Question

How do you calculate `K_"eq"` from `DeltaG^@`?

Answer 1

You'd start from the expression for the change in Gibbs' free energy, `DeltaG`, relative to a reference, `DeltaG^@`, at standard pressure and a convenient temperature:

`DeltaG = DeltaG^@ + RTlnQ`

where:

• `Q` is the reaction quotient for the current state of the reaction.
• `R` and `T` are known from the ideal gas law.
• `RTlnQ` describes the shift in the free energy in reference to standard pressure and the chosen temperature (usually `25^@ "C"` for convenience).

At chemical equilibrium, the reaction has no tendency to shift in either direction, so the change in Gibbs' free energy is zero, i.e.

`DeltaG = 0`

Thus, with `Q = K` as well at equilibrium,

`color(blue)(DeltaG^@ = -RTlnK_(eq))`

And usually the other kind of calculation of this kind is to solve for `K_(eq)`.

`-(DeltaG^@)/(RT) = ln K_(eq)`

`=> color(blue)(K_(eq) = "exp"(-(DeltaG^@)/(RT))`,

where `"exp"(x) = e^x`.