# Introductory Chemistry Online/Chemical Reactions
## Chapter 5. Chemical Reactions

In Chapter 2, we learned that chemical changes result in the transformation of one chemical substance into a different substance having a new set of chemical and physical properties. The transformation of one substance into another is called a chemical reaction and is described using a chemical equation. In this chapter we will learn how to write and balance simple chemical equations. We will learn the basic types of chemical reactions and we will learn how to predict the products that are likely to be formed when these reactions occur. We will examine a special type of chemical reaction in which one of the products has low solubility in water and precipitates from solution. Understanding the basic rules of solubility is simple and again will allow us to predict when this type of reaction is likely to be observed. Finally, we will address the energetics of chemical reactions, laying a fundamental background for the study of reaction rates and equilibrium later in the course.

## 5.1 Chemical Changes and Chemical Reactions

In Chapter 2, we classified changes in our environment utilizing the concepts of physical and chemical changes. We said that a physical change alters the appearance of a substance without changing its molecular structure. Ice melts, water evaporates and mountains are slowly weathered into dust. All of these change the characteristics of substances, but they do not alter its basic structure. A chemical change, however, results in the transformation of one molecular substance into another. Gasoline burns, reacting with oxygen in the atmosphere, generating light, heat, and converting the carbon-based molecules into carbon dioxide gas and water vapor. When substances combine like this and undergo chemical changes, we say that a chemical reaction has occurred. Some chemical reactions are quite evident, like the burning of gasoline, and involve the production of heat or light. In other types of chemical reactions, gases are evolved, color changes occur and clear solutions become cloudy, with the ultimate formation of an insoluble substance (a precipitate). Chemical changes can also be quite obscure and their occurrence can only be detected by sophisticated chemical analysis.
Sometimes chemical changes occur spontaneously, others require the input of energy (heat) in order to occur. Chemical reactions can occur rapidly, like the explosive reaction of sodium metal in the presence of water, and others occur very slowly, like the rusting of iron or the tarnish that slowly develops on some metal surfaces exposed to air. In this chapter we will learn to represent chemical reactions using chemical equations. We will learn to balance these equations, explore types of reactions and learn to predict products from simple reactions. Central to all of this is the concept of the chemical equation.

## 5.2 Chemical Equations

The processes that occur during a chemical change can be represented using a chemical equation. In a chemical equation, the chemical formulas for the substance or substances that undergo the chemical reaction (the reactants) and the formulas for the new substance or substances that are formed (the products) are both shown, and are linked by an arrow. The arrow in a chemical equation has the properties of an “equals sign” in mathematics, and because of this, in a chemical equation, there must be the same number and types of atoms on each side of the arrow.
Reactants → Products
As an example of a chemical reaction, consider the reaction between solid carbon and oxygen gas to form carbon dioxide gas. This chemical equation for this reaction can be written as shown below. Figure 5.1
C (s) + O$_2$ (g) → CO$_2$ (g)
In this equation, we have used (s) and (g) to represent the physical state of the reactants at the time of the reaction (solid and gas). Other abbreviations that are often used include (l) for liquid and (aq) to indicate that the reactant or product is dissolved in aqueous solution.
As we inspect this equation we see that there is one carbon atom on each side of the arrow and that there are two oxygen atoms on each side. An equation in which there are the same number and types of atoms on both sides of the arrow is referred to as balanced. As you write chemical equations, it is important to remember those elements that naturally occur as diatomic molecules (Table 1.1). Remember that when you include these as reactants or products, remember to indicate that they are diatomic by using the subscript “2”.
Table 1.1 Common Diatomic Elements :
| Element | Chemical Formula |
| Hydrogen | H$_2$ |
| Oxygen | O$_2$ |
| Nitrogen | N$_2$ |
| Fluorine | F$_2$ |
| Chlorine | Cl$_2$ |
| Bromine | Br$_2$ |
| Iodine | I$_2$ |
Exercise 5.1 Writing Chemical Equations.

