The rate of a gaseous reaction triples when temperature is increased from $\displaystyle 17^{\circ} \mathrm{C}$ to $\displaystyle 27^{\circ} \mathrm{C}$. Calculate the energy of activation for this reaction. [Given : $\displaystyle 2.303 \mathrm{R}=19 \cdot 15 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \log 3=0 \cdot 48$ ]

\[\begin{aligned} & \log \frac{k_{2}}{k_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{2 \cdot 303 \mathrm{R}}\left[\frac{1}{\mathrm{T}_{1}}-\frac{1}{\mathrm{T}_{2}}\right] \text { or } \frac{\mathrm{E}_{\mathrm{a}}}{2 \cdot 303 \mathrm{R}}\left[\frac{\mathrm{~T} 2-\mathrm{T} 1}{\mathrm{T}_{1} \mathrm{T} 2}\right] \\ & \log \frac{3 k_{1}}{k_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{19 \cdot 15}\left[\frac{1}{290}-\frac{1}{300}\right] \\ & 0 \cdot 48=\frac{\mathrm{E}_{\mathrm{a}}}{19 \cdot 15}\left[\frac{10}{290 \times 300}\right] \\ & \mathrm{E}_{\mathrm{a}}=\frac{0 \cdot 48 \times 19 \cdot 15 \times 290 \times 300}{10} \\ & \mathrm{E}_{\mathrm{a}}=79970 \mathrm{Jmol}^{-1} \mathrm{or}^{2} 9.970 \mathrm{kJmol}^{-1} \end{aligned} \]