1. 1
   0.500 g of magnesium powder is added to 100 cm³ of 0.500 mol dm⁻³ sulfuric acid at 20 °C: Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g). A conductivity probe, calibrated against concentration, shows that the concentration of sulfuric acid falls to 0.420 mol dm⁻³ in the first 20.0 s. The experiment is then repeated at 30 °C and, separately, with 1.00 mol dm⁻³ sulfuric acid at 20 °C.

   (a)

   What is the mean rate of consumption of sulfuric acid over the first 20.0 s?

   [1 mark]

   • A2.10 × 10⁻² mol dm⁻³ s⁻¹
   • B2.50 × 10² mol dm⁻³ s⁻¹
   • C4.00 × 10⁻³ mol dm⁻³ s⁻¹
   • D8.00 × 10⁻² mol dm⁻³ s⁻¹

   (b)

   How does the Maxwell–Boltzmann distribution for the solution particles at 30 °C compare with that at 20 °C?

   [1 mark]

   • AThe peak is higher and at a lower energy; the area under the curve is unchanged.
   • BThe peak is lower and at a higher energy; the area under the curve is larger.
   • CThe peak is at the same energy but lower; the area under the curve is smaller.
   • DThe peak is lower and at a higher energy; the area under the curve is unchanged.

   (c)

   Explain, in terms of collision theory, why using 1.00 mol dm⁻³ sulfuric acid at 20 °C increases the initial rate.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The acid-catalysed reaction of propanone with iodine is CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + H⁺(aq) + I⁻(aq). Initial rates were measured at 25 °C. Experiment 1: [CH₃COCH₃] = 0.40, [I₂] = 0.0020, [H⁺] = 0.20 mol dm⁻³, rate = 2.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 2: as experiment 1 but [CH₃COCH₃] = 0.80 mol dm⁻³, rate = 5.6 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: as experiment 1 but [I₂] = 0.0040 mol dm⁻³, rate = 2.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 4: as experiment 1 but [H⁺] = 0.40 mol dm⁻³, rate = 5.6 × 10⁻⁶ mol dm⁻³ s⁻¹.

   (a)

   Propanone and H⁺ are now used in large excess, so only [I₂] changes significantly during the reaction. Which describes how [I₂] changes with time?

   [1 mark]

   • A[I₂] falls in a straight line until the iodine is used up.
   • B[I₂] falls along a curve with a constant half-life.
   • C[I₂] falls along a curve whose half-life gets longer as the reaction proceeds.
   • D[I₂] stays constant because iodine does not appear in the rate equation.

   (b)

   What is the initial rate when [CH₃COCH₃] = 1.20, [I₂] = 0.0060 and [H⁺] = 0.10 mol dm⁻³?

   [1 mark]

   • A1.4 × 10⁻⁶ mol dm⁻³ s⁻¹
   • B4.2 × 10⁻⁶ mol dm⁻³ s⁻¹
   • C8.4 × 10⁻⁶ mol dm⁻³ s⁻¹
   • D1.3 × 10⁻⁵ mol dm⁻³ s⁻¹

   (c)

   Deduce the rate equation, and calculate the rate constant, k, including its units.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The decomposition 2N₂O₅(g) → 4NO₂(g) + O₂(g) is first order, and its rate constant k (in s⁻¹) was measured at several temperatures. A plot of ln k against 1/T, with T in kelvin, is a straight line with a gradient of −1.24 × 10⁴ K and an intercept on the ln k axis of 31.0. Use R = 8.31 J K⁻¹ mol⁻¹.

   (a)

   Determine the activation energy of the reaction, in kJ mol⁻¹.

   [3 marks]

   (b)

   Determine the Arrhenius factor, A, with units, and calculate the rate constant at 318 K.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   Below 500 K, carbon monoxide reacts with nitrogen dioxide: NO₂(g) + CO(g) → NO(g) + CO₂(g), ΔH = −226 kJ mol⁻¹. Experiments show that doubling [NO₂] at constant [CO] makes the initial rate four times larger, whereas doubling [CO] at constant [NO₂] leaves the initial rate unchanged. Two mechanisms are proposed. Mechanism 1: a single step, NO₂ + CO → NO + CO₂. Mechanism 2: step 1, NO₂ + NO₂ → NO₃ + NO (slow); step 2, NO₃ + CO → NO₂ + CO₂ (fast).

   (a)

   Deduce the rate equation and the overall order of reaction, and evaluate the two proposed mechanisms.

   [6 marks]

   (b)

   Describe, in words, the energy profile for mechanism 2. In your answer, distinguish between NO₃ and a transition state, and comment on the molecularity of each step and why a termolecular mechanism for this reaction would be unlikely.

   [6 marks]

   Total for question 4: 12 marks

End of questions