`DeltaH and DeltaS` for the reaction
`Ag_(2)O_((s))rarr2Ag_((s))+1/2O_(2_((g))) " are "30.56 kJ mol^(-1)`
and `66.0 Jk^(-1) mol^(-1)` respectively. Calculate the temperature at which the free energy for this reaction will be zero. What will be the direction of reaction at this temperature and at temperature below this and why ?
Given: `DeltaH=30.56kJ mol^(-1)=30560 J mol^(-1)`
`DeltaS=66.0JK^(-1)mol^(-1)`
`DeltaG=0`

Explanation

Given: `DeltaH=30.56 "kJ mol"^(-1)`
`=30560 J mol^(-1)`
`DeltaS=6.66 xx 10^(-3) kJ K^(-1) "mol"^(-1)`
T=? At which `DeltaG=0`
`DeltaG=DeltaH-TDeltaS`
`0=DeltaH-TDeltaS`
`T=(DeltaH)/(DeltaS)`
`T=("30.56 kJ mol"^(-1))/(6.66xx10^(-3) kJ K^(-1) "mol"^(-1))`
T=4589 K
(i) At 4589 K , `DeltaG=0` , the reaction is in equilibrium .
(ii)At temperature below 4598 K, `DeltaH > T DeltaS`
`DeltaG=DeltaH-T DeltaS > 0`, the reaction in the forward direction, is non-spontaneous. In other words the reaction occurs in the backward direction.