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Find the equilibrium constant `K_(p)` in `log K_(p)` if the standard free energy change of a reaction `DeltaG^(@)=-115kJ` at 298K isA. 2.303B. 13.83C. 2.016D. 20.16 |
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Answer» Correct Answer - D `DeltaG^(@)=-115xx10^(3)J,T=298K` and `R=8.314JK^(-1)mol^(-1)` `-DeltaG^(@)=2.303RTlog_(10)K_(p)` `-(-115xx10^(3))=2.303xx8.314xx298xxlog_(10)K_(p)` `log_(10)K_(p)=(115000)/(2.303xx8.314xx298)` `log_(10)K_(p)=20.6` |
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