Predict the effect of increasing pressure on N_2 + 3H_2 2NH_3. [2 marks]

AQA AQA 2022-style practice: In the equilibrium N_2(g) + 3H_2(g) 2NH_3(g), a 2.0\ dm^3 vessel at equilibrium contains 0.40\ mol N_2, 0.60\ mol H_2 and 0.80\ mol NH_3. Calculate K_c and state its units.

Convert amounts to concentrations by dividing by the 2.0\ dm^3 volume: [N_2] = 0.20, [H_2] = 0.30, [NH_3] = 0.40\ mol dm^-3. K_c = [NH_3]^2[N_2][H_2]^3 = (0.40)^20.20 × (0.30)^3 = 0.160.20 × 0.027 = 0.160.0054 = 29.6. Units: (mol dm^-3)^2(mol dm^-3)(mol dm^-3)^3 = (mol dm^-3)^-2 = dm^6\,mol^-2, so K_c = 29.6\ dm^6\,mol^-2. Markers reward converting moles to concentrations, the correct expression, the value, and deriving the units.

EnglandChemistrySyllabus dot point

How do reversible reactions settle into balance, and how do we measure that balance?

Dynamic equilibrium, Le Chatelier's principle and the effect of changing concentration, pressure and temperature, the role of a catalyst, and the equilibrium constant Kc and its calculation.

A focused answer to AQA A-Level Chemistry 3.1.6, covering dynamic equilibrium, Le Chatelier's principle, the effects of concentration, pressure, temperature and catalysts, and writing and calculating the equilibrium constant Kc.

Generated by Claude Opus 4.811 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
Dynamic equilibrium
Le Chatelier's principle
The equilibrium constant Kc
Try this

What this dot point is asking

AQA wants you to describe dynamic equilibrium, apply Le Chatelier's principle to predict the effect of changing concentration, pressure and temperature (and to explain the role of a catalyst), and to write an expression for $K_c$ , calculate it from equilibrium amounts, and deduce its units.

Dynamic equilibrium

Le Chatelier's principle

Concentration: adding a reactant shifts equilibrium to the right (towards products) to remove it; removing a product also shifts it right.
Pressure (gases): increasing pressure shifts equilibrium towards the side with fewer moles of gas. If both sides have equal moles, pressure has no effect.
Temperature: increasing temperature shifts equilibrium in the endothermic direction; decreasing it shifts towards the exothermic direction.
Catalyst: speeds up forward and reverse reactions equally, so it has no effect on the position of equilibrium; it only reaches equilibrium faster.

The equilibrium constant Kc

For the reaction $aA + bB \rightleftharpoons cC + dD$ :

Only temperature changes $K_c$ . Changing concentration or pressure shifts the position of equilibrium, but the system adjusts until the ratio returns to the same $K_c$ value; a catalyst does not change $K_c$ (it only reaches equilibrium faster). A large $K_c$ (much greater than 1) means the equilibrium lies far to the right and products dominate; a small $K_c$ (much less than 1) means reactants dominate. This links to industrial compromise conditions, such as the Haber process, where a moderate temperature is chosen: a lower temperature would give a higher equilibrium yield of ammonia (the forward reaction is exothermic, so a lower temperature raises $K_c$ ) but the rate would be too slow to be economic, so a compromise of around 450 degrees Celsius is used alongside an iron catalyst and high pressure.

The units of $K_c$ are found by substituting $\text{mol dm}^{-3}$ for each concentration term and cancelling. They are not fixed: they depend on the difference between the total powers on the top and bottom of the expression, so they must be worked out for each equilibrium rather than memorised.

Worked example: calculating Kc

For $H_2 + I_2 \rightleftharpoons 2HI$ , at equilibrium $[H_2] = 0.20$ , $[I_2] = 0.20$ and $[HI] = 1.60\ \text{mol dm}^{-3}$ . Calculate $K_c$ and state its units.

Step 1: Write the expression for $K_c$

Products go on top and reactants go on the bottom, each concentration raised to its balancing number from the equation. For this 1:1:2 equilibrium, the two reactants each appear to the power 1 and the product appears to the power 2.

K_c = \frac{[HI]^2}{[H_2][I_2]}

Step 2: Substitute the equilibrium concentrations

Replace each bracket with the given equilibrium concentration values. Square the $[HI]$ term because its coefficient in the equation is 2.

K_c = \frac{(1.60)^2}{0.20 \times 0.20} = \frac{2.56}{0.040}

Step 3: Evaluate and derive the units

Carrying out the division gives the numerical value. To find the units, substitute $\text{mol dm}^{-3}$ for each concentration term and cancel: here $(\text{mol dm}^{-3})^2$ on top and $(\text{mol dm}^{-3})^2$ on the bottom cancel completely, leaving no units.

K_c = 64 \quad (\text{no units})

Final answer: $K_c = 64$ with no units for this equilibrium.

Try this

Q1. State the effect of a catalyst on the position of equilibrium. [1 mark]

Cue. No effect; it only speeds up reaching equilibrium.

Q2. Predict the effect of increasing pressure on $N_2 + 3H_2 \rightleftharpoons 2NH_3$ . [2 marks]

Cue. Shifts right (4 moles of gas to 2 moles), increasing the yield of ammonia.

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 marksFor an exothermic forward reaction at equilibrium, explain the effect on the position of equilibrium and on

K_c

of increasing the temperature.

Show worked answer →

Increasing the temperature shifts the equilibrium in the endothermic direction to oppose the change. For an exothermic forward reaction, the endothermic direction is the reverse, so the position of equilibrium shifts to the left and the yield of product falls.

Because the equilibrium shifts left, the equilibrium concentrations of products fall and reactants rise, so $K_c$ decreases.

Markers reward the shift in the endothermic direction, the leftward shift, and the decrease in $K_c$ (temperature is the only factor that changes $K_c$ ).

AQA 20224 marksIn the equilibrium

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

, a

2.0\ \text{dm}^3

vessel at equilibrium contains

0.40\ \text{mol}

\text{N}_2

0.60\ \text{mol}

\text{H}_2

and

0.80\ \text{mol}

\text{NH}_3

. Calculate

K_c

and state its units.

Show worked answer →

Convert amounts to concentrations by dividing by the $2.0\ \text{dm}^3$ volume: $[\text{N}_2] = 0.20$ , $[\text{H}_2] = 0.30$ , $[\text{NH}_3] = 0.40\ \text{mol dm}^{-3}$ .

$K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \dfrac{(0.40)^2}{0.20 \times (0.30)^3} = \dfrac{0.16}{0.20 \times 0.027} = \dfrac{0.16}{0.0054} = 29.6$ .

Units: $\dfrac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3} = (\text{mol dm}^{-3})^{-2} = \text{dm}^6\,\text{mol}^{-2}$ , so $K_c = 29.6\ \text{dm}^6\,\text{mol}^{-2}$ .

Markers reward converting moles to concentrations, the correct expression, the value, and deriving the units.

Related dot points

Sources & how we know this

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