A series of experiments is carried out with compounds C and D - AQA - A-Level Chemistry - Question 3 - 2017 - Paper 2

Using the Arrhenius equation:

$ext{ln} k = -\frac{E_a}{RT} + ext{ln} A$

From Question 3.1, we found:

$k = 2.81 \times 10^{-2} \text{ dm}^{3} \text{ mol}^{-1} \text{ s}^{-1}$

Thus, we calculate:

1. Find ln k:

$ext{ln} (2.81 \times 10^{-2}) = -3.58$

2. Substitute known values into the Arrhenius equation:

$-3.58 = -\frac{E_a}{8.31 \times 298} + 16.9$

3. Solve for $E_a$:

Rearranging gives:

$E_a = -\frac{(-3.58 - 16.9)(8.31)(298)}{1}$

Calculating:

First, calculate $(-3.58 - 16.9) = -20.48$.

Now:

$E_a = 20.48 \times 8.31 \times 298 \approx 51 \text{ kJ mol}^{-1}$

Thus, the activation energy is approximately 51 kJ mol⁻¹.