The rate of a reaction doubles when its temperatures changes from $300 \mathrm{~K}$ to $310 \mathrm{~K}$. The activation energy of such a reaction will be $\left(R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}ight.$ and $\log 2=0.301$ )

Question

Aicharya · Accepted Answer

We have $\quad k=A \mathrm{e}^{-E_{\mathrm{a}} / R T} \Rightarrow \ln \left(k / k^{\circ}ight)=\ln \left(A / k^{\circ}ight)-E_{\mathrm{a}} / R T$
$$
\Rightarrow \quad \ln \left(k_{2} / k_{1}ight)=\frac{E_{\mathrm{a}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}ight)=\frac{E_{\mathrm{a}}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}ight) \Rightarrow E_{\mathrm{a}}=\frac{R T_{1} T_{2}}{T_{2} T_{1}}\left[2.303 \log \frac{k_{2}}{k_{1}}ight]
$$
Substituting the given values we get
$$
E_{\mathrm{a}}=\left[\frac{\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}ight)(300 \mathrm{~K})(310 \mathrm{~K})}{(310 \mathrm{~K}-300 \mathrm{~K})}ight](2.303 \log 2)=53.599 \mathrm{~J} \mathrm{~mol}^{-1}=53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$ *

The rate of a reaction doubles when its temperatures changes from $300 \mathrm{~K}$ to $310 \mathrm{~K}$.…

Solution