Rate-determining step and reaction mechanismsAQA A-Level Chemistry: Subtopic test

1. 1
   The reaction NO₂(g) + CO(g) → NO(g) + CO₂(g) has the experimentally determined rate equation rate = k[NO₂]². A two-step mechanism is proposed. Step 1 (slow): 2NO₂ → NO₃ + NO. Step 2 (fast): NO₃ + CO → NO₂ + CO₂.

   (a)

   Which statement about the mechanism is consistent with the rate equation?

   [1 mark]

   • AOne NO₂ molecule and one CO molecule collide in the rate-determining step
   • BTwo NO₂ molecules take part in the slow step and CO reacts in a later, faster step
   • CThe slow step must be the last step of the mechanism
   • DCO is zero order, so it does not take part in the reaction

   (b)

   Which species is an intermediate in the mechanism?

   [1 mark]

   • ANO
   • BNO₂
   • CNO₃
   • DCO₂

   (c)

   Explain why CO does not appear in the rate equation even though it is a reactant.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   In acidic solution, propanone reacts with iodine: CH₃COCH₃ + I₂ → CH₃COCH₂I + HI. The experimentally determined rate equation is rate = k[CH₃COCH₃][H⁺]. The reaction is zero order with respect to iodine. Hydrogen ions are not used up.

   (a)

   Which statement about the rate-determining step is correct?

   [1 mark]

   • APropanone and H⁺ are involved in the rate-determining step or in steps before it, and iodine reacts only after it
   • BIodine is involved in the rate-determining step, because it is a reactant in the overall equation
   • COnly H⁺ ions are involved, because they are a catalyst
   • DThe rate-determining step involves all three species

   (b)

   The concentrations of propanone and iodine are both doubled at constant [H⁺] and constant temperature. By what factor does the rate change?

   [1 mark]

   • A1
   • B4
   • C8
   • D2

   (c)

   Explain why H⁺ appears in the rate equation although it does not appear in the overall equation.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Two halogenoalkanes are hydrolysed by aqueous sodium hydroxide at constant temperature. For 2-bromo-2-methylpropane, (CH₃)₃CBr, the reaction is first order with respect to the halogenoalkane and zero order with respect to hydroxide ions. For bromoethane, CH₃CH₂Br, the reaction is first order with respect to the halogenoalkane and first order with respect to hydroxide ions.

   (a)

   Use the orders of reaction for 2-bromo-2-methylpropane to explain what they show about its mechanism.

   [3 marks]

   (b)

   Use the orders of reaction for bromoethane to deduce what they show about the rate-determining step. Describe how the mechanism for the reaction fits these orders.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The reaction 2NO₂(g) + F₂(g) → 2NO₂F(g) has the experimentally determined rate equation rate = k[NO₂][F₂]. A two-step mechanism is proposed. Step 1 (slow): NO₂ + F₂ → NO₂F + F. Step 2 (fast): NO₂ + F → NO₂F.

   (a)

   Evaluate whether the proposed mechanism is consistent with the experimental rate equation, and explain why a one-step mechanism in which two NO₂ molecules and one F₂ molecule collide would be less likely.

   [6 marks]

   (b)

   In one experiment the initial rate is 4.0 × 10⁻⁵ mol dm⁻³ s⁻¹ when [NO₂] = 0.010 mol dm⁻³ and [F₂] = 0.020 mol dm⁻³. Calculate the rate constant, with units. Calculate the rate when [NO₂] = 0.030 mol dm⁻³ and [F₂] = 0.050 mol dm⁻³. State the effect on the rate of doubling [NO₂] while halving [F₂], and explain why a catalyst that speeds up only step 2 would not change the rate.

   [6 marks]

   Total for question 4: 12 marks

End of questions