b
\(\ln k =10-\frac{69( KJ )}{ RT } \cdots \cdots( i )\)

\(k=A e^{-E_{a} / R T}\)

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

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.

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

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

\(T _{2}=307.7 \,K\)