Structured Questions
Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.
Question 1: Initial Rates Method & Calculations
8 marksThe reaction between peroxodisulfate(VIII) ions (S₂O₈²⁻) and iodide ions (I⁻) is represented by:
S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)
Initial rates data gathered at constant temperature:
• Exp 1: [S₂O₈²⁻] = 0.010 mol dm⁻³; [I⁻] = 0.015 mol dm⁻³; Initial Rate = 3.00 × 10⁻⁵ mol dm⁻³ s⁻¹
• Exp 2: [S₂O₈²⁻] = 0.020 mol dm⁻³; [I⁻] = 0.015 mol dm⁻³; Initial Rate = 6.00 × 10⁻⁵ mol dm⁻³ s⁻¹
• Exp 3: [S₂O₈²⁻] = 0.010 mol dm⁻³; [I⁻] = 0.030 mol dm⁻³; Initial Rate = 6.00 × 10⁻⁵ mol dm⁻³ s⁻¹
(a) Determine the order of reaction with respect to S₂O₈²⁻ and I⁻. Explain your reasoning. [4]
(b) Write the rate equation for the reaction, and calculate the value of the rate constant, \(k\), stating its units. [4]
Show Mark Scheme
(a)
- Order with respect to S₂O₈²⁻ is 1 [1]
- Comparing Experiment 1 and 2, [I⁻] remains constant, [S₂O₈²⁻] is doubled, and the initial rate doubles [1]
- Order with respect to I⁻ is 1 [1]
- Comparing Experiment 1 and 3, [S₂O₈²⁻] remains constant, [I⁻] is doubled, and the initial rate doubles [1]
(b)
- Rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) [1]
- \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{3.00 \times 10^{-5}}{0.010 \times 0.015}\) [1]
- \(k = \mathbf{0.20}\) [1]
- Units: mol⁻¹ dm³ s⁻¹ [1]
Question 2: Mechanisms & Rate-Determining Step
6 marksThe reaction between nitrogen dioxide and carbon monoxide is represented by:
NO₂(g) + CO(g) → NO(g) + CO₂(g)
The rate equation is determined experimentally to be: \(\text{Rate} = k[\text{NO}_2]^2\).
(a) Explain why this reaction cannot occur in a single step. [2]
(b) Suggest a two-step mechanism that is consistent with the rate equation, identifying the rate-determining step. [4]
Show Mark Scheme
(a)
- If the reaction occurred in a single step, the rate equation would be \(\text{Rate} = k[\text{NO}_2][\text{CO}]\) (reflecting the stoichiometry of the reactants) [1]
- However, CO is zero order in the experimental rate equation, meaning it does not participate in the rate-determining step [1]
(b)
- Step 1 (Slow / Rate-determining step): NO₂ + NO₂ → NO₃ + NO [2] (1 mark for reactants matching rate equation [NO₂]²; 1 mark for correct products NO₃ and NO)
- Step 2 (Fast step): NO₃ + CO → NO₂ + CO₂ [1]
- Adding both steps gives the overall equation: NO₂ + CO → NO + CO₂ [1]
Question 3: Arrhenius Equation Calculations
8 marks(a) Rearrange the Arrhenius equation (\(k = A e^{-E_a / RT}\)) to show how activation energy (\(E_a\)) can be determined from a graph of \(\ln k\) against \(1/T\). [2]
(b) A plot of \(\ln k\) against \(1/T\) gave a straight line with a gradient of −1.45 × 10⁴ K. Calculate the activation energy (\(E_a\)) of the reaction in kJ mol⁻¹. (\(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)). [3]
(c) The pre-exponential factor, \(A\), has a value of 2.50 × 10¹¹ s⁻¹. Calculate the rate constant, \(k\), at a temperature of 320 K. [3]
Show Mark Scheme
(a)
- \(\ln k = \ln A - \frac{E_a}{RT}\) [1]
- Comparing to \(y = mx + c\), where \(y = \ln k\) and \(x = 1/T\), the gradient \(m = \mathbf{-\frac{E_a}{R}}\) [1]
(b)
- \(\text{Gradient} = -\frac{E_a}{R} \Rightarrow E_a = -\text{gradient} \times R\) [1]
- \(E_a = -(-1.45 \times 10^4) \times 8.314 = 120,553\text{ J mol}^{-1}\) [1]
- \(E_a = \mathbf{121\text{ kJ mol}^{-1}}\) (accept 120.6 kJ mol⁻¹) [1]
(c)
- \(k = A e^{-E_a / RT}\)
- \(k = 2.50 \times 10^{11} \times e^{-120,553 / (8.314 \times 320)}\) [1]
- \(k = 2.50 \times 10^{11} \times e^{-45.31}\) [1]
- \(k = \mathbf{5.25 \times 10^{-9}\text{ s}^{-1}}\) (accept 5.1 × 10⁻⁹ to 5.4 × 10⁻⁹ s⁻¹) [1]