SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 8

Chemical Reactions and Thermochemistry

These skills are usually tested on the SAT Subject Test in Chemistry. You should be able to …

•   Identify the driving force for these four major types of chemical reactions, and write balanced equations for each: combination (or synthesis), decomposition (or analysis), single replacement, and double replacement.

•   Explain hydrolysis using a balanced equation.

•   Identify and explain graphically enthalpy changes in exothermic and endothermic reactions.

•   Use Hess’s Law to show the additivity of heats of reactions.

•   Calculate enthalpy from bond energies.

This chapter will review and strengthen these skills. Be sure to do the Practice Exercises at the end of the chapter.

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Know the four types of chemical reactions:

1.   Combination or synthesis

2.   Decomposition or analysis

3.   Single replacement

4.   Double replacement

The many kinds of reactions you may encounter can be placed into four basic categories: combination, decomposition, single replacement, and double replacement. The first type, combination, can also be called synthesis. This means the formation of a compound from the union of its elements. Some examples of this type are:

Zn(s) + S(s) → ZnS(s)

2H2(g) + O2(g) → 2H2O(g)

C(s) + O2(g) → CO2(g)

The second type of reaction, decomposition, can also be referred to as analysis. This means the breakdown of a compound to release its components as individual elements or other compounds. Some examples of this type are:

2H2O() → 2H2(g) + O2(g) (electrolysis of water)

C12H22O11(s) → 12C(s) + 11H2O()

2HgO(s) → 2Hg(s) + O2(g)

The third type of reaction is called single replacement or single displacement. This type can best be shown by examples in which one substance is displacing another. Some examples are:

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

The last type of reaction is called double replacement or double displacement because there is an actual exchange of “partners” to form new compounds. Some examples are:

PREDICTING REACTIONS

One of the most important topics of chemistry deals with the reasons why reactions take place. Taking each of the above types of reactions, let us see how a prediction can be made concerning how the reaction gets the driving force to make it occur.

1. Combination (Also Known as Synthesis)

The best source of information to predict a chemical combination is the heat of formation table. A heat of formation table gives the number of kilojoules evolved or absorbed when a mole (gram-formula mass) of the compound in question is formed by the direct union of its elements. In this book a positive number indicates that heat is absorbed and a negative number that heat is evolved. It makes some difference whether the compounds formed are in the solid, liquid, or gaseous state. Unless otherwise indicated (g = gas, Il_7-4480_SATChem_Ch8_0002_004.gif = liquid), the compounds are in the solid state. The values given are in kilojoules, 4.18 joules is the amount of heat needed to raise the temperature of 1 gram of water 1 degree on the Kelvin scale. The symbol ΔH is used to indicate the heat of formation.

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–ΔH = exothermic reaction

H = endothermic reaction

If the heat of formation is a large number preceded by a minus sign, the combination is likely to occur spontaneously and the reaction is exothermic. If, on the other hand, the number is small and negative or is positive, heat will be needed to get the reaction to proceed at any noticeable rate. Some examples are:

EXAMPLE 1: Zn(s) + S(s) → ZnS(s) + 202.7 kJ   ΔH = −202.7 kJ

This means that 1 mole of zinc (65 grams) reacts with 1 mole of sulfur (32 grams) to form 1 mole of zinc sulfide (97 grams) and releases 202.7 kilojoules of heat.

EXAMPLE 2: Mg(s) + O2(g) → MgO(s) + 601.6 kJ   ΔH = −601.6 kJ/mol

indicates that the formation of 1 mole of magnesium oxide requires 1 mole of magnesium and  mole of oxygen with the release of 601.6 kJ of heat. Notice the use of the fractional coefficient for oxygen. If the equation had been written with the usual whole-number coefficients, 2 moles of magnesium oxide would have been released.

Since, by definition, the heat of formation is given for the formation of 1 mole, this latter thermal equation shows 2 × (−601.6) kJ released.

EXAMPLE 3: H2(g) +  O2(g) → H2O() + 285.8 kJ   ΔH = −285.8 kJ

In combustion reactions the heat evolved when 1 mole of a substance is completely oxidized is called the heat of combustion of the substance, so, in this equation:

ΔH is the heat of combustion of carbon. Because the energy of a system is conserved during chemical activity, the same equation could be arrived at by adding the following equations:

2. Decomposition (Also Known as Analysis)

The prediction of decomposition reactions uses the same source of information, the heat of formation table. If the heat of formation is a high exothermic (ΔH is negative) value, the compound will be difficult to decompose since this same quantity of energy must be returned to the compound. A low heat of formation indicates decomposition would not be difficult, such as the decomposition of mercuric oxide with ΔH = −90.8 kJ/mole:

2HgO(s) → 2Hg(s) + O2(g) (Priestley’s method of preparation)

A high positive heat of formation indicates extreme instability of a compound, which can explosively decompose.

3. Single Replacement

A prediction of the feasibility of this type of reaction can be based on a comparison of the heat of formation of the original compound and that of the compound to be formed. For example, in a reaction of zinc with hydrochloric acid, the 2 moles of HCl have ΔH = 2 × −92.3 kJ.

EXAMPLE 1:

and the zinc chloride has ΔH = −415.5 kJ. This comparison leaves an excess of 230.9 kJ of heat given off, so the reaction would occur.