## 5.3 Balancing Chemical Equations

In another example of a chemical reaction, sodium metal reacts with chlorine gas to form solid sodium chloride. An equation describing this process is shown below.Figure 5.2
Na (s) + Cl$_2$ (g) → NaCl (s)
Inspection of this equation, however, shows that, while there is one sodium atom on each side of the arrow, there are two chlorine atoms in the reactants and only one in the products. This equation is not balanced, and is therefore not a valid chemical equation. In order to balance this equation, we must insert coefficients (not subscripts) in front of the appropriate reactants or products so that the same number and types of atoms appear on both sides of the equation. Because chlorine is diatomic, there are two chlorines in the reactants and there must also be two chlorines in the products. In order to accomplish this, we place the coefficient “2” in front of the product, NaCl. Now we are balanced for chlorine, but there are two atoms of sodium in the products and only one shown in the reactants. To resolve this, we need to place the coefficient “2” in front of the sodium in the reactant, to give the equation shown below.
2 Na (s) + Cl$_2$(g) → 2 NaCl (s)
In this equation, there are two sodiums in the reactants, two sodiums in the products, two chlorines in the reactants and two chlorines in the products; the equation is now balanced.
There are many different strategies that people use in order to balance chemical equations. The simplest methods, where you examine and modify coefficients in some systematic order, is generally called “balancing by inspection”. These methods are generally useful for most simple chemical equations, although mathematical algorithms are often necessary for highly complex reactions. One version of the “inspection” method that we will use in this section can be called the “odd-even” approach. Looking at the first equation that we wrote for the sodium-chlorine reaction, we note that there are an odd number of chlorines in the products and an even number of chlorines in the reactants. The first thing we did in balancing this equation was to insert the multiplier “2” in front of the product (NaCl) so that there were now an even number of chlorines on both sides of the equation. Once we did that, we simply had to balance the other element (Na) which was “odd” on both sides, and the equation was easily balanced. When you are using this approach with more complicated equations, it is often useful to begin by balancing the most complex molecule in the equation first (the one with the most atoms), and focus on the element in this compound that is present in the greatest amount.
Another example where the “odd-even” approach works well is the decomposition of hydrogen peroxide to yield water and oxygen gas, as shown below.
H$_2$O$_2$ (aq) → O$_2$ (g) + H$_2$O (l)
As we inspect this equation, we note that there are an even number of oxygen atoms in the reactants and an odd number of oxygens in the products. Specifically, water has only one oxygen (in the products) and the number of oxygen atoms in the products can be made “even” by inserting the coefficient “2” in front of H$_2$O. Doing this (shown below) we note that we now have four hydrogens in the products and only two in the reactants.
H$_2$O$_2$ (aq) → O$_2$ (g) + 2 H$_2$O (l)
Balancing the hydrogens by inserting “2” in front of H$_2$O$_2$ in the reactants gives us an equation with four hydrogens on both sides on four oxygens on both sides; the equation is now balanced.
2 H$_2$O$_2$ (aq) → O$_2$ (g) + 2 H$_2$O (l)
Example 5.1 Balancing Chemical Equations.
Exercise 5.2 Balancing Chemical Equations.
Write a balanced chemical equation for the reactions given below:

