(\Page {1}\). What is the free energy change for the process shown here under the specified conditions? = 25 °C, \(P_{\ce{N2}}=\mathrm{0.870\: atm}\), \(P_{\ce{H2}}=\mathrm{0.250\: atm}\), and \(P_{\ce{NH3}}=\mathrm{12.9\: atm}\) \[\ce{2NH3}(g)⟶\ce{3H2}(g)+\ce{N2}(g) \hspace{20px} ΔG°=\mathrm{33.0\: kJ/mol} \nonumber \] Equation \ref{eq10A} relates free energy change to standard free energy change and reaction quotient and may be used directly: \[\begin{align*} ΔG&=ΔG°+RT\ln Q \\[4pt] &=\mathrm{33.0\:\dfrac{kJ}{mol}+\left(8.314\:\dfrac{J}{mol\: K}×298\: K×\ln\dfrac{(0.250^3)×0.870}{12.9^2}\right)}\\[4pt] &=\mathrm{9680\:\dfrac{J}{mol}\:or\: 9.68\: kJ/mol} \end{align*} \nonumber \] Since the computed value for Δ is positive, the reaction is nonspontaneous under these conditions. The reaction will proceed in the reverse direction to reach equilibrium. Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions? Δ = −47 kJ; yes, the reaction proceeds in the forward direction, as written, to reach equilibrium. For a system at equilibrium, = and Δ = 0, and the Equation \ref{eq10A} may be written as \[ \underbrace{0=ΔG°+RT\ln K}_{\text{at equilibrium}} \nonumber \] \[ΔG°=−RT\ln K \label{eq4A} \] or alternatively \[K=e^{−\frac{ΔG°}{RT}} \label{eq4B} \] This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table \(\Page {1}\). Given that the standard free energies of formation of Ag ( ), Cl ( ), and AgCl( ) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, , for AgCl. The reaction of interest is the following: \[\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ag+,Cl- ]} \nonumber \] The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products: \[ \begin{align*} ΔG° =ΔG^\circ_{298} &=[ΔG^\circ_\ce{f}(\ce{Ag+}(aq))+ΔG^\circ_\ce{f}(\ce{Cl-}(aq))]−[ΔG^\circ_\ce{f}(\ce{AgCl}(s))] \\[4pt] &=[77.1\: kJ/mol−131.2\: kJ/mol]−[−109.8\: kJ/mol] \\[4pt] &=55.7\: kJ/mol \end{align*} \nonumber \] The equilibrium constant for the reaction may then be derived from its standard free energy change via Equation \ref{eq4B}: \[\begin{align*} K_\ce{sp}&=e^{−\dfrac{ΔG°}{RT}}=\exp\left(−\dfrac{ΔG°}{RT}\right) \\[4pt] &=\mathrm{\exp\left(−\dfrac{55.7×10^3\:J/mol}{8.314\:J/mol⋅K×298.15\:K}\right)}\\&=\mathrm{\exp(−22.470)=e^{−22.470}=1.74×10^{−10}} \end{align*} \nonumber \] Use the thermodynamic data provided in Tables T1 or T2 to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C. \[\ce{NO}_{2(g)} \rightleftharpoons \ce{N_2O}_{4(g)} \nonumber \] K = 6.9 To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of ), equilibrium is established when the system’s free energy is minimized (Figure \(\Page {3}\)). If a system is present with reactants and products present in nonequilibrium amounts ( ≠ ), the reaction will proceed spontaneously in the direction necessary to establish equilibrium. Gibbs free energy ( ) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for Δ indicates that the process will proceed in the forward direction to reach equilibrium; a positive Δ indicates that the process will proceed in the reverse direction to reach equilibrium ; and a Δ of zero indicates that the system is at equilibrium. A negative value for Δ means that the reaction is product-favored at equilibrium.  A positive value for Δ means that the reaction is reactant-favored at equilibrium. If Δ equals 0 (a rare occurrence), the reaction has roughly equal amounts of reactants and products at equilibrium.A number of approaches to the computation of free energy changes are possible.

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

Diastereomers are stereoisomers that are not related as object and mirror image and are not . Unlike enatiomers which of each other and , diastereomers are of each other and . Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities. They have stereocenters. It is easy to mistake between diasteromers and enantiomers. For example, we have four steroisomers of 3-bromo-2-butanol. The four possible combination are SS, RR, SR and RS (Figure 1). One of the molecule is the enantiomer of its mirror image molecule and diasteromer of each of the other two molecule (SS is enantiomer of RR and diasteromer of RS and SR). SS's mirror image is RR and they are not superimposable, so they are enantiomers. RS and SR are not mirror image of SS and are not superimposable to each other, so they are diasteromers. Figure 1 Tartaric acid, C H O is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (figure 2). (R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius. Figure 2 To identify meso, meso compound is superimposed on its mirror image, and has an internal plane that is symmetry (figure 3). Meso-tartaric acid is achiral and optically unactive. Identify which of the following pair is enantiomers, diastereomers or meso compounds.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/15%3A_Benzene_and_Aromaticity

After you have completed Chapter 15, you should be able to In Chapter 3, we identified an aromatic compound as being a compound which contains a benzene ring (or phenyl group). It is now time to define aromaticity in a more sophisticated manner. In this chapter, we discuss the stability of benzene and other aromatic compounds, explaining it in terms of resonance and molecular orbital theory. You will study the nomenclature of aromatic compounds and the Hückel (4 + 2) rule for predicting aromaticity. The chapter concludes with a brief summary of the spectroscopic properties of arenes.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/16%3A_Thermodynamics/16.1%3A_Spontaneity

In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system. Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A is one that occurs naturally under certain conditions. A , on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze. The spontaneity of a process is correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure \(\Page {1}\)). As another example, consider the conversion of diamond into graphite (Figure \(\Page {2}\)). \[\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1} \] The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be but under ambient conditions. As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure \(\Page {3}\)). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero. \[ \begin{align} w&=−PΔV \\[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align} \] Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process. \[ \begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \\[4pt] &=0+0=0 \label{Eq3}\end{align} \] The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask). Now consider two objects at different temperatures: object X at temperature and object Y at temperature , with > (Figure \(\Page {4}\)). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y. \[q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4} \] From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a . As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy. Describe how matter and energy are redistributed when the following spontaneous processes take place: Describe how matter and energy are redistributed when you empty a canister of compressed air into a room. This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings. Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging. 

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/07%3A_Solids_and_Liquids/7.05%3A_Changes_of_State

