The activation energy for a reaction is $9.0 \mathrm{k} \mathrm{cals} \mathrm{mole}^{-1}$. The incre

Question

The activation energy for a reaction is $9.0 \mathrm{k} \mathrm{cals} \mathrm{mole}^{-1}$. The increase in the rate constant when its temperature is increased from 298 K to 308 K is\

Professor PC Thomas Classes · Answer

$\log \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{~T}_{2}}-\frac{1}{\mathrm{~T}_{1}}ight)$\ $=-\frac{9000}{2.303 	imes 2}\left(\frac{1}{308}-\frac{1}{298}ight)=0.213$\ $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\operatorname{antilog} 0.213=1.63$.\ The increase is 63%\