Chemical Equilibrium

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam. Chemical equilibrium is the state reached in a reversible reaction when the forward reaction and reverse reaction proceed at identical rates, leaving macroscopic concentrations unchanged — a dynamic equilibrium, because molecular-level processes keep running. The must-know formula is the equilibrium constant expression for a homogeneous system:

Kc expression

$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$

for the reaction aA + bB ⇌ cC + dD, where each concentration is raised to its stoichiometric coefficient. For gas-phase equilibria, Kp uses partial pressures and converts via Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants) and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Le Chatelier’s principle: an imposed change in concentration, temperature, or pressure is counteracted by a shift in equilibrium position. A catalyst does not shift equilibrium. NABTEB pointers: write Kc with products over reactants, omit pure solids/liquids, and remember that only temperature changes the numerical value of Kc.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Reversible reactions and the dynamic state

A reversible reaction is one in which products can react to re-form reactants (e.g. N₂ + 3H₂ ⇌ 2NH₃). At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Although concentrations appear constant to an observer, the reactions continue in both directions — hence the term dynamic equilibrium. Equilibrium can only be established in a closed system at constant temperature.

The equilibrium constant Kc

For a reversible reaction at equilibrium, the ratio of product concentrations to reactant concentrations (each raised to its stoichiometric coefficient) is a constant at a given temperature. This is the equilibrium law (Guldberg–Waage, 1864):

Kc for aA + bB ⇌ cC + dD

$$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

A large Kc (> 1, often >> 1) means products dominate at equilibrium; a small Kc (< 1) means reactants dominate. Kc depends only on temperature. Changing concentration shifts the position of equilibrium but not Kc itself.

Kp and its link to Kc

For gas-phase equilibria, Kp is built from partial pressures: P_A, P_B, … in atm. The two constants are related by:

Conversion between Kp and Kc

$$K_p = K_c(RT)^{\Delta n}$$

where Δn = (sum of gaseous-product moles) − (sum of gaseous-reactant moles). If Δn = 0, Kp = Kc numerically.

Le Chatelier’s principle

The system responds to an external stress (change in concentration, temperature, or pressure) by shifting in the direction that opposes the change:

Concentration: adding a reactant shifts equilibrium right; adding a product shifts it left.
Pressure (gaseous equilibria, Δn ≠ 0): increasing pressure shifts towards the side with fewer moles of gas.
Temperature: raising T favours the endothermic direction; lowering T favours the exothermic direction. Temperature actually changes the value of Kc.
Catalyst: speeds up forward and reverse reactions equally — no shift in position, no change in Kc.

Reaction quotient Q

At any moment, Q is computed with current (non-equilibrium) concentrations in the Kc expression. Equilibrium is reached when Q = Kc.

NABTEB question patterns

Expect short-answer items asking you to (a) write Kc expressions for given equations, (b) calculate Kc from a concentration table (ICE method), (c) predict direction of shift after a stated change, and (d) state how Kc varies with temperature for an exothermic reaction.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Heterogeneous equilibria

In a heterogeneous equilibrium more than one phase is present (e.g. CaCO₃(s) ⇌ CaO(s) + CO₂(g)). Pure solids and pure liquids have essentially constant activities and are omitted from Kc. Thus for the example above, Kc = [CO₂], and the amount of CaCO₃ or CaO present does not appear. NABTEB frequently tests this omission — including a solid or liquid in the expression is one of the highest-frequency errors.

Worked example — Kp → Kc (ammonia synthesis)

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K, the partial pressures at equilibrium give Kp. Convert to Kc using Kp = Kc(RT)^Δn. Here Δn = 2 − (1 + 3) = −2. With R = 0.0821 L·atm·mol⁻¹·K⁻¹ and T = 500 K:

Substitution

$$K_c = K_p / (RT)^{\Delta n} = K_p / [(0.0821)(500)]^{-2}$$

So Kc = Kp · (41.05)² ≈ Kp · 1685. Notice that Kc is numerically much larger than Kp here, reflecting the Δn penalty for reactants gaining mole count.

Common mistakes (worth memorising)

Writing Kc as [reactants]/[products] instead of the reverse.
Raising a substance that has no coefficient written (coefficient is implicitly 1) — students forget [N₂] and [H₂] must each be raised to the power 1.
Stating that a catalyst shifts equilibrium — it only shortens the time to reach the same equilibrium.
Applying Le Chatelier to pressure of a reaction where Δn = 0; nothing happens.
Believing Kc changes when concentrations are altered — only temperature alters Kc.

Connections to adjacent NABTEB topics

Chemical equilibrium links directly to ionic equilibria (Ka, Kw at 25 °C = 1.0 × 10⁻¹⁴, Ksp for sparingly soluble salts), buffer solutions, and electrode potentials (E°cell = 0 when Q = K). The Nernst equation used in electrochemistry is the thermodynamic cousin of the equilibrium expression.

Practice prompts

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), write Kc and Kp, then state the effect of (a) doubling [O₂] and (b) compressing the system to half its volume.
The equilibrium constant Kp for the dissociation of PCl₅ at 250 °C is 1.78 atm. Given Δn = +1, calculate Kc and explain in one sentence why a chemist would prefer Kc when working in solution-linked processes.

Content adapted based on your selected roadmap duration. Switch tiers using the selector above.

Chemical Equilibrium