EXAMPLE 2:

In this next example, −928.4 − (−771.4) = −157.0 kJ, which is the excess to be given off as the reaction occurs:

Another simple way of predicting single replacement reactions is to check the relative positions of the two elements in the activity series below. If the element that is to replace the other in the compound is higher on the chart, the reaction will occur. If it is below, there will be no reaction.

Some simple examples of this are the following reactions.

In predicting the replacement of hydrogen by zinc in hydrochloric acid, reference to the activity series shows that zinc will replace hydrogen. This reaction would occur:

In fact, most metals in the activity series would replace hydrogen in an acid solution. If a metal such as copper were chosen, no reaction would occur.

Cu(s) + HCl(aq) → no reaction

The determination of these replacements using a quantitative method is covered in Chapter 12.

TIP 

Know how to use this activity series to predict reactions.

4. Double Replacement

For double replacement reactions to go to completion, that is, proceed until the supply of one of the reactants is exhausted, one of the following conditions must be present: (1) an insoluble precipitate is formed, (2) a nonionizing substance is formed, or (3) a gaseous product is given off.

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Know these three conditions for going to completion.

1. To predict the formation of an insoluble precipitate, you should have some knowledge of the solubilities of compounds. Table 9 gives some general solubility rules.

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Know the general solubility rules.

(A table of solubilities could also be used as reference.)

An example of this type of reaction is given in its complete ionic form.

The silver ions combine with the chloride ions to form an insoluble precipitate, silver chloride. If the reaction had been like this:

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Notice that the ions “switch” partners.

merely a mixture of the ions would have been shown in the final solution.

2. Another reason for a reaction of this type to go to completion is the formation of a nonionizing product such as water. This weak electrolyte keeps its component ions in molecular form and thus eliminates the possibility of reversing the reaction. All neutralization reactions are of this type.

(H++Cl)+(Na++OH) → H2O(l)+Na++Cl

This example shows the ions of the reactants, hydrochloric acid and sodium hydroxide, and the nonelectrolyte product water with sodium and chloride ions in solution. Since the water does not ionize to any extent, the reverse reaction cannot occur.

The third reason for double displacement to occur is the evolution of a gaseous product. An example of this is calcium carbonate reacting with hydrochloric acid:

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

3. Another example of a compound that evolves a gas in sodium sulfite with an acid is:

Na2SO3(aq)+ 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g)

In general, acids with carbonates or sulfites are good examples of this type of equation.

Hydrolysis Reactions

Hydrolysis reactions are the opposite of neutralization reactions. In hydrolysis the salt and water react to form an acid and a base. For example, if sodium chloride is placed in solution, this reaction occurs to some degree:

(Na+ + Cl) + H2O(l) → (Na+ + OH) + (H+ + Cl)

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Salt interacts with water.

In this hydrolysis reaction the same number of hydrogen ions and of hydroxide ions is released so that the solution is neutral. This occurs because sodium hydroxide is a strong base and hydrochloric acid is a strong acid. (There are charts of relative acid and base strengths to use as references.) Because they are both classified as strong, sodium hydroxide and hydrochloric acid essentially exist as ions in solutions. Therefore, the NaCl solution has an excess of neither hydrogen nor hydroxide ions, and it will test neutral. Thus, the salt of a strong acid and a strong base forms a neutral solution when dissolved in water. However, if Na2COis dissolved, we have:

(2Na+ + CO32−) + 2H2O() → (2Na+ + 2OH) + H2CO3

The H2COis written as a single entity because it is a slightly ionized acid, or, in other words, a weak acid. Since the hydroxide ions are free in the solution, the solution is basic. Notice that here it was the salt of a strong base and a weak acid that formed a basic solution. This generalization is true for this type of salt.

If we use ZnCl2, which is the salt of a strong acid and a weak base, the reaction will be:

(Zn2+ + 2Cl) + H2O(l) → (2H+ + 2Cl) + Zn(OH)2

In this case the hydroxide ions are held in the weakly ionizing compound while the hydrogen ions are free to make the solution acidic. In general, then, the salt of a strong acid and a weak base forms an acid solution by hydrolysis.

The fourth possibility is that of a salt of a weak acid and weak base dissolving in water. An example would be ammonium carbonate, (NH4)2CO3, which is the salt of a weak base and a weak acid. The hydrolysis reaction would be:

(NH4)2CO3 + 2H2O(l) → 2NH4OH + H2CO3

Both the ammonium hydroxide, NH4OH, and the carbonic acid, H2CO3, are written as nonionized compounds because they are classified as a weak base and a weak acid, respectively. Therefore, a salt of a weak acid and a weak base forms a neutral solution since neither hydrogen ion nor hydroxide ion will be present in excess.

Entropy

In many of the preceding predictions of reactions, we used the concept that reactions will occur when they result in the lowest possible energy state.

There is, however, another driving force to reactions that is related to their state of disorder or of randomness. This state of disorder is called entropy. A reaction is also driven, then, by a need for a greater degree of disorder. An example is the intermixing of gases in two connected flasks when a valve is opened to allow the two previously isolated gases to travel between the two flasks. Because temperature remains constant throughout the process, the total heat content cannot have changed to a lower energy level, and yet the gases will become evenly distributed in the two flasks. The system has thus reached a higher degree of disorder or entropy.

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Entropy is a measure of the degree of disorder.