12. Which process has the highest activation energy?

A. The first step of the forward reaction

B. The first step of the reverse reaction

C. The second step of the forward reaction

D. The second step of the reverse reaction

13. Which of the following components of the reaction mechanism will NEVER be present in the reaction vessel when the reaction coordinate is at point B?

A. Reactants

B. Products

C. Intermediates

D. Catalysts

14. Consider the following two reactions:

3A + 2B 3C + 4D (Reaction 1)

4D + 3C 3A + 2B (Reaction 2)

If Keq for reaction 1 is equal to 0.1, what is Keq for reaction 2?

A. 0.1

B. 1

C. 10

D. 100

15. Which of the following statements would best describe the experimental result if the temperature of the following theoretical reaction were decreased?

A + B C + D H = -1.12 kJ/mol

A. [C] + [D] would increase.

B. [A] + [B] would increase.

C. H would increase.

D. H would decrease.

16. Compound A has a Ka of approximately 10-4. Which of the following compounds is most likely to react with a solution of compound A?

A. HNO2

B. NO2

C. NH3

D. N2O5

17. The following system obeys second-order kinetics.

2NO2 NO3 + NO (slow)

NO3 + CO NO2 + CO2 (fast)

What is the rate law for this reaction?

A. Rate = k [NO2] [CO].

B. Rate = k [NO2]2 [CO].

C. Rate = k [NO2] [NO3].

D. Rate = k [NO2]2.

18. The potential energy diagram shown represents four different reactions.

Assuming identical conditions, which of the reactions displayed on the energy diagram proceeds the fastest?

A. A

B. B

C. C

D. D

Small Group Questions

1. Because catalysts affect reaction rate, why don’t we include their concentrations in reaction rate laws?

2. Do zero-order reactions contradict the collision theory of chemical kinetics?

3. How can Le Châtelier’s principle be used to manipulate reactions?

Explanations to Practice Questions

1. D

Based on the information given in the question, the rate is first-order with respect to the concentration of the first reactant; when the concentration of that reactant doubles, the rate also doubles. Because the reaction is third-order, the sum of the exponents in the rate law must be equal to 3. Therefore, the rate law is defined as follows:

Rate = k [reactant 1]1 [reactant 2]2

So the concentration of reactant 2 must be squared in order to write a rate law that represents a third-order reaction. When the concentration of reactant 2 is multiplied by ½, the rate will be multiplied by (½)2 = ¼.

2. A

Before you try to answer this question, you should draw a potential energy diagram for the system.

If the activation energy of the forward reaction is greater than the activation energy of the reverse reaction, then the products must have a higher enthalpy than the reactants. The overall energy of the system is higher at the end than it was in the beginning. The net enthalpy change is positive, indicating an endothermic reaction. Spontaneity is correlated with Gibbs free energy (G).

3. A

Carbon dioxide gas evolves and leaves the bottle, which decreases the total pressure of the reactants. Le Châtelier’s principle explains that a decrease in pressure shifts the equilibrium so as to increase the number of moles of gas present. This particular reaction will shift to the left, thereby decreasing the amount of carbonic acid and increasing the amount of carbon dioxide and water. Oxygen and nitrogen are not highly reactive and are unlikely to combine spontaneously with CO2 or H2CO3.

4. D

H is always positive for an endothermic reaction (meaning that option I is correct), and G is always negative for a spontaneous reaction (meaning that option II is correct). Based on these two facts, S can be determined by the free energy equation: G = H-TS, which can be rewritten as: = H-G. If H is positive and G is negative, H-G must be positive. This means that is positive. T (the temperature of the system in Kelvin) is always positive, so S must also be positive. Therefore, option III is also correct. (A), (B), and (C) all include true but incomplete statements.

5. C

Recall that the Ksp of a salt is equal to the product of the concentrations of each of the salt’s ions in saturated solution. Each molecule of this salt dissociates into three A+ ions and one B3- ion, so Ksp is equal to [A+] [A+] [A+] [B3-] = [A+]3[B3-]. The molar solubility of the salt is equal to the total number of moles of the salt in solution. The molar solubility of the salt is 10 M. Thus, the A+ ion is present in a concentration of 30 M (3 A+ ions per molecule), and the B3- ion is present in a concentration of 10 M (one B3- ion per molecule). These values can be substituted into the Ksp expression to yield Ksp = [30 M]3[10 M] = (27 × 103) (101) = 27 × 104 = 2.7 × 105.

6. B

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