What is the activation energy for a reaction, if its rate doubles when the temperature is raised from 20^oC to 35^oC ? $\left(\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\right)$

Given, initial temperature ${\mathrm{T}}_1=20+273=293\mathrm{K}$
Final temperature, $\begin{array}{r}{\mathrm{T}}_2=35+273=308\mathrm{K}\\ \mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\end{array}$
Since, rate becomes double on raising temperature ${\mathrm{r}}_2=2{\mathrm{r}}_1\Rightarrow \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=2$
As rate constant, $\mathrm{k}\propto \mathrm{r}\Rightarrow \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=2$
From Arrhenius equation, we know that $\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right)$
                                                      $\log 2=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{308-293}{293\times 308}\right]$
                                                   $0.3010=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{15}{293\times 308}\right]$
                                              ${\mathrm{E}}_{\mathrm{a}}=\frac{0.3010\times 2.303\times 8.314\times 293\times 308}{15}$
                                                  $=34673.48\mathrm{J}{\mathrm{mol}}^{-1}=34.7\mathrm{kJ}{\mathrm{mol}}^{-1}$