A reaction has a rate constant of 5.4 × 10⁻⁴ s⁻¹ at 285 K and 6.8 × 10⁻³ s⁻¹ at 310 K…

Step 1. The ratio. $\dfrac{k_2}{k_1} = \dfrac{6.8 \times 10^{-3}}{5.4 \times 10^{-4}} = 12.6$.

Step 2. Its logarithm. $\log 12.6 = 1.10$.

Step 3. The temperature part. $\dfrac{1}{285\ \text{K}} - \dfrac{1}{310\ \text{K}} = 2.830 \times 10^{-4}\ \text{K}^{-1}$.

Step 4. The activation energy. $E_a = \dfrac{2.303 \times 8.314\ \text{J/(mol K)} \times 1.10}{2.830 \times 10^{-4}\ \text{K}^{-1}} = 74\,400\ \text{J/mol}$.