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

From arrhenius equation,
`K=A.e^(-Ea)/(RT)`
so, `k_(1)=A.e^(-E_(a_(1)) // RT)` ...(1)
`k_(2)=A.e^(-E_(a_(2)) // RT)` ...(2)
On dividing equation `(2) // (1) implies (k_(2))/(k_(1))=e^(((E_(a_(1))-E_(a_(2))))/(RT))`
In `((k_(2))/(k_(1)))=(E_(a_(1))-E_(a_(2)))/(RT)=(10,000)/(8.314 xx 300)=4`