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 terms in the context of this topic. A given substance will exist in the form of a solid, liquid, or gas, depending on the temperature and pressure. In this unit, we will learn what common factors govern the preferred state of matter under a particular set of conditions, and we will examine the way in which one phase gives way to another when these conditions change. Earlier in the morning, the droplets of water in Figure \(\Page {1}\) were tiny crystals of ice, but even though the air temperature is still around 0°C and will remain so all day, the sun's warmth has rendered them into liquid form, bound up by surface tension into reflective spheres. By late afternoon, most of the drops will be gone, their H O molecules now dispersed as a tenuous atmospheric gas. Solid, liquid, and gas — these are the basic three states, or , in which the majority of small-molecule substances can exist. At most combinations of pressure and temperature, only one of these phases will be favored; this is the phase that is most under these conditions. A proper explanation of why most substances have well-defined melting and boiling points needs to invoke some principles of thermodynamics and quantum mechanics. A full explanation of this would go beyond the scope of what most students who see this lesson are familiar with, but the following greatly over-simplified explanation should convince you that it is something more than black magic. All atoms and molecules at temperatures above absolute zero possess thermal energy that keeps them in constant states of motion. A fundamental law of nature mandates that this energy tends to spread out and be shared as widely as possible. Within a single molecular unit, this spreading and sharing can occur by dispersing the energy into the many allowed states of motion (translation, vibration, rotation) of the molecules of the substance itself. There are a huge number of such states, and they are quantized, meaning that they all require different amounts of thermal energy to come into action. Temperature is a measure of the intensity of thermal energy, so the higher the temperature, the greater will be the number of states that can be active, and the more extensively will the energy be dispersed among these allowed states. In solids, the molecular units are bound into fixed locations, so the kinds of motion (and thus the number of states) that can be thermally activated is relatively small. Because the molecules of solids possess the lowest potential energies, solids are the most stable states at low temperatures. At the other extreme are gas molecules which are not only free to vibrate and rotate, but are in constant translational motion. The corresponding number of quantum states is hugely greater for gases, providing a nearly-endless opportunity to spread energy. But this can only happen if the temperature is high enough to populate this new multitude of states. Once it does, the gaseous state wins out by a landslide. Escaping tendency is more formally known as . Bear in mind also that changes in state always involve changes in enthalpy and internal energy. In much the same way that tea spreads out from a tea bag into the larger space of the water in which it is immersed, molecules that are confined within a phase (liquid, solid, or gas) will tend to spread themselves (and the thermal energy they carry with them) as widely as possible. This fundamental law of nature is manifested in what we will call the of the molecules from the phase. The escaping tendency is a quantity of fundamental importance in understanding all chemical equilibria and transformations. We need not define the term in a formal way at this point. What is important for now is how we can observe and compare escaping tendencies. Think first of a : what property of the gas constitutes the best measure of its tendency to escape from a container? It does not require much reflection to conclude that the greater the of the gas, the more frequently will its molecules collide with the walls of the container and possibly find their way through an opening to the outside. What about liquids and solids? Although we think of the molecules of condensed phases as permanently confined within them, these molecules still possess some thermal energy, and there is always a chance that one that is near the surface will occasionally fly loose and escape into the space outside the solid or liquid. We can observe the tendency of molecules to escape into the gas phase from a solid or liquid by placing the substance in a closed, evacuated container connected to a manometer for measuring gas pressure (Figure \(\Page {2}\)). If we do this for water (Figure \(\Page {3}\)), the partial pressure of water in the vapor space will initially be zero ( ). Gradually, will rise as molecules escape from the substance and enter the vapor phase. But at the same time, some of the vapor molecules will "escape" back into the liquid phase ( ). But because this latter process is less favorable (at the particular temperature represented here), continues to rise. Eventually a balance is reached between the two processes ( ), and eventually stabilizes at a fixed value that depends on the substance and on the temperature and is known as the , or simply as the “vapor pressure” of the liquid or solid. Note carefully that if the container is left open to the air, it is unlikely that many of the molecules in the vapor phase will return to the liquid phase. They will simply escape from the entire system and the partial pressure of water vapor will never reach ; the liquid will simply without any kind of equilibrium ever being achieved. The escaping tendency of molecules from a phase always increases with the temperature; therefore the vapor pressure of a liquid or solid will be greater at higher temperatures. As Figure \(\Page {4}\) shows, the variation of the vapor pressure with the temperature is not linear. It's important that you be able to interpret vapor pressure plots such as the three shown here. Take special note of how boiling points can be found from these plots. You will recall that the is the temperature at which the liquid is in equilibrium with its vapor at a partial pressure of 1 atm (760 torr). Thus the intercepts of each curve with the blue dashed 760-torr line indicate the normal boiling points of each liquid. Similarly, you can easily estimate the boiling points these liquids would have in Denver, Colorado where the atmospheric pressure is 630 torr by simply constructing a horizontal line corresponding to this pressure. The is the temperature at which the liquid is in equilibrium with its vapor at a partial pressure of 1 atm. This is when the vapor pressure is at atmospheric pressure. The great importance of H O in our world merits a more detailed look at its vapor pressure properties. Because the vapor pressure of water varies greatly over the range of temperatures in which the liquid can exist (Figure \(\Page {5}\)). The larger plot in Figure \(\Page {5}\) covers the lowest temperatures, while the inset shows the complete range of pressure values. Note particularly that The vapor pressure of water at 22°C is about 20 torr, or around 0.026 atm (2.7 kPa). This is the partial pressure of H O that will be found in the vapor space within a closed container of water at this temperature; the air in this space is said to be with water vapor. Humid air is sometimes described as "heavy", but this is misleading; the average molar mass of dry air is 29, but that of water is only 18, so humid air is actually less dense. The feeling of "heaviness" probably relates to the reduced ability of perspiration to evaporate in humid air. In ordinary air, the partial pressure of water vapor is normally less than its saturation or equilibrium value. The ratio of the partial pressure of H O in the air to its (equilibrium) vapor pressure at any given temperature is known as the . Water enters the atmosphere through evaporation from the ocean and other bodies of water, and from water-saturated soils. The resulting vapor tends to get dissipated and diluted by atmospheric circulation, so the relative humidity rarely reaches 100 percent. When it does and the weather is warm, we are very uncomfortable because vaporization of water from the skin is inhibited; if the air is already saturated with water, then there is no place for our perspiration to go, other than to drip down our face. Because the vapor pressure increases with temperature, a parcel of air containing a fixed partial pressure of water vapor will have a larger relative humidity at low temperatures than at high temperatures. Thus when cold air enters a heated house, its water content remains unchanged but the relative humidity drops. In climates with cold winters, this promotes increased moisture loss from house plants and from mucous membranes, leading to wilting of the former and irritation of the latter. The vapor pressure of water is 3.9 torr at –2°C and 20 torr at 22°C. What will be the relative humidity inside a house maintained at 22°C when the outside air temperature is –2°C and the relative humidity is 70%? At 70 percent relative humidity, the partial pressure of the –2° air is (0.7 × 3.9 torr) = 2.7 torr. When this air enters the house, its relative humidity will be (2.7 torr)/(20 torr) = 0.14 or . In the evening, especially on clear nights, solid objects (even spider webs!) lose heat to the sky more rapidly than does the air. It is often important to know what temperature such objects must drop to so that atmospheric moisture will condense out on them (Figure \(\Page {1}\)). The is the temperature at which the relative humidity is 100 percent — that is, the temperature at which the vapor pressure of water becomes equal to its partial pressure at a given [higher] temperature and relative humidity. For water to condense directly out of the atmosphere as rain, the air must be the dew point, but this is not of itself the only requirement for the formation of rain, as we will see shortly. Many solid salts incorporate water molecules into their crystal lattices; the resulting compounds are known as . These solid hydrates possess definite vapor pressures that correspond to an equilibrium between the hydrated and anhydrous compounds and water vapor. For example Strontium chloride hexahydrate: \[ \ce{SrCl2 \cdot 6H2O(s) \rightarrow SrCl2(s) + 6H2O(g)} \label{\(\Page {1}\)}\] The vapor pressure of this hydrate is 8.4 torr at 25°C. Only at this unique partial pressure of water vapor can the two solids coexist at 25°C. If the partial pressure of water in the air is greater than 8.4 torr, a sample of anhydrous SrCl will absorb moisture from the air and change into the hydrate. In fact, when fully hydrates, water is responsible for 40% of the hydrate's mass. What will be the relative humidity of air in an enclosed vessel containing solid SrCl ·6H O at 25°C? What fraction of the vapor pressure of water at this temperature (23.8 torr) is the vapor pressure of the hydrate (8.4 torr)? Expressed in percent, this is the relative humidity. If the partial pressure of H O in the air is less than the vapor pressure of the hydrate, the latter will tend to lose moisture and revert back to its anhydrous form. This process is sometimes accompanied by a breakup of the crystal into a powdery form, an effect known as . Evaporation and boiling of a liquid, and condensation of a gas (vapor) are such ordinary parts of our daily life that we hardly give them a thought. Every time we boil water to make a pot of tea and see the cloud of steam above the teapot, we are observing this most common of all phase changes. How can we understand these changes in terms of vapor pressure? Figure \(\Page {6}\) plots the vapor-pressure as a function of temperature can represent water or any other liquid. When we say this is a vapor-pressure plot, we mean that each point on the curve represents a combination of temperature and vapor pressure at which the liquid (green) and the vapor (blue) can coexist. Thus at the normal boiling point, defined as the temperature at which the vapor pressure is 1 atm, the state of the system corresponds to the point labeled . Suppose that we select an arbitrary point at a temperature and pressure at which only the gaseous state is stable. We then decrease the temperature so as to move the state point toward point in the liquid region. When the state point falls on the vapor pressure line, the two phases can coexist and we would expect some liquid to condense. Once the state point moves to the left of the vapor pressure line, the substance will be entirely in the liquid phase. This is supposedly what happens when "steam" (actually tiny water drops) forms above a pot of boiling water. The reverse process should work the same way: starting with a temperature in the liquid region, nothing happens until we reach the vapor pressure line, at which point the liquid begins to change into vapor. At higher temperatures, only the vapor remains. This is the , but it is not complete. The is that a vapor will generally condense to a liquid at the boiling point (also called the condensation point or dew point), and a liquid will generally boil at its boiling point. The reason for the discrepancy is that the vapor pressure, as we normally use the term and as it is depicted by the liquid-vapor line on the phase diagram, refers to the partial pressure of vapor in equilibrium with a liquid whose surface is reasonably flat, as it would be in a partially filled container. In a drop of liquid or in a bubble of vapor within a liquid, the surface of the liquid is not flat, but curved. For drops or bubbles that are of reasonable size, this does not make much difference, but these drops and bubbles must grow from smaller ones, and these from tinier ones still. Eventually, one gets down to the primordial drops and bubbles having only a few molecular dimensions, and it is here that we run into a problem: this is the problem of — the formation and growth of the first tiny drop (in the vapor) or of a bubble (in a liquid). The vapor pressure of a liquid is determined by the attractive forces that act over a 180° solid angle at the surface of a liquid. In a very small drop, the liquid surface is curved in such a way that each molecule experiences fewer nearest-neighbor attractions than is the case for the bulk liquid. The outermost molecules of the liquid are bound to the droplet less tightly, and the drop has a larger vapor pressure than does the bulk liquid. If the vapor pressure of the drop is greater than the partial pressure of vapor in the gas phase, the drop will evaporate. Thus it is highly unlikely that a droplet will ever form within a vapor as it is cooled. A bubble, like a drop, must start small and grow larger, but there is a diffculty here that is similar to the one with bubbles. A bubble is a hole in a liquid; molecules at the liquid boundary are curved inward, so that they experience nearest-neighbor attractions over a solid angle greater than 180°. As a consequence, the vapor pressure of the liquid facing into a bubble is always less than that of the bulk liquid at the same temperature. When the bulk liquid is at its boiling point (that is, when its vapor pressure is 1 atm), the pressure of the vapor within the bubble will be than 1 atm, so the bubble will tend to collapse. Also, since the bubble is formed within the liquid, the hydrostatic pressure of the overlaying liquid will add to this effect. For both of these reasons, a liquid will not boil until the temperature is raised slightly above the boiling point, a phenomenon known as . Once the boiling begins, it will continue to do so at the liquid's proper boiling point. These plots show how, in the case of water, the vapor pressure of a very small bubble or drop varies with its radius of curvature; the quantity being plotted is the ratio of the actual vapor pressure to , the vapor pressure of a flat surface. If the tiniest of drops are destined to self-destruct, why do vapors ever condense (e.g., why does it rain)? In the region of the atmosphere where rain forms there are large numbers of solid particles, mostly of microscopic size. Some of these are particles of salt produced by evaporation of spray from the ocean surface. Many condensation nuclei are of biological origin; these include bacteria, spores, and particles of ammonium sulfate. There is volcanic and meteor dust, and of course there is dust and smoke due to the activities of humans. These particles tend to adsorb water vapor, and some may even dissolve to form a droplet of concentrated solution. In either case, the vapor pressure of the water is reduced below its equilibrium value, thus stabilizing the aggregate until it can grow to self-sustaining size and become fog, rain, or snow. This, by the way, is why fog is an irritant to the nose and throat; each fog droplet carries within it a particle of dust or (in air polluted by the burning of sulfur-containing fossil fuels) a droplet of sulfuric acid, which it effciently deposits on your sensitive mucous membranes. If you own a car which is left outside on a foggy night, you may have noticed how dirty the windshield is in the morning. What is the difference between the evaporation and boiling of a liquid? When a liquid evaporates at a temperature below its boiling point, the molecules that enter the vapor phase do so directly from the surface. When a liquid boils, bubbles of vapor form in the interior of the liquid, and are propelled to the surface by their lower density (buoyancy). As they rise, the diminishing hydrostatic pressure causes the bubbles to expand, reducing their density (and increasing their buoyancy) even more. But as we explained above, getting that first bubble to form and survive is often sufficiently difficult that liquids commonly superheat before they begin to boil. If you have had experience in an organic chemistry laboratory, you probably know this as “bumping”, and have been taught to take precautions against it. In large quantities, superheated liquids can be very dangerous, because the introduction of an impurity (such as release of an air bubble from the container surface) or even a mechanical disturbance can trigger nucleation and cause boiling to occur suddenly and almost explosively ( \(\Page {1}\)). Many people have been seriously burned after attempting to boil water in a microwave oven, or after having added powdered material such as instant coffee to such water. When water is heated on a stove, the bottom of the container superheats only the thin layer of water immediately in contact with it, producing localized "microexplosions" that you can hear just before regular smooth boiling begins; these bubbles quickly disperse and serve as nucleation centers for regular boiling. In a microwave oven, however, the energy is absorbed by the water itself, so that the entire bulk of the water can become superheated. If this happens, the slightest disturbance can produce an explosive flash into vapor. Some solids have such high vapor pressures that heating leads to a substantial amount of direct vaporization even before the melting point is reached. This is the case for solid iodine, for example. I melts at 115°C and boils at 183°C, is easily sublimed at temperatures around 100°C. Even ice has a measurable vapor pressure near its freezing point, as evidenced by the tendency of snow to evaporate in cold dry weather. There are other solids whose vapor pressure overtakes that of the liquid before melting can occur. Such substances sublime without melting; a common example is solid carbon dioxide (“Dry Ice”) at 1 atm (see the CO phase diagram below). The temperatures and pressures at which a given phase of a substance is stable (that is, from which the molecules have the lowest escaping tendency) is an important property of any substance. Because both the temperature and pressure are factors, it is customary to plot the regions of stability of the various phases in - coordinates, as in this generic for a hypothetical substance. Because pressures and temperatures can vary over very wide ranges, it is common practice to draw phase diagrams with non-linear or distorted coordinates. This enables us to express a lot of information in a compact way and to visualize changes that could not be represented on a linearly-scaled plot. It is important that you be able to interpret a phase diagram, or alternatively, construct a rough one when given the appropriate data. Take special note of the following points: The best way of making sure you understand a phase diagram is to imagine that you are starting at a certain temperature and pressure, and then change just one of these parameters, keeping the other constant. You will be traversing a horizontal or vertical path on the phase diagram, and there will be a change in state every time your path crosses a line. Of special importance is the horizontal path (shown by the blue line on the diagram above) corresponding to a pressure of 1 atmosphere; this line defines the of a substance. Notice the following features for the phase diagram of water (Figure \(\Page {11}\)): Dry ice, solid carbon dioxide, is widely used as a refrigerant and the phase diagram in Figure \(\Page {12}\) shows why it is “dry”. The triple point pressure is at 5.11 atm, so below this pressure, liquid CO cannot exist; the solid can only sublime directly to vapor. Gaseous carbon dioxide at a partial pressure of 1 atm is in equilibrium with the solid at 195K (−79 °C, ); this is the of carbon dioxide. The surface temperature of dry ice will be slightly less than this, since the partial pressure of CO in contact with the solid will usually be less than 1 atm. Notice also that the critical temperature of CO is only 31°C. This means that on a very warm day, the CO in a fire extinguisher will be entirely vaporized; the vessel must therefore be strong enough to withstand a pressure of 73 atm. This view of the carbon dioxide phase diagram employs a logarithmic pressure scale and thus encompasses a much wider range of pressures, revealing the upper boundary of the fluid phase (liquid and supercritical). (CO above its critical temperature) possesses the solvent properties of a liquid and the penetrating properties of a gas; one major use is to remove caffeine from coffee beans. Elemental iodine, I , forms dark gray crystals that have an almost metallic appearance. It is often used in chemistry classes as an example of a solid that is easily sublimed; if you have seen such a demonstration or experimented with it in the lab, its phase diagram might be of interest. The most notable feature of iodine's phase behavior is the small difference (less than a degree) between the temperatures of its triple point and melting point . Contrary to the impression many people have, there is nothing really special about iodine's tendency to sublime, which is shared by many molecular crystals including ice and naphthalene ("moth balls".) The vapor pressure of iodine at room temperature is really quite small — only about 0.3 torr (40 Pa).The fact that solid iodine has a strong odor and is surrounded by a purple vapor in a closed container is mainly a consequence of its strong ability to absorb green light (this leaves blue and red which make purple) and the high sensitivity of our noses to its vapor. Sulfur exhibits a very complicated phase behavior that has puzzled chemists for over a century; what you see here is the greatly simplified phase diagram shown in most textbooks. The difficulty arises from the tendency of S molecules to break up into chains (especially in the liquid above 159°C) or to rearrange into rings of various sizes (S to S ). Even the vapor can contain a mixture of species S through S . The phase diagram of sulfur contains a new feature: there are two solid phases, and . The names refer to the crystal structures in which the S molecules arrange themselves. This gives rise to , indicated by the numbers on the diagram. ulfur (\(S_8\) (hint there are several correct answers)? When rhombic sulfur (the stable low-temperature phase) is heated slowly, it changes to the monoclinic form at 114°C, which then melts at 119°. But if the monoclinic form is heated rapidly the molecules do not have time to rearrange themselves, so the rhombic arrangement persists as a metastable phase until it melts at 119-120°. We tend to think of the properties of substances as they exist under the conditions we encounter in everyday life, forgetting that most of the matter that makes up our world is situated inside the Earth, where pressures are orders of magnitude higher (Figure \(\Page {15}\)). Geochemists and planetary scientists need to know about the phase behavior of substances at high temperatures and pressures to develop useful models to test their theories about the structure and evolution of the Earth and of the solar system. What ranges of temperatures and pressures are likely to be of interest — and more importantly, are experimentally accessible? Figure \(\Page {16}\) shows several scales (all of which, please note, are logarithmic) that cover respectively the temperature range for the universe; the low temperatures of importance to chemistry (note the green line indicating the temperatures at which liquid water can exist); the higher temperatures, showing the melting and boiling points of several elements for reference. The highest temperatures that can be produced in the laboratory are achieved (but only for very short time intervals) by light pulses from laser or synchrotron radiation. The study of is limited by the laws of physics that prohibit reaching absolute zero. But the fact that there is no limit to how close one can approach 0 K has encouraged a great deal of creative experimentation. The study of matter at high pressures is not an easy task. The general techniques were pioneered between 1908-1960 by P.W. Bridgeman of Harvard University, whose work won him the 1946 Nobel Prize in physics. The more recent development of the diamond anvil cell has greatly extended the range of pressures attainable and the kinds of observations that can be made. Shock-wave techniques have made possible the production of short-lived pressures in the tPa range. High pressure laboratory studies have revealed that many molecular substances such as hydrogen and water change to solid phases having melting points well above room temperature at very high pressures; there is a solid form of ice that remains frozen even at 100°C. At still higher pressures, many of these substances become metals. It is believed that much of the inner portion of the largest planets consists of metallic hydrogen — and, in fact, that all substances can become metallic at sufficiently high pressures. Figure \(\Page {19}\): Phase diagram of carbon: diamond and graphite Graphite is the stable form of solid carbon at low pressures; diamond is only stable above about 10 atm. But once it is in this form, the rate at which diamond converts back to graphite is immeasurably slow under ordinary environmental conditions; there is simply not enough thermal energy available to break all of those carbon-carbon bonds. So the diamonds we admire in jewelry and pay dearly for are said to be To transform graphite into diamond at a reasonable rate, a pressure of 200,000 atm and a temperature of about 4000 K would be required.Since no apparatus can survive these conditions, the process, known as is (HPHT) is carried out commercially at 70,000 atm and 2300 K in a solution of molten nickel, which also acts as a catalyst. Traces of Ni in the finished product serve to distinguish synthetic diamonds from natural ones. However, most synthetic diamonds are too small (only a few millimeters) and too flawed for gem quality, and are used mainly to fabricate industrial grinding and cutting tools. Figure \(\Page {20}\): Phase diagram of carbon: diamond and graphite More recently, thin diamond films have become important for engineering applications and semiconductor fabrication. These are most commonly made by condensation of gaseous carbon onto a suitable substrate (chemical vapor deposition, CVD). The conditions under which synthetic diamonds are made are depicted on the above phase diagram from Bristol University. Helium is unique in that quantum phenomena, which normally apply only to tiny objects such as atoms and electrons, extend to and dominate its macroscopic properties. A glance at the phase diagram of \(\ce{^4He}\) reveals some of this quantum weirdness. The main points to notice are: The low mass of the He atoms and their close confinement in the solid provides them with a very high (the principle in action!) that allows them to vibrate with such amplitude that they overcome the dispersion forces that would otherwise hold the solid together, thus keeping the atoms too separated to form a solid. Only by applying a high pressure (25 atm) can this effect be overcome. We usually need quantum theory only to describe the properties of tiny objects such as electrons, but with liquid He-II, it extends to the macroscopic scale of the bulk liquid. \(\ce{^4He}\) atoms (99.99+ percent of natural helium) are , which means that at low temperatures, they can all occupy the same quantum state (all other normal atoms are and are subject to the ). Objects that occupy the same quantum state all possess the same momentum. Thus when one atom moves, they all move together. In a sense, this means that the entire bulk of liquid He-II acts as a single entity. This property, known as , gives rise to a number of remarkable effects, most notably: In liquid helium-II, only about 10% of the atoms are in such a state, but it is enough to give the liquid some of the weird properties of a quantum liquid. He exhibits similar properties, but owing to its low natural abundance, it was not extensively studied until the 1940s when large amounts became available as a byproduct of nuclear weapons manufacture. Finally, in 1996, its superfluidity was observed at a temperature of 2 nK. Although He atoms are fermions, only those that pair up (and thus assume the properties of bosons) give rise to superfluidity. Water, like most other substances, exhibits many solid forms at higher pressures (Figure \(\Page {22}\)). So far, fifteen distinct ice phases have been identified; these are designated by Roman numerals ice-I through ice-XV. Ice-I can exist in two modifications; the crystal lattice of ice-Ic is cubic, while that of ice-lh is hexagonal. The latter corresponds to the ordinary ice that we all know. It's interesting to note that several high-pressure phases of ice can exist at temperatures in excess of 100°C.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.07%3A_Corrosion

