Rate Equation Calculations Revision Notes | A Level AQA Chemistry

Worked example: Calculating k from experimental rate data

The following data were obtained for the reaction A + B → C at constant temperature.\n\nExperiment 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹\nExperiment 2: [A] = 0.20 mol dm⁻³, [B] = 0.10 mol dm⁻³, rate = 4.0 × 10⁻³ mol dm⁻³ s⁻¹\nExperiment 3: [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹\n\nDetermine the rate equation and calculate k with units.

1. Compare experiments 1 and 2: [B] is constant, [A] doubles from 0.10 to 0.20 mol dm$^{-3}$.\n$\frac{\text{rate}_2}{\text{rate}_1} = \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2$\n$\frac{[A]_2}{[A]_1} = \frac{0.20}{0.10} = 2$\nRate doubles when [A] doubles → first order in A. — [M1] Doubling [A] doubles the rate, so the order with respect to A is 1.
2. Compare experiments 1 and 3: [A] is constant, [B] doubles from 0.10 to 0.20 mol dm$^{-3}$.\n$\frac{\text{rate}_3}{\text{rate}_1} = \frac{2.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 1$\n[B] doubles but rate is unchanged → zero order in B. — [M1] No change in rate when [B] doubles, so the order with respect to B is 0.
3. rate $= k[A]^1[B]^0 = k[A]$ — [B1] The rate equation is first order overall. [B] does not appear because its order is zero.
4. $k = \frac{\text{rate}}{[A]} = \frac{2.0 \times 10^{-3}}{0.10} = 2.0 \times 10^{-2}$ — [M1] Rearrange rate = k[A] for k, then substitute values from any experiment.
5. $k = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}$ — [A1] For a first-order reaction, k has units of s⁻¹. Always derive units — never guess.

Answer: Rate equation: rate = k[A]. Rate constant k = 2.0 × 10⁻² s⁻¹

Worked example: Calculating activation energy from two k values at different temperatures

The rate constant for a reaction is 3.46 × 10⁻³ mol⁻¹ dm³ s⁻¹ at 300 K and 2.71 × 10⁻² mol⁻¹ dm³ s⁻¹ at 350 K. Calculate the activation energy Ea in kJ mol⁻¹. (R = 8.314 J mol⁻¹ K⁻¹)

1. At $T_1 = 300$ K: $\ln k_1 = \ln A - \frac{E_a}{R T_1}$ At $T_2 = 350$ K: $\ln k_2 = \ln A - \frac{E_a}{R T_2}$ — Writing both equations sets up the subtraction that eliminates ln A.
2. $\ln k_2 - \ln k_1 = -\frac{E_a}{R T_2} + \frac{E_a}{R T_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ — [M1] This rearrangement removes the unknown ln A, leaving only Ea to find.
3. $\ln k_1 = \ln(3.46 \times 10^{-3}) = -5.666$ $\ln k_2 = \ln(2.71 \times 10^{-2}) = -3.608$ $\ln k_2 - \ln k_1 = -3.608 - (-5.666) = 2.058$ — [M1] Use a calculator for ln values. Keep at least 3 significant figures throughout.
4. $\frac{1}{T_1} = \frac{1}{300} = 3.333 \times 10^{-3}$ K$^{-1}$ $\frac{1}{T_2} = \frac{1}{350} = 2.857 \times 10^{-3}$ K$^{-1}$ $\frac{1}{T_1} - \frac{1}{T_2} = 3.333 \times 10^{-3} - 2.857 \times 10^{-3} = 4.76 \times 10^{-4}$ K$^{-1}$ — [M1] Always use temperatures in Kelvin. Converting 300 K and 350 K is not needed here as they are already in K.
5. $E_a = \frac{(\ln k_2 - \ln k_1) \times R}{\frac{1}{T_1} - \frac{1}{T_2}} = \frac{2.058 \times 8.314}{4.76 \times 10^{-4}}$ $E_a = \frac{17.11}{4.76 \times 10^{-4}} = 35940$ J mol$^{-1}$ — [M1] Numerator: 2.058 × 8.314 = 17.11 J mol⁻¹. Denominator: 4.76 × 10⁻⁴ K⁻¹. Division gives J mol⁻¹.
6. $E_a = \frac{35940}{1000} = 35.9$ kJ mol$^{-1}$ — [A1] Divide by 1000 to convert J mol⁻¹ to kJ mol⁻¹. Give to 3 significant figures to match the data.

