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Calculate the standard free energy chang...

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)`.

Text Solution

Verified by Experts

`DeltaS^(Theta) = 2S^(Theta) (HI) -S^(Theta) (H_(2))+S^(Theta)(I_(2))]`
`= 165.3 J K^(-1) mol^(-1)`
`DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`= 51.9 xx 10^(3) - 298 (165.3) =2.641 kJ mol^(-1)`
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