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

1. 1
   Nitrogen dioxide reacts with carbon monoxide: NO₂(g) + CO(g) → NO(g) + CO₂(g). At 500 K the rate equation is rate = k[NO₂]². A proposed mechanism has two steps, and the first step is the slower.

   (a)

   Which particles are involved in the rate-determining step?

   [1 mark]

   • AOne NO₂ and one CO
   • BOne NO₂ only
   • COne CO only
   • DTwo NO₂

   (b)

   The first step is 2NO₂ → NO₃ + NO. Which equation is the second step of a mechanism consistent with the rate equation and the overall equation?

   [1 mark]

   • ANO₃ + CO → NO₂ + CO₂
   • BNO₃ + NO → 2NO₂
   • CNO₃ + CO → NO + CO₂
   • DNO₃ → NO₂ + O

   (c)

   Explain why CO does not appear in the rate equation.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The hydrolysis of two bromoalkanes by aqueous sodium hydroxide was studied at constant temperature. For 2-bromo-2-methylpropane, doubling the concentration of the bromoalkane doubled the initial rate, but doubling the concentration of hydroxide ions had no effect on the initial rate. For 1-bromobutane, doubling the concentration of either the bromoalkane or the hydroxide ions doubled the initial rate.

   (a)

   What is the rate equation for the hydrolysis of 2-bromo-2-methylpropane?

   [1 mark]

   • Arate = k[bromoalkane][OH⁻]
   • Brate = k[OH⁻]
   • Crate = k[bromoalkane]
   • Drate = k[bromoalkane]²

   (b)

   Which of the following correctly identifies the mechanism for 1-bromobutane and gives the reason?

   [1 mark]

   • ASN1, because a carbocation intermediate forms
   • BSN2, because both reagents are in the rate-determining step
   • CSN1, because the rate is first order in hydroxide ions
   • DSN2, because it is a tertiary bromoalkane

   (c)

   Explain why the data show that 2-bromo-2-methylpropane is hydrolysed by an SN1 mechanism.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The acid-catalysed reaction of propanone with iodine is CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq). Initial rates were measured at constant temperature. Experiment 1: [CH₃COCH₃] = 1.00 mol dm⁻³, [I₂] = 0.0100 mol dm⁻³, [H⁺] = 0.50 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 2: [CH₃COCH₃] = 2.00 mol dm⁻³, [I₂] = 0.0100 mol dm⁻³, [H⁺] = 0.50 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: [CH₃COCH₃] = 1.00 mol dm⁻³, [I₂] = 0.0200 mol dm⁻³, [H⁺] = 0.50 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 4: [CH₃COCH₃] = 1.00 mol dm⁻³, [I₂] = 0.0100 mol dm⁻³, [H⁺] = 1.00 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹.

   (a)

   Deduce the order of reaction with respect to each of propanone, iodine and hydrogen ions, and write the rate equation.

   [3 marks]

   (b)

   Use the rate equation to suggest a possible mechanism for the reaction. Identify what is in the rate-determining step, and explain why iodine is not.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The overall reaction between nitrogen dioxide and fluorine is 2NO₂(g) + F₂(g) → 2NO₂F(g). Its rate equation is rate = k[NO₂][F₂]. At a certain temperature, when [NO₂] = 0.020 mol dm⁻³ and [F₂] = 0.010 mol dm⁻³, the initial rate is 4.0 × 10⁻⁶ mol dm⁻³ s⁻¹. Two students propose mechanisms. Student X suggests a single step in which two NO₂ molecules and one F₂ molecule collide. Student Y suggests two steps: step 1 (slow) NO₂ + F₂ → NO₂F + F, then step 2 (fast) NO₂ + F → NO₂F.

   (a)

   Evaluate which of the two mechanisms is more likely to be correct, using the rate equation and the overall equation.

   [6 marks]

   (b)

   Calculate the rate constant, with units. Calculate the initial rate when [NO₂] = 0.040 mol dm⁻³ and [F₂] = 0.0050 mol dm⁻³. Explain, using Student Y's mechanism, why only one of the two NO₂ molecules used affects the rate.

   [6 marks]

   Total for question 4: 12 marks

End of questions