Answer: Ea = 35.9 kJ mol⁻¹

Worked example: Finding k at a new temperature using the Arrhenius equation

A reaction has an activation energy of 75 000 J mol⁻¹. At 298 K, k = 1.50 × 10⁻⁴ s⁻¹. Calculate k at 320 K. (R = 8.314 J mol⁻¹ K⁻¹)

1. $\ln k_2 - \ln k_1 = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ — This is the working form derived by subtracting the Arrhenius equation at T₁ from that at T₂.
2. $\frac{E_a}{R} = \frac{75000}{8.314} = 9021$ K — 75 000 ÷ 8.314 = 9021 K. Keep this value unrounded for the next step.
3. $\frac{1}{298} = 3.3557 \times 10^{-3}$ K$^{-1}$ $\frac{1}{320} = 3.1250 \times 10^{-3}$ K$^{-1}$ $\frac{1}{T_1} - \frac{1}{T_2} = 3.3557 \times 10^{-3} - 3.1250 \times 10^{-3} = 2.307 \times 10^{-4}$ K$^{-1}$ — [M1] T₁ = 298 K (the temperature where k is known), T₂ = 320 K (the target temperature).
4. $\ln k_2 - \ln k_1 = 9021 \times 2.307 \times 10^{-4} = 2.081$ — 9021 × 2.307 × 10⁻⁴: multiply 9021 × 2.307 = 20820, then × 10⁻⁴ = 2.082. (Slight rounding gives 2.081–2.082.)
5. $\ln k_1 = \ln(1.50 \times 10^{-4}) = -8.805$ $\ln k_2 = -8.805 + 2.081 = -6.724$ — [M1] Add the calculated difference to ln k₁ to get ln k₂.
6. $k_2 = e^{-6.724} = 1.20 \times 10^{-3}$ s$^{-1}$ — [A1] Use e^x on your calculator. The units of k₂ are the same as k₁ (s⁻¹ for a first-order reaction).

Answer: k at 320 K = 1.20 × 10⁻³ s⁻¹

Worked example: Calculating Ea from the gradient of a ln k vs 1/T graph

A graph of ln k against 1/T gives a straight line passing through the points (3.10 × 10⁻³ K⁻¹, −4.2) and (3.50 × 10⁻³ K⁻¹, −7.0). Calculate Ea in kJ mol⁻¹. (R = 8.314 J mol⁻¹ K⁻¹)

1. $\text{gradient} = \frac{\Delta(\ln k)}{\Delta(1/T)} = \frac{-7.0 - (-4.2)}{3.50 \times 10^{-3} - 3.10 \times 10^{-3}}$ $= \frac{-2.8}{0.40 \times 10^{-3}} = \frac{-2.8}{4.0 \times 10^{-4}}$ — [M1] Always use two points that are far apart on the line to minimise reading error. Show the triangle clearly.
2. $\text{gradient} = \frac{-2.8}{4.0 \times 10^{-4}} = -7000$ K — −2.8 ÷ (4.0 × 10⁻⁴) = −2.8 ÷ 0.0004 = −7000. Units: (dimensionless) ÷ K⁻¹ = K.
3. $\text{gradient} = -\frac{E_a}{R}$ $-7000 = -\frac{E_a}{8.314}$ $E_a = 7000 \times 8.314 = 58198$ J mol$^{-1}$ — [M1] Multiply both sides by −R. 7000 × 8.314 = 58 198 J mol⁻¹.
4. $E_a = \frac{58198}{1000} = 58.2$ kJ mol$^{-1}$ — [A1] Divide by 1000. Round to 3 significant figures to match the precision of the graph data.

Answer: Ea = 58.2 kJ mol⁻¹