For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?

Explanation

We have `log K_(p_(2))-log K_(p_(1))=(triangleH^(0))/(2.303 R) xx (T_(2)-T_(1))/(T_(2) T_(1)) ………….("Eqn. 11")`
As we know from Eqn. (11) that if on increasing temperature, `K_(p)` increases `triangleH^(0)` becomes positive i.e, the reaction is endothermic. Thus from the given data, we see that the reaction is endothermic.