Which of the following expressions is correct?

Options

A. `DeltaG=DeltaG^(Θ)+RT 1n Q`
B. `DeltaG^(Θ)=-RT1nK_(eq)`
C. `K_(eq)=e^(-DeltaG^(Θ)//RT`
D. All of these

Answer

D

Explanation

`DeltaG=DeltaG^(Θ)+RT 1nQ`
At equilibrium, `DeltaG=0` and `Q=K_(eq)`
`:. 0=DeltaG^(Θ)+RT 1n K_(eq)`
`DeltaG^(Θ)=-RT1nK_(eq)`
`1n K_(eq)=(-DeltaG^(Θ))/(RT)`
Taking the antilog of both sides, we get
`K_(eq)=e^(-DeltaG^(Θ)//RT)`