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InterviewSolution
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`Delta G^(@)` for a reaction is `46.06 kcal mol^(-)`. `K_(P)` for the reaction at `300 K` isA. `10^(-22.22)`B. `10^(-8)`C. `10^(-44.55)`D. `10^(-35.54)` |
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Answer» Correct Answer - D According to thermodynamics, `Delta G^(@) = - 2.303 RT log k_(P)` We have `Delta G^(@) = 46.06 kcal mol^(-1)` `= (46.06) (1000) (4.184) J mol^(-1)` `R = 8.314 J K^(-1) mol^(-1)` `T = 300 K` `:. Log K_(p) = (Delta G^(@))/(-2.303 RT)` `= ((46.06)(1000)(4.184))/(-(2.303)(8.314)(300))` `= (192715.04)/(5744.14)` `= 33.54` or `K_(P) = 10^(-33.54)` |
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