The standard free energy change of a reaction is `DeltaG^(@)=-115` at 298K. Calculate the equilibrium constant `K_(P)` in log `K_(P).(R=8.314JK^(-1)mo
The standard free energy change of a reaction is `DeltaG^(@)=-115` at 298K. Calculate the equilibrium constant `K_(P)` in log `K_(P).(R=8.314JK^(-1)mol^(-1))`
A. 20.16
B. 2.303
C. 2.016
D. 13.83
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Correct Answer - A
`log" "K_(P)=(-DeltaG^(@))/(2.303RT)`
`=(-(-115xx1000))/(2.303xx8.313xx298)`
`=20.16`
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