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

AQA AQA 2018-style practice: An equilibrium mixture contains 2\ mol N_2, 6\ mol H_2 and 2\ mol NH_3 at a total pressure of 200\ kPa. Calculate the partial pressure of ammonia.

First find the total moles of gas: 2 + 6 + 2 = 10\ mol. The mole fraction of ammonia is 210 = 0.20. Partial pressure = mole fraction × total pressure = 0.20 × 200 = 40\ kPa. Markers reward the total moles, the mole fraction, and the partial pressure of 40\ kPa.

AQA AQA 2022-style practice: For the equilibrium N_2(g) + 3H_2(g) 2NH_3(g), an equilibrium mixture at a total pressure of 300\ kPa contains 1.0\ mol N_2, 1.0\ mol H_2 and 2.0\ mol NH_3. Calculate K_p and give its units.

Total moles = 1.0 + 1.0 + 2.0 = 4.0\ mol. Partial pressures (mole fraction × total): p(N_2) = (1.0)/(4.0) × 300 = 75\ kPa, p(H_2) = 75\ kPa, p(NH_3) = (2.0)/(4.0) × 300 = 150\ kPa. K_p = p(NH_3)^2p(N_2)\,p(H_2)^3 = (150)^275 × (75)^3 = 2250075 × 421875 = 7.1 × 10^-4. Units: kPa^2kPa × kPa^3 = kPa^-2, so K_p = 7.1 × 10^-4\ kPa^-2. Markers reward all partial pressures, the correct expression, the value, and the derived units.

EnglandChemistrySyllabus dot point

How do we express equilibrium for gas-phase reactions using partial pressures?

Mole fractions and partial pressures, the equilibrium constant Kp written in terms of partial pressures, calculating Kp, and the effect of changing conditions on Kp.

A focused answer to AQA A-Level Chemistry 3.1.10, covering mole fractions and partial pressures, writing and calculating the equilibrium constant Kp for gaseous equilibria, and the effect of changing conditions on Kp.

Generated by Claude Opus 4.810 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
Mole fractions and partial pressures
The Kp expression
Calculating Kp
Effect of changing conditions
Try this

What this dot point is asking

AQA wants you to calculate mole fractions and partial pressures, write the expression for $K_p$ in terms of partial pressures, calculate $K_p$ with correct units, and explain how changing temperature, pressure or a catalyst affects $K_p$ .

Mole fractions and partial pressures

Mole fractions are useful because they avoid having to know the volume or temperature: as long as the gases share the same container, the ratio of their moles equals the ratio of their partial pressures. This is why $K_p$ problems start by finding total moles, then each mole fraction, then each partial pressure.

The Kp expression

For the gaseous equilibrium $aA + bB \rightleftharpoons cC + dD$ :

Calculating Kp

Find equilibrium moles of each gas, then the total moles, then each mole fraction, then each partial pressure, then substitute into the $K_p$ expression. Work out the units by cancelling the pressure units from the expression.

A typical question gives starting moles and the amount reacted, so you build an "initial, change, equilibrium" (ICE) table in moles first. The change column follows the stoichiometry: if $x$ mol of one reactant is used, the others change in proportion to their balancing numbers. Only after finding equilibrium moles do you convert to partial pressures, because mole fractions need the total at equilibrium, not at the start. A frequent slip is to forget that the total number of moles of gas can change during the reaction (for example $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ goes from 4 moles to 2), so the total used for mole fractions must be the equilibrium total.

The link between $K_p$ and $K_c$ is conceptual at A-level: both describe the same equilibrium, and both depend only on temperature. $K_p$ is preferred for gas-phase equilibria because partial pressures are easier to measure than concentrations, and it is the natural quantity for industrial gas reactions such as the Haber and Contact processes.

Worked example: calculating Kp

For $2SO_2 + O_2 \rightleftharpoons 2SO_3$ , equilibrium partial pressures are $p(SO_2) = 40$ , $p(O_2) = 20$ and $p(SO_3) = 80\ \text{kPa}$ . Calculate $K_p$ and state its units.

