Rate Equations Exam Practice
Exam-style practice questions with detailed mark schemes and examiner tips.
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 (S2O8^2-) and iodide ions (I^-) is represented by:
S2O8^2-(aq) + 2I^-(aq) → 2SO4^2-(aq) + I2(aq)
Initial rates data gathered at constant temperature:
• Exp 1: [S2O8^2-] = 0.010 mol dm^-3; [I^-] = 0.015 mol dm^-3; Initial Rate = 3.00 x 10^-5 mol dm^-3 s^-1
• Exp 2: [S2O8^2-] = 0.020 mol dm^-3; [I^-] = 0.015 mol dm^-3; Initial Rate = 6.00 x 10^-5 mol dm^-3 s^-1
• Exp 3: [S2O8^2-] = 0.010 mol dm^-3; [I^-] = 0.030 mol dm^-3; Initial Rate = 6.00 x 10^-5 mol dm^-3 s^-1
(a) Determine the order of reaction with respect to S2O8^2- 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 S2O8^2- is 1 [1]
- Comparing Experiment 1 and 2, [I^-] remains constant, [S2O8^2-] is doubled, and the initial rate doubles [1]
- Order with respect to I^- is 1 [1]
- Comparing Experiment 1 and 3, [S2O8^2-] remains constant, [I^-] is doubled, and the initial rate doubles [1]
(b)
- Rate equation: Rate = k [S2O8^2-] [I^-] [1]
- k = Rate / ([S2O8^2-] [I^-]) = 3.00 × 10^-5 / (0.010 × 0.015) [1]
- k = 0.20 [1]
- Units: mol^-1 dm^3 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: Rate = k [NO2]^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 Rate = k [NO2] [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^(-Ea / RT)) to show how activation energy (Ea) 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 x 10^4 K. Calculate the activation energy (Ea) of the reaction in kJ mol^-1. (R = 8.314 J K^-1 mol^-1). [3]
(c) The pre-exponential factor, A, has a value of 2.50 x 10^11 s^-1. Calculate the rate constant, k, at a temperature of 320 K. [3]
Show Mark Scheme
(a)
- ln k = ln A − (Ea / RT) [1]
- Comparing to y = mx + c, where y = ln k and x = 1/T, the gradient m = -Ea / R [1]
(b)
- Gradient = -Ea / R ⇒ Ea = -Gradient × R [1]
- Ea = -(-1.45 × 10^4) × 8.314 = 120,553 J mol^-1 [1]
- Ea = 121 kJ mol^-1 (accept 120.6 kJ mol^-1) [1]
(c)
- k = A × e^(-Ea / RT)
- k = 2.50 × 10^11 × e^(-120,553 / (8.314 × 320)) [1]
- k = 2.50 × 10^11 × e^(-45.31) [1]
- k = 5.25 × 10^-9 s^-1 (accept 5.1 × 10^-9 to 5.4 × 10^-9 s^-1) [1]