\(k=\frac{2 \cdot 303}{t} \log \frac{a}{a-x}\)
\(\begin{aligned}
\therefore \quad k_{298} &=\frac{2 \cdot 303}{t_1} \log \frac{a}{a-0 \cdot 10 a} \\
&=\frac{2 \cdot 303}{t_1} \log \frac{10}{9} \\
&=\frac{2.303}{t_1}(0.0458)=\frac{0.1055}{t_1}
\end{aligned}\)
\(\begin{aligned} t_1 &=\frac{0-1055}{k_{298}} ...(I)\\ k_{308} &=\frac{2 \cdot 303}{t_2} \log \frac{a}{a-0 \cdot 25 a} \\ &=\frac{2 \cdot 303}{t_2} \log \frac{4}{3} \\ &=\frac{2 \cdot 303}{t_2}(0 \cdot 125) \\ k_{308} &=\frac{0 \cdot 2879}{t_2} \\ \therefore \quad t_2 &=\frac{0 \cdot 2879}{k_{308}} ..(ii) \end{aligned}\)
As given \(t_1=t_2, \therefore\) comparing equation \((i)\) and equation (ii), we get
\(\begin{aligned}
\frac{0 \cdot 1055}{k_{298}} &=\frac{0.2879}{k_{308}} \\
\frac{k_{308}}{k_{298}} &=2.7289
\end{aligned}\)
Using Arrhenius equation,
\(\begin{gathered}
\log \frac{k_{308}}{k_{298}}=\frac{E_a}{2.303 \mathrm{R}}\left(\frac{T_2-T_1}{T_1 T_2}\right) \\
\log (2 \cdot 7289)=\frac{E_a}{2 \cdot 303 \times 8.314} \times\left(\frac{308-298}{298 \times 308}\right)
\end{gathered}\)
\(0.4360=\frac{E_a \times 10}{2.303 \times 8.314 \times 298 \times 308}\)
\(\therefore \quad E_a=\frac{0.4360 \times 2.303 \times 8.314 \times 298 \times 308}{10}\)
\(=76.623 \mathrm{kJmol}^{-1}\)
Calculation of \(k_{318}\) :
\(\log k=\log A-\frac{E_a}{2 \cdot 303 R T}\)
\(=\log \left(4 \times 10^{10}\right)-\frac{76623 \mathrm{Jmol}^{-1}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 318 \mathrm{~K}}\)
\(\log k=10.6021-12.5843=-1.9822\)
\(k=\) Antilog \((-1.9822)\)
\(=\) Antilog \((\overline{2} \cdot 0178)=1 \cdot 042 \times 10^{-2} \mathrm{~s}^{-1}\).