Write the K_c expression for N_2 + 3H_2 2NH_3. [1 mark]

State the units of K_c for H_2 + I_2 2HI. [1 mark]

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How do we quantify the position of an equilibrium?

The equilibrium constants Kc and Kp, writing and evaluating expressions, units, the effect of changing conditions, and the role of a catalyst.

A focused answer to WJEC A-Level Chemistry Unit 3, covering the equilibrium constants Kc and Kp, writing expressions and determining units, calculating values from equilibrium amounts, and how temperature, pressure and catalysts affect them.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

WJEC wants you to write $K_c$ and $K_p$ expressions, determine their units, calculate values from equilibrium amounts, and explain how temperature, pressure and a catalyst affect both the position of equilibrium and the value of the constant.

The answer

The equilibrium constants

Units

Calculating Kc

Kc with an ICE table

For $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$ , $0.100$ mol $\text{N}_2\text{O}_4$ is placed in a $2.00$ dm $^3$ flask and $40$ percent dissociates. Find $K_c$ .

step 1 Find amounts at equilibrium

Dissociated: $0.40 \times 0.100 = 0.0400$ mol $\text{N}_2\text{O}_4$ , forming $0.0800$ mol $\text{NO}_2$ . Remaining $\text{N}_2\text{O}_4 = 0.0600$ mol.

step 2 Convert to concentrations

$[\text{N}_2\text{O}_4] = 0.0600/2.00 = 0.0300$ , $[\text{NO}_2] = 0.0800/2.00 = 0.0400$ mol dm $^{-3}$ .

step 3 Substitute

$K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{0.0400^2}{0.0300} = \dfrac{0.00160}{0.0300} = 0.0533$ .

step 4 Units

$\dfrac{(\text{mol dm}^{-3})^2}{\text{mol dm}^{-3}} = \text{mol dm}^{-3}$ , so $K_c = 0.0533$ mol dm $^{-3}$ .

What changes the constant

Only temperature changes the value of $K_c$ or $K_p$ . Changing concentration or pressure shifts the position of equilibrium but the constant stays the same. A catalyst changes neither the position nor the value; it only speeds attainment.

Position versus value of the constant

A frequent source of confusion is the difference between shifting the position of equilibrium and changing the value of the constant. Changing concentration or pressure shifts the position (more or less product forms) but the system returns to the same $K_c$ or $K_p$ , so the constant is unchanged. Only temperature changes the value, because it alters the balance of the forward and reverse rates differently. For an exothermic forward reaction, raising the temperature shifts the position toward reactants and lowers $K$ ; for an endothermic one, raising the temperature raises $K$ .

Partial pressures for Kp

To use $K_p$ , first find the mole fraction of each gas (its moles divided by the total moles of gas), then multiply by the total pressure to get its partial pressure. Substituting these partial pressures into the $K_p$ expression and working out the units (powers of Pa) gives the constant. This mole-fraction route is the standard method for gas-phase equilibrium calculations, and care with the total number of gas moles is where most marks are lost.

Examples in context

The Contact process. Sulfur dioxide oxidation to sulfur trioxide is run at a moderate temperature to keep $K_p$ favourable (the forward reaction is exothermic) while a vanadium(V) oxide catalyst secures a workable rate. The Haber process. A compromise temperature balances a higher $K_p$ at low temperature against an acceptable rate, the classic industrial equilibrium optimisation.

Try this

Q1. Write the $K_c$ expression for $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ . [1 mark]

Cue. $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ .

Q2. State which single factor changes the value of $K_c$ . [1 mark]

Cue. Temperature.

Q3. State the units of $K_c$ for $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ . [1 mark]

Cue. No units (equal moles of gas on both sides).

Q4. State the effect of increasing pressure on the value of $K_p$ . [1 mark]

Cue. No effect; pressure shifts the position but does not change $K_p$ .

Q5. State how you find the partial pressure of a gas in a mixture. [1 mark]

Cue. Multiply its mole fraction by the total pressure.

Exam-style practice questions

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

WJEC 20195 marksFor the equilibrium

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

, 1.00 mol of

\text{H}_2

and 1.00 mol of

\text{I}_2

were mixed in a

1.00\ \text{dm}^3

vessel. At equilibrium 1.56 mol of

\text{HI}

had formed. Calculate

K_c

Show worked answer →

Set up an ICE table. Initial: $\text{H}_2 = 1.00$ , $\text{I}_2 = 1.00$ , $\text{HI} = 0$ .

Change: $1.56$ mol HI forms, and the ratio is $1:1:2$ , so $0.78$ mol each of $\text{H}_2$ and $\text{I}_2$ are used.

Equilibrium: $\text{H}_2 = 1.00 - 0.78 = 0.22$ mol, $\text{I}_2 = 0.22$ mol, $\text{HI} = 1.56$ mol. Volume is $1.00$ dm3, so these are also concentrations.

$K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{1.56^2}{0.22 \times 0.22} = \dfrac{2.43}{0.0484} = 50.3$ (no units, as the powers cancel).

Markers reward the ICE table, correct equilibrium amounts, the expression, and a value with no units.

WJEC 20213 marksState and explain the effect of increasing the temperature on the value of Kc for an exothermic forward reaction.

Show worked answer →

For an exothermic forward reaction, raising the temperature shifts the equilibrium in the endothermic (reverse) direction by Le Chatelier's principle.

This reduces the amount of product and increases the amount of reactant at equilibrium.

Because $K_c$ depends on these equilibrium concentrations, $K_c$ decreases as temperature rises.

Markers reward the shift toward reactants, and the conclusion that $K_c$ decreases (temperature is the only factor that changes $K_c$ ).

Related dot points

Sources & how we know this

WJEC A-level Chemistry specification — WJEC (2015)