$k=A e^{-E_{a} / R T}$

$\ln k=\ln A-\frac{E_{a}}{R T} \cdots \cdots(i i)$

Comparison of equation $(i)$ and $(ii)$

$E_{a}$ is $69$

The rate constants in Arrhenius equation at two different

temperatures are expressed as stated below

$\ln k_{1}=\ln A-\frac{E_{ a }}{ RT _{1}}$

$\ln k_{2}=\ln A-\frac{E_{ a }}{ RT _{2}}$

Therefore, the Arrhenius equation for two different temperatures is expressed as shown below.

$\ln k_{2}-\ln k_{1}=\left(\ln A-\frac{E_{ a }}{ RT _{2}}\right)-\left(\ln A-\frac{E_{ a }}{ RT _{1}}\right)$

This equation is simplified as shown below.

$\ln \frac{k_{2}}{k_{1}}=\frac{E_{2}}{ R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$

$\ln \frac{k_{2}}{k_{1}}=\frac{E_{ a }}{ R }\left(\frac{ T _{2}- T _{1}}{ T _{1} T _{2}}\right)$ or $\log \frac{k_{2}}{k_{1}}=\frac{E_{ a }}{2.303 R }\left(\frac{ T _{2}- T _{1}}{ T _{1} T _{2}}\right)$

Substitute the values in the above equation as follows.

$\log \frac{2 k_{1}}{k_{1}}=\frac{69}{2.303 \times 8.314}\left(\frac{ T _{2}-300}{300 T _{2}}\right)$

$\log 2=\frac{69 \times 10^{3}}{2.303 \times 8.314}\left(\frac{ T _{2}-300}{300 T _{2}}\right)$

$T _{2}=307.7 \,K$