Question

In the presence of a catalyst, the rate of a reaction grows to the extent of `10^(5)` times at `298 K`. Hence, the catalyst must have lowered `E_(a)` by
A. `25 kJ mol^(-1)`
B. `20 kJ mol^(-1)`
C. `10 kJ mol^(-1)`
D. `28.5 kJ mol^(-1)`

Answer 1

Correct Answer - D
`(k_(cat))/(k) = "antilog"((Delta E_(a))/(2.303 RT))`
`log.(k_(cat))/(k) = (DeltaE_(a))/(2.303 RT)`
`log 10^(5) = (Delta E_(a))/(2.303xx8.314xx 10^(3)xx 298)`
`Delta E_(a) = 5 xx 2.303 xx 8.314 xx 298 xx 10^(-3)`
`= 28.53 kJ mol^(-1)`