Find the equilibrium constant `K_(p)` in `log K_(p)` if the standard free energy change of a reaction `DeltaG^(@)=-115kJ` at 298K is
Find the equilibrium constant `K_(p)` in `log K_(p)` if the standard free energy change of a reaction `DeltaG^(@)=-115kJ` at 298K is
A. 2.303
B. 13.83
C. 2.016
D. 20.16
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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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