What is dynamic equilibrium?

The two key words are "closed" (nothing enters or leaves) and "dynamic" (the reactions never stop, they just balance). For H_2(g) + I_2(g) 2HI(g), once equilibrium is reached the amount of each gas is fixed, but individual molecules are constantly forming and decomposing.

0.50\ mol of ethanoic acid and 0.50\ mol of ethanol reach equilibrium with 0.33\ mol of ester in a fixed volume V. Write the K_c expression and show that K_c is dimensionless. [3 marks]

Pearson Edexcel Edexcel 2021-style practice: Hydrogen and iodine were mixed and allowed to reach equilibrium at constant temperature: H_2(g) + I_2(g) 2HI(g). At equilibrium a 1.00\ dm^3 vessel contained 0.020\ mol H_2, 0.020\ mol I_2 and 0.160\ mol HI. Calculate K_c and state its units.

Write the K_c expression, substitute equilibrium concentrations, and evaluate. Expression: K_c = [HI]^2[H_2][I_2] (1). Concentrations (volume = 1.00\ dm^3): [HI] = 0.160, [H_2] = 0.020, [I_2] = 0.020\ mol dm^-3 (1). Substitute: K_c = (0.160)^2(0.020)(0.020) = 0.02560.0004 = 64 (2). Units: the powers cancel (2 - 1 - 1 = 0), so K_c has no units (1). Markers reward the correct expression, correct substitution, the value 64, and the recognition that K_c is dimensionless here.

EnglandChemistrySyllabus dot point

Where does a reversible reaction settle, and how can we shift it?

Dynamic equilibrium, Le Chatelier's principle, the effects of concentration, pressure, temperature and catalysts on the position of equilibrium, and the meaning of the equilibrium constant Kc.

An Edexcel 9CH0 Topic 10 answer covering dynamic equilibrium, Le Chatelier's principle, the effects of changing conditions, and the equilibrium constant Kc with worked calculations.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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What this topic is asking

Edexcel Topic 10 wants you to explain what a dynamic equilibrium is, apply Le Chatelier's principle to predict how changing conditions shifts the position of equilibrium, and write and use the equilibrium constant $K_c$ . You must be able to deduce the direction of a shift, state whether $K_c$ changes, and calculate $K_c$ from equilibrium amounts including its correct units.

The answer

Dynamic equilibrium

The two key words are "closed" (nothing enters or leaves) and "dynamic" (the reactions never stop, they just balance). For $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ , once equilibrium is reached the amount of each gas is fixed, but individual molecules are constantly forming and decomposing.

Le Chatelier's principle

Concentration. Increasing a reactant shifts the position towards products; removing a product also drives the reaction forward.
Pressure (gases only). Increasing total pressure shifts towards the side with fewer gas moles. If both sides have equal moles, pressure has no effect on position.
Temperature. Increasing temperature shifts in the endothermic direction. For an exothermic forward reaction, heating reduces the yield of products.
Catalyst. A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but the position and $K_c$ are unchanged.

The equilibrium constant Kc

For the general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ :

A large $K_c$ ( $K_c \gg 1$ ) means products are favoured at equilibrium; a small $K_c$ ( $K_c \ll 1$ ) means reactants dominate. Only a change of temperature changes the value of $K_c$ . Concentration, pressure and catalysts alter the position of equilibrium but leave $K_c$ unchanged, because the system re-adjusts until the same ratio is restored.

The units of $K_c$ depend on the equation. Work out the overall power: (sum of product powers) minus (sum of reactant powers), and attach that many factors of $\text{mol dm}^{-3}$ . For the Haber process $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , the power is $2 - (1 + 3) = -2$ , so units are $\text{mol}^{-2}\,\text{dm}^{6}$ .

Calculating Kc from initial amounts

For $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ , $0.100\ \text{mol}$ $\text{H}_2$ and $0.100\ \text{mol}$ $\text{I}_2$ are placed in a $2.00\ \text{dm}^3$ flask. At equilibrium $0.160\ \text{mol}$ $\text{HI}$ has formed. Find $K_c$ .

step 1: Set up an ICE table in moles

Start: $\text{H}_2 = 0.100$ , $\text{I}_2 = 0.100$ , $\text{HI} = 0$ . Formation of $0.160\ \text{mol}$ $\text{HI}$ uses $0.080\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ (1:1:2 stoichiometry).

step 2: Equilibrium moles

$\text{H}_2 = 0.100 - 0.080 = 0.020\ \text{mol}$ ; $\text{I}_2 = 0.020\ \text{mol}$ ; $\text{HI} = 0.160\ \text{mol}$ .

step 3: Convert to concentrations (divide by $2.00\ \text{dm}^3$ )

$[\text{H}_2] = 0.010$ , $[\text{I}_2] = 0.010$ , $[\text{HI}] = 0.080\ \text{mol dm}^{-3}$ .

step 4: Substitute

K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.080)^2}{(0.010)(0.010)} = \frac{0.0064}{0.0001} = 64

step 5: Units

Power $= 2 - 1 - 1 = 0$ , so $K_c = 64$ with no units.

