Chemical equilibrium

Many reactions do not run to completion. Mix nitrogen and hydrogen over a hot iron catalyst and some ammonia forms, but the reaction stops while plenty of nitrogen and hydrogen remain. Seal some brown nitrogen dioxide in a flask and its colour fades to a steady intermediate shade as some of it forms colourless dinitrogen tetroxide. These systems have reached chemical equilibrium. This lesson explains what equilibrium is, how the equilibrium constant describes it, how to calculate equilibrium concentrations, how equilibrium responds to change (Le Châtelier's principle), and how K connects to the free energy of lesson 3. Lessons 8 to 10 apply these ideas to solubility, acids and bases.

Dynamic equilibrium

Equilibrium is dynamic, not static. Consider N₂O₄(g) ⇌ 2 NO₂(g). Start with pure N₂O₄: the forward reaction begins at its maximum rate and slows as N₂O₄ is used up; the reverse reaction starts at zero and speeds up as NO₂ accumulates. Eventually the two rates become equal. From then on, both reactions continue, but each molecule of NO₂ formed is matched by one consumed, so concentrations no longer change. The double arrow ⇌ signals a reaction that can reach equilibrium.

Equilibrium can be approached from either direction. Starting with pure NO₂ instead, at the same temperature and total amount of nitrogen and oxygen, the system reaches the same equilibrium composition. This shows that equilibrium is a property of the system, not of its history.

The equilibrium constant

In 1864 the Norwegian chemists Cato Guldberg and Peter Waage found that at equilibrium a particular ratio of concentrations is constant at a given temperature, whatever the starting amounts. For a general reaction a A + b B ⇌ c C + d D, the law of mass action defines the equilibrium constant in terms of concentrations:

Products go in the numerator and reactants in the denominator, each raised to the power of its coefficient. Strictly, the terms are activities, dimensionless ratios relative to the standard state, so K has no units; for dilute solutions and ideal gases, activities are approximated by molar concentrations (relative to 1 M) or partial pressures (relative to 1 bar). Pure solids and liquids have activity 1 and are left out of the expression, because their concentrations do not change. For example, for CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_c = [CO₂].

For gas reactions, an equilibrium constant can also be written with partial pressures, K_p. The two are related through the ideal gas law, where Δn is the change in moles of gas (products minus reactants) and R = 0.08314 L·bar/(mol·K) when pressures are in bar.

What the size of K means

The magnitude of K shows how far a reaction goes. When K is very large (say 10¹⁰), products dominate at equilibrium and the reaction effectively goes to completion; the combustion of fuels and the neutralization of strong acids are examples. When K is very small (10⁻¹⁰), almost no products form; nitrogen and oxygen in air barely react at room temperature (K_c for N₂ + O₂ ⇌ 2 NO is about 10⁻³⁰ at 25 °C). When K is near 1, significant amounts of both reactants and products are present.

Equilibrium constants obey simple rules when equations are manipulated, much like the Hess's law rules for ΔH. Reversing an equation inverts K (K′ = 1/K). Multiplying all coefficients by n raises K to the power n. Adding equations multiplies their K values.

The reaction quotient

The reaction quotient, Q, has the same form as K but uses the concentrations at any moment, not necessarily at equilibrium. Comparing Q with K predicts which way a reaction will shift.

Calculating equilibrium concentrations: ICE tables

The most common equilibrium problem gives the starting concentrations and K and asks for the equilibrium concentrations. An ICE table organizes the work: a row for Initial concentrations, a row for the Change (written in terms of an unknown x and the coefficients), and a row for Equilibrium concentrations (initial plus change). Substituting the equilibrium row into the K expression gives an equation for x.

In general the ICE equation is a polynomial in x, and only one root is physically sensible: it must keep every concentration positive. The code below solves a general case numerically and checks the answer.

import numpy as np

# PCl5(g) <=> PCl3(g) + Cl2(g), Kc = 0.0420 at 250 C. Start with 0.100 M PCl5 only.
Kc, c0 = 0.0420, 0.100
# Kc = x^2 / (c0 - x)  ->  x^2 + Kc*x - Kc*c0 = 0
roots = np.roots([1, Kc, -Kc * c0])
x = [r.real for r in roots if 0 < r.real < c0][0]
print(f"roots: {np.round(roots.real, 5)}; physical root x = {x:.5f} M")
print(f"[PCl5] = {c0 - x:.4f} M, [PCl3] = [Cl2] = {x:.4f} M")
print(f"check K = {x * x / (c0 - x):.4f}; percent dissociated = {100 * x / c0:.1f} %")

# the small-x approximation x = sqrt(Kc*c0) and the 5 % test
x_approx = (Kc * c0) ** 0.5
print(f"small-x estimate = {x_approx:.4f} M ({100 * x_approx / c0:.0f} % of c0: approximation fails)")

roots: [-0.08912  0.04712]; physical root x = 0.04712 M
[PCl5] = 0.0529 M, [PCl3] = [Cl2] = 0.0471 M
check K = 0.0420; percent dissociated = 47.1 %
small-x estimate = 0.0648 M (65 % of c0: approximation fails)

