Write the expression for K_c for 2NO_2(g) N_2O_4(g). [1 mark]

OCR OCR 2020-style practice: For the equilibrium H_2(g) + I_2(g) 2HI(g), an equilibrium mixture in a 1.0\ 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 K_c and state its units.

(a) K_c = [HI]^2[H_2][I_2] (1). (b) Concentrations equal moles here (volume 1.0\ dm^3). K_c = (1.60)^2(0.20)(0.20) = 2.560.040 = 64 (1)(1). The units cancel ((mol dm^-3)^2(mol dm^-3)^2), so K_c has no units (1). Markers reward the correct expression, the substituted values, the value 64, and the recognition that the units cancel.

OCR OCR 2022-style practice: For N_2O_4(g) 2NO_2(g) at equilibrium the total pressure is 100\ kPa and the mole fraction of NO_2 is 0.40. (a) Find the partial pressures. (b) Write and evaluate K_p.

(a) Mole fraction of N_2O_4 = 1 - 0.40 = 0.60. Partial pressures: p(NO_2) = 0.40 × 100 = 40\ kPa and p(N_2O_4) = 0.60 × 100 = 60\ kPa (1)(1). (b) K_p = p(NO_2)^2p(N_2O_4) = (40)^260 = 160060 = 27\ kPa (1)(1). Markers reward both partial pressures from the mole fractions, the correct K_p expression, the value, and the units kPa.

EnglandChemistrySyllabus dot point

How do we quantify the position of equilibrium, and what changes its value?

The equilibrium constant Kc in terms of concentrations and Kp in terms of partial pressures and mole fractions, calculations from equilibrium amounts, and the effect of temperature and catalysts on the constant.

An OCR H432 module 5 answer on equilibrium constants: writing and calculating Kc from concentrations and Kp from partial pressures and mole fractions, and the effect of temperature and catalysts on the value of the constant.

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
The equilibrium constant Kc
The equilibrium constant Kp
What changes the constant
Examples in context
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What this topic is asking

OCR specification point 5.1.2 wants you to write the equilibrium constant $K_c$ in terms of concentrations and $K_p$ in terms of partial pressures, calculate them from equilibrium amounts (using mole fractions for $K_p$ ), deduce their units, and state how temperature changes the constant while a catalyst does not. This puts numbers on the Le Chatelier ideas of Module 3.

The equilibrium constant Kc

To calculate $K_c$ : find the equilibrium amounts (often from an initial-change-equilibrium table), convert to concentrations by dividing by the volume, substitute into the expression, and work out the units by cancelling.

The equilibrium constant Kp

The mole fraction of a gas is its moles divided by the total moles of gas. The partial pressures must add up to the total pressure.

Calculating Kp from mole fractions

Find $K_p$ for $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ given equilibrium amounts of $1.0\ \text{mol}$ $\text{N}_2$ , $3.0\ \text{mol}$ $\text{H}_2$ and $4.0\ \text{mol}$ $\text{NH}_3$ at a total pressure of $200\ \text{kPa}$ .

step 1: Total moles and mole fractions

Total $= 1.0 + 3.0 + 4.0 = 8.0\ \text{mol}$ . Mole fractions: $\text{N}_2 = 0.125$ , $\text{H}_2 = 0.375$ , $\text{NH}_3 = 0.500$ .

step 2: Partial pressures

$p(\text{N}_2) = 0.125 \times 200 = 25\ \text{kPa}$ , $p(\text{H}_2) = 0.375 \times 200 = 75\ \text{kPa}$ , $p(\text{NH}_3) = 0.500 \times 200 = 100\ \text{kPa}$ .

step 3: Substitute into Kp

K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)\,p(\text{H}_2)^3} = \frac{(100)^2}{(25)(75)^3} = \frac{10000}{25 \times 421875} \approx 9.5 \times 10^{-4}\ \text{kPa}^{-2}.

step 4: Units

Two pressures on top, four on the bottom, so the units are $\text{kPa}^{2-4} = \text{kPa}^{-2}$ .

What changes the constant

Examples in context

Example 1. The Haber process compromise. A high pressure raises the equilibrium yield of ammonia (fewer gas moles on the product side) but does not change $K_p$ ; the moderate temperature used is a compromise between a higher yield (favoured by low temperature, since the forward reaction is exothermic) and a workable rate.

Example 2. Why $K$ alone does not give the rate. A large $K_c$ means the equilibrium lies well towards products, but says nothing about how fast equilibrium is reached, which is why a separate catalyst is used in industrial processes.

Try this

Q1. Write the expression for $K_c$ for $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$ . [1 mark]

Cue. $K_c = \dfrac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}$ .

Q2. State the only factor that changes the value of $K_p$ . [1 mark]

Cue. Temperature.

Exam-style practice questions

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

OCR 20204 marksFor the equilibrium

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

, an equilibrium mixture in a

1.0\ \text{dm}^3

vessel contains

0.20\ \text{mol}

\text{H}_2

0.20\ \text{mol}

\text{I}_2

and

1.60\ \text{mol}

\text{HI}

. (a) Write the expression for

K_c

. (b) Calculate

K_c

and state its units.

Show worked answer →

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

(b) Concentrations equal moles here (volume $1.0\ \text{dm}^3$ ). $K_c = \dfrac{(1.60)^2}{(0.20)(0.20)} = \dfrac{2.56}{0.040} = 64$ (1)(1). The units cancel ( $\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2}$ ), so $K_c$ has no units (1).

Markers reward the correct expression, the substituted values, the value $64$ , and the recognition that the units cancel.

OCR 20224 marksFor

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

at equilibrium the total pressure is

100\ \text{kPa}

and the mole fraction of

\text{NO}_2

0.40

. (a) Find the partial pressures. (b) Write and evaluate

K_p

Show worked answer →

(a) Mole fraction of $\text{N}_2\text{O}_4 = 1 - 0.40 = 0.60$ . Partial pressures: $p(\text{NO}_2) = 0.40 \times 100 = 40\ \text{kPa}$ and $p(\text{N}_2\text{O}_4) = 0.60 \times 100 = 60\ \text{kPa}$ (1)(1).

(b) $K_p = \dfrac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \dfrac{(40)^2}{60} = \dfrac{1600}{60} = 27\ \text{kPa}$ (1)(1).

Markers reward both partial pressures from the mole fractions, the correct $K_p$ expression, the value, and the units $\text{kPa}$ .

Related dot points

Sources & how we know this

OCR A-Level Chemistry A (H432) specification — OCR (2015)