If reaction `A` and `B` are given with same temperature and same concentration but rate of `A` is double than `B`. Pre exponential factor is same for both the reaction then difference in activation energy `E_(A) - E_(B)` is ?

Options

A. `-RTln2`
B. `RT ln2`
C. 2RT
D. `(RT)/(2)`

Answer

A

Explanation

`(r_(A))/(r_(B))=(A_(1)e^(-E_(A))//RT)/(A_(2)e^(-E_(B)//RT))`
`(2)/(1)=(e^(-E_(A))//RT)/(e^(-E_(B)//RT))`
`In2=E_(B)-E_A//RT`
`E_(B)-E_(A)=RTln2`
`E_(A)-E_(B)=-RTln2`