Both $\Delta G^\circ$ and K describe how far a reaction goes, so they must be linked. The link is logarithmic, so you will solve log equations with it.
Key idea
$\Delta G = \Delta G^\circ + RT\ln Q$, and at equilibrium $\Delta G = 0$ and $Q = K$, so
$$\Delta G^\circ = -RT\ln K, \qquad K = e^{-\Delta G^\circ/RT}$$
with $R = 8.314\ \text{J/(mol·K)}$ and $\Delta G^\circ$ in J/mol. Negative $\Delta G^\circ$ means $K > 1$.
Why it works
$\Delta G^\circ$ is the free-energy change when everything is in its standard state ($Q = 1$). For any other mixture, $RT\ln Q$ corrects it: too much product (large $Q$) makes $\Delta G$ less negative. As the reaction runs, $Q$ moves toward $K$ and $\Delta G$ moves toward zero. When $\Delta G$ reaches zero there is no more drive either way: that is equilibrium. Setting $\Delta G = 0$ and $Q = K$ gives the relation.
Worked examples
Worked example: K from ΔG°
For $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, $\Delta G^\circ = -32.8\ \text{kJ}$ at $298\ \text{K}$. Find $K$.
- $\dfrac{-\Delta G^\circ}{RT} = \dfrac{32\,800}{(8.314)(298)} = 13.24$.
- $K = e^{13.24} = 5.6 \times 10^{5}$.
Answer: $K = 5.6 \times 10^{5}$.
Worked example: ΔG° from K
Acetic acid has $K_a = 1.8 \times 10^{-5}$ at $298\ \text{K}$. Find $\Delta G^\circ$ for its ionization.
- $$\Delta G^\circ = -RT\ln K = -(8.314)(298)\ln(1.8 \times 10^{-5})$$
- $\ln(1.8 \times 10^{-5}) = -10.93$, so $\Delta G^\circ = +2.71 \times 10^4\ \text{J} = +27.1\ \text{kJ/mol}$. Positive, as it must be for $K < 1$.
Answer: $+27.1\ \text{kJ/mol}$.
Worked example: Away from standard conditions
For the ammonia reaction at 298 K, find $\Delta G$ when $P_{\mathrm{NH_3}} = 0.10\ \text{atm}$ and $P_{\mathrm{N_2}} = P_{\mathrm{H_2}} = 1.0\ \text{atm}$.
- $Q = \dfrac{(0.10)^2}{(1.0)(1.0)^3} = 0.010$.
- $$\Delta G = -32.8\ \text{kJ} + (0.008314)(298)\ln(0.010) = -32.8 - 11.4 = -44.2\ \text{kJ}$$ With little product present, the drive forward is even stronger.
Answer: $-44.2\ \text{kJ}$.
| $\Delta G^\circ$ (kJ/mol, 298 K) | K | At equilibrium |
|---|---|---|
| −40 | $1 \times 10^{7}$ | essentially all products |
| −10 | 57 | mostly products |
| 0 | 1 | comparable amounts |
| +10 | 0.018 | mostly reactants |
| +40 | $1 \times 10^{-7}$ | essentially all reactants |
Combining $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ with $\Delta G^\circ = -RT\ln K$ gives $\ln K = -\dfrac{\Delta H^\circ}{RT} + \dfrac{\Delta S^\circ}{R}$. For an exothermic reaction, $K$ falls as $T$ rises; for an endothermic one, it grows. This is the quantitative form of Le Chatelier’s rule for temperature.
For reactions in solution, $K$ uses concentrations; for gases, partial pressures. Either way the activities in $Q$ and $K$ are pure numbers, so the logarithm makes sense.
Common mistake
Thinking $\Delta G^\circ$ is zero at equilibrium. It is $\Delta G$ that reaches zero; $\Delta G^\circ$ is a fixed number for the reaction at a given temperature, and it sets where equilibrium lies. Check: if $\Delta G^\circ$ were zero, $K$ would be exactly 1.