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Two reactions R(2) and R(2) have identic...

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))`

A

4

B

8

C

12

D

6

Text Solution

Verified by Experts

The correct Answer is:
A
From Arrhenius equation `k=Ae^(-E_(a)//RT)`
Since In `((k_(2))/(k_(1)))=(1)/(RT) [(E_(a))_(1) -(E_(a))_(2)]=(1xx10xx10^(3))/(8.314xx300)=4`
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