An important aspect of the use of some metals, particularly of iron, is the possibility of corrosion. It is estimated that about one-seventh of all iron production goes to replace the metal lost to corrosion. Rust is apparently a hydrated form of iron(III)oxide. The formula is approximately Fe O •\(\tfrac{\text{3}}{\text{2}}\)H O, although the exact amount of water is variable. (Note that this is about halfway between iron(III) hydroxide, Fe(OH) or ½{Fe O •3H O], and anhydrous Fe O ). Rusting requires both oxygen and water, and it is usually sped up by acids, strains in the iron, contact with less-active metals, and the presence of rust itself. In addition, observation of a rusted object, such as an iron nail from an old wooden building, shows that rust will deposit in one location (near the head of the nail) while the greatest loss of metallic iron will occur elsewhere (near the point). These facts suggest that the mechanism of rusting involves a . The half-equations involved are \[\text{2Fe}(S) \rightarrow \text{2Fe}^{2+}(aq) + \text{4}e^-\label{1} \] \[\text{4}e^- + \text{4H}^+(aq) + \text{O}_2(g) \rightarrow \text{2H}_2\text{O}\label{2} \] yielding the full reaction: \[\text{2Fe}(s) + \text{4H}^+(aq) + \text{O}_2(g) \rightarrow \text{2Fe}^{2+}(aq) + \text{2H}_2\text{O}\label{3} \] Once Fe (aq) is formed, it can migrate freely through the aqueous solution to another location on the metal surface. At that point the iron can precipitate: \[\text{4Fe}(s) + \text{O}_2(g) + \text{7 H}_2\text{O}(l) \rightarrow \text{2Fe}_2\text{O}_3 \cdot \frac{3}{2} \text{H}_2\text{O}(s) + \text{8H}^+(aq) \nonumber \] Hydrogen ions liberated by this reaction are then partially consumed by Equation \(\ref{2}\). The electrons required for half-equation \(\ref{2}\) are supplied from Equation \(\ref{1}\) via metallic conduction through the iron or by ionic conduction if the aqueous solution contains a significant concentration of ions. Thus iron rusts faster in contact with salt water than in fresh. The mechanism proposed in the preceding paragraph implies that some regions of the iron surface become , i.e., that reduction of oxygen to water occurs there. Other locations are anodic; oxidation of Fe to Fe occurs. The chief way in which such regions may be set up depends on restriction of oxygen supply, because oxygen is required for the cathodic reaction shown in Equation \(\ref{2}\). In the case of the iron nail, for example, rust forms near the head because more oxygen is available. Most of the loss of metal takes place deep in the wood, however, near the point of the nail. At this location Equation \(\ref{1}\) but not \(\ref{2}\) can occur. A similar situation occurs when a drop of moisture adheres to an iron surface (Figure \(\Page {1}\)). Pitting occurs near the center of the drop, while hydrated iron(III) oxide deposits near the edge. A second way in which anodic and cathodic regions may be set up involves the presence of a second metal which has a greater attraction for electrons (is less easily oxidized) than iron. Such a metal can drain off electrons left behind in the iron when Fe dissolves. This excess of electrons makes the less-active metal an ideal site for Equation \(\ref{2}\), and so a cell is set up at the intersection of the metals. Rust may actually coat the surface of the less-active metal while pits form in the iron. The most important technique for rust prevention is simply to exclude water and oxygen by means of a protective coating. This is the principle behind oiling, greasing, painting, or metal plating of iron. The coating must be complete, however, or rusting may be accelerated by exclusion of oxygen from part of the surface. This is especially true when iron is coated with a less-active metal such as tin. Even a pinhole in the coating on a tin can will rust very quickly, since the tin becomes cathodic due to its larger electrode potential and to the oxygen exclusion from the iron beneath. A second technique involves bringing the iron object in contact with a more active metal. This is called because the more active metal donates electrons to the iron, strongly inhibiting Equation \(\ref{1}\). Both cathodic protection and a surface coating are provided by , a process in which zinc is plated onto steel electrolytically or by dipping in the molten metal. Like many other metals, zinc is self-protective—it reacts with oxygen and carbon dioxide from air to form an adherent impervious coating of zinc hydroxycarbonate, Zn (OH) CO . Should there be a scratch in the zinc plate, the iron still cannot rust because zinc will be preferentially oxidized. The hydroxycarbonate formed will then cover the opening, preventing further contact of oxygen with the iron or zinc. A third technique applies to situations (such as an automobile radiator) where aqueous solutions are in contact with the iron. Corrosion inhibitors include chromate salts and organic compounds such as tribntylamine, (C H ) N. Chromates apparently form an impervious coating of FeCrO ( ) as soon as any iron is oxidized to iron(II). Tributylamine, a derivative of ammonia, reacts with organic acids formed by decomposition of antifreeze at the high temperatures of an automobile engine. The tributylammonium salts produced are insoluble and coat the inside of the cooling system. Thus tributylamine neutralizes acid which would accelerate corrosion and provides a surface coating as well.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles

