Question

`DeltaH` and `DeltaS` for the reaction:
`Ag_(2)O(s) rarr 2Ag(s) +(1//2)O_(2)(g)`
are `30.56 kJ mol^(-1)` and `66.0 J JK^(-1) mol^(-1)` respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be favoured above or below this temperature.

Answer 1

We know that, `DeltaG = DeltaH - T DeltaS`
At equilibrum, `DeltaG = 0`
so that `0 = DeltaH - T DeltaS`
or `T = (DeltaH)/(DeltaS)`
Given that `DeltaH = 30.56 kJ mol^(-1) = 30560 J mol^(-1)`
`DeltaS = 66.0 J K^(-1) mol^(-1)`
`T = (30560)/(66) = 463 K`
Above this temperature, `DeltaG` will be negative and the process will be spontaneous if forward direction.