The specific reaction rate of a chemical reaction at 303K and 273 K are respectively $2.45\times {10}^{-5 }$and $1.62\times {10}^{-5 }$. What energy of activation of the reaction ?  

Its given that

 $$\begin{gathered} {\mathrm{T}}_1=303\mathrm{K} \mathrm{and} {\mathrm{T}}_2=273\mathrm{K}, \\ {\mathrm{K}}_1=2.45\times {10}^{-5} \mathrm{and} {\mathrm{K}}_1=1.62\times {10}^{-5} \end{gathered}$$

Now we know that the relation between 

$$\begin{gathered} K_{1 },K_{2 },E_a, T_1, T_{2 }is \\ \log \frac{K_2}{K_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ Put values of cons\tan ts in above expression, \\ \log \frac{1.62\times {10}^{-5}}{2.45\times {10}^{-5}}=\frac{E_a}{2.303\times 8.314}\left[\frac{1}{303}-\frac{1}{273}\right] \\ on solving we get , \\ {E}_a=9484.57 J \\ {E}_a= 9.48457 kJ \\ hence option 4 is correct . \end{gathered}$$