The rate constants `k_(1)` and `k_(2)` of two reactions are in the ratio `2:1`. The corresponding energies of acativation of the two reaction will be related by

Options

A. `E_(1) gt E_(2)`
B. `E_(1) lt E_(2)`
C. `E_(1)=E_(2)`
D. `E_(1)=2E_(2)`

Answer

B

Explanation

`k_(1)//k_(2)= 2:1 or k_(1)=2k_(1), i.e., k_(1) gt k_(2)`. Uisng Arrhenius eqaution,
`k=Ae^(-E_(a)//RT)`
Therefore, if `k` is high `E_(a)` is low. Hence, `E_(2) gt E_(1)`