Kinetics Free Response Questions for AP Chemistry

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By — McGraw-Hill Professional
Updated on Feb 2, 2011

Review the following concepts if necessary:


You have 15 minutes for this question. You may use a calculator and the tables at the back of the book.

    2 ClO2(aq) + 2 OH (aq) → ClO3 (aq) + ClO2(aq) + H2O(l)

A series of experiments were conducted to study the above reaction. The initial concentrations and rates are reported in the table below.

    1. Determine the order of the reaction with respect to each reactant. Make sure you explain your reasoning.
    2. Give the rate law for the reaction.
  2. Determine the value of the rate constant, making sure the units are included.
  3. Calculate the initial rate of disappearance of ClO2 in experiment 1.
  4. The following has been proposed as a mechanism for this reaction.
      Step 1: ClO2 + ClO2 → Cl2O4
      Step 2: Cl2O4 + OH– → ClO3 – + HClO2
      Step 3: HClO2 + OH– → ClO2 – + H2O
  5. Which step is the rate–determining step? Show that this mechanism is consistent with both the rate law for the reaction and with the overall stoichiometry.

Answers and Explanations

    1. This part of the problem begins with a generic rate equation: Rate = k[ClO2]m[OH ]n. The values of the exponents, the orders, must be determined. It does not matter which is done first. If you want to begin with ClO2, you must pick two experiments from the table where its concentration changes but the OH concentration does not change. These are experiments 1 and 3. Experiment 3 has twice the concentration of ClO2 as experiment 1. This doubling of the ClO2 concentration has quadrupled the rate. The relationship between the concentration (× 2) and the rate (× 4 = × 22) indicates that the order for ClO2 is 2 (= m). Using experiments 1 and 2 (only the OH concentration changes), we see that doubling the concentration simply doubles the rate. Thus, the order for OH is 1 (= n). Give yourself 1 point for each order you got correct.
    2. Inserting the orders into the generic rate law gives: Rate = k[ClO2]2 [OH ]1,which is usually simplified to: Rate = k[ClO2]2[OH ]. Give yourself 1 point if you got this correct.
  2. Any one of the three experiments may be used to calculate the rate constant. If the problem asked for an average rate constant, you would need to calculate a value for each of the experiments and then average the values.
  3. The rate law should be rearranged to: k = Rate/[ClO2]2[OH ]. Then the appropriate values are entered into the equation. Using experiment 1 as an example:

      k = (0.166 mol/L min)/[(0.020 M)2(0.030 M)]
      = 1.3833 × 104 M/M3 min = 1.4 × 104/M2 min

    The answer could also be reported as 1.4 × 104 L2/mol2 min. You should not forget that M = mol/L.

    Give yourself 1 point for the correct numerical value. Give yourself 1 point for the correct units.

  4. The coefficients from the equation say that for every mole of ClO3 that forms, 2 mol of ClO2 reacted. Thus, the rate of ClO2 is twice the rate of ClO3. Do not forget that since ClO3 is forming, it has a positive rate, and since ClO2 is reacting it has a negative rate. This gives:
  5. Rearranging and inserting the rate from experiment 1 gives:

      Δ[ClO]/Δt = –2(0.166 mol/L min)] =–8.332 mol/L min

    Give yourself 2 points if you got the entire answer correct. You get only 1 point if the sign or units are missing.

  6. The rate–determining step must match the rate law. One approach is to determine the rate law for each step in the mechanism. This gives:
      Step 1: Rate = k[ClO2]2
      Step 2: Rate = k[Cl2O4][OH ] = k[Cl2O]2[OH ]
      Step 3: Rate = k[HClO2][OH ] = k[ClO2][OH ]2

For steps 2 and 3, the intermediates must be replaced with reactants. Step 2 gives a rate law matching the one derived in part a. Give yourself 1 point if you picked step 2, or if you picked a step with a rate law that matches a wrong answer for part a. Give yourself 1 more point if you explained the substitution of reactants for intermediates.

To see if the stoichiometry is correct, simply add the three steps together and cancel the intermediates (materials that appear on both sides of the reaction arrow).

    Step 1: ClO2 + ClO2 → Cl2O4
    Step 2: Cl2O4 + OH → ClO3 + HClO2
    Step 3: HClO2 + OH → ClO2 + H2O
    Total: 2 ClO2 + Cl2O4 + 2 OH + HClO2 → Cl2O4 + ClO3 + HClO2 + ClO2 + H2O

After removing the intermediates (Cl2O4 and HClO2):

    2 ClO2 + 2 OH → ClO3, + ClO2 + H2O

As this matches the original reaction equation, the mechanism fulfills the overall stoichiometry requirement. Give yourself 1 point if you have done this.

The total is 10 points for this question. Subtract one point if any answer has an incorrect number of significant figures.

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