SAT Chemistry Chemical Reactions and Thermochemistry - Predicting Reactions

SAT Chemistry Chemical Reactions and Thermochemistry - 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)
A good 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. For chemists, 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, ℓ, = 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 unit on the Kelvin scale. The symbol ΔH_ƒ, is used to indicate the heat of formation.
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_ƒ = -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) + 1/2 O₂ (g) MgO(s) + 601.6 kj    ΔH_ƒ = -601.6 kj/mol

indicates that the formation of 1 mole of magnesium oxide requires 1 mole of magnesium and 1/2 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.
2Mg(s) + O₂(g) → 2MgO(s) + 2 (+601.6) kj
Since, by definition, the heat of formation is given for the formation of 1 mole, this latter thermal equation shows 2 x (-601.6) kj released.

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 relatively low and negative heat of formation indicates decomposition would not be difficult, such as the decomposition of mercuric oxide with ΔH_ƒ = -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_ƒ = 2 x -92.3 kj.

Example 1 ______________________________________

Zn(s) + 2HCl(aq) —> ZnCl₂(aq) + H2(g)               Note: ΔH_ƒ = 0 for elements
2 x -92.3 kj
-184.6 kj
and the zinc chloride has ΔH_ƒ = -415.5 kj. This comparison leaves an excess of 230.9 kj of heat given off, so the reaction would likely occur.

Example 2 ______________________________________
In this next example, -928.4 kj - (-771.4) kj = -157.0 kj, which is the excess to be given off as the reaction occurs:
Fe(s) + CuSO₄(aq) —> FeSO₄(aq) + Cu(s)
-771.4 kj         - 928.4 kj

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:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
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, viewing them as the result of a transfer of electrons, is covered in Chapter 12.


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.
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.

(A table of solubilities could also be used as reference.)
An example of this type of reaction is given in its complete ionic form.
(K⁺ + Cl^-) + (Ag⁺ + NO₃^-) → AgCl (s) + (K⁺ + NO₃^-)

The silver ions combine with the chloride ions to form an insoluble precipitate, silver chloride. If the reaction had been like this:
(H⁺ + Cl^-) + (Na⁺ + NO₃^-) -> K⁺ + NO₃^- + Na⁺ + Cl^-
 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. Salt interacts with water.

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^-)

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 on pages 246 and 247 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 Na₂CO₃ is dissolved, we have:
(2Na⁺ + CO₃^2- ) + 2H₂O(1) → (2Na⁺ + 2Cl^-) + H₂CO₃

The H₂CO₃ is 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(1) → 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, a more fundamental driving force to reactions that is related to their state of disorder or of randomness. This measure of disorder is called entropy. A reaction is ultimately 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 fl asks. 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. A quantitative treatment of entropy is given on page 236.