$\begin{array}{1 1}(a)\;20.23KJ\\(b)\;21.9KJ\\(c)\;22.97KJ\\(d)\;23.97KJ\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$K=Ae^{-E/RT}$

Applying Arrehenius equation

At two point we will get

$\log\large\frac{K_2}{K_1}=\frac{E_a}{2.303R}\bigg[\large\frac{1}{T_1}-\frac{1}{T_2}\bigg]$

Here $\large\frac{K_2}{K_1}$$=10$

$T_1=300K$

$T_2=400K$

So $\log 10=\large\frac{E_a}{2.303\times 8.314}$$\bigg[\large\frac{1}{300}-\frac{1}{400}\bigg]$

$E_a=\large\frac{1\times 2.303\times 8.314\times 300\times 400}{100}$

$E_a=22.97KJ$

Hence (c) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...