MCAT General Chemistry Practice Test 6: Kinetics

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One way to determine a rate law is to look at the slowest elementary step in a reaction mechanism. The rate law is equal to the rate constant times the initial concentrations of the reactants in the slowest step raised to the power of their coefficients in the balanced equation. If a chemical appears in the rate law raised to the X power, we say the reaction is X order for that chemical.

In cases where the slow step is not the first step, the rate law will likely depend on the concentration of intermediate species. This is experimentally inconvenient, since the concentration of intermediates is not as straightforward to control as starting materials. As such, rate laws are often rewritten substituting terms consisting solely of starting materials, when possible. For example, consider the decomposition of nitramide:

O2 NNH2 (aq) → N2O(g) + H2O(l)

Reaction 1

This reaction consists of three elementary steps (shown below), with step 2 as the slow step.

Step 1 (fast equilibrium):
O2NNH2(aq) O2NNH-(aq) + H+(aq)

Step 2 (slow):
O2NNH-(aq) → N2O(g) + OH-(aq)

Step 3 (fast):
H+(aq) + OH-(aq) → H2O(l)

One could write a valid rate law for this reaction of the form:

rate = k[O2NNH-]

However, the inclusion of the intermediate term is not ideal. The fast equilibrium in Step 1 allows the substitution of [O2NNH-] according to the equilibrium condition:

Solving for [O2NNH-], and substituting into the rate law gives an equally valid expression, detailing how the rate may be altered by varying the concentration of starting material and the pH of the reaction mixture:

The mechanism for this reaction consists of three elementary steps. The first step is an equilibrium reaction with a significant back reaction and the last step is an equilibrium reaction lying so far to the right that we consider it to go to completion:

1. What is the order of the decomposition of nitramide in water with respect to H+?

  • A. Negative first order
  • B. One half order
  • C. First order
  • D. Second order

2. If Step 1 were simply a fast reaction and not a fast equilibrium, what would be the expected rate law for the decomposition of nitramide in water?

  • A. Rate = k[O2NNH2]
  • B. Rate = k[O2NNH-]
  • C. Rate = k[H+][OH-]
  • D. Rate =

3. If separately synthesized Na+[O2NNH-] were added to a reaction in progress (assuming total solubility of the salt), what effect would this have on the rate?

  • A. No reaction would be observed.
  • B. The rate of the reaction would decrease.
  • C. The rate of the reaction would increase.
  • D. It is impossible to tell without experimental data.

4. If the [H+] goes up by a factor of four, the reaction rate will

  • A. increase by a factor of four.
  • B. increase by a factor of two.
  • C. decrease by a factor of two.
  • D. decrease by a factor of four.

5. Considering Step 1 in isolation, if a known amount of O2NNH2 is dissolved in water, which of the following plays a role in determining how fast the reaction reaches equilibrium?

  • A. The pH of the solution
  • B. The reaction temperature
  • C. The magnitude of the equilibrium constant
  • D. The stability of O2NNH2 compared to O2NNH--and H+

6. What is true regarding the enthalpy and entropy changes for Step 3 of the mechanism?

  • A. ∆H > 0,S > 0
  • B. ∆H > 0,S < 0
  • C. ∆H < 0,S > 0
  • D. ∆H < 0,S < 0

7. Which of the following best describes how a catalyst might speed up a reaction?

  • A. It raises the temperature of the reaction mixture.
  • B. It raises the activation energy, making it easier to overcome the energy barrier.
  • C. It increases the rate of the slowest step.
  • D. It increases the frequency of collisions between molecules.