The small-x approximation

When K is small compared with the initial concentrations, x is small and the term (c₀ − x) can be replaced by c₀, avoiding the quadratic. The approximation is accepted if x turns out to be less than 5 % of c₀, a test you should always check. In the PCl₅ example K is not small enough: the approximation gives x equal to 65 % of c₀, far beyond 5 %, and the quadratic is required. The approximation becomes essential for weak acids and bases in lesson 9, where K values are often 10⁻⁵ or smaller.

Le Châtelier's principle

In 1884 the French chemist Henri Le Châtelier stated a principle that predicts how equilibria respond to disturbances: if a system at equilibrium is disturbed, it shifts in the direction that partly counteracts the disturbance. It is a qualitative summary of what comparing Q and K shows quantitatively.

Changing a concentration. Adding a reactant (or removing a product) makes Q < K, and the reaction shifts forward. Adding a product shifts it in reverse. Industrial processes exploit this by continuously removing a product; ammonia plants cool the gas stream to condense ammonia out, letting more form.

Changing pressure by changing volume. Compressing a gas mixture increases all partial pressures. The system shifts toward the side with fewer moles of gas, which reduces the pressure. For N₂ + 3 H₂ ⇌ 2 NH₃ (4 mol of gas becoming 2), high pressure favours ammonia, which is why Haber–Bosch plants run at about 150 to 300 bar. Adding an inert gas at constant volume changes no partial pressure and has no effect.

Changing temperature. Temperature is different: it changes the value of K itself. Treat heat as a product of an exothermic reaction and a reactant of an endothermic one. Raising the temperature shifts an exothermic reaction in reverse (K falls) and an endothermic one forward (K rises). The N₂O₄ ⇌ 2 NO₂ reaction is endothermic as written, so a sealed tube of the gas turns darker brown when warmed and paler when chilled in ice, a classic demonstration.

Adding a catalyst. A catalyst speeds forward and reverse reactions equally, so it does not shift equilibrium; it only gets there faster.

Free energy and the equilibrium constant

Lesson 3 noted that ΔG for a reaction depends on composition. The link is exact.

At equilibrium, ΔG = 0 and Q = K, which gives the second equation. It connects the thermodynamic tables of lesson 3 to equilibrium compositions. A negative ΔG° means K > 1 and products are favoured; a positive ΔG° means K < 1. Because the relationship is logarithmic, each 5.7 kJ/mol of ΔG° at 25 °C changes K by a factor of 10.

from math import exp, log

R, T = 8.314, 298.15
for dG0 in (-40, -11.4, -5.7, 0, 5.7, 11.4, 40):   # kJ/mol
    K = exp(-dG0 * 1000 / (R * T))
    print(f"dG0 = {dG0:6.1f} kJ/mol  ->  K = {K:.2e}")

# Haber-Bosch at 298 K from dG0f(NH3) = -16.4 kJ/mol
dG0 = 2 * -16.4
K = exp(-dG0 * 1000 / (R * T))
print(f"N2 + 3 H2 <=> 2 NH3 at 298 K: dG0 = {dG0:.1f} kJ, K = {K:.1e}")

dG0 =  -40.0 kJ/mol  ->  K = 1.02e+07
dG0 =  -11.4 kJ/mol  ->  K = 9.94e+01
dG0 =   -5.7 kJ/mol  ->  K = 9.97e+00
dG0 =    0.0 kJ/mol  ->  K = 1.00e+00
dG0 =    5.7 kJ/mol  ->  K = 1.00e-01
dG0 =   11.4 kJ/mol  ->  K = 1.01e-02
dG0 =   40.0 kJ/mol  ->  K = 9.82e-08
N2 + 3 H2 <=> 2 NH3 at 298 K: dG0 = -32.8 kJ, K = 5.6e+05

At room temperature, ammonia synthesis has a large K, about 6 × 10⁵, yet it does not happen without a catalyst because it is far too slow. At the 700 K or so of an industrial reactor, the exothermic reaction's K is far smaller, but the rate is acceptable. The plant operates at high pressure to push the equilibrium back toward ammonia and recycles the unreacted gases. This interplay between thermodynamics, kinetics and Le Châtelier's principle is the classic case study of industrial chemistry, and the Haber–Bosch process still produces most of the nitrogen fertilizer used on Canadian and American farms.

The van ’t Hoff equation

How much K changes with temperature depends on ΔH°. Combining ΔG° = ΔH° − TΔS° with ΔG° = −RT ln K gives the van ’t Hoff equation, which has the same form as the Arrhenius equation and the Clausius–Clapeyron equation from CHEM-101.