## 5.4 Classifying Chemical Reactions

The reactions we have examined in the previous sections can be classified into a few simple types. Organizing reactions in this way is useful because it will assist us in predicting the products of unknown reactions. There are many different classifications of chemical reactions, but here we will focus on the following types: synthesis, decomposition, single replacement and double replacement. In addition, we will see that some of these reactions involve changes in the oxidation numbers of the reactants and products; these will be referred to as oxidation-reduction, or “redox” reactions.
The first type of reaction we will consider is a synthesis reaction (also called a combination reaction). In a synthesis reaction, elements or compounds undergo reaction and combine to form a single new substance. The reaction of sodium metal with chlorine gas to give sodium chloride is an example of a synthesis reaction where both reactants are elements.
2 Na (s) + Cl$_2$ (g) → 2 NaCl (s)
In this reaction, sodium metal and chlorine gas have combined to yield (synthesize) the more complex molecule, sodium chloride. Another example of a synthesis reaction, where compounds are involved as reactants, is the reaction of the organic molecule ethene (a carbon-based molecule – covered in more depth in Chapter 14) with HBr to yield the organic molecule bromoethane.
In this example, two molecular compounds (the organic compound, ethene, and hydrogen bromide) have combined to yield (synthesize) the new molecule, bromoethane. In a similar manner, synthesis reaction can also involve elements reacting with compounds.
In decomposition reactions, a single compound will break down to form two or more new substances. The substances formed can be elements, compounds, or a mixture of both elements and compounds. Two simple examples of decomposition reactions are shown below.
Cu$_2$S (s) → 2 Cu (s) + S (s)
CaCO$_3$ (s) → CaO (s) + CO$_2$ (g)
In a single-replacement reaction (also called a single-displacement reaction) an element and a compound will react so that their elements are switched. In other words, an element will typically displace another element from within a compound. As a general rule, metals will replace metals in compounds and non-metals will typically replace non-metals. An example of a single replacement reaction is shown below.
Zn (s) + CuCl$_2$ (s) → ZnCl$_2$ (s) + Cu (s)
In this example, elemental zinc has displaced the metal, copper, from copper(II) chloride to form zinc chloride and elemental copper. In the reactants, zinc was elemental and in the products, it is present within the compound, zinc chloride. Likewise, copper was present in a compound in the reactants and is elemental in the products.
In another example, iron metal will react with an aqueous solution of copper sulfate to give copper metal and iron(II) sulfate.
Fe (s) + CuSO$_4$ (aq) → Cu (s) + FeSO$_4$ (aq)
In this reaction, elemental iron replaces copper in a compound with sulfate anion and elemental copper metal is formed; metal replaces metal. The tendency of metals to replace other metals in single-replacement reactions is often referred to as an activity series. In the activity series shown below, any metal will replace any other metal which is to the right of itself in the series. Thus, iron will replace copper, as shown above, but copper metal would not replace iron from iron sulfate (iron is more active than copper). The activity series is useful in predicting whether a given single-replacement reaction will occur or not. Note that hydrogen is also included in the table, although it is not generally considered a “metal”. Within this table, metals which occur before hydrogen will react with acids to form hydrogen gas and metal salts. Copper, silver, mercury and gold are less active than hydrogen and will not liberate hydrogen from simple acids. The metals in Groups I & II of the periodic table (Li, Na, K, Rb, Cs, Ca, Sr and Ba) are so reactive that they will directly react with water to liberate hydrogen gas and form metal hydroxides. These are generally referred to as “active metals”.
The activity series is useful in predicting whether a given single-replacement reaction will occur or not. Note that hydrogen is also included in the table, although it is not generally considered a “metal”. Within this table, metals which occur before hydrogen will react with acids to form hydrogen gas and metal salts. Copper, silver, mercury and gold are less active than hydrogen and will not liberate hydrogen from simple acids. The metals in Groups I & II of the periodic table (Li, Na, K, Rb, Cs, Ca, Sr and Ba) are so reactive that they will directly react with water to liberate hydrogen gas and form metal hydroxides. These are generally referred to as “active metals”. Figure 5.3
A double-replacement reaction (or double-displacement) two ionic compounds in aqueous solution switch anions and form two new compounds. In order for a chemical reaction to occur in a double-replacement reaction, one of the new compounds that is formed must be insoluble in water, forming a solid precipitate (or a gas, to be covered in Chapter 6). If both of the new compounds which are formed are water-soluble, then no reaction has occurred. In the reactants, there were two cations and two anions in solution, and in the products there are the same two cations and two anions, in the same solution; nothing has happened. An example of a double-replacement reaction is shown below.
BaCl$_2$ (aq) + Na$_2$SO$_4$ (aq) → BaSO$_4$ (s) + 2 NaCl (aq)
In this reaction, solid barium sulfate is formed as a precipitate. This is a chemical change and this is a valid chemical reaction. In order to predict whether a double-replacement reaction will occur or not you need to understand rules for predicting solubility of ionic compounds. These rules will be covered in a later section, but are not related to the activity series discussed above. Figure 5.4
Table 5.2 Activity Series for Common Metals :
| Li > | K > | Ba > | Sr > | Ca > |
| Na > | Mg > | Al > | Mn > | Zn > |
| Fe > | Cd > | Co > | Ni > | Sn > |
| Pb > | H > | Cu > | Ag > | Hg > |
| Au |

