AQA GCSE Chemistry Chemistