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

b) According to Arrhenius equation,
`R=Ae^(Ea//RT)`
For reaction, `R_(1), k_(1)=Ae^(-Ea//RT)`
In `k_(1)=In A-(E_(a1))/(RT)`
for reaction, `R_(2),k_(2)=Ae^(-Ea2//RT)`
In `k_(2) = In A-(Ea2)/(RT)`
`(E_(a1))/(RT) - (E_(a2))/(RT)`
`=(10 xx 10^(3)J mol^(-1))/((8.314JK^(-1)mol^(-1)) xx (300K))=4`