Step 1: Write the expression for $K_p$

Products go on top and reactants go on the bottom, each partial pressure raised to its balancing number from the equation. $SO_3$ has coefficient 2, $SO_2$ has coefficient 2, and $O_2$ has coefficient 1.

K_p = \frac{(p_{SO_3})^2}{(p_{SO_2})^2\,(p_{O_2})}

Step 2: Substitute the equilibrium partial pressures

Replace each partial pressure symbol with the given value and carry out the arithmetic.

K_p = \frac{(80)^2}{(40)^2 \times 20} = \frac{6{,}400}{32{,}000}

Step 3: Evaluate and derive the units

Completing the division gives the numerical value. To find the units, track what happens to the kPa units: $\text{kPa}^2$ on top divided by $(\text{kPa}^2 \times \text{kPa})$ on the bottom leaves $\text{kPa}^{-1}$ .

K_p = 0.20\ \text{kPa}^{-1}

Final answer: $K_p = 0.20\ \text{kPa}^{-1}$ .

Effect of changing conditions

When the total pressure is increased, the individual partial pressures rise, but the system responds by shifting towards the side with fewer gas moles until the ratio in the $K_p$ expression returns to the same value. This is why pressure changes the position of equilibrium without changing $K_p$ . A catalyst speeds the forward and reverse reactions equally, so it has no effect on either the position or $K_p$ , only on how quickly equilibrium is reached.

Try this

Q1. Write the expression for $K_p$ for $N_2 + 3H_2 \rightleftharpoons 2NH_3$ . [1 mark]

Cue. $K_p = \dfrac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$ .

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 AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20183 marksAn equilibrium mixture contains

2\ \text{mol}

\text{N}_2

6\ \text{mol}

\text{H}_2

and

2\ \text{mol}

\text{NH}_3

at a total pressure of

200\ \text{kPa}

. Calculate the partial pressure of ammonia.

Show worked answer →

First find the total moles of gas: $2 + 6 + 2 = 10\ \text{mol}$ .

The mole fraction of ammonia is $\dfrac{2}{10} = 0.20$ .

Partial pressure $=$ mole fraction $\times$ total pressure $= 0.20 \times 200 = 40\ \text{kPa}$ .

Markers reward the total moles, the mole fraction, and the partial pressure of $40\ \text{kPa}$ .

AQA 20225 marksFor the equilibrium

\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})

, an equilibrium mixture at a total pressure of

300\ \text{kPa}

contains

1.0\ \text{mol}

\text{N}_2

1.0\ \text{mol}

\text{H}_2

and

2.0\ \text{mol}

\text{NH}_3

. Calculate

K_p

and give its units.

Show worked answer →

Total moles $= 1.0 + 1.0 + 2.0 = 4.0\ \text{mol}$ . Partial pressures (mole fraction $\times$ total): $p(\text{N}_2) = \frac{1.0}{4.0} \times 300 = 75\ \text{kPa}$ , $p(\text{H}_2) = 75\ \text{kPa}$ , $p(\text{NH}_3) = \frac{2.0}{4.0} \times 300 = 150\ \text{kPa}$ .

$K_p = \dfrac{p(\text{NH}_3)^2}{p(\text{N}_2)\,p(\text{H}_2)^3} = \dfrac{(150)^2}{75 \times (75)^3} = \dfrac{22500}{75 \times 421875} = 7.1 \times 10^{-4}$ .

Units: $\dfrac{\text{kPa}^2}{\text{kPa} \times \text{kPa}^3} = \text{kPa}^{-2}$ , so $K_p = 7.1 \times 10^{-4}\ \text{kPa}^{-2}$ .

Markers reward all partial pressures, the correct expression, the value, and the derived units.

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Sources & how we know this

AQA A-level Chemistry (7405) specification — AQA (2015)