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`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» 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. |
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