Equilibrium constant, `K` changes with temperature. At `300` K, equilibrium constant is `25` and at `400` K it is `10`. Hence, backward reaction will have energy of activation

Options

A. equal to that of forward reaction
B. less than that of forward reaction
C. greater than that of forward reaction
D. given values are not sufficient to explain given statement.

Answer

C

Explanation

`log.(K_(2))/(K_(1))=(DeltaH)/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`DeltaH=(2.303xxRT_(1)T_(2))/((T_(2)-T_(1)))log. (K_(2))/(K_(1))`
`=(2.303xx8.314xx300xx400)/(400-300)log. (10)/(25)`
`=(2.303xx8.314xx300xx400)/(100)log. (10)/(25)lt0`
As `DeltaH` is -ve , the reaction is exothermic.

Hence, `E_(a)` ( backward) is more than `E_(a)` ( forward).