question_answer

A first order reaction at 298 K can be expressed by generalisation log \[K(In\,{{S}^{-1}})=17.64-\frac{2.5\times {{10}^{4}}}{T},\] then activation energy \[({{K}_{a}})\] of the reaction will be [K =Rate constant of reaction; T= Kelvin temperature]

A)  \[479\text{ }kJ\text{ }mo{{l}^{-1}}\]      

B)  \[100\text{ }kJ\text{ }mo{{l}^{-1}}\]

C)  \[50\,kJ\,\,mo{{l}^{-1}}\]                            

D)  \[25\text{ }kJ\text{ }mo\]

Correct Answer: A

Solution :