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. 8
B. 12
C. 6
D. 4

Answer

D

Explanation

We know `k_(2)=Ae^(-E_(a)//RT)`
`(k_(2))/(k_(1))=e^((E_(a)^(1)-E_(b)^(2))//RT)`
`ln.(k_(2))/(k_(1))=(E_(a)^(1)-E_(b)^(2))/(RT)`
`=(10xx1000)/(8.314xx300)`
`~~4`