Using Arrhenius equation, \(k=A e^{-E_a / R T}\)
\(\ln k=\ln A-\frac{E_a}{R T}\)
\(\quad \log k=\log A-\frac{E_a}{2 \cdot 303 R T}\) ...(i)
\(\log k=14 \cdot 34-1.25 \times 10^4 \mathrm{~K} / T\) {given} (ii)
By compairing equations (i) and (ii), we get
\(\Rightarrow \frac{E_a}{2 \cdot 303 R T}=\frac{1.25 \times 10^4}{T}\)
\(E_a=\frac{2.303 R T \times 1.25 \times 10^4}{T}\)
\(=2 \cdot 303 \times\left(8 \cdot 314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \times 1.25 \times 10^4 \mathrm{~K}\)
\(=239 \cdot 34 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
For \(t_{1 / 2}=256 \mathrm{~min}\),
\(\begin{aligned}
k &=\frac{0-693}{t_{1 / 2}}=\frac{0 \cdot 693}{256 \mathrm{~min}}=\frac{0 \cdot 693}{256 \times 60 \mathrm{~s}} \\
&=4.51 \times 10^{-5} \mathrm{~s}^{-1} \\
k &=4.51 \times 10^{-5} \mathrm{~s}^{-1}
\end{aligned}\)
Putting this value of \(k\) in equation \((i)\) we get,
\(\begin{aligned}
&\log \left(4.51 \times 10^{-5}\right)=14.34-\frac{1.25 \times 10^4 \mathrm{~K}}{T} \\
&(-5+0.6542)=14.34-\frac{1.25 \times 10^4 \mathrm{~K}}{T} \\
&\frac{1.25 \times 10^4 \mathrm{~K}}{T}=18.6858 \\
&T=669 \mathrm{~K} .
\end{aligned}\)