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

Jul 18, 2017

You'd start from the expression for the change in Gibbs' free energy, $\Delta G$, relative to a reference, $\Delta {G}^{\circ}$, at standard pressure and a convenient temperature:

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$

where:

• $Q$ is the reaction quotient for the current state of the reaction.
• $R$ and $T$ are known from the ideal gas law.
• $R T \ln Q$ describes the shift in the free energy in reference to standard pressure and the chosen temperature (usually ${25}^{\circ} \text{C}$ 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.

$\Delta G = 0$

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

$\textcolor{b l u e}{\Delta {G}^{\circ} = - R T \ln {K}_{e q}}$

And usually the other kind of calculation of this kind is to solve for ${K}_{e q}$.

$- \frac{\Delta {G}^{\circ}}{R T} = \ln {K}_{e q}$

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

where $\text{exp} \left(x\right) = {e}^{x}$.