Calculate the standard free energy change for the reaction:
`H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1)`
Given: `S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1)`,
`S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1)`
and `S^(Theta) (HI) = 206.8 J K^(-1) mol^(-1)`.

Explanation

`Delta^(Theta)S = sum S^(Theta) ("products") -sum S^(Theta) ("Reactant")`
`=2S_(HI)^(Theta)-(S_(H_(2))^(Theta)+S_(I_(2))^(Theta))`
`=2 xx 206.3 - 130.6 - 116.7`
`= 165.3 JK^(-1) mol^(-1)`
`DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`= 51.9 kJ -298 xx 165.3 xx 10^(-3)kJ`
`= 51.9 - 49.25 = 2.6406 kJ mol^(-1) = 2640.6 J mol^(-1)`
`DeltaG = +ve`. Reaction is not feasible.