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-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.