InterviewSolution
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InterviewSolution
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The standard Gibbd energies `(Delta_(f)S^(Theta))` for the formation of `SO_(2)(g)` and `SO_(3)(g)` are `-300.0` and `-371.0 kJ mol^(-1)` at `300K`, respectively. Calculate `DeltaG` and equilibrium constant for the following reaction at `300K`: |
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Answer» `2SO_(2)(g) +O_(2)(g) hArr 2SO_(2)(g)` `Delta_(r)G^(Theta) = 2Delta_(f)G^(Theta) (SO_(3)) - 2Delta_(f)G^(Theta) (SO_(2)) -Delta_(f)G^(Theta) (O_(2))` `= 2(-371) -2(-300)-0` `= -742 +600 =- 142 kJ mol^(-1)` Now `log K = (Delta_(r)G)/(2.303 RT)` `Delta_(r)G^(Theta) =- 142 kJ mol^(-1), R = 8.314 xx 10^(-3) kJ mol^(-1)K^(-1)`, `T = 300K` `log K = (-142)/(2.303 xx 8.314 xx 10^(-3) xx 300) = 24.72` `:. K = Antilog (24.72) = 5.248 xx 10^(24)` |
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