Unit 5: Kinetics

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

1. Question 1Foundational

   What does adding a catalyst do to the activation energy and to the enthalpy change of a reaction?

   1. AIt lowers both the activation energy and the enthalpy change
   2. BIt raises the activation energy and lowers the enthalpy change
   3. CIt leaves the activation energy unchanged and lowers the enthalpy change
   4. DIt lowers the activation energy and leaves the enthalpy change unchanged

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   Answer: D. It lowers the activation energy and leaves the enthalpy change unchanged

   A catalyst supplies an alternative pathway with a lower energy barrier, so more collisions have enough energy to react and the rate increases. Enthalpy change depends only on the energies of the reactants and products, which the catalyst does not alter, so delta H is unchanged. Because the barrier drops for both directions equally, a catalyst also cannot shift the position of equilibrium.

   Key insight: Catalysts change the path, never the starting and ending energies.

2. Question 2Foundational

   In a multistep reaction mechanism, which step determines the rate law of the overall reaction?

   1. AThe slowest step
   2. BThe fastest step
   3. CThe first step, whatever its speed
   4. DThe final step, whatever its speed

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   Answer: A. The slowest step

   The slowest elementary step is the bottleneck, so the overall reaction can proceed no faster than that step allows, which is why it is called the rate-determining step. Choosing the first step is tempting because many mechanisms happen to place the slow step first, but that is a coincidence rather than a rule. A proposed mechanism is only acceptable if its steps sum to the overall equation and its slow step reproduces the experimental rate law.

   Key insight: A mechanism must pass two tests: the steps add up, and the slow step matches the measured rate law.

3. Question 3Intermediate

   In an initial-rate experiment, doubling the concentration of reactant A while holding every other concentration constant makes the reaction rate four times as fast. What is the order of the reaction with respect to A?

   1. AFirst order
   2. BZero order
   3. CSecond order
   4. DThird order

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   Answer: C. Second order

   Rate is proportional to the concentration of A raised to the order n, so doubling the concentration multiplies the rate by 2 raised to n. Since the rate quadrupled, 2 to the n equals 4 and n is 2. First order is the tempting answer, but that would only double the rate.

   Key insight: Compare the factor by which concentration changes with the factor by which rate changes, then solve for the exponent.

4. Question 4Intermediate

   A first-order reaction has a half-life of 20. minutes. What fraction of the original reactant remains after 60. minutes?

   1. A1/8
   2. B1/6
   3. C1/4
   4. D1/3

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   Answer: A. 1/8

   Sixty minutes is three half-lives, and each half-life leaves half of what was there before, so the fraction remaining is one half cubed, or 1/8. Answering 1/3 divides by the number of half-lives instead of halving repeatedly. The half-life of a first-order reaction does not depend on starting concentration, so this result holds for any initial amount.

   Key insight: Count half-lives, then raise one half to that power; concentration decays geometrically, not linearly.

5. Question 5Advanced

   A reaction has a forward activation energy of 80. kJ/mol and an overall enthalpy change of -30. kJ/mol. What is the activation energy of the reverse reaction?

   1. A30 kJ/mol
   2. B50 kJ/mol
   3. C80 kJ/mol
   4. D110 kJ/mol

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   Answer: D. 110 kJ/mol

   Both directions pass through the same transition state, so the reverse barrier is measured from the products up to that peak. The products sit 30 kJ/mol below the reactants, so the climb is 80 plus 30, or 110 kJ/mol. Answering 50 kJ/mol subtracts instead of adds, which would only be correct if the reaction were endothermic.

   Key insight: Reverse activation energy equals forward activation energy minus delta H, so an exothermic reaction always has the larger reverse barrier.