Do not assume K is dimensionless; work out the units from the expression each time, because they depend on the stoichiometry.

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

WJEC Eduqas Eduqas 2019-style practice: For H_2(g) + I_2(g) 2HI(g), at equilibrium a 1.00\ dm^3 vessel contains 0.20\ mol H_2, 0.20\ mol I_2 and 1.60\ mol HI. (a) Write the expression for K_c. (b) Calculate its value and state the units.

(a) K_c = [HI]^2[H_2][I_2] (1). (b) Concentrations (in a 1.00\ dm^3 vessel) are [H_2] = 0.20, [I_2] = 0.20, [HI] = 1.60\ mol dm^-3. K_c = (1.60)^20.20 × 0.20 = 2.560.040 = 64 (3). The units cancel, so K_c has no units (1).

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How do we quantify the position of an equilibrium with Kc and Kp, and what changes their value?

The equilibrium constants Kc and Kp, writing their expressions, calculating their values and units, and the effect of temperature, concentration, pressure and catalysts on the constant and the position of equilibrium.

An Eduqas A-Level Chemistry PI5.1 answer on the equilibrium constants Kc and Kp, writing and calculating their expressions and units, and the effect of changing conditions on K and on the position of equilibrium.

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 topic is asking
Writing Kc and Kp
Calculating Kc
What changes the equilibrium constant
Linking to Le Chatelier
Examples in context
Try this

What this topic is asking

Eduqas topic PI5.1 makes equilibrium quantitative through the equilibrium constants $K_c$ (in terms of concentration) and $K_p$ (in terms of partial pressure). You write their expressions, calculate their values and units, and explain how (and whether) changing temperature, concentration, pressure or adding a catalyst affects the constant and the position of equilibrium. It builds on the qualitative Le Chatelier treatment of topic C2.1.

Writing Kc and Kp

The units of $K$ are found by substituting the units of each term; they may cancel (as in the $\text{H}_2/\text{I}_2/\text{HI}$ system) or leave a net unit.

Calculating Kc

Calculating Kc from equilibrium amounts

$1.00\ \text{mol}$ of ethanoic acid and $1.00\ \text{mol}$ of ethanol react in a fixed volume $V$ ; at equilibrium $0.67\ \text{mol}$ of ester has formed. Find $K_c$ for $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{ester} + \text{H}_2\text{O}$ .

step 1: Find equilibrium moles

Ester $= 0.67$ , water $= 0.67$ ; acid $= 1.00 - 0.67 = 0.33$ ; ethanol $= 0.33$ mol.

step 2: Write the expression

$K_c = \dfrac{[\text{ester}][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$ .

step 3: Substitute (the volume cancels)

Because there are equal moles of gas/solution on each side, $V$ cancels: $K_c = \dfrac{0.67 \times 0.67}{0.33 \times 0.33} = \dfrac{0.449}{0.109} = 4.1$ (no units).

What changes the equilibrium constant

Linking to Le Chatelier

The constant and Le Chatelier's principle are consistent: when you change a concentration, the system shifts so that the reaction quotient returns to $K$ . When you change temperature, $K$ itself changes, which is why temperature is the only factor that alters the constant.

Examples in context

Example 1. The Contact process. $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ is exothermic, so a low temperature gives a large $K_p$ and high yield, but a moderate temperature and a vanadium(V) oxide catalyst are used to reach equilibrium fast enough, a yield-versus-rate compromise.

Example 2. Esterification equilibrium. Because $K_c$ for esterification is only a few, the reaction does not go to completion; removing water (or using excess alcohol) shifts the position to improve the yield without changing $K_c$ .

Try this

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

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

Q2. State the only factor that changes the value of the equilibrium constant, and explain why pressure does not. [2 marks]

Cue. Only temperature changes $K$ ; changing pressure shifts the position of equilibrium so the reaction quotient returns to the same $K$ , leaving the constant unchanged.

Exam-style practice questions

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

Eduqas 20195 marksFor

\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 contains

0.20\ \text{mol}

\text{H}_2

0.20\ \text{mol}

\text{I}_2

and

1.60\ \text{mol}

HI. (a) Write the expression for

K_c

. (b) Calculate its value and state the units.

Show worked answer →

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

(b) Concentrations (in a $1.00\ \text{dm}^3$ vessel) are $[\text{H}_2] = 0.20$ , $[\text{I}_2] = 0.20$ , $[\text{HI}] = 1.60\ \text{mol dm}^{-3}$ . $K_c = \dfrac{(1.60)^2}{0.20 \times 0.20} = \dfrac{2.56}{0.040} = 64$ (3). The units cancel, so $K_c$ has no units (1).

Eduqas 20214 marksThe reaction

2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})

is exothermic. (a) Write the expression for

K_p

. (b) State and explain the effect on the value of

K_p

of increasing the temperature.

Show worked answer →

(a) $K_p = \dfrac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}$ (2).

(b) Increasing the temperature decreases $K_p$ (1). The forward reaction is exothermic, so raising the temperature shifts the equilibrium in the endothermic (backward) direction, lowering the proportion of products and so decreasing the equilibrium constant (1).

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Sources & how we know this

WJEC Eduqas GCE A Level Chemistry specification (from 2015) — WJEC Eduqas (2015)