$K=A e^{-E_a / R T}$

After taking In both side

$$\ln \mathrm{K}=\ln \mathrm{A}-\frac{E_a}{R T}$$

$$\ln K_1=\ln A-\frac{E_a}{R T_1}$$ at temp. $T_1$ .... (i)

$$\ln K_2=\ln A-\frac{E_a}{R T_2}$$ at temp. $T_2$ .... (ii)

$$\begin{aligned} & \text { (ii) }- \text { (i) } \\ & \ln K_2-\ln K_1=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ & \ln \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{1}{500}-\frac{1}{700}\right] \\ & \ln \frac{0.14}{0.04}=\frac{E_a}{R}\left[\frac{700-500}{500 \times 700}\right] \\ & \ln \frac{14}{4}=\frac{E_a}{R}\left[\frac{200}{500 \times 700}\right] \\ & \log 3.5=\frac{E_a}{2.303 \times R}\left[\frac{1}{250 \times 7}\right] \\ & 0.5441=\frac{E_a}{2.303 \times 8.31}\left[\frac{1}{250 \times 7}\right] \\ & E_a=0.5441 \times 8.31 \times 250 \times 7 \times 2.303 \\ & =0.5441 \times 83.1 \times 25 \times 7 \times 2.303 \\ & =18222.65 \\ & \approx 18219 \mathrm{~J} \end{aligned}$$