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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

12

B

6

C

4

D

8

Text Solution

Verified by Experts

The correct Answer is:
c
`k_(1) = Ae^(-E_(a_(1)) //RT) , k_(2) = Ae^(-(Ea_(1) - 10) //RT)`
Ln `((k_(2))/(k_(1))) = (10)/(RT) = (10)/(8.314 xx 10^(-3) xx 300) = 4`
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