Kinetics Multiple Choice Review Questions for AP Chemistry (page 2)

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By — McGraw-Hill Professional
Updated on Feb 9, 2011

Answers and Explanations

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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)
  6. C—The value will be decreased by one-half for each half-life. Using the following table:
  7. Four half-lives = 4(200 s) = 800 s

  8. C—This is the definition of the activation energy.
  9. B—The friction supplies the energy needed to start the reaction. The energy needed to start the reaction is the activation energy.
  10. 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].
  11. 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.
  12. 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.
  13. 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.
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