1. Writing Kc
For a homogeneous equilibrium aA + bB ⇌ cC + dD, write the equilibrium constant in terms of equilibrium concentrations:
Square brackets mean concentration in mol dm⁻³. Products go above reactants. Each balancing coefficient becomes a power, not a multiplier in front of a concentration. Write the symbolic expression before inserting numbers. Concentrations must belong to the same equilibrium mixture at the stated temperature.
| Equation | Kc expression |
|---|---|
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | [NH₃]² / ([N₂][H₂]³) |
| N₂O₄(g) ⇌ 2NO₂(g) | [NO₂]² / [N₂O₄] |
| H₂(g) + I₂(g) ⇌ 2HI(g) | [HI]² / ([H₂][I₂]) |
The iodine example assumes a temperature at which all species are gases. Do not label a halogen as a separate pure liquid and then apply the homogeneous gas expression to it. A homogeneous liquid reaction mixture, such as esterification, is different: its dissolved reactants and products—including water—appear in the concentration expression used here. There is no rule that “all liquids are omitted”.
2. Deriving units
Replace each bracket with mol dm⁻³ and apply the same powers. Cancel before simplifying. If the sum of product coefficients minus the sum of reactant coefficients is Δν, the AQA concentration-based units are (mol dm⁻³)Δν.
| Equilibrium | Power difference | Units |
|---|---|---|
| Ammonia formation | 2 − (1 + 3) = −2 | (mol dm⁻³)⁻² = dm⁶ mol⁻² |
| N₂O₄ dissociation | 2 − 1 = 1 | mol dm⁻³ |
| Hydrogen iodide formation | 2 − (1 + 1) = 0 | No units |
“No units” is the appropriate result when the concentration units cancel in these AQA calculations. Always derive units for the equation as written; reversing or scaling the equation changes the expression.
3. Initial, change, equilibrium
Start with a balanced equation. Make an initial/change/equilibrium table in moles. Use the coefficients to relate the amounts reacting or forming, then subtract from initial reactants and add to initial products. Finally divide every equilibrium amount by the vessel volume in dm³.
The ratio 1:3:2 relates changes. It does not say that equilibrium amounts must be in that ratio. Any product present initially must be included in the final amount. Check that no calculated amount is negative and keep unrounded values until the final answer.
4. When volume cancels
Substitute [X] = n(X)/V into the expression. A common volume cancels only when the sums of coefficients in numerator and denominator are equal. Counting the number of different substances is not enough.
5. Finding a missing concentration or amount
Rearrange algebraically first. Take a square root if the unknown concentration is squared. Check whether the question asks for concentration or moles; convert using n = cV if necessary.
6. What changes Kc?
| Change | Equilibrium composition | Kc |
|---|---|---|
| Change temperature | Usually changes | Changes: heating increases Kc for an endothermic forward reaction and decreases it for an exothermic one |
| Change concentration | Shifts to re-establish equilibrium | Unchanged at fixed temperature |
| Compress gases | May shift if gas coefficients differ | Unchanged at fixed temperature |
| Add catalyst | Unchanged at equilibrium | Unchanged |
A larger Kc indicates a more product-favoured equilibrium for the specified equation and concentration convention. It does not mean a faster reaction, nor does it give a product amount without the initial composition. Compare values for the same equation.
| Equation change | New constant | Example |
|---|---|---|
| Reverse | 1/Kc | If Kc = 0.250 mol dm⁻³, reverse Kc = 4.00 dm³ mol⁻¹ |
| Multiply every coefficient by r | Kc raised to r | Doubling an equation squares its Kc and its units |
Quick checks
Original Finesse practice with indicative worked answers, not official AQA mark allocations.
Q1. Write Kc and its units for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).Show answer
Kc = [SO₃]² / ([SO₂]²[O₂]). Unit power = 2 − 3 = −1, so units are dm³ mol⁻¹.
Q2. Initially 0.600 mol H₂, 0.400 mol I₂ and no HI are present. At equilibrium 0.200 mol I₂ has reacted. Find Kc for H₂(g) + I₂(g) ⇌ 2HI(g).Show answer
Changes: −0.200 mol H₂, −0.200 mol I₂, +0.400 mol HI. Equilibrium amounts: 0.400, 0.200, 0.400 mol.
The common volume cancels. Kc = 0.400² / (0.400 × 0.200) = 2.00, no units.
Q3. For N₂O₄(g) ⇌ 2NO₂(g), Kc = 0.125 mol dm⁻³ and [N₂O₄] = 0.0800 mol dm⁻³. Find the moles of NO₂ in 2.50 dm³.Show answer
[NO₂] = √(0.125 × 0.0800) = 0.100 mol dm⁻³. n = 0.100 × 2.50 = 0.250 mol.
Q4. Kc for a forward equation is 0.0400 dm⁶ mol⁻². Give Kc for the reverse equation and its units.Show answer
Reverse Kc = 1 / 0.0400 = 25.0 mol² dm⁻⁶. Both the numerical value and the units are reciprocated.
Q5. Cooling a mixture increases Kc. Deduce the sign of ΔH for the forward reaction and explain whether compression at that same temperature changes Kc.Show answer
Cooling favours products, so the forward reaction is exothermic (ΔH negative): it releases heat and opposes cooling. Compression does not change Kc at fixed temperature, even if the composition shifts.
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.6 Equilibria revision guide (N. Goalby, Oct 2025 upload) — Kc expressions, units, mole tables, rearrangement and temperature effects; phase labels and early rounding corrected.
- AQA 7405 specification — 3.1.6 Chemical equilibria, Le Chatelier's principle and Kc — 3.1.6.2: equilibrium constant Kc for homogeneous systems.
- AQA 7404/1 mark scheme, June 2023 — Q08.1–Q08.4: equilibrium amounts, expressions, temperature and rearrangement.
- AQA 7404/1 examiner report, June 2023 — Q08: coefficient ratios, expressions and algebra.
- AQA 7404/1 mark scheme, June 2022 — Q07.4–Q07.5: concentrations, final moles and reciprocal constant.
- AQA 7404/1 examiner report, June 2022 — Q07.4–Q07.5: completing the calculation requested.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.