\(\ln k\left(s^{-1}\right)=14.34-\frac{1.25 \times 10^{4}}{T} K \ldots(i)\)

We know that,

\(\ln k\left(s^{-1}\right)=\ln A-\frac{E_{a}}{R T} \ldots( ii )\)

On comparing equation \((i)\) and \((ii)\), we get

\(\frac{E_{a}}{R}=1.25 \times 10^{4} \,K\)

\(\therefore E_{a}=1.25 \times 10^{4} K \times R\) \(cal\) \(K^{-1} mol ^{-1}\)

\(=1.25 \times 10^{4} K \times 2\) \(cal\) \(K^{-1} mol ^{-1}\)

\(\left[\because R=2 \operatorname{cal} K^{-1} mol ^{-1}\right]\)

\(=2.50 \times 10^{4}\) \(cal\) \(mol ^{-1}\)