`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

`DeltaG = 0 ` at 463K . At this temp, the reaction will be in equilibrium . Above this temperature , the reaction will be spontaneous in the forward direction . Below this temp, the reaction will be non-spontaneous or it wll proceed in the bakcward direction.

Explanation

When `DeltaG = 0, T = (DeltaH )/(DeltaS) = ( 30560J mol^(-1))/( 66JK^(-1)mol^(-1))= 463.0K`
`DeltaG = Delta H - T DeltaS `. Above 463K `, DeltaG = -ve` ( because `DeltaH ` and `Delta S` both are `+ve`) . Below 463K, `DeltaG = + ve`.