Equilibrium

What this chapter is about

This chapter introduces the idea that many chemical reactions do not go to completion. Instead, they reach a state called equilibrium, where the forward and reverse reactions occur at equal rates and the concentrations of reactants and products remain constant over time. You learn how to describe this balance quantitatively using the equilibrium constant, and how factors such as concentration, pressure and temperature shift the position of equilibrium according to Le Chatelier's principle.

The chapter also extends equilibrium ideas to reactions in aqueous solution, particularly the dissociation of acids, bases and salts. You study pH, the ionic product of water, buffer solutions and the solubility product. These concepts explain everyday phenomena — why lemon juice is sour, why antacids relieve acidity, and why some salts dissolve easily while others precipitate.

By the end you should be able to write equilibrium-constant expressions, calculate Kc and Kp, predict the direction of a reaction using the reaction quotient Q, apply Le Chatelier's principle, work out pH of strong and weak acids or bases, and decide whether a precipitate will form from ion concentrations.

Key ideas

• Dynamic equilibrium: At equilibrium the forward and reverse reactions continue, but their rates are equal, so macroscopic properties (concentration, colour, pressure) stay constant.
• Equilibrium constant (Kc and Kp): For aA + bB ⇌ cC + dD, Kc = [C]^c [D]^d / [A]^a [B]^b. A large K means products dominate; a small K means reactants dominate.
• Relation Kp = Kc (RT)^Δn: Here Δn = (moles of gaseous products) − (moles of gaseous reactants), R = 8.314 J mol⁻¹ K⁻¹, T in kelvin.
• Reaction quotient Q: Calculated the same way as K but with current (non-equilibrium) concentrations. If Q < K the reaction proceeds forward; if Q > K it proceeds backward.
• Le Chatelier's principle: When a system at equilibrium is disturbed, it shifts to counteract the change. Adding reactant pushes equilibrium toward products; raising temperature favours the endothermic direction.
• Ionic product of water: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. Pure water has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol L⁻¹.
• pH and pOH: pH = −log₁₀[H⁺]; pOH = −log₁₀[OH⁻]; pH + pOH = 14 at 25 °C.
• Solubility product (Ksp): For a sparingly soluble salt MX₂ ⇌ M²⁺ + 2 X⁻, Ksp = [M²⁺][X⁻]². If the ionic product exceeds Ksp, precipitation occurs.

Formulas and facts to remember

1. Kc = [products]^coefficients / [reactants]^coefficients (each concentration in mol L⁻¹).
2. Kp = Kc (RT)^Δn, with Δn = sum of gaseous product coefficients minus sum of gaseous reactant coefficients.
3. pH = −log₁₀[H⁺]; pOH = −log₁₀[OH⁻]; pH + pOH = 14 (at 25 °C).
4. Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.
5. For a weak acid HA with dissociation constant Ka, degree of dissociation α ≈ √(Ka / C) when α is small.
6. Henderson–Hasselbalch equation for a buffer: pH = pKa + log₁₀([salt]/[acid]).
7. Ksp for AgCl ⇌ Ag⁺ + Cl⁻ is Ksp = [Ag⁺][Cl⁻]; precipitation begins when ionic product > Ksp.

Worked examples

Example 1 — Calculating Kc

Nitrogen and hydrogen react: N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g). At equilibrium in a 1.0 L vessel at 500 K, the mixture contains 0.10 mol N₂, 0.30 mol H₂ and 0.20 mol NH₃. Find Kc.

Step 1: Write concentrations (mol L⁻¹): [N₂] = 0.10, [H₂] = 0.30, [NH₃] = 0.20.

Step 2: Write the expression: Kc = [NH₃]² / ([N₂][H₂]³).

Step 3: Substitute: Kc = (0.20)² / (0.10 × (0.30)³) = 0.040 / (0.10 × 0.027) = 0.040 / 0.0027 ≈ 14.8.

So Kc ≈ 15 (dimensionless when unit conventions are followed).

Example 2 — Finding pH of a weak acid

A 0.020 mol L⁻¹ solution of a weak acid HX has Ka = 1.8 × 10⁻⁵. Calculate its pH.

Step 1: Use α ≈ √(Ka / C) = √(1.8 × 10⁻⁵ / 0.020) = √(9.0 × 10⁻⁴) = 0.030.

Step 2: [H⁺] = C × α = 0.020 × 0.030 = 6.0 × 10⁻⁴ mol L⁻¹.

Step 3: pH = −log₁₀(6.0 × 10⁻⁴) = −(−3.22) ≈ 3.2.

The solution is acidic but much less so than a strong acid of the same molarity.

Example 3 — Predicting precipitation

Will a precipitate form when 50 mL of 2.0 × 10⁻³ mol L⁻¹ AgNO₃ is mixed with 50 mL of 4.0 × 10⁻⁴ mol L⁻¹ NaCl? Ksp of AgCl = 1.8 × 10⁻¹⁰.

Step 1: After mixing, total volume = 100 mL. Concentrations halve.

[Ag⁺] = 1.0 × 10⁻³ mol L⁻¹; [Cl⁻] = 2.0 × 10⁻⁴ mol L⁻¹.

Step 2: Ionic product Q = [Ag⁺][Cl⁻] = 1.0 × 10⁻³ × 2.0 × 10⁻⁴ = 2.0 × 10⁻⁷.

Step 3: Compare: Q (2.0 × 10⁻⁷) > Ksp (1.8 × 10⁻¹⁰). Precipitation of AgCl will occur.

Common mistakes

• Forgetting to raise concentrations to the power of their coefficients → always use exponents from the balanced equation.
• Using moles instead of molar concentrations in the Kc expression → divide by volume in litres first.
• Thinking equilibrium means concentrations are equal → equilibrium means they are constant, not necessarily equal.
• Ignoring the sign of Δn when relating Kp and Kc → count only gaseous species and keep track of whether Δn is positive or negative.
• Assuming pH = 7 means neutral regardless of temperature → pH of pure water changes with temperature because Kw changes.

Quick revision

• Equilibrium is dynamic: forward rate = reverse rate; concentrations stay constant.
• K large → products favoured; K small → reactants favoured.
• Le Chatelier: the system opposes any external change.
• pH = −log₁₀[H⁺]; neutral water at 25 °C has pH 7.
• Precipitation occurs when ionic product exceeds Ksp.
• Buffers resist pH change; use Henderson–Hasselbalch to calculate pH.