It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol^$-$1. [nearest integer]

(Given ln10 = 2.3, R = 8.3 J K^$-$1 mol^$-$1, log 2 = 0.30)

$T_{1}=300 \mathrm{~K}$ <br/><br/> (Rate constant) <br/><br/> $\mathrm{K}_{2}=2 \mathrm{~K}_{1}$, on increase temperature by $9 \mathrm{~K}$ <br/><br/> $\mathrm{T}_{2}=309 \mathrm{~K}$ <br/><br/> $\mathrm{Ea}=?$ <br/><br/> $\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{2.3 \mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{2} \cdot \mathrm{T}_{1}}\right]$ <br/><br/> $\log 2=\frac{\mathrm{Ea}}{2.3 \times 8.3}\left[\frac{9}{309 \times 300}\right]$ <br/><br/> $\mathrm{Ea}=\frac{0.3 \times 309 \times 300 \times 2.3 \times 8.3}{9}$ <br/><br/> $=58988.1 \mathrm{~J} / \mathrm{mole}$ <br/><br/> $\simeq 59 \mathrm{~kJ} / \mathrm{mole}$