Question

The rate of a reaction $A$ doubles on increasing the temperature from $300K$ to $310K$. By how much, the temperature of reaction $B$ should be increased from $300K$ so that rate doubles if activation energy of the reaction $B$ is twice to that of reaction $A$.

• A. $9.84K$
• B. $304.92K$
• C. $2.45K$
• D. $19.67K$

Answer 1

Answer: (B)

$2=\frac{Eq}{R}\left\{\frac{1}{300}-\frac{1}{310}\right\}...\left(i\right)$
$2=e^2\frac{Ea}{R}\left\{\frac{1}{300}-\frac{1}{T}\right\}...\left(ii\right)$
$\frac{2Ea}{R}\left\{\frac{1}{300}-\frac{1}{T}\right\}=\frac{E_a}{R}\left\{\frac{1}{300}-\frac{1}{310}\right\}$
$\frac{1}{300}+\frac{1}{310}=\frac{2}{T}\Rightarrow T=\frac{300\times 310}{610}\times 2$
$=304.92$