For the equilibrium, `PCI_(5)(g) hArr PCI_(3)(g) +CI_(2)(g) at 25^(@)C K_(c) =1.8 xx 10^(-7)` Calculate `DeltaG^(Theta)` for the reaction `(R = 8.314 J K^(-1) mol^(-1))`.

For the equilibrium, `PCI_(5)(g) hArr PCI_(3)(g) +CI_(2)(g) at 25^(@)C

1.	For the equilibrium, `PCI_(5)(g) hArr PCI_(3)(g) +CI_(2)(g) at 25^(@)C K_(c) =1.8 xx 10^(-7)` Calculate `DeltaG^(Theta)` for the reaction `(R = 8.314 J K^(-1) mol^(-1))`.
Answer» We know, `DeltaG^(Theta) =- 2.303 RT log_(10)K_(c)` `=- 2.303 xx 8.314 xx 298 log (1.8 xx 10^(-7))` `= 38484 J mol^(-1) = 38.484 kJ mol^(-1)`

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For the equilibrium, `PCI_(5)(g) hArr PCI_(3)(g) +CI_(2)(g) at 25^(@)C K_(c) =1.8 xx 10^(-7)` Calculate `DeltaG^(Theta)` for the reaction `(R = 8.314 J K^(-1) mol^(-1))`.

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