Question

The rate of a reaction doubles when its temperature changes from $300K$ to $310K$. Activation energy of such a reaction will be:
$(R=8.314J{K}^{-1}mo{l}^{-1}$ and $lo{g}_{10}2=0.301)$

• A. $58.5kJmo{l}^{-1}$
• B. $60.5kJmo{l}^{-1}$
• C. $53.6kJmo{l}^{-1}$
• D. $48.6kJmo{l}^{-1}$

Answer 1

Answer: (C)

$53.6kJmo{l}^{-1}$

$\because \frac{r_2}{r_1}=\frac{k_2}{k_1}=2$
Now, $2.303lo{g}_{10}\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$2.303lo{g}_{10}\left(2\right)=\frac{E_a}{8.314}\times \left[\frac{310-300}{310\times 300}\right]$
$\therefore E_a=\frac{2.303\times 0.301\times 8.314\times 310\times 300}{10}=53598.6J$$=53.6kJmo{l}^{-1}$