Lesson 5 · 30 min

Free energy and equilibrium

ΔG° = −RT ln K: one table of thermodynamic data yields the equilibrium constant of ammonia synthesis, the ionic product of water and the solubility product of silver chloride. The van ’t Hoff equation says how K moves with temperature; comparing Q with K says which way a reaction goes.

In this lesson

Use ΔG = ΔG° + RT ln Q to show how ΔG changes with composition
Compute equilibrium constants from thermodynamic data with ΔG° = −RT ln K
Explain with the van ’t Hoff equation how temperature changes K
Compare Q with K to decide which way a mixture will react

The acid–base unit’s $K_\mathrm{w} = 1.0\times10^{-14}$ and the ammonium ion’s $\mathrm{p}K_\mathrm{a} = 9.25$ came from countless pH measurements. They can also be calculated: the same data table that gave reaction enthalpies lists every substance’s standard Gibbs energy of formation. One formula turns that table into the ionic product of water, the solubility product of silver chloride and the equilibrium constant of ammonia synthesis at any temperature — and explains why plants make ammonia at 700 K, where its equilibrium constant is tiny.

Predict first

At some temperature a reaction has ΔG° = 0. What is its equilibrium constant K?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

K = 0: nothing happens K = 1 K is infinite: the reaction goes to completion Cannot tell; it depends on the starting amounts

ΔG depends on composition

ΔG° describes one special mixture: every species in its standard state. A real reaction mixture has some other composition, and its reaction Gibbs energy gets a composition-dependent correction:

\Delta G = \Delta G^\circ + RT\ln Q

The reaction quotient $Q$ is written just like the equilibrium constant — the product of product activities over the product of reactant activities, each raised to its coefficient — but with present values. Gas activities are approximately $p/p^\circ$ (bar), solutes $c/c^\circ$ (mol/L), pure solids and liquids 1.

As the reaction proceeds, reactants are used up, products accumulate, $Q$ grows and ΔG rises; when ΔG reaches 0 there is no net push left — that is equilibrium, the bottom of the curve of G against the extent of reaction (see the story The downhill road of a reaction).

\Delta G^\circ = -RT\ln K \qquad K = e^{-\Delta G^\circ/RT}

At equilibrium ΔG = 0 and Q = K; substitute and this follows. At 298 K every 5.71 kJ/mol of ΔG° changes K tenfold.

$K$: thermodynamic equilibrium constant (dimensionless)
$R$: 8.314 J/(K·mol)
$T$: thermodynamic temperature / K

ΔG° and K at 298 K
ΔG° / (kJ/mol)	K	At equilibrium
−57.1	$10^{10}$	almost all products
−11.4	100	products dominate
0	1	comparable amounts
+11.4	0.01	reactants dominate
+57.1	$10^{-10}$	almost all reactants

Source: K = exp(−ΔG°/RT) with R = 8.314 J/(K·mol), T = 298.15 K.

ΔG° = −RT ln K · van ’t Hoff · Q versus K

Equilibrium constants from data

Reaction Temperature T · 298 K

$\ce{H2O(l)} \rightleftharpoons \ce{H+(aq)} + \ce{OH-(aq)}$

ΔH°: 55.8 kJ
ΔS°: −80.7 J/K
ΔG°(T) = ΔH° − TΔS°: 79.9 kJ
Equilibrium constant K = e^−ΔG°/RT: K = 9.95 × 10⁻¹⁵ · lg K = −14.00

Reactants dominate (K ≪ 1) · Endothermic: K rises on heating

Your mixture (activities)

\ce{H+(aq)}

1 concentration / M

\ce{OH-(aq)}

1 concentration / M

Pure solids and liquids have activity 1 and drop out of Q.

Reaction quotient Q: 1
ΔG = ΔG° + RT ln Q: 79.9 kJ

Q > K: the reaction runs backward (←)

Start with the autoionization of water: from ΔfG° values, K = 1.0 × 10⁻¹⁴ — exactly Kw. Then try silver chloride dissolving (Ksp ≈ 1.8 × 10⁻¹⁰) and the ammonium ion giving up a proton (pKa ≈ 9.26). Adjust the concentrations under “Your mixture”: when Q is below K the reaction runs to the right, when above, to the left.

