Kinetics Multiple Choice Review Questions for AP Chemistry (page 2)
Review the following concepts if necessary:
- Rates of Reaction for AP Chemistry
- Integrated Rate Laws for AP Chemistry
- Reaction Mechanisms for AP Chemistry
- Activation Energy and Catalysts for AP Chemistry
You have 15 minutes. You may not use a calculator. You may use the periodic table at the back of the book. For each question, circle the letter of your choice.
- A reaction follows the rate law: Rate = k[A]2. Which of the following plots will give a straight line?
- 1/[A] versus 1/time
- [A]2 versus time
- 1/[A] versus time
- ln[A] versus time
- [A] versus time
- For the following reaction: NO2(g) + CO(g) → NO(g) + CO2(g), the rate law is: Rate = k[NO2]2. If a small amount of gaseous carbon monoxide (CO) is added to a reaction mixture that was 0.10 molar in NO2 and 0.20 molar in CO, which of the following statements is true?
- Both k and the reaction rate remain the same.
- Both k and the reaction rate increase.
- Both k and the reaction rate decrease.
- Only k increases, the reaction rate remains the same.
- Only the reaction rate increases; k remains the same.
- The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains after 28 seconds?
- 0.250 mg
- 0.125 mg
- 0.0625 mg
- 0.375 mg
- 0.500 mg
- The slow rate of a particular chemical reaction might be attributed to which of the following?
- a low activation energy
- a high activation energy
- the presence of a catalyst
- the temperature is high
- the concentration of the reactants are high
- The steps below represent a proposed mechanism for the catalyzed oxidation of CO by O3.
- Step 1: NO2(g) + CO(g) → NO(g) + CO2(g)
- Step 2: NO(g) + O3(g) → NO2(g) + O2(g)
- CO2 and O2
- NO and CO2
- NO2 and O2
- NO and O2
- NO2 and CO2
- The decomposition of ammonia to the elements is a first–order reaction with a half–life of 200 s at a certain temperature. How long will it take the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm?
- 200 s
- 400 s
- 800 s
- 1000 s
- 1200 s
- The energy difference between the reactants and the transition state is
- the free energy
- the heat of reaction
- the activation energy
- the kinetic energy
- the reaction energy
- The purpose of striking a match against the side of a box to light the match is
- to supply the free energy for the reaction
- to supply the activation energy for the reaction
- to supply the heat of reaction
- to supply the kinetic energy for the reaction
- to catalyze the reaction
- The table below gives the initial concentrations and rate for three experiments.
- Rate = k[CO]
- Rate = k[CO]2[Cl2]
- Rate = k[Cl2]
- Rate = k[CO][Cl2]2
- Rate = k[CO][Cl2]
- The reaction (CH3)3CBr(aq) + H2O(l) → (CH3)3COH(aq) + HBr(aq) follows the rate law: Rate = k[(CH3)3CBr]. What will be the effect of decreasing the concentration of (CH3)3CBr?
- The rate of the reaction will increase.
- More HBr will form.
- The rate of the reaction will decrease.
- The reaction will shift to the left.
- The equilibrium constant will increase.
- When the concentration of H+(aq) is doubled for the reaction H2O2(aq) + 2 Fe2+(aq) + 2 H+(aq) → 2 Fe3+(aq) + 2 H2O(g), there is no change in the reaction rate. This indicates
- the H+ is a spectator ion
- the rate-determining step does not involve H+
- the reaction mechanism does not involve H+
- the H+ is a catalyst
- the rate law is first order with respect to H+
- The mechanism below has been proposed for the reaction of CHCl3 with Cl2.
- Step l: Cl2 2 Cl(g) fast
- Step 2: Cl(g) + CHCl3(g) → CCl3(g) + HCl(g) slow
- Step 3: CCl3(g) + Cl(g) → CCl4(g) fast
- Rate = k[Cl2]
- Rate = k[CHCl3][Cl2]
- Rate = k[CHCl3]
- Rate = k[CHCl3]/[Cl2]
- Rate = k[CHCl3][Cl2]1/2
What are the overall products of the catalyzed reaction?
The reaction is CO(g) + Cl2(g) → COCl2(g). What is the rate law for this reaction?
Which of the following rate laws is consistent with this mechanism?
Answers and Explanations
- C—The "2" exponent means this is a second-order rate law. Second-order rate laws give a straight-line plot for 1/[A] versus t.
- A—The value of k remains the same unless the temperature is changed or a catalyst is added. Only materials that appear in the rate law, in this case NO2, will affect the rate. Adding NO2 would increase the rate, and removing NO2 would decrease the rate. CO has no effect on the rate.
- B—The half-life is 0.693/k = 0.693/0.049 s–1 = 14 s. The time given, 28 s, represents two half-lives. The first half-life uses one-half of the isotope, and the second half-life uses one-half of the remaining material, so only one-fourth of the original material remains.
- B—Slow reactions have high activation energies. High activation energies are often attributed to strong bonds within the reactant molecules. All the other choices give faster rates.
- A—Add the two equations together: NO2(g) + CO(g) + NO(g) + O3(g) → NO(g) + CO2(g) + NO2(g) + O2(g) Then cancel identical species that appear on opposite sides: CO(g) + O3(g) → CO2(g) + O2(g)
- C—The value will be decreased by one-half for each half-life. Using the following table:
- C—This is the definition of the activation energy.
- B—The friction supplies the energy needed to start the reaction. The energy needed to start the reaction is the activation energy.
- E—Beginning with the generic rate law: Rate = k[CO]m[Cl2]n, it is necessary to determine the values of m and n (the orders). Comparing Experiments 2 and 3, the rate doubles when the concentration of CO is doubled. This direct change means the reaction is first order with respect to CO. Comparing Experiments 1 and 3, the rate doubles when the concentration of Cl2 is doubled. Again, this direct change means the reaction is first order. This gives: Rate = k[CO]1[Cl2]1 = k[CO][Cl2].
- C—The compound appears in the rate law, and so a change in its concentration will change the rate. The reaction is first order in (CH3)3CBr, so the rate will change directly with the change in concentration of this reactant. There is no equilibrium arrow, so the reaction is not in equilibrium. If the reaction were in equilibrium were in equilibrium, D would also be true.
- B—All substances involved, directly or indirectly, in the rate-determining step will change the rate when their concentrations are changed. The ion is required in the balanced chemical equation, so it cannot be a spectator ion, and it must appear in the mechanism. Catalysts will change the rate of a reaction. Since H+ does not affect the rate, the reaction is zero order with respect to this ion.
- E—The rate law depends on the slow step of the mechanism. The reactants in the slow step are Cl and CHCl3 (one of each). The rate law is first order with respect to each of these. The Cl is half of the original reactant molecule Cl2. This replaces the [Cl] in the rate law with [Cl2]1/2. Do not make the mistake of using the overall reaction to predict the rate law.
Four half-lives = 4(200 s) = 800 s
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