Kc Expressions and Calculations Revision Notes | A Level AQA Chemistry

1. Why Kc Matters — The Equilibrium Snapshot

Picture a reversible reaction that has reached equilibrium. The concentrations of reactants and products are no longer changing — but they are not necessarily equal. Kc captures that exact ratio in a single number, telling you precisely how far the reaction has gone.

A large Kc means the equilibrium position lies to the right — products dominate. A small Kc means reactants dominate. That one number encodes the whole story of where the reaction sits at a given temperature.

2. Constructing a Kc Expression

The Kc expression is deduced directly from the balanced equation for the reversible reaction. For a general homogeneous equilibrium:

The rule is consistent: products always go on the numerator, reactants on the denominator. Each concentration is written in square brackets — the AQA notation $[X]$ means the equilibrium concentration of species X in mol dm⁻³.

Now let's apply this to a real reaction. Consider the formation of hydrogen iodide:

$\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$Products on top, reactants on the bottom, each raised to its stoichiometric coefficient: $K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$

3. Units of Kc — Deriving Them from the Expression

The units of Kc are not fixed — they depend entirely on the stoichiometry of the reaction. You derive them by substituting the units of concentration (mol dm⁻³) into the expression and cancelling.

The key quantity is $\Delta n$, the change in the total number of moles of gas (or, for homogeneous liquid-phase reactions, total moles of solute): $\Delta n = \text{moles of products} - \text{moles of reactants}$ in the balanced equation. When $\Delta n = 0$, Kc is dimensionless. When $\Delta n eq 0$, the units are (mol dm⁻³)$^{\Delta n}$.

4. Calculating Kc from Equilibrium Concentrations

When equilibrium concentrations are given directly, calculating Kc is a three-step process: write the expression, substitute the values, and evaluate — including units. The worked example below uses the esterification equilibrium, a reaction you will encounter across multiple topics.

Worked example: Calculating Kc from Given Equilibrium Concentrations

At a fixed temperature, the following equilibrium is established in a homogeneous system: $\text{CH}_3\text{COOH(l)} + \text{C}_2\text{H}_5\text{OH(l)} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5\text{(l)} + \text{H}_2\text{O(l)}$ At equilibrium: $[\text{CH}_3\text{COOH}] = 0.500$ mol dm⁻³, $[\text{C}_2\text{H}_5\text{OH}] = 0.500$ mol dm⁻³, $[\text{CH}_3\text{COOC}_2\text{H}_5] = 0.333$ mol dm⁻³, $[\text{H}_2\text{O}] = 0.333$ mol dm⁻³. Calculate Kc.

1. $K_c = \dfrac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$ — Products over reactants. All stoichiometric coefficients are 1, so no powers are needed. [M1]
2. $K_c = \dfrac{0.333 \times 0.333}{0.500 \times 0.500}$ — Substitute each equilibrium concentration directly. [M1]
3. Numerator: $0.333 \times 0.333 = 0.111$ Denominator: $0.500 \times 0.500 = 0.250$ — Working each product separately reduces arithmetic errors. Using 0.111 here keeps the arithmetic consistent with the quoted data.
4. $K_c = \dfrac{0.111}{0.250} = 0.444$ — Divide numerator by denominator. [A1]
5. $\Delta n = (1+1) - (1+1) = 0$, so units cancel: Kc has no units. — When Δn = 0, Kc is dimensionless. State this explicitly for full marks. [B1]

Answer: $K_c = 0.444$ (no units)

5. Finding Equilibrium Concentrations Before Calculating Kc

Exam questions rarely hand you equilibrium concentrations directly. More often, you are given initial amounts and told how much reacts — or you are given equilibrium amounts and a volume. The key step is always to convert moles to concentrations by dividing by the volume of the reaction mixture.

A structured approach using an ICE table (Initial, Change, Equilibrium) keeps the working organised and reduces errors. Here is a full example using the decomposition of dinitrogen tetroxide, a reaction that appears frequently in AQA papers.

Worked example: Kc Calculation from Initial Moles and Equilibrium Data

0.200 mol of $\text{N}_2\text{O}_4$ is placed in a 2.00 dm³ container and allowed to reach equilibrium at constant temperature: $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$ At equilibrium, 0.0800 mol of $\text{N}_2\text{O}_4$ remains. Calculate Kc.

