The rate of a certain biochemical reaction at physiological temperature (T) occurs `10^(6)` times faster with enzyme than without. The change in the reaction energy upon adding enzyme is :

Options

A. `+ RT`
B. `-RT`
C. `+6(2.303) RT`
D. `-6(2.303) RT`

Answer

D

Explanation

`therefore k=Ae^(-Ea//RT)" ….(1)"`
`therefore 10^(6) xx k =Ae^((-E_(a))_("cat/Rt"))" ....(2)"`
Dividing equation (2) by (1),
`10^(6) =e^(E_(a)-(E_(a))_("cat/RT")) rArr 6xx" In 10"=(E_(a) -(E_(a))_("cat"))/(RT)`
`rArr Delta E_(a) =E_(a) - (E_(a))_("cat") =6xx" In 10 "xxRT rArr (E_(a))_("cat")-E_(a) =-6xx2.303 xx RT`