Examples in context

Example 1. The Haber process. Ammonia is made by $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$ , exothermic forward. A low temperature would favour a high yield (the forward reaction is exothermic) but would be far too slow, so industry compromises at about $450\ ^\circ\text{C}$ with an iron catalyst. High pressure (around $200\ \text{atm}$ ) shifts the position towards ammonia because the product side has fewer gas moles ( $2$ versus $4$ ). Removing ammonia by liquefaction keeps pulling the position to the right.

Example 2. The Contact process. Sulfur trioxide is made by $2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$ , exothermic forward. A moderate temperature of about $450\ ^\circ\text{C}$ and a vanadium(V) oxide catalyst give a fast reaction with a high equilibrium yield. Pressure is kept only slightly above atmospheric because the position already lies well to the right, illustrating that conditions are chosen for the best balance of rate, yield and cost.

Try this

Q1. State Le Chatelier's principle. [2 marks]

Cue. If a system at equilibrium is subjected to a change, the position of equilibrium shifts to oppose that change.

Q2. For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (exothermic forward), predict and explain the effect of increasing temperature on the yield of ammonia. [2 marks]

Cue. Yield falls; the equilibrium shifts in the endothermic (reverse) direction to oppose the rise in temperature, and $K_c$ decreases.

Q3. $0.50\ \text{mol}$ of ethanoic acid and $0.50\ \text{mol}$ of ethanol reach equilibrium with $0.33\ \text{mol}$ of ester in a fixed volume $V$ . Write the $K_c$ expression and show that $K_c$ is dimensionless. [3 marks]

Cue. $K_c = \frac{[\text{ester}][\text{H}_2\text{O}]}{[\text{acid}][\text{alcohol}]}$ ; the $V$ terms and powers cancel ( $2 - 2 = 0$ ), so $K_c$ has no units.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksFor the equilibrium

\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})

, which is endothermic in the forward direction, explain the effect on the position of equilibrium and on

K_c

of (a) increasing the temperature and (b) increasing the total pressure.

Show worked answer →

Award marks for linking each change to Le Chatelier's principle and to $K_c$ .

(a) Increasing temperature shifts the position to the right (1) because the system opposes the rise by absorbing heat in the endothermic forward direction (1). Because temperature is the only factor that changes $K_c$ , $K_c$ increases (1).

(b) Increasing total pressure shifts the position to the left (1) because that side has fewer gas moles (1 mol versus 2 mol), so the system reduces the pressure. $K_c$ is unchanged because pressure does not alter $K_c$ (1).

Common loss of marks: stating that pressure changes $K_c$ , or failing to give the gas mole count that justifies the direction.

Edexcel 20215 marksHydrogen and iodine were mixed and allowed to reach equilibrium at constant temperature:

\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})

. At equilibrium a

1.00\ \text{dm}^3

vessel contained

0.020\ \text{mol}

\text{H}_2

0.020\ \text{mol}

\text{I}_2

and

0.160\ \text{mol}

\text{HI}

. Calculate

K_c

and state its units.

Show worked answer →

Write the $K_c$ expression, substitute equilibrium concentrations, and evaluate.

Expression: $K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ (1).

Concentrations (volume $= 1.00\ \text{dm}^3$ ): $[\text{HI}] = 0.160$ , $[\text{H}_2] = 0.020$ , $[\text{I}_2] = 0.020\ \text{mol dm}^{-3}$ (1).

Substitute: $K_c = \dfrac{(0.160)^2}{(0.020)(0.020)} = \dfrac{0.0256}{0.0004} = 64$ (2).

Units: the powers cancel ( $2 - 1 - 1 = 0$ ), so $K_c$ has no units (1).

Markers reward the correct expression, correct substitution, the value $64$ , and the recognition that $K_c$ is dimensionless here.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)