When you have completed Chapter 24, you should be able to Amines are the first nitrogen-containing compounds that we study in detail in this course. We begin the chapter with an explanation of the differences in structure among primary, secondary and tertiary amines. We explain the nomenclature of aliphatic and arylamines, and examine the structure and bonding of these compounds, relating these features to their physical properties and basicity. We describe the use of amines to resolve racemic mixtures of chiral carboxylic acids. Amines may be prepared by a number of different synthetic methods. We describe each of these methods and assess the relative merits of each. After a description of the reactions of aliphatic amines, we devote sections to a discussion of the use of tetraalkylammonium salts as phase-transfer agents. The chapter concludes with a summary of the spectroscopic properties of amines.

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

are catalysts that drive reaction rates forward. Most catalysts, but not all, are made up of amino acid chains called proteins that accelerate the rate of reactions in chemical systems. The functionality of a catalyst depends on how the proteins are folded, what they bind to, and what they react with. For protein-based catalysts, amino acid polarization lies at the core of catalytic activity. In chemistry, a catalyst is a chemical that drives a reaction forward. Catalysts lower the , which is the amount of energy required for reactants to form products (Figure 1). Catalysts also lower the kinetic barrier, which is needed to drive a reaction forward and backward. A certain amount of energy contained in the molecules is required when the two molecules react together to form a product. If the two molecules do not have enough energy to react, then no product is produced. By lowering the , a catalyst allows the molecules to gain sufficient energy to overcome the barrier and form products. Catalysts increase the rates of the forward and backward reaction ( to denote the rate of the forward reaction and to denote the rate of the backward reaction). convert reactants to products, whereas convert products to reactants. Typically, the conversion of products to reactants requires more energy, measured in Gibbs energy. Recall that the difference in energy between the products and the reactants is measured as ΔG (Gibbs energy). It is very important to note that catalysts do not change the free energy, G, they simply affect the speed of the reaction. Catalysts are very beneficial in biological systems because they drive individual reactions forward. Our bodies are a vast combination of redox reactions. Heat may drive a reaction forward. For example, when you catch a fever, your body raises its temperature to drive reactions forward in your body, dissipating energy in the form of heat. Note that the body raises its temperature to drive reactions forward. This extra energy drives your immune system forward to get rid of the germs faster. Often, life does not want whole systems to be driven forward; instead, the biological system merely wants to produce a little extra product or a small amount of excess reactant of one reaction. It would be a waste of energy to constantly have our body hotter than needed, and we would probably die much faster. Therefore, our body uses . These reaction specific catalysts are required to keep our body alive. In this section, we will talk about the chemistry of inorganic and organic biological catalysts, also called , and how their composition is evaluated in medicine. What makes an amino acid polar or nonpolar? What level of polarity affects an amino acid's hydrophobicity or hydrophilic characteristics? This is the basic structure of an : Figure 3 Notice that the basic structure carries an amine group (NH ), and a carboxyl functional group (CO H). The general formula for an amino acid is H N COOH, which denotes the order in which the hydrogen and carbon atoms are bonded. The 20 amino acids have this same general structure. What makes them different is the R-linked side chain. Images of the 20 different amino acids. Amino acids differ in their electronegativity in the R groups, causing differences in their hydrophobicity. The side chains denote whether an amino acid is: Recall that the more electronegative an R side chain is compared with its amine and carboxyl, the more polar the amino acid. In general, side chains with hydrocarbon alkyl groups (C H ), or are . Examples: , Leucine, Isoleucine The number of alkyl groups affects the polarity. The more C H groups, the more nonpolar. What makes an amino acid more polar? What makes an amino acid basic? What makes an amino acid acidic? Enzyme and Substrate Chemistry can be described biologically. Enzymes provide the particular substrate with an active site, which forms an enzyme-substrate complex, which is necessary for its catalyst properties and the formation of products. The Rate of an Enzyme-Substrate reaction is proportionally related to the concentrations of both the enzyme and the substrate. As the concentration of either decreases or increases, so does the reaction rate. However, there are certain exceptions to this rule of proportionality. The reaction rate disregards or is independent of the concentration of the substrate when it is very high. Thus, the rate of the reaction for an enzyme-catalyzed reaction with a high substrate concentration follows a zero rate equation: Rate of Reaction=K. In terms of a normal proportional reaction rate to the concentration level, the rate equation is considered as first order. Homogeneous catalysts interact with the reactants in the same phase (i.e: turning a substrate into a product at a faster rate). The homogeneous catalysts do not change their current states, unlike heterogeneous catalysts. By "states" we mean the phase state, which indicates either solid, liquid, or gas. If homogeneous catalysts are solid, then they will remain solid after the reaction is completed; the same is true for liquid and gas homogeneous catalysts. Although many biological enzymes are heterogeneous, there are some homogeneous enzymes that remain in their same state after the reaction, such as in an immunoassay (EIA). Heterogeneous catalysts are catalysts that speed up the rate of reactions by allowing them to occur on a solid surface. An example of a heterogeneous catalyst is a clay DNA polymer scaffolding, where the DNA's individual purines and pyrimidines link together on a clay surface to enable more secure bonding. Enzymes are composed of many amino acids that react with substrates in biological chemistry. Enzymes exist to drive the rates of reactions forward in our bodies. Without enzymes, products would not form quickly enough for our body to actually process the energy that we need. The basic reaction for any enzyme-substrate complex is this: \[ \text{Step 1}\;\;\;\;\; E+S \rightleftharpoons ES \] The enzyme-substrate complex bound together is an intermediate in a reaction, denoted by ]. \[ \text{Step 2}\;\;\;\;\; ES \rightarrow E + P \] where P stands for products, E for enzyme, and S for substrate. The rate determining step for an enzyme-substrate reaction is always the second step in which [ES] is converted into the product. The reason for this is because once the enzyme performs its duty, it is free to do more work. Once an enzyme can do more work after conversion, a reaction can go faster. The rates of enzyme-substrate reactions oscillate between and . Initially, a reaction will be first order because it will depend on the amount of substrate added. When the maximum amount of active sites are consumed, the rate of an enzyme-substrate reaction maximizes, becoming a zero order reaction in which the rate of reaction is constant. A reaction is typically denoted graphically by an asymptote, which indicates the rate limit of the reaction. When an enzyme-substrate reaction tends toward zero order, the only way to make a reaction speed up is to add more enzyme, therefore adding more active sites. If more enzyme is added to a zero order maximized reaction, then the reaction will go back to first order until either all of the active sites are taken once again or all of the substrate is converted into product, leaving an excess of empty active sites. The rate of production of the product depends on the velocity (V) of a reaction. For any enzyme-substrate reaction to go forward, the rate of product formation (or decomposition of ES) must equal the rate of formation of ES. If ES only forms and does not decompose into product, then the enzyme is not working. Enzymes that do not work are discussed later, and may be a result of faulty RNA translation from DNA, which causes the active site on an enzyme to be malformed. Because the total concentration of an enzyme can never be accurately measured, biochemists like to use \(E_0\) to denote the sum of unbound enzyme versus bound enzyme ES. A convenient constant to use when relating the rates of the forward versus reverse reactions of enzyme chemistry is \(K_M\). This constant can be derived by dividing the rates of formation of unbound E (k and k ) by the bound ES (k ). \[ K_M=\dfrac{k_{-1}+k_2}{k_1} \] where Note that this constant is always changing due to fluctuations in the rates of the forward and reverse reactions due to the concentrations of the enzyme or substrate. The equation for velocity can then be understood. \[V=\dfrac{k_2[E_0,S]}{K_M+[S]}\] If \([S]\) is low, then the value of \(K_M\) will be large, and the reaction rate on the concentration of the substrate. This reaction will be a first order reaction because there is enough enzyme to drive the reaction forward at a relatively fast rate. \[K_M>>[S]\] If the concentration of the substrate is high and all of the active sites are taken, then \(K_M\) will be sufficiently less than [S], and the reaction will tend toward its maximum rate. More enzyme will need to be added to drive this reaction faster, and the reaction will become zero order and attain an asymptote. \[K_M<<[S]\] The thermodynamics of a biological reaction are crucial. Your body temperature stays at a constant 97.5 to 98.8 degrees Fahrenheit because a higher body temperature can cause certain proteins to denature in your body. A lowering of this range will cause reactions to slow down, which also may cause death. However, slight changes above this set homeostasis can drive all of the reactions forward in your body, causing you to burn more energy, which partially escapes in the form of heat. Therefore, when you have a fever, your mother or father may have felt your forehead to see if you were warmer than usual. are the parts of enzymes that are substrate-specific. Certain enzymes will only bind to certain substrates because of a site resembling a lock-key on the surface of the enzyme. We will be taking a look at a very common enzyme family called serine protease as an example of how active site chemistry works. The serine protease family is an important enzyme for digestion, blood clotting, and fertilization. They are also the enzymes that catalyse peptide bond cleavage by attacking the carbonyl bond. Serine proteases are most famous for their specificity for substrates. They contain disulphide linkages (S--S) to keep their shape. Charged side chains are found on the outside of the enzyme, interacting with the solvent unless involved in catalysis. Let us use an enzyme called trypsin in the serine protease family. Trypsin's active site has two domains, with the active site between the two. At the center of each domain is a barrel structure. Polar regions of the structure are well hydrated. Trypsin's active site contains the amino acid sequence , His 57, (Aspartic Acid, Histidine, and Serine respectively). The numbers correspond with the actual sequence and position of the amino acids. These amino acids are found on loop regions of the two domains, and represent the charge relay system for the active site. The specificity pocket is also found in the loops of the two domains. These two domains of barrel structures are important because they provide a scaffold on which the specific amino acid bonds can interact to form the substrate-specific active site. The connection between the domains is less tight at the active site and may allow more rigid movements within the domains that may contribute to catalysis. These rigid body movements are a fundamental part of enzyme catalysis. We still have much to research on serine proteases because not much is understood about their crucial chemistry. His 57 and Asp 102 are supposed to fix the Ser 195 to a state capable of reacting with the incoming and to stabilize any intermediate formed during catalysis. His 57 acts as a strong base, abstracting the alcoholic proton of Ser 195 and moves it to the amine leaving group. The negative end of Asp 102 cancels out the positive charge developed by His 57 during the transition state. Then, the hydrolysis (adding of H O) of the acyl-enzyme releases the product. (At this link, do not pay attention to the actual reaction, just pay attention to the highlighted intermediate of acyl-enzyme. This reaction is a ) The actual reaction mechanism. ... These catalysts drive the reaction forward 1,000,000 times faster than the reaction without a catalyst. 1) What is a catalyst? A catalyst is a compound that speeds up a reaction. 2) What is an enzyme? An enzyme is a biological catalyst that speeds up reactions and interactions between molecules in biological systems. 3) What is the name of reactants that enter a substrate to form products at a faster rate? The name of the reactants that enter a substrate to form products at a faster time are called substrates. 4) In which place on the enzyme does the substrate bind (to that enzyme, specifically) to give us products at a faster rate? The name of the place on which the substrate binds is the active site of the enzyme.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.5%3A_Collision_Theory

We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates. is based on the following postulates: We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen: \[\ce{2CO}(g)+\ce{O2}(g)⟶\ce{2CO2}(g) \nonumber \] Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient quantity, the reaction is spontaneous at high temperature and pressure. The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules: \[\ce{CO}(g)+\ce{O2}(g)⟶\ce{CO2}(g)+\ce{O}(g) \nonumber \] Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure \(\Page {1}\). In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms \(\ce{(O=C=O)}\). This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction. If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. Every reaction requires a certain amount of activation energy for it to proceed in the forward direction, yielding an appropriate activated complex along the way. As Figure \(\Page {2}\) demonstrates, even a collision with the correct orientation can fail to form the reaction product. In the study of reaction mechanisms, each of these three arrangements of atoms is called a proposed or . In most circumstances, it is impossible to isolate or identify a transition state or activated complex. In the reaction between carbon monoxide and oxygen to form carbon dioxide, activated complexes have only been observed spectroscopically in systems that utilize a heterogeneous catalyst. The gas-phase reaction occurs too rapidly to isolate any such chemical compound. Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.   The minimum energy necessary to form a product during a collision between reactants is called the y (\(E_a\)). The kinetic energy of reactant molecules plays an important role in a reaction because the energy necessary to form a product is provided by a collision of a reactant molecule with another reactant molecule. (In single-reactant reactions, activation energy may be provided by a collision of the reactant molecule with the wall of the reaction vessel or with molecules of an inert contaminant.) If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly: Only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, the fraction of molecules possessing the necessary kinetic energy will be large; most collisions between molecules will result in reaction, and the reaction will occur rapidly. Figure \(\Page {3}\) shows the energy relationships for the general reaction of a molecule of \(A\) with a molecule of \(B\) to form molecules of \(C\) and \(D\): \[A+B⟶C+D \nonumber \] The figure shows that the energy of the transition state is higher than that of the reactants \(A\) and \(B\) by an amount equal to \(E_a\), the activation energy. Thus, the sum of the kinetic energies of \(A\) and \(B\) must be than to reach the transition state. After the transition state has been reached, and as \(C\) and \(D\) begin to form, the system loses energy until its total energy is lower than that of the initial mixture. This lost energy is transferred to other molecules, giving them enough energy to reach the transition state. The forward reaction (that between molecules \(A\) and \(B\)) therefore tends to take place readily once the reaction has started. In Figure \(\Page {3}\), \(ΔH\) represents the difference in enthalpy between the reactants (\(A\) and \(B\)) and the products (\(C\) and \(D\)). The sum of \(E_a\) and \(ΔH\) represents the activation energy for the reverse reaction: \[C+D⟶A+B \nonumber \]   We can use the to relate the activation energy and the rate constant, , of a given reaction: \[k=Ae^{−E_a/RT} \label{Arrhenius} \] In this equation, Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor is related to the rate at which collisions having the correct occur. The exponential term, \(e^{−E_a/RT}\), is related to the fraction of collisions providing adequate to overcome the activation barrier of the reaction. At one extreme, the system does not contain enough energy for collisions to overcome the activation barrier. In such cases, no reaction occurs. At the other extreme, the system has so much energy that every collision with the correct orientation can overcome the activation barrier, causing the reaction to proceed. In such cases, the reaction is nearly instantaneous. The Arrhenius equation (Equation \ref{Arrhenius}) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of \(E_a\) results in a smaller value for \(e^{−E_a/RT}\), reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller \(E_a\) has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of \(e^{−E_a/RT}\), a larger rate constant, and a faster rate for the reaction. An increase in temperature has the same effect as a decrease in activation energy. A larger fraction of molecules has the necessary energy to react (Figure \(\Page {4}\)), as indicated by an increase in the value of \(e^{−E_a/RT}\). The rate constant is also directly proportional to the frequency factor, \(A\). Hence a change in conditions or reactants that increases the number of collisions with a favorable orientation for reaction results in an increase in \(A\) and, consequently, an increase in \(k\). A convenient approach to determining \(E_a\) for a reaction involves the measurement of \(k\) at different temperatures and using of an alternate version of the Arrhenius equation that takes the form of linear equation: \[\begin{align*} \ln k&=\left(\dfrac{−E_a}{R}\right)\left(\dfrac{1}{T}\right)+\ln A\\ y&=mx+b \end{align*} \nonumber \] Thus, a plot of \(\ln k\) versus \(\dfrac{1}{T}\) gives a straight line with the slope \(\dfrac{-E_\ce{a}}{R}\), from which may be determined. The intercept gives the value of \(\ln A\). This is sometimes call an Arrhenius Plot. The variation of the rate constant with temperature for the decomposition of HI( ) to H ( ) and I ( ) is given here. What is the activation energy for the reaction? \[\ce{2HI}(g)⟶\ce{H2}(g)+\ce{I2}(g) \nonumber \] Values of \(\dfrac{1}{T}\) and ln are: Figure \(\Page {5}\) is a graph of ln versus \(\dfrac{1}{T}\). To determine the slope of the line, we need two values of ln , which are determined from the line at two values of \(\dfrac{1}{T}\) (one near each end of the line is preferable). For example, the value of ln determined from the line when \(\dfrac{1}{T}=1.25×10^{−3}\) is −2.593; the value when \(\dfrac{1}{T}=1.78×10^{−3}\) is −14.447. The slope of this line is given by the following expression: Thus: \[ \begin{align*} E_\ce{a} &=\mathrm{−slope×\mathit R=−(−2.2×10^4\:K×8.314\: J\: mol^{−1}\:K^{−1})} \\[4pt] &=\mathrm{1.8×10^5\:J\: mol^{−1}} \end{align*} \nonumber \] In many situations, it is possible to obtain a reasonable estimate of the activation energy without going through the entire process of constructing the Arrhenius plot. The Arrhenius equation: \[\ln k=\left(\dfrac{−E_\ce{a}}{R}\right)\left(\dfrac{1}{T}\right)+\ln A \nonumber \] can be rearranged as shown to give: \[\dfrac{Δ(\ln k)}{Δ\left(\dfrac{1}{T}\right)}=−\dfrac{E_\ce{a}}{R} \nonumber \] or \[\ln\dfrac{k_1}{k_2}=\dfrac{E_\ce{a}}{R}\left(\dfrac{1}{T_2}−\dfrac{1}{T_1}\right) \nonumber \] This equation can be rearranged to give a one-step calculation to obtain an estimate for the activation energy: \[E_\ce{a}=−R\left( \dfrac{\ln k_2−\ln k_1}{\left(\dfrac{1}{T_2}\right)−\left(\dfrac{1}{T_1}\right)}\right ) \nonumber \] Using the experimental data presented here, we can simply select two data entries. For this example, we select the first entry and the last entry: After calculating \(\dfrac{1}{T}\) and ln , we can substitute into the equation: \[E_\ce{a}=\mathrm{−8.314\:J\:mol^{−1}\:K^{−1}\left(\dfrac{−3.231−(−14.860)}{1.28×10^{−3}\:K^{−1}−1.80×10^{−3}\:K^{−1}}\right)} \nonumber \] and the result is = 185,900 J/mol. This method is very effective, especially when a limited number of temperature-dependent rate constants are available for the reaction of interest. The rate constant for the rate of decomposition of N O to NO and O in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K: \[\ce{2N2O5}(g)⟶\ce{4NO}(g)+\ce{3O2}(g) \nonumber \] Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition. 113,000 J/mol Chemical reactions require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant and its activation energy, temperature, and dependence on collision orientation.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Lavoisier

So what happened to turn alchemy, which was like magical potion-brewing in Harry Potter, into the science of chemistry? It was . Careful, careful measurement of quantities, such as masses, volumes, densities, temperatures, pressures. An early hero of measurement was Antoine Lavoisier. He was one of the first true chemical scientists. He conducted careful experiments, and tried to draw no conclusions except those required by his data. He said fact, idea, and word should be as closely connected as possible: that you can't improve your language without improving your thinking, and you can't improve your thinking without improving your language. So he pioneered a systematic chemical that is essentially what we use today. Remarkably, if you read his text, written in 1789, intended to introduce chemistry to beginners, much of it is still perfectly understandable and even correct by modern standards. Lavoisier first describes the : gases, liquids and solids. He points out when a solid material is heated, it tends to expand, becoming first a liquid, which takes up a constant volume, but can be poured, unlike a solid. More heating, and it becomes a gas, which he describes as elastic because it will expand or compress to different volumes depending on the pressure. Unlike the Greek philosophers, he understood that this is a , not a , and he has a good submicroscopic-scale intuition of what's happening: the particles of the material don't change, they just get further apart. He recognized the following as elements: oxygen, nitrogen, hydrogen, sulfur, phosphorus, chlorine and fluorine (although he did not know their elemental forms), carbon, iron, copper, silver, gold, mercury, lead, tin, antimony, arsenic, bismuth, cobalt, manganese, molybdenum, nickel, platinum, tungsten, and zinc. He burned sulfur and phosphorus and charcoal (carbon) and made careful observations, often using the bell jar over a bucket of mercury as shown in the drawing from his book, Figure 1. This is an example of a chemical change or chemical , in which chemicals turn into different chemicals. If you light the sulfur in the dish labeled D under the bell jar of air, it burns until it goes out leaving some extra sulfur. The air remaining in the jar is no longer good for breathing. If you put a mouse in the jar, it will die, just as the flame did. This demonstrates the concept of . The reaction or burning stopped when it ran out of oxygen, leaving primarily nitrogen (and a few trace other gases) in the jar. Priestley, another scientist, showed him how to prepare pure oxygen gas, and he used this to do many burning experiments as well. Lavoisier was obsessed with measurement. He developed elaborate apparatus for measuring everything. He would burn phosphorus, as shown in Figure 1, and observe the formation of a white flaky . The phosphorus (the in this case) wasn't water soluble, but the product was, so he collected the product very carefully, separating it from the unreacted phosphorus by washing with water. After drying, he could measure how much phosphorus had burned, how much oxygen had been consumed (because he knew the density of oxygen gas), and how much product had formed. He found that the mass of product was the sum of the masses of reactant consumed, in every experiment. This is the (which, actually, some earlier alchemists and chemists had also used). He also observed that the phosphorus has no taste, but the product, which he called phosphoric acid, is sour. He knew from these experiments that in many cases elements combine in only certain proportions, and also that oxygen can combine with sulfur, phosphorus, etc in two different ratios. He gave us the terminology we still use today: sulfuric acid is composed of sulfur and more oxygen, sulfurous acid is composed of sulfur and less oxygen. -ous means less oxygen; -ic means more oxygen. See the page for details. Lavoisier paid close attention to and . For instance, in the experiment we just described, he measured the volume of gas in the bell jar, before and after the reaction, but noted that after the reaction, you must wait until the temperature returns to what it was when you measured originally. If the gas is hot when you measure its volume after the reaction, it will have expanded, and your standard density will not apply. This would introduce a systematic error into the measurements: each time you perform the experiment, you will think that there is more gas leftover than there actually is, and your measurement won't be accurate. If the average result of your experiment is near the correct value, it is accurate. However, if your experiment gives very different numbers each time, even if the average is correct and the experiment is accurate, it is not precise. Precision is the difference between meeting "around 2 o'clock" and meeting "at 3 minutes and 27 seconds before 2 pm." Precision is how specific you are, how much detail you use. Lavoisier also helped develop the system of units (kg, L, m) that are currently in use in Korea and many other countries. Overall, while he didn't do very many original experiments that nobody else had done before, he did his experiments very carefully, so they were as accurate and precise as possible, and then he thought about them clearly and created words to describe the chemicals and ideas that helped make everything clearer. If you read a chemistry textbook written before Lavoiser, you will be very confused because the names for chemicals would be based on history (and would sound like they came from Harry Potter), rather than being based on what the chemicals are. If you read a chemistry text written after Lavoiser, you will recognize the language as similar to what we use today. describes how close a measured value is to the actual value. describes how well a group of measured values agree with each other. The states that matter can neither be created nor destroyed by a chemical or physical process. This results in the sum of the masses of reactant consumed in any experiment is equal to the mass of product. Chemical changes involve changing a substance's chemical identity such that new substances are formed. Physical changes involve altering a substance without changing its chemical identity. Combustion and rusting are two examples of chemical processes while boiling and melting are examples of physical processes. Chemical involve turning , chemicals that get consumed in the process of chemical change, into , chemicals produced through the process of chemical change that have a different composition from the reactants. A determines, or limits, the amount of product that can be produced from a chemical reaction.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.11%3A_Yogurt

Yogurt has been around for several millennia. The mythological story about the discovery of yogurt suggests that sheepherders stored their milk in bags made of the intestinal gut of the animals. The intestines contain natural enzymes that cause the milk to curdle and sour. This soured milk lasted longer so they continued making it. Today, the FDA defines yogurt as a milk product fermented by two bacterial strains: a lactic acid producing bacteria: and . Yogurt Production Process: In step 4, yogurt cultures are added to milk. These bacteria are lactic acid fermenters; they use enzymes to produce energy (ATP) from lactose. Yogurt is often tart. This flavor is often attributed to the presence of lactic acid. However, there are also a number of carbonyl compounds like acetoin, diacetyl and acetaldehyde that also contribute to the tangy yogurt flavor. During yogurt fermentation, acetaldehyde could be produced from lactose metabolism as a result of pyruvate decarboxylation. However, the primary source of acetaldehyde in these bacteria is from the conversion of threonine (amino acid) into acetaldehyde and glycine. Many yogurt bacteria lack the enzyme, alcohol dehydrogenase. Both and produce diacetyl which provides a distinctive “buttery” flavor to yogurt (and other fermented milk products). Acetoin is the reduced form of diacetyl and it complements the diacetyl with a mild creamy flavor. Propose a mechanism. Yogurt cultures in the intestinal tract have been shown to release the enzyme lactase which continues to break down lactose in the dairy product. This makes yogurt edible for people who are lactose-intolerant. To give their products a longer shelf life, manufacturers often fermentation. This kills the live cultures. What will happen to lactase if the yogurt has been heat-treated after fermentation? and are the two main bacteria used for creating yogurt. However, these strains do not survive the gastrointestinal tract. They are destroyed by the acidity of the stomach and the enzymes of the pancreas. It has become common to add ‘probiotic’ bacterial strains to yogurt such as , There is evidence that these bacteria will make it to the intestine intact. When probiotics are added to foods, the food industry often also adds ingredients known as prebiotics, such as inulin, which will, after digestion, aid in the growth of the probiotics in the colon. Kefir is a carbonated fermented milk drink. The microbes involved in the production of kefir are a symbiotic culture of lactic acid bacteria and yeasts embedded in a matrix of proteins, lipids, and polysaccharides, ‘kefir grains’. Kefir Production Process: During the first fermentation, lactic acid bacteria are responsible for the conversion of the lactose present in milk into lactic acid, which results in a pH decrease and milk preservation. Similar to yogurt, the flavor of kefir is often attributed to diacetyl and acetoin (both of which contribute a "buttery" flavor), acetaldehyde, and related carbonyl products. Non-lactose fermenting yeast and acetic acid bacteria (AAB) also participate in the process. Propionibacteria further break down some of the lactic acid into propionic acid (these bacteria also carry out the same fermentation in Swiss cheese). Other kefir microbial constituents include lactose-fermenting yeasts such as , as well as strains of yeast that do not metabolize lactose, including and The lactose-fermenting yeast break the lactose down into ethanol and carbon dioxide resulting in a carbonated taste. Ethanol concentration is typically low, usually 0.2-0.3%. Zourari, Accolas, & Desmazeaud, Metabolism and Biochemical Characteristics of Yogurt Bacteria, A Review. (1), pp.1-34. (Available in Canvas)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.08%3A_Ideal_vs._Real_Solutions

Make sure you thoroughly understand the following essential ideas: The popular liquor vodka consists mainly of ethanol (ethyl alcohol) and water in roughly equal portions. Ethanol and water both have substantial vapor pressures, so both components contribute to the total pressure of the gas phase above the liquid in a closed container of the two liquids. One might expect the vapor pressure of a solution of ethanol and water to be directly proportional to the sums of the values predicted by for the two liquids individually, but in general, this does not happen. The reason for this can be understood if you recall that Raoult's law reflects a single effect: the smaller proportion of vaporizable molecules (and thus their reduced escaping tendency) when the liquid is diluted by otherwise "inert" (non-volatile) substance. There are some solutions whose components follow Raoult's law quite closely. An example of such a solution is one composed of hexane C H and heptane C H . The total vapor pressure of this solution varies in a straight-line manner with the mole fraction composition of the mixture. Note that the mole fraction scales at the top and bottom run in opposite directions, since by definition, \[\chi_{hexane} = 1 – \chi_{heptane}\] If this solution behaves ideally, then is the sum of the Raoult's law plots for the two pure compounds: \[P_{total} = P_{ heptane } + P_{ hexane }\] An is one whose vapor pressure follows Raoult's law throughout its range of compositions. Experience has shown solutions that approximate ideal behavior are composed of molecules having very similar structures. Thus hexane and heptane are both linear hydrocarbons that differ only by a single –CH group. This provides a direct clue to the underlying cause of non-ideal behavior in solutions of volatile liquids. In an ideal solution, the interactions are there, but they are all energetically identical. Thus in an ideal solution of molecules A and B, A—A and B—B attractions are the same as A—B attractions. This is the case only when the two components are chemically and structurally very similar. The ideal solution differs in a fundamental way from the definition of an , defined as a hypothetical substance that follows the ideal gas law. The kinetic molecular theory that explains ideal gas behavior assumes that the molecules occupy no space and that intermolecular attractions are totally absent. The definition of an ideal gas is clearly inapplicable to liquids, whose volumes directly reflect the volumes of their component molecules. And of course, the very ability of the molecules to form a condensed phase is due to the attractive forces between the molecules. So the most we can say about an ideal solution is that the attractions between its all of its molecules are identical — that is, A-type molecules are as strongly attracted to other A molecules as to B-type molecules. Ideal solutions are perfectly democratic: there are no favorites. Real solutions are more like real societies, in which some members are "more equal than others." Suppose, for example, that unlike molecules are more strongly attracted to each other than are like molecules. This will cause A–B pairs that find themselves adjacent to each other to be energetically more stable than A–A and B–B pairs. At compositions in which significant numbers of both kind of molecules are present, their tendencies to escape the solution — and thus the vapor pressure of the solution, will fall below what it would be if the interactions between all the molecules were identical. This gives rise to a from Raoult's law. The chloroform-acetone system, illustrated above, is a good example. Conversely, if like molecules of each kind are more attracted to each other than to unlike ones, then the molecules that happen to be close to their own kind will be stabilized. At compositions approaching 50 mole-percent, A and B molecules near each other will more readily escape the solution, which will therefore exhibit a higher vapor pressure than would otherwise be the case. It should not be surprising molecules as different as benzene and \(CS_2\) should interact more strongly with their own kind, hence the positive deviation illustrated here. You will recall that all gases approach ideal behavior as their pressures approach zero. In the same way, as the mole fraction of either component approaches unity, the behavior of the solution approaches ideality. This is a simple consequence of the fact that at these limits, each molecule is surrounded mainly by its own kind, and the few A-B interactions will have little effect. Raoult's law is therefore a limiting law: \[P_i = \lim_{x_i \rightarrow 0} P^o \chi_i\] it gives the partial pressure of a substance in equilibrium with the solution more and more closely as the mole fraction of that substance approaches unity.

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

Polyethylene is probably the polymer you see most in daily life. Polyethylene is the most popular in the world. This is the polymer that makes grocery bags, shampoo bottles, children's toys, and even bullet proof vests. For such a versatile material, it has a very simple structure, the simplest of all commercial polymers. A molecule of polyethylene is nothing more than a long chain of carbon atoms, with two hydrogen atoms attached to each carbon atom. That's what the picture at the top of the page shows, but it might be easier to draw it like the picture below, only with the chain of carbon atoms being many thousands of atoms long. Sometimes it's a little more complicated. Sometimes some of the carbons, instead of having hydrogens attached to them, will have long chains of polyethylene attached to them. This is called branched, or low-density polyethylene, or LDPE. When there is no branching, it is called linear polyethylene, or HDPE. Linear polyethylene is much stronger than branched polyethylene, but branched polyethylene is cheaper and easier to make. Linear polyethylene is normally produced with molecular weights in the range of 200,000 to 500,000, but it can be made even higher. Polyethylene with molecular weights of three to six million is referred to as ultra-high molecular weight polyethylene, or UHMWPE. UHMWPE can be used to make fibers which are so strong they replaced Kevlar for use in bullet proof vests. Large sheets of it can be used instead of ice for skating rinks. ( )

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Lipids/Steroids

One major class of lipids is the steroids, which have structures totally different from the other classes of lipids. The main feature of steroids is the ring system of three cyclohexanes and one cyclopentane in a fused ring system as shown below. There are a variety of functional groups that may be attached. The main feature, as in all lipids, is the large number of carbon-hydrogens which make steroids non-polar. Steroids include such well known compounds as cholesterol, sex hormones, birth control pills, cortisone, and anabolic steroids. The best known and most abundant steroid in the body is cholesterol. Cholesterol is formed in brain tissue, nerve tissue, and the blood stream. It is the major compound found in gallstones and . Cholesterol also contributes to the formation of deposits on the inner walls of blood vessels. These deposits harden and obstruct the flow of blood. This condition, known as atherosclerosis, results in various heart diseases, strokes, and high blood pressure. Much research is currently underway to determine if a correlation exists between cholesterol levels in the blood and diet. Not only does cholesterol come from the diet, but cholesterol is synthesized in the body from carbohydrates and proteins as well as fat. Therefore, the elimination of cholesterol rich foods from the diet does not necessarily lower blood cholesterol levels. Some studies have found that if certain unsaturated fats and oils are substituted for saturated fats, the blood cholesterol level decreases. The research is incomplete on this problem. are also steroids. The primary male hormone, testosterone, is responsible for the development of secondary sex characteristics. Two female sex hormones, progesterone and estrogen or estradiol control the ovulation cycle. Notice that the male and female hormones have only slight differences in structures, but yet have very different physiological effects. Testosterone promotes the normal development of male genital organs ans is synthesized from cholesterol in the testes. It also promotes secondary male sexual characteristics such as deep voice, facial and body hair. Estrogen, along with progesterone regulates changes occurring in the uterus and ovaries known as the menstrual cycle. For more details see . Estrogen is synthesized from testosterone by making the first ring aromatic which results in mole double bonds, the loss of a methyl group and formation of an alcohol group. The adrenocorticoid hormones are products of the adrenal glands ("adrenal" means jacent to the (kidney). The most important mineralocrticoid is , which regulates the reabsorption of sodium and chloride ions in the kidney tubules and increases the loss of potassium ions. Aldosterone is secreted when blood sodium ion levels are too low to cause the kidney to retain sodium ions. If sodium levels are elevated, aldosterone is not secreted, so that some sodium will be lost in the urine. Aldosterone also controls swelling in the tissues. Cortisol, the most important glucocortinoid, has the function of increasing glucose and glycogen concentrations in the body. These reactions are completed in the liver by taking fatty acids from lipid storage cells and amino acids from body proteins to make glucose and glycogen. In addition, cortisol and its ketone derivative, , have the ability to inflammatory effects. Cortisone or similar synthetic derivatives such as prednisolone are used to treat inflammatory diseases, rheumatoid arthritis, and bronchial asthma. There are many side effects with the use of cortisone drugs, so there use must be monitored carefully. Thumbnail: Ball-and-stick model of the cholesterol molecule, a compound essential for animal life that forms the membranes of animal cells. (Public Domain; ).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.09%3A_Corrosion_Gallery

The nails are immersed in agar which forms a moist solid gel. The agar also contains phenolphthalein and hexacyanoiron(III) Fe(CN ) which forms a deep blue color ("prussian blue") in the presence of Fe . The blue colors are clearly associated with those parts of the nail that have been stressed, thus faciliting the anodic release of Fe from the metal. The the pink color shows the cathodic regions that have been made alkaline by the reaction O + 2 H O + 4 → 4 OH This clearly shows the separation between the anodic and cathodic processes in corrosion. [Illustration from U of West Indies: link] If you live in the older part of a city where the mains are 50-100 years old, the water you drink may well have passed through a pipe in this condition! Severe corrosion like this is more common in areas where the water is acidic. Such water comes from mountain snowmelt and runoff, and usually acquires its acidity from dissolved atmospheric carbon dioxide. Waters from rivers, lakes, and especially groundwaters from wells have usually been in sufficiently long contact with carbonate-containing sediments to have been neutralized. Water-works administrators like to make the water slightly alkaline and slightly supersaturated in calcium carbonate in order to maintain a thin coating of solid carbonate on the interior of the pipe which acts to protect it from corrosion. All large concrete structures contain steel reinforcing bars ("rebars") that help ensure structural integrity under varying load conditions and especially during earthquakes. Intrusion of water, even in the form of fog or mists, can lead to serious corrosion damage, as seen in this picture of this column which supports a highway overpass. The picture shows two steel structural members joined by cast iron flanges which have been bolted together. For some reason, one of the pieces has become more anodic than the other, leading to extensive corrosion of the upper part. This gas pipe was buried in a red soil that contained iron pyrites (FeS.) The bacterium derives its energy by oxidizing Fe to the more soluble Fe , transferring the electrons to O . It also oxidizes the sulfur, producing sulfuric acid. The resulting chemical cocktail has eaten a hole into the pipe.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Hydrocarbons/Alkanes

Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula \(C_nH_{2n+2}\) and can be subdivided into the following three groups: the , , and . Alkanes are also . Alkanes are the simplest and least reactive hydrocarbon species containing only carbons and hydrogens. They are commercially very important, being the principal constituent of gasoline and lubricating oils and are extensively employed in organic chemistry; though the role of pure alkanes (such as hexanes) is delegated mostly to solvents. The distinguishing feature of an alkane, making it distinct from other compounds that also exclusively contain carbon and hydrogen, is its lack of unsaturation. That is to say, it contains no double or triple bonds, which are highly reactive in organic chemistry. Though not totally devoid of reactivity, their lack of reactivity under most laboratory conditions makes them a relatively uninteresting, though very important component of organic chemistry. As you will learn about later, the energy confined within the carbon-carbon bond and the carbon-hydrogen bond is quite high and their rapid oxidation produces a large amount of heat, typically in the form of fire.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/13%3A_Synthesis_of_Vitamin_B

The total synthesis of Vitamin B was accomplished in 1973 by a grand collaboration between R. B. Woodward’s group at Harvard University (USA) and A. Eschenmoser’s group, Swiss Federal Institute of Technology (ETH), Zürich, Switzerland. It took about twelve years and more than two dozen senior scientists to complete this gigantic task. The achievement is variously eulogized by organic chemists – monumental achievement in the annals of organic synthetic chemistry; a breakthrough; a mile stone in organic synthesis; unrivaled even after 40 years; a task no less adventurous than concurring Mount Everest. At the time the adventure started it was the most formidable challenge in synthetic chemistry, which few would have dared. The announcement of its synthesis marked the coming of age of synthetic organic chemistry. Woodward and Eschenmoser worked in close collaboration and in competition during this historic pace. In the process, they achieved not only an astounding synthesis, but also opened several new fields for future investigations. Woodward Hoffmann Rule is the most famous of the offshoots. While studying this synthesis, a student should also ponder over the thoroughness in their planning of all aspects of the scheme and initiation of suitable basic studies well in advance to facilitate the main scheme at appropriate junctures. This long introduction is just in tune with the length of the scheme, time taken and magnificent achievements. shows the structure of Vit and the main structural features / challenges of the complex molecule. In his inimitable style, R. B. Woodward draws attention to these challenges in the beginning of his lectures and to the fact that it took close to 50 years to establish the structure of Vit by chemical degradation and finally by X-Ray diffraction studies by Dorothy Hodgkin in 1956. One should note that most of the chemical degradation studies, remarkable as they might be, happened during the early years of the twentieth century, when modern organic chemistry was in its making. Spectroscopic data, if any, were conspicuous by their absence and most of the degradation and synthetic chemistry were too harsh by modern (meaning 1960s) standards for this delicate molecule. Hence, this was a daunting scenario. This meant that a whole lot of synthetic chemistry had to be reinvented to suit these challenges. Such aspects were thoroughly planned in and contingency plans were put in place well in advance. Discerning students could see a glimpse of this planning in this short presentation. The ideas on retroanalysis, as we understand today, were in the developing stages during the sixties. The retroanalyses we use here are later-day additions by other scientists, based on the facts (lectures, papers etc.,) published by the two groups. All such citations are included at the end of the write-up. The synthetic target was identified as Cobyric Acid, because this compound was a natural product and had been converted to Vit by Bernhauer K., et.al., (1960) . Hence total synthesis of Cobyric Acid would amount to a formal synthesis of Vit . Cobyric acid had seven carboxylic acid side chains, out of which four were propionic acid moieties, one on each heterocyclic ring. The main challenge was to differentiate the propionic acid chain on the D ring from the other acetic acid and propionic acid moieties. It was therefore decided that this odd acid moiety would be masked as nitrile . That still leaves a daunting task of differentiation, which we could address later. It was first decided to view the molecule as made up of two halves – the . The first disconnection was at the A/B ring junction at the methylene bridge . Cleavage of the second bridge at C/D ring junctions gave the Eastern Half as in charge of Eschenmoser’s group at Zurich and the Western Half as in charge of Woodward’s group at Harvard (US). This western half has a formidable array of on an eight-carbon frame. Note that the stereocentres were planned on the basis of known stereoselectivities and the six membered rings were built to provide the propionic acid chains . The nitrogen for the A ring came from an indole, whose benzene ring gave the side chains for the A ring. The D ring nitrogen came via a Beckmann rearrangement. Corrnorsterone was the key intermediate (corner stone) that held all the stereocentres and chains on the Western Half. The required enantiopure 1,2,3-trimethylyclopentene unit came from camphorquinone as shown in . Through another convergent synthesis, a five membered ring was fused to indole at C2 – C3 bond and resolved as shown in . The (+)- enantiomer was actually needed for target synthesis. The useless (-)- enentiomer was used as a model compound (for this was "just about the only kind of model study which we regard as wholly reliable” – RBW). Fragments and were combined and then processed to Corrnorsterone as shown in . Note that the Beckmann ring expansion process set in motion a cascade of reaction, leading to a Claisen condensation and D ring formation, all in one step. The six membered imide carbonyl was also cleaved and placed an acetate chain on the D ring. In spite of this elaborate planning and execution, the process yielded a mixture of epimers at propionic acid side chain on the A ring, the required isomer being a minor component of the mixture. The major undesired product does not cleave at the amide bond due to unfavourable steric compression at the developing side chains . This unfavorable steric problem was however soon solved. On hydrolysis under strong base conditions, the amide ring opened and the propionic acid side chain isomerised to the less strained isomer. This could be then be acidified and esterified to β-corrnorsterone, with a recovery of 90% of the desired isomer . This correct isomer was treated with a mixture of methanol and thiophenol under acid conditions . This set two processes in motion. The thiophenol attacked the ketone and activated this center, while the methanol oxygen attacked the amide bond leading to an ester and a thioenol ether. Ozonolysis of the thioenol ether at – 90°C cleaved the olefin unit to the aldehyde-thioester compound . An interesting new chemistry evolved here. While thioesters are less reactive to acid hydrolysis and showed comparable reactivity with oxygen nucleophiles, nitrogen nucleophiles were unique. The thioester reacted much faster then normal oxyesters to give amides. Thus, the thioester was exclusively cleaved to amide with ammonia, leaving three methyl esters untouched. The aldehyde moiety was then selectively converted to alcohol and then mesylated under mixed anhydride / pyridine conditions. This sequence also converted the amide moiety to nitrile. The mesylate was then converted to bromide to give the key intermediate cyanobromide. The Zurich group was simultaneously working on the synthesis of the Eastern Half named . The fragments for B and C rings were planned via., a single intermediate . The synthesis started with a Diels-Alder reaction to secure two asymmetric centers properly and the racemate was resolved. The pure enantiomer was followed through the scheme to obtain the B ring segment. The same intermediate yielded the C ring fragment as well . To connect two such fragments Eschenmoser had developed two sulphide contraction procedures. The mechanisms of the processes are shown below. Using the oxidative coupling / sulfur extrusion procedure, they coupled the B and C rings as shown in . Woodward’s group also developed a new synthesis for the C ring starting from (+)-Camphorquinone. The scheme is shown in . Though great care had been spared for the stereocentres at all points, the problem of stereoisomers could not be avoided. The crystalline thiodextrollin that was synthesized was actually a mixture of two stereoisomers at the propionic acid moiety of the B ring. Though they had purified the mixture at this point, it was of no value because this stereocentre was due for further disturbances at later stages. The mixture was taken forward for the first coupling at C / D bridge. After considerable effort that lasted over a year, they were first coupled at the Southern end using the alkylation / extrusion procedure . Note that the first product of alkylation was a thioether Type I, which readily isomerised to thioether Type II. The product was named Cyanocorrigenolide. This isomerisation disturbed the stereocentre at the C ring. The next phase was the formation of A / B bridge. The C / D coupled compound was first treated with phosphorous pentasulphide followed by trimethyloxonium fluoroborate (M OB ) . This procedure replaced the oxygens on the A and B rings by sulfur and finally to the S-methyl derivative. Dimethylamine in methnol cleaved the thiolactone ring selectively to a dimethylacetamide chain and a terminal olefin. The olefin was rather unstable. This was to be converted immediately to the cobalt complex. This procedure was not easy. Under several conditions, the cobalt metal ion catalyzed further reactions leading to extensive “destruction” of the compound. After several experiments it was observed that cobalt chloride or iodide in THF was unique for smooth cobaltation. The complexation process brought the A and B rings in close proximity. A base catalysed reaction then enabled formation of the bridge and removal of the sulfur moiety the best condition being DBN catalysed cyclization. Note that the overall transformation interfered with the asymmetric center at C ring. Nonetheless, the A / B bridge was finally in place. The Zurich group also came up with an alternate Zn-complex procedure along the same lines. All these manipulations were indeed harsh to the (three) epimerisable propionic acid chains on the A, B and C rings. The final product was purified by TLC (“plate chromatography”) and critically analyzed by HPLC (a new chromatographic tool at that time). The UV chromophores in all the products were of great help in this chromatography. The synthesis has now two major milestones to pass. There are three active methine bridges in the molecule. The bridge at B / C rings had to be ‘protected’ from methylation. This was achieved by an oxidative lactone formation reaction at the B ring (I2, AcOH) . This new quaternary center and the existing quaternary carbon at C12 together exerted a steric congestion around C10. The chloromethyl ether entered the C5 and C15 centres exclusively. Ra-Ni hydrogenolysis cleaved the thioethers and the lactone ring in one step. This step was followed by esterification. Conc. sulfuric acid converted the nitrile to amide .At this stage, they faced the difficult task of selective cleavage of amide, in the presence of six esters in the molecule. After extensive parallel experimentations, the Harvard group rediscovered an efficient selective cleavage of the amide moiety in reagent. The Zurich group also came out with a “diabolically cleaver scheme” for selective hydrolysis of the amide group in the presence of ester groups. This scheme is shown in . However the former was preferred due to its simplicity and better yields. Nonetheless, Eschenmoser’s solution stands testimony to human ingenuity. The very last step of this long synthesis posed a major problem that needs special mention. Amidation of esters with ammonia was the only remaining step for the final assault on the synthesis of Cobyric acid. A closer look reveals that the task may not be that easy. The ester moieties (in particular the acetate units) were in crowed environments. Parallel to these developments, model studies on very similar molecules were in progress. Based on these studies, when the hexamethylester 1 acid was treated with ammonia in ethylene glycol at 75°C for 30 hours, the product obtained was not cobyric acid but a pseudocobyric acid, whose structure was established as dehydrocobyric acid. This product could be purified only by HPLC. This was one among the products obtained by earlier workers. But they had no idea about such a complication. This mystery took several critical studies to solve. Throughout all these studies, great care was exercised to see that the solvents were well deoxygenated before use. Oxygen was strictly avoided from the reaction atmosphere. The source for this oxidative cyclization was attributed to cobalt, which was suspected to oxidatively bond to the C9 position, facilitating the amide anion to cyclise. This unwanted reaction took a long time to solve. Several studies were aimed at ammonialysis under reducing conditions and also to open the lactam ring under reduction conditions. Finally a few milligrams of ammonium acetate were added to the reaction mixture to prevent the formation of amide anion, which was suspected to be the culprit for cyclization . This trick helped. The reaction was complete within 10 hours, with cobyric acid as the only product in good yields. This cobyric acid was identical in all respects, particularly in HPLC, with the natural product, thus ending this long journey for the formal synthesis of Vitamin . This extraordinary adventure in organic synthesis is noted for several most significant achievements. Even after half a century this Vit synthesis remains unrivaled and continues to inspire generations of chemists.