## 5.5 Oxidation and Reduction in Chemical Reactions

The oxidation number of an element represents the total number of electrons which have been removed (a positive oxidation state) or added (a negative oxidation state) to get the element into its present state. The term oxidation describes the loss of electrons by an element and an increase in oxidation state; the term reduction describes the gain of electrons and a decrease in oxidation state. Oxidation numbers for elements in compounds can be calculated using a simple set of rules, which are reproduced below in Table 5.3.
* Table 5.3 Rules for Assigning Oxidation Numbers
* 1. The oxidation number of an element in the free state is zero.
* 2. A monoatomic ion will have an oxidation number that is equal to its charge.
* 3. In compounds with metals, hydrogen will be –1, otherwise it will always be +1.
* 4. Oxygen, within a compound, will generally have an oxidation number of –2.
* 5. Halogens will be –1, except in compounds with oxygen.
* 6. Sulfur will generally be –2, except in compounds with oxygen.
* 7. In a molecular compound, the most electronegative element is assigned a negative oxidation number.
In many chemical reactions, the oxidation number of elements change. Consider the synthesis reaction shown below. In the reactants, carbon and oxygen are both elements and their oxidation numbers are zero (Rule 1). In the product, oxygen will have an oxidation number of –2 (Rule 4), therefore, carbon in CO$_2$ must have an oxidation number of +4 in order to balance the four negative charges on the oxygens. During this reaction, the oxidation number of carbon has changed from zero in the reactants to +4 in the products and the oxidation number of oxygen has changed from zero to –2. This is an example of a redox reaction; a chemical reaction in which the oxidation numbers of elements change on going from reactants to products.
C (s) + O$_2$ (g) → CO$_2$ (g)
In a redox reaction, the element that “loses electrons” is said to be oxidized and will have an increase in its oxidation number. In the example above, the oxidation number of carbon increases from zero to +4; it has “lost electrons” and has been oxidized. The element that “gains electrons” in a redox reaction is said to be reduced and will have a decrease in its oxidation number. In the reaction above, the oxidation number of oxygen has decreased from zero to –2; it has “gained electrons” and has been reduced.
Example 5.2 Recognizing Redox Reactions
Exercise 5.3 Recognizing Redox Reactions
For each of the reactions given below, calculate the oxidation number of each of the elements in the reactants and the products and determine if the reaction involves oxidation-reduction. If it is a redox reaction, identify the elements that have been oxidized and reduced.

## 5.6 Predicting Products from Chemical Reactions

Part of the lure of chemistry is that things don’t always work out the way you expect. You plan a reaction, anticipate the products and, quite often, the results astound you! The exercise, then, is trying to figure out what was formed, why, and whether your observation leads to other useful generalizations. The first step in this process of discovery is anticipating or predicting the products which are likely to be formed in a given chemical reaction. The guidelines we describe here will accurately predict the products of most classes of simple chemical reactions. As your experience in chemistry grows, however, you will begin to appreciate the unexpected! Figure 5.5
In simple synthesis reactions involving reaction of elements, such as aluminum metal reacting with chlorine gas, the product will be a simple compound containing both elements. In this case, it is easiest to consider the common charges that the elements adopt as ions and build your product accordingly. Aluminum is a Group III element and will typically form a +3 ion. Chlorine, being Group VII, will accept one electron and form a monoanion. Putting these predictions together, the product is likely to be AlCl$_3$. In fact, if aluminum metal and chlorine gas are allowed to react, solid AlCl$_3$ is the predominant product.
2 Al (s) + 3 Cl$_2$ (g) → 2 AlCl$_3$ (s)
The synthesis reaction involving the non-metals hydrogen gas and bromine can be approached similarly. The product will contain both elements. Hydrogen, Group I, has one valence electron and will form one covalent bond. Bromine, Group VII, has seven valence electrons and will form one covalent bond. The likely product is therefore HBr, with one covalent bond between the hydrogen and the bromine.
H$_2$ (g) + Br$_2$ (g) < → 2 HBr (g)
For a single-replacement reaction, recall that (in general) metals will replace metals and non-metals will replace non-metals. For the reaction between lead(IV) chloride and fluorine gas, the fluorine will replace the chlorine, leading to a compound between lead and fluorine and the production of elemental chlorine. The lead can be viewed as a “spectator” in the reaction and the product is likely to be lead(IV) fluoride. The complete equation is shown below.
PbCl$_4$ (s) + 2 F$_2$ (g) → PbF$_4$ (s) + 2 Cl$_2$ (g)
In single-replacement reactions in which metals (or carbon or hydrogen) are expected to replace metals, first you should check the activity series to see if any reaction is anticipated. Remember that metals can only replace metals that are less active than themselves (to the right in the Table). If the reaction is predicted to occur, use the same general guidelines that we used above. For example, solid iron reacting with aqueous sulfuric acid (H$_2$SO$_4$). In this reaction the question is whether iron will displace hydrogen and form hydrogen gas. Consulting the activity series, we see that hydrogen is to the right of iron, meaning that the reaction is expected to occur. Next, we reason that iron will replace hydrogen, leading to the formation of iron sulfate, where the sulfate is the “spectator” ion. The formation of hydrogen gas requires a change in oxidation number in the hydrogen of +1 to zero. Two hydrogen atoms must therefore be reduced (a decrease in oxidation number) and the two electrons required for the reduction must come from the iron. The charge on the iron is therefore most likely to be +2 (it starts off at zero and donates two electrons to the hydrogens). The final product is therefore most likely iron(II) sulfate. The complete equation is shown below.
Fe (s) + H$_2$SO$_4$ (aq) → FeSO$_4$ (aq) + H$_2$ (g)
Decomposition reactions are the most difficult to predict, but there are some general trends that are useful. For example, most metal carbonates will decompose on heating to yield the metal oxide and carbon dioxide.
NiCO$_3$ (s) → NiO (s) + CO$_2$ (g)
Metal hydrogen carbonates also decompose on heating to give the metal carbonate, carbon dioxide and water.
2 NaHCO$_3$ (s) → Na$_2$CO$_3$ (s) + H$_2$O (g) + CO$_2$ (g)
Finally, many oxygen-containing compounds will decompose on heating to yield oxygen gas and “other compounds”. Identifying these compounds and building an understanding of why and how they are formed is one of the challenges of chemistry. Some examples:
H$_2$O$_2$ (aq) → O$_2$ (g) + H$_2$O (l)
2 HgO (s) → O$_2$ (g) + 2 Hg (l)
2 KClO$_3$ (s) → 3 O$_2$ (g) + 2 KCl (s)
The potential products in double-replacement reactions are simple to predict; the anions and cations simply exchange. Remember, however, that one of the products must precipitate, otherwise no chemical reaction has occurred. For the reaction between lead(II) nitrate and potassium iodide, the products are predicted to be lead(II) iodide and potassium nitrate. No redox occurs, and the product, lead iodide, precipitates from the solution as a bright yellow solid. The question of how do you predict this type of solubility trend is addressed in the next section.
Pb(NO$_3$)$_2$ (aq) + 2 KI(aq)< → PbI$_2$ (s) + 2 KNO$_3$ (aq)

