Two reactions `R_(2) and R_(2)` have identical pre - exponential factors. Activations enery of `R_(1)` exceeds that of `R_(2)` by 10 kJ `mol_(-1)` . If `k_(1) and k_(2)` are rate constants for rate constants for reactions `R_(1) and R_(2)`
respectively at 300k , then In `(k_(2)/k_(1))`is equal to `(R=8.314 J mol^(-1)K^(-1))`

Options

A. 6
B. 4
C. 8
D. 12

Answer

B

Explanation

`k_1=Ae^(-E_(a_(1))//RT), K_2=Ae^(-E_(a_(2))//RT)`
`:." "k_2/k_1=e^(1/(RT)(E_(a_(1))-E_(a_(2)))`
or `In k_2/k_1=(E_(a_(1))-E_(a_(2)))/(RT)=(10xx10^3)/(8.314xx300)=4`