How temperature moves K: the van ’t Hoff equation

Put $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ into $-RT\ln K$ and divide by $-RT$ :

\ln K = -\frac{\Delta H^\circ}{R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{R}

A plot of ln K against 1/T is a straight line of slope $-\Delta H^\circ/R$ . For an exothermic reaction (ΔH° < 0) the slope is positive: raise T, 1/T falls, and K falls; an endothermic reaction does the opposite. This is the quantitative version of Le Chatelier’s rule for temperature. Conversely, K at two temperatures is enough to find ΔH°.

\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

The two-temperature form of the van ’t Hoff equation (assuming ΔH° does not change with temperature).

$K_1, K_2$: equilibrium constants at T₁ and T₂
$\Delta H^\circ$: standard reaction enthalpy / (J/mol)

Worked example

Ammonia synthesis at 298 K and 700 K

$\ce{N2(g) + 3H2(g)} \rightleftharpoons \ce{2NH3(g)}$ : ΔH° = −91.88 kJ, ΔS° = −198.1 J/K. Find K at 298 K and at 700 K, and explain why industry works at about 700 K.

Q versus K: which way will it go?

Put ΔG° = −RT ln K back into ΔG = ΔG° + RT ln Q and you get a very handy form:

\Delta G = RT\ln\frac{Q}{K}

$Q < K$ : ΔG < 0, the forward reaction runs and products increase;
$Q > K$ : ΔG > 0, the reverse reaction runs;
$Q = K$ : equilibrium.

For example, $\ce{N2O4} \mathrel{\ce{<=>}} \ce{2NO2}$ has K ≈ 0.15 at 298 K. A flask with $p(\ce{NO2}) = 0.20$ bar and $p(\ce{N2O4}) = 0.50$ bar has $Q = 0.20^2/0.50 = 0.08 < K$ , so more N₂O₄ will dissociate and the brown colour deepens.

Concept check

At 298 K, a solution with $[\ce{H+}] = 1.0\times10^{-3}$ M is mixed with an equal volume with $[\ce{OH-}] = 1.0\times10^{-3}$ M (each diluted to 5.0 × 10⁻⁴ M). For $\ce{H2O} \mathrel{\ce{<=>}} \ce{H+} + \ce{OH-}$ (Kw = 1.0 × 10⁻¹⁴):

Q < Kw, so water ionizes further Q > Kw, so the reverse reaction (neutralisation) runs Q = Kw, already at equilibrium Cannot tell

Try the calculation

For $\ce{AgCl(s)} \rightleftharpoons \ce{Ag+(aq) + Cl-(aq)}$ , ΔG° = +55.67 kJ/mol at 298.15 K. Find the solubility product Ksp.

Key takeaways

ΔG = ΔG° + RT ln Q = RT ln(Q/K): forward when Q < K, backward when Q > K, equilibrium when Q = K.
ΔG° = −RT ln K: at 298 K every 5.71 kJ/mol is a factor of 10 in K.
Kw, Ksp, Ka and other constants follow from ΔfG° data.
Van ’t Hoff: ln K against 1/T is a straight line of slope −ΔH°/R; heating an exothermic reaction lowers K.

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Data and references

P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topics 6A–6B (chemical equilibrium, response to temperature).
D. D. Wagman et al., “The NBS tables of chemical thermodynamic properties”, J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982) — ΔfG° of aqueous ions.
M. W. Chase, NIST-JANAF Thermochemical Tables, 4th ed., J. Phys. Chem. Ref. Data Monograph 9 (1998) — ammonia at high temperature.
D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016), Appendix F (Ksp of AgCl 1.8 × 10⁻¹⁰).
V. Smil, Enriching the Earth: Fritz Haber, Carl Bosch, and the Transformation of World Food Production, MIT Press (2001).