For the reaction `Ag_(2)O(s) rarr 2Ag(s) + (1)/(2) O_(2)(g) , DeltaH ` is `30.56kJ mol^(-1)` and `Delta S ` is `66JK^(-1) mol^(-1)` at one atmosphere pressure. Calculate the temperature at which `Delta G` for it will be zero. What will be the direction of the reaction at this temperature and at temperature above or below this temperature and why ?

For the reaction `Ag_(2)O(s) rarr 2Ag(s) + (1)/(2) O_(2)(g) , DeltaH `

1.	For the reaction `Ag_(2)O(s) rarr 2Ag(s) + (1)/(2) O_(2)(g) , DeltaH ` is `30.56kJ mol^(-1)` and `Delta S ` is `66JK^(-1) mol^(-1)` at one atmosphere pressure. Calculate the temperature at which `Delta G` for it will be zero. What will be the direction of the reaction at this temperature and at temperature above or below this temperature and why ?
Answer» Correct 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. 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`.

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