A-Level · Chemistry · AQA · Mark scheme decoded

AQA A-Level Chemistry: Equilibrium Constant K in Homogeneous Systems — mark scheme explained

Machine-verifiedchecked against the AQA A-Level Chemistry specificationlast verified 3 July 2026

The short answer

In physical chemistry, understanding equilibrium is crucial for predicting the behavior of chemical systems. The equilibrium constant, denoted as K , is a fundamental concept that helps us quantify the extent to which a reversible reaction proceeds.

The question

For the reaction N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g), write the equilibrium constant expression and calculate K c if [N 2 ] = 0.5 mol dm -3 , [H 2 ] = 1.5 mol dm -3 , and [NH 3 ] = 0.8 mol dm -3 .

[Paraphrased for study — not reproduced from any exam paper.]

4 marks

Mark scheme, decoded

How the examiner actually awards the marks on this topic.

Gradora's own decode of the marking approach — not the exam board's published mark scheme.

How marks are awarded

For calculation questions, marks are typically awarded for correct substitution of values into formulas, accurate arithmetic, and providing the final answer with appropriate units. For conceptual questions, marks are given for clear and concise explanations that demonstrate a deep understanding of the topic.

What the command words demand

Calculate: Perform a numerical calculation using given data and appropriate formulas.
Explain: Provide a detailed description of the concept, including relevant principles and relationships.
Determine: Find or derive a specific value or quantity based on given information.
Predict: Use known principles to forecast how changes in conditions will affect the system.

Model answer

A full-mark response to the question above, worked through step by step.

Timing: Allocate approximately 5-7 minutes per question to ensure you have enough time to carefully read the problem, perform calculations, and check your work.

Write the equilibrium constant expression for the reaction.1 mark
K c = [NH 3 ] 2 / [N 2 ][H 2 ] 3
Substitute the given equilibrium concentrations into the expression.1 mark
K c = (0.8) 2 / (0.5)(1.5) 3
Calculate the value of K c .2 marks
(0.8) 2 = 0.64(1.5) 3 = 3.375K c = 0.64 / (0.5 × 3.375) = 0.64 / 1.6875 ≈ 0.38

Final answer: 0.38

Work through every step correctly and you earn all 4 marks.

Another worked example

For the reaction CO(g) + H 2 O(g) ⇌ CO 2 (g) + H 2 (g), if K c = 10 at 500 K, predict the value of K c at 600 K given that ΔH° = -41.2 kJ mol -1 .

6 marks

Use the Van't Hoff equation to relate K c at two different temperatures.1 mark
ln( K 2 / K 1 ) = -ΔH° / R (1/T 2 - 1/T 1 )
Substitute the given values into the equation.1 mark
K 1 = 10, T 1 = 500 K, T 2 = 600 K, ΔH° = -41.2 kJ mol -1 , R = 8.314 J K -1 mol -1
Convert ΔH° to joules.1 mark
ΔH° = -41.2 × 1000 = -41200 J mol -1
Calculate the right-hand side of the equation.2 marks
-ΔH° / R = -(-41200) / 8.314 = +4955.5 K1/600 - 1/500 = 0.001667 - 0.002 = -0.000333 K -1+4955.5 × (-0.000333) ≈ -1.65
Solve for K 2 .1 mark
ln( K 2 / 10) = -1.65K 2 / 10 = e -1.65 ≈ 0.19K 2 ≈ 10 × 0.19 ≈ 1.9Check: the reaction is exothermic, so raising the temperature should DECREASE K — consistent with K 2 1 .

Final answer: 1.9

Work through every step correctly and you earn all 6 marks.

Common mistakes

Using the wrong units for concentrations.
Why it happens: Students often forget to use mol dm -3 as the unit for concentrations in the equilibrium constant expression.
Fix: Always ensure that concentrations are in mol dm -3 when substituting into the equilibrium constant expression.
Forgetting to use stoichiometric coefficients as exponents in the equilibrium constant expression.
Why it happens: The stoichiometric coefficients can be easily overlooked, leading to incorrect expressions and calculations.
Fix: Always include the stoichiometric coefficients as exponents in the equilibrium constant expression.
Misinterpreting the effect of temperature on K for endothermic and exothermic reactions.
Why it happens: Students may confuse the effects of temperature on K for endothermic and exothermic reactions, leading to incorrect predictions.
Fix: Remember that increasing temperature increases K for endothermic reactions and decreases K for exothermic reactions.
Incorrectly applying the Van't Hoff equation.
Why it happens: Students may make arithmetic errors or use incorrect values when applying the Van't Hoff equation to predict changes in K with temperature.
Fix: Double-check your calculations and ensure you are using the correct values for ΔH°, R, T 1 , and T 2 .
Believing that changes in concentration or catalysts affect K.
Why it happens: Students may incorrectly think that changing the concentration of reactants or products or adding a catalyst will change the value of K.
Fix: Understand that changes in concentration only shift the position of equilibrium, and catalysts do not affect K; they speed up both forward and reverse reactions equally.
Failing to explain why changes in concentration or catalysts do not affect K clearly.
Why it happens: Students may struggle to articulate the reasons why changes in concentration or the addition of a catalyst do not affect K, leading to vague or incorrect explanations.
Fix: Practice explaining that changes in concentration shift the position of equilibrium but do not change K, and catalysts speed up both forward and reverse reactions equally without affecting K.

Where the marks go

The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.

Question type	What you’re asked to do	Marks
Calculate Kc	Write the Kc expression and calculate its value with correct units from given concentrations.	4
Van't Hoff Equation	Predict Kc at a new temperature using the Van't Hoff equation and given enthalpy change.	6
Calculate Kc	Substitute equilibrium concentrations into the Kc expression and calculate its value with units.	3
Van't Hoff Equation	Predict a new equilibrium constant at a different temperature using the enthalpy change.	6
Total across these question types		19

Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.