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Two reactions , R(1) and R(2) have iden...

Two reactions , `R_(1) and R_(2)` have identical pre-exponential factors . Activation energy of `R_(1)` exceeds that of `R_(2)` by `10"kJ.mol"^(-1)`. If `k_(1) and k_(2)` are rate constants for reaction `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

6

B

4

C

8

D

12

Text Solution

Verified by Experts

The correct Answer is:
B
`k_(1)=A.e^((-E_(a_(1)))/(RT)),k_(2)=A.e^((-E_(a_(2)))/(RT))`
`therefore (k_(2))/(k_(1))=e^((E_(a_(1))-E_(a_(2)))//RT)`
or, In`(k_(2))/(k_(1))=(E_(a_(1))-E_(a_(2)))/(RT)=(10xx10^(3))/(8.314xx300)=4`
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