Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20°C\text{to}35°C$? $(R=8.314Jmo{l}^{-1}{K}^{-1})$ $(\text{Given:}log2=0.3010)$

• A. $15.1kJmo{l}^{-1}$
• B. $342kJmo{l}^{-1}$
• C. $269kJmo{l}^{-1}$
• D. $34.7kJmo{l}^{-1}$

Answer 1

The correct option is D $34.7kJmo{l}^{-1}$

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$

$\frac{r_2}{r_1}=\frac{K_2}{K_1}=2$
$\Rightarrow K_2=2K_1$

$\log 2=\frac{Ea}{2.303\times 8.314}(\frac{1}{293}-\frac{1}{308})$
$\Rightarrow log2=\frac{E_a}{2.303\times 8.314}\left(\frac{308-293}{293\times 308}\right)$
$\Rightarrow log2=\frac{E_a}{2.303\times 8.314}\left(\frac{15}{293\times 308}\right)$
$\Rightarrow Ea=34.7kJmo{l}^{-1}$