InterviewSolution
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InterviewSolution
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Calculate equilibrium constant for the reaction: `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g) at 25^(@)C` Given: `Delta_(f)G^(Theta) SO_(3)(g) = - 371.1 kJ mol^(-1)`, `Delta_(f)G^(Theta)SO_(2)(g) =- 300.2 kJ mol^(-1)` and `R = 8.31 J K^(-1) mol^(-1)` |
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Answer» Step I: Calculation of `DeltaG^(Theta)` for the reaction `DeltaG^(Theta) = sum Delta_(f)G^(Theta) (p) - sum Delta_(f)G^(Theta) (r )` `=[2mol xx Delta_(f)G^(Theta)SO_(3)(g)]` `-[2mol xx Delta_(f)G^(Theta) SO_(2),(g) +Delta_(f)G^(Theta) O_(2)(g)]` `=[2mol xx 371.1 kJ mol^(-1)]` `-[2mol xx (-300.2 kJ mol^(-1))]` `=- 742.2 +600.4 =- 141.8 kJ` Step II: Calculation of equilibrium constant `(K)` `DeltaG^(Theta) =- 2.303 RT log K` `DeltaG^(Theta) = - 141.8 kJ =- 141800 J, T = 25 +273 = 298K`, `R = 8.31 JK^(-1) mol^(-1)` `logK =- (DeltaG^(@))/(2.303 RT)` `=(-) ((-141800J))/(2.303xx(8.31JK^(-1)mol^(-1)xx298K)) = 24.864` `K = Antilog (24.864) = 7.31 xx 10^(24)` |
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