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`Delta G^(@)` for a reaction is `46.06 kcal mol^(-)`. `K_(P)` for the reaction at `300 K` isA. `10^(-8)`B. `10^(-22.22)`C. `10^(-33.55)`D. none of these |
Answer» Correct Answer - C `DeltaG^(@)=46-.06 kcal mol^(-1)` `=46.06xx1000xx4.814 J mol^(-1)` `DeltaG^(@)=-RT ln K_(P)=-2.303 RT log K_(P)` `46.04xx1000xx4.184` `=-2.303xx8.31xx300 log K_(P)` or `log K_(P)=-33.55` or `K_(P)=10^(-33.55)` |
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