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

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

JEE MainJEE Main 2017Chemical Kinetics

Solution:

$2=\frac{Eq}{R}\left\{\frac{1}{300}-\frac{1}{310}\right\} ...\left(i\right)$
$2=e^{2} \frac{Ea}{R}\left\{\frac{1}{300}-\frac{1}{T}\right\} ...\left(ii\right)$
$\frac{2Ea}{R}\left\{\frac{1}{300}-\frac{1}{T}\right\}=\frac{E_{a}}{R}\left\{\frac{1}{300}-\frac{1}{310}\right\}$
$\frac{1}{300}+\frac{1}{310}=\frac{2}{T} \Rightarrow T=\frac{300\times310}{610}\times2$
$=304.92$