### JEE Main On 8 April 2017 Question 6

##### Question: The rate of a reaction A doubles on increasing the temperature from 300 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 Online 08-04-2017]

#### Options:

A) 4.92 K

B) 9.84 K

C) 19.67 K

D) 2.45 K

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

- $ 2=\frac{Eq}{R}{ \frac{1}{300}-\frac{1}{310} } $ ?(i) $ 2=e^{2}\frac{Ea}{R}{ \frac{1}{300}-\frac{1}{T} } $ ?(ii) $ \frac{2Ea}{R}{ \frac{1}{300}-\frac{1}{T} }=\frac{E _{a}}{R}{ \frac{1}{300}-\frac{1}{310} } $ $ \frac{1}{300}+\frac{1}{310}=\frac{2}{T}\Rightarrow T=\frac{300\times 310}{610}\times 2 $ $ =304.92-300K $

$ =4.92 $