Rate Equations Exam Practice: AQA A-Level

The 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]

Examiner Tip: Always state which experiments you are comparing in your reasoning. If your calculation of the rate constant uses data from one experiment, check it against another to verify the value.

The 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]

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(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]

Examiner Tip: The reactants in the rate-determining step must match the species in the rate equation. The intermediate species (NO3 in this case) must cancel out when adding the steps together.

(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]

Examiner Tip: Pay close attention to standard units. Ea must be in J mol^-1 when substituting into the Arrhenius exponent, not kJ mol^-1. Convert back to kJ mol^-1 only when specifically requested.

Topic 3.1.9: Rate Equations Exam Practice

Structured Questions

Question 1: Initial Rates Method & Calculations

Question 2: Mechanisms & Rate-Determining Step

Question 3: Arrhenius Equation Calculations