`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.

Explanation

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