According to the Arrhenius equation, $k = A e^{-E_a/RT}$ Taking the natural logarithm on both sides yields: $\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A$ Comparing this with the equation of a straight line, $y = mx + c$, the plot of $\ln k$ versus $1/T$ gives a straight line with slope $m = -\frac{E_a}{R}$. Given slope $= -5 \times 10^3 \text{ K}$ $-\frac{E_a}{R} = -5 \times 10^3 \text{ K}$ $E_a = 5 \times 10^3 \times R$ $E_a = 5 \times 10^3 \text{ K} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} = 41570 \text{ J mol}^{-1} = 41.57 \text{ kJ mol}^{-1} \approx 41.5 \text{ kJ mol}^{-1}$.