1. Initial moles: $\text{N}_2\text{O}_4 = 0.200$, $\text{NO}_2 = 0$ Change: $\text{N}_2\text{O}_4$ decreases by $x$, $\text{NO}_2$ increases by $2x$ Equilibrium moles: $\text{N}_2\text{O}_4 = 0.200 - x$, $\text{NO}_2 = 2x$ — The stoichiometry tells us that for every mole of N₂O₄ that decomposes, 2 moles of NO₂ form.
2. $0.200 - x = 0.0800$ $x = 0.200 - 0.0800 = 0.120$ mol — x is the number of moles of N₂O₄ that decomposed.
3. $\text{NO}_2 = 2x = 2 \times 0.120 = 0.240$ mol — Use the stoichiometric ratio: 2 moles of NO₂ per mole of N₂O₄ decomposed.
4. $[\text{N}_2\text{O}_4] = \dfrac{0.0800}{2.00} = 0.0400$ mol dm⁻³ $[\text{NO}_2] = \dfrac{0.240}{2.00} = 0.120$ mol dm⁻³ — Divide each equilibrium mole value by the volume (2.00 dm³). This is the step most students forget. [M1]
5. $K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.120)^2}{0.0400}$ — Products over reactants; NO₂ has a stoichiometric coefficient of 2, so it is squared. [M1]
6. $(0.120)^2 = 0.120 \times 0.120 = 0.0144$ — Expand the square explicitly to avoid arithmetic errors.
7. $K_c = \dfrac{0.0144}{0.0400} = 0.360$ mol dm⁻³ — $\Delta n = 2 - 1 = +1$, so units are mol dm⁻³. [A1] for value, [B1] for units.

Answer: $K_c = 0.360$ mol dm⁻³

6. Working Backwards — Finding an Equilibrium Concentration from Kc

AQA also tests the reverse calculation: you are given Kc and some equilibrium concentrations, and asked to find a missing one. The method is algebraic — rearrange the Kc expression to make the unknown concentration the subject.

Worked example: Finding a Missing Equilibrium Concentration

For the equilibrium $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$, Kc = 49.0 (no units) at a certain temperature. At equilibrium, $[\text{H}_2] = 0.100$ mol dm⁻³ and $[\text{I}_2] = 0.100$ mol dm⁻³. Calculate the equilibrium concentration of HI.

1. $K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ — Products over reactants; HI has coefficient 2, so it is squared.
2. $49.0 = \dfrac{[\text{HI}]^2}{0.100 \times 0.100}$ — Substitute Kc and the two known concentrations. [M1]
3. $0.100 \times 0.100 = 0.0100$ — Work out the denominator before rearranging.
4. $[\text{HI}]^2 = 49.0 \times 0.0100 = 0.490$ — Multiply both sides by the denominator. [M1]
5. $[\text{HI}] = \sqrt{0.490} = 0.700$ mol dm⁻³ — $\sqrt{0.490} = 0.700$ (verify: $0.700 \times 0.700 = 0.490$ ✓). [A1]

Answer: $[\text{HI}] = 0.700$ mol dm⁻³

7. Interpreting Kc and the Effect of Temperature

The numerical value of Kc tells you where the equilibrium position lies. A Kc much greater than 1 means the equilibrium lies to the right — products are favoured. A Kc much less than 1 means the equilibrium lies to the left — reactants are favoured.

Temperature is the only factor that changes the value of Kc. Increasing temperature shifts the equilibrium in the endothermic direction, which changes the ratio of product to reactant concentrations and therefore changes Kc. Changing concentration or pressure shifts the equilibrium position but leaves Kc unchanged — the system adjusts concentrations to restore the same ratio.

8. Exam Technique and Mark-Winning Strategies

AQA Kc questions follow predictable patterns. Recognising the structure of each question type before you start writing saves time and secures method marks even if your arithmetic slips.

For 'calculate' questions, the mark scheme awards marks at three distinct stages: writing the correct Kc expression, substituting correct equilibrium concentrations (in mol dm⁻³, not moles), and reaching the correct numerical answer with correct units. Losing the units mark is one of the most common avoidable errors at A Level.

• Write the Kc expression before substituting any numbers — this secures the method mark even if later arithmetic is wrong.
• Convert moles to concentrations (divide by volume in dm³) before substituting — never substitute raw mole values.
• Derive units every time by substituting mol dm⁻³ into the expression and cancelling; do not assume Kc is dimensionless.
• When squaring a concentration, write out the multiplication explicitly (e.g. $(0.120)^2 = 0.120 \times 0.120 = 0.0144$) to avoid errors.
• State that Kc only changes with temperature if asked to explain the effect of a concentration or pressure change on Kc.

Across all five specification requirements for this topic, the thread connecting them is the same: the balanced equation is the starting point for everything. The expression follows from the equation, the units follow from the expression, and the calculation follows from substituting equilibrium concentrations — always in mol dm⁻³ — into that expression at constant temperature.