Does temperature change K?

Yes — K depends on temperature through the van't Hoff equation ln(K₂/K₁) = −(ΔH/R)(1/T₂ − 1/T₁). Concentration changes, pressure changes, and catalysts do not.

Why are pure solids and liquids excluded from K?

Their activities are defined as unity in standard thermodynamics, so they do not appear. Their amount only matters for whether equilibrium can be established at all.

When is K_p numerically equal to K_c?

Only when Δn (moles of gaseous products − reactants) is zero, e.g. H₂ + I₂ ⇌ 2 HI.

What is the difference between Q and K?

K is the value of the reaction quotient at equilibrium. Q is computed at any instant; comparing Q to K predicts the direction of approach.

Why does the Haber process use moderate temperature instead of low T?

The forward reaction is exothermic, so low T favours yield (Le Chatelier) but is kinetically slow. 450 °C is a compromise that gives reasonable yield in reasonable time with an Fe catalyst.

For the reaction N₂ + 3 H₂ ⇌ 2 NH₃, K_p and K_c are related by:

K_p = K_c(RT)⁻². Δn = 2 − 4 = −2, so K_p = K_c(RT)⁻².

Adding a catalyst to an equilibrium system:

Has no effect on K. A catalyst reduces activation energy for both forward and reverse, leaving K (and the equilibrium position) unchanged.

Increasing pressure on N₂ + 3 H₂ ⇌ 2 NH₃ at constant T:

Shifts right. There are 4 mol of gas on the left and 2 on the right; higher P favours fewer moles, shifting right.

For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_p equals:

P(CO₂). Pure solids have unit activity and do not appear; only the partial pressure of CO₂ enters K_p.

If Q < K for a reaction, the system will:

Shift right (toward products). Q < K means there are too few products relative to equilibrium; the reaction proceeds forward.

Chemical Equilibrium MDCAT 2026 — Notes, Key Concepts & MCQs (Chemistry)

Chemical Equilibrium

Chemical Equilibrium averages 3 MCQs per MDCAT paper — Le Chatelier's principle, K_c vs K_p, and ICE-table calculations are core MDCAT material.

Chemical Equilibrium is a Chemistry chapter on the official PMDC MDCAT 2026 syllabus, contributing roughly 2 MCQs to the 45-MCQ Chemistry section. Mastering the core concepts below typically secures the full chapter weightage.

Dynamic equilibrium

A reversible reaction reaches equilibrium when the forward and reverse rates are equal — concentrations are then constant in time, but molecules continue interconverting. For a A + b B ⇌ c C + d D, the equilibrium constant K_c = [C]^c[D]^d / [A]^a[B]^b. K_c depends only on temperature; it is unaffected by initial concentrations, the presence of a catalyst, or the addition of inert species at constant volume. Atkins Chapter 6 stresses that the position of equilibrium is one thing, the value of K is another.

K_c, K_p, and the relationship between them

For gas-phase reactions, K_p = K_c(RT)^Δn, where Δn is moles of gaseous products minus reactants. For N₂ + 3 H₂ ⇌ 2 NH₃, Δn = 2 − 4 = −2; K_p = K_c(RT)⁻². When Δn = 0 (e.g. H₂ + I₂ ⇌ 2 HI), K_p = K_c. Pure liquids and solids do not appear in K expressions because their activities are essentially 1 — so for CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_p = P(CO₂).

Le Chatelier's principle — qualitative shifts

When a stress is applied to an equilibrium, the system shifts to partially relieve it. Increasing reactant concentration shifts forward; raising pressure shifts toward fewer moles of gas; raising temperature shifts in the endothermic direction. The Haber process N₂ + 3 H₂ ⇌ 2 NH₃, ΔH = −92 kJ, illustrates the practical compromise: high pressure favours NH₃, but low temperature (favoured thermodynamically) is too slow kinetically, so industry uses 450 °C and 200 atm with an Fe catalyst — the classic FSc XI Chapter 7 case study.

ICE-table calculations

For 2 SO₂ + O₂ ⇌ 2 SO₃ with K_c = 280 at 1000 K, starting with [SO₂] = [O₂] = 1.0 M and [SO₃] = 0: setting up Initial-Change-Equilibrium with x = mol/L of O₂ consumed gives [SO₃]² / ([SO₂]²[O₂]) = (2x)² / ((1−2x)²(1−x)) = 280. Solve numerically (or by approximation when 2x is comparable to 1, requiring a quadratic). The reaction quotient Q has the same form as K but with non-equilibrium concentrations: Q < K shifts forward, Q > K reverse.

Catalysts, common-ion effect, and applications

A catalyst speeds both forward and reverse reactions equally, so it reaches equilibrium faster but does not change K or the equilibrium position. The common-ion effect explains why adding NaCl decreases AgCl solubility: K_sp = [Ag⁺][Cl⁻] is fixed, so increasing [Cl⁻] forces [Ag⁺] down. Industrial equilibria — Haber (NH₃), Contact (SO₃), Ostwald (HNO₃) — are the recurrent MDCAT case studies: identify the conditions, justify them with Le Chatelier.

Key Concepts

Worked MCQs

Frequently Asked Questions

How Chemical Equilibrium Is Tested

MDCAT questions on Chemical Equilibrium are a mix of recall (definitions, classifications), application (predict outcomes, interpret diagrams), and basic numerical/analytical reasoning. PMDC papers from 2020–2025 emphasized the concepts above; older UHS papers (2008–2019) tested them too, with slight variations in question framing.