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. `-RT ln 2`
B. `RT ln 2`
C. `2RT`
D. `(RT)/(2)`

Answer

A

Explanation

`(r_(A))/(r_(B)) = (A_(1)e^(-E_(A)//RT))/(A_(2)e^(-E_(a)//RT))`
`(2)/(1) = (e^(-E_(A)//RT))/(e^(-E_(B)//RT))`
`ln2 = E_(B) - E_(A)//RT`
`E_(B) - E_(A) - RTln2`
`E_(A) - E_(B) = -RTln2`