## 5.7 Predicting Solubility Trends

The solubility of many simple ionic compounds can be predicted by applying the set of rules shown below.
* Table 5.5 Solubility Trends for Ionic Compounds
* 1. Salts of the alkali metal ions and the ammonium ion, Li$^+$, Na$^+$, K$^+$, and NH$_4$$^+$ are almost always soluble.
* 2. Virtually all metal nitrates and metal acetates are soluble.
* 3. Metal halides are generally soluble, except for salts of Ag$^+$, Pb$^{2+}$, Cu$^+$ and Hg$^+$.
* 4. Metal sulfates are generally soluble, except for salts of Ba$^{2+}$, Pb$^{2+}$ and Ca$^{2+}$.
* 5. With exception of the alkali metal ions and ammonium (Rule 1), the following salts are generally insoluble: metal carbonates (CO$_3$$^{2-}$), metal phosphates (PO$_4$$^{3-}$) and metal chromates (CrO$_4$$^{2-}$).
* 6. Metal hydroxides and metal sulfides are generally insoluble, except for those covered by Rule 1 and Ca$^{2+}$, Sr$^{2+}$ and Ba$^{2+}$.
Applying these rules to the reaction between lead nitrate and potassium iodide, the reactants are both soluble (Rule 1 and Rule 2). In the products, potassium nitrate will be soluble (Rule 2) and lead iodide will be insoluble, based on Rule 3.
Example 5.3 Predicting Solubility
Mixing each of the following salt solutions results in the formation of a precipate. In each case, identify the insoluble salt.
Exercise 5.4 Predicting Solubility
For each of the ionic compounds given below, determine whether or not the compound will be soluble in water, according to the trends given above.

## 5.8 The Energetics of Chemical Reactions

Many of the chemical reactions that we have discussed in this chapter occur with the generation of significant amounts of light and heat. A prime example is the synthesis reaction between zinc and sulfur, described by the equation shown below.
Zn (s) + S (s) → ZnS (s)
Initially, both elements are present as fine powders. They are mixed (carefully) and the mixture is stable sitting of the laboratory bench. When the mixture is touched with a heated metal rod, however, a violent reaction occurs (the reaction is termed exothermic; producing heat) and zinc sulfide is formed as the product. The reaction is obviously favorable, so why does it need heat to start the reaction? This concept can be understood by considering a reaction coordinate diagram for a simple one-step reaction (Figure 5.8). Figure 5.8
In a reaction coordinate diagram, the y-axis corresponds to energy. The initial and final “energy wells” represent the ground-state energies of the reactants and products, and the path connecting them describes the energy changes that occur in the course of the reaction.
Looking at the reaction coordinate diagram for the zinc sulfide reaction, the reactants sit at an initial energy level that is characteristic and unique for each element or compound. Likewise, zinc sulfide sits at a lower overall energy level; that means that the conversion from the elements to the compound is favorable and that heat is liberated during the reaction. If the energy level of the products was higher than the reactants, the reaction would be unfavorable and heat would be absorbed during the reaction (the reaction is said to be endothermic; consuming heat).
Why, then, does the zinc sulfide reaction need energy input before the reaction begins? The answer lies in the curved path that connects the reactants and products in the reaction coordinate diagram. In order for the zinc-sulfur mixture to react, enough energy must be put into the system so that the energy level of the reactants equals the highest hill in the diagram. Once that point is reached, the reactants can “tumble down” the energy hill and form the more stable products (with the evolution of all of the excess energy as heat, light, etc.).
The top of the energy hill in a reaction coordinate diagram is called the transition state (or activated complex). In modern chemical theory, the transition state is the energy maximum corresponding to the processes of bond-making and bond-breaking. The energy required to go from the reactants to the transition state is the activation energy. Reactions that occur with little requirement for heat simply have a small activation energy. The magnitude of the activation energy controls the rate of a reaction and the difference in energy between the reactants and the products controls the equilibrium distribution of the products and reactants in a reversible reaction. We will return to these concepts when we address reaction rates and equilibrium later in the book.

## Study Points

* The processes that occur during a chemical change can be represented using a chemical equation. In a chemical equation, the chemical formulas for the substance or substances that undergo the chemical reaction (the reactants) and the formulas for the new substance or substances that are formed (the products) are both shown, and are linked by an arrow. The arrow in a chemical equation has the properties of an “equals sign” in mathematics, and because of this, in a chemical equation, there must be the same number and types of atoms on each side of the arrow.
* A chemical equation in which the same number and types of atoms appear on each side of the arrow is called balanced. In order to balance an equation, insert coefficients in front of the appropriate reactants or products until there are the same number and types of atoms on both sides of the arrow.
* In a synthesis reaction, elements or compounds undergo reaction and combine to form a single new substance.
* In a decomposition' reaction, a single compound will break down to form two or more new substances. The substances formed can be elements, compounds, or a mixture of both. '
* In a single-replacement (single-displacement) reaction, an element and a compound will react so that their elements are switched. As a general rule, metals will replace metals in compounds and non-metals will typically replace non-metals.
* A double-replacement (double-displacement) reaction, two ionic compounds in aqueous solution switch anions and form two new compounds. In order for a chemical reaction to occur, one of the new compounds that is formed must be insoluble in water, forming a solid precipitate or a gas.
* The oxidation number of an element represents the total number of electrons which have been removed (a positive oxidation state) or added (a negative oxidation state) to get the element into its present state. The term oxidation describes the loss of electrons by an element and an increase in oxidation state; the term reduction describes the gain of electrons and a decrease in oxidation state.
* A chemical reaction in which oxidation numbers undergo a change is called a redox reaction. In a redox reaction, the element that “loses electrons” is said to be oxidized and will have an increase in its oxidation number. The element that “gains electrons” in a redox reaction is said to be reduced and will have a decrease in its oxidation number.
* In a simple synthesis reaction involving reaction of elements, the product will be a compound containing both elements. Write the product considering the common charges that the elements adopt as ions or the number of bonds that the elements typically form in molecules.
* In a simple single-replacement reaction, (in general) metals (including carbon and hydrogen) will tend to replace metals and that non-metals will replace non-metals.
* In a double-replacement reaction, the anions and cations simply of the two compounds simply exchange. In order for a reaction to occur, one of the products must precipitate, otherwise no chemical reaction has occurred. Changes in oxidation numbers do not occur in double-replacement reactions.
* Solubility trends can be predicted using a simple set of rules shown in Table 5.5; you should review there rules, remembering to apply them in order.
* The energy required to initiate a chemical reaction is called the activation energy. The greater the activation energy, the slower, or less favorable a reaction will be. The magnitude of the activation energy is directly linked to the rate of a chemical reaction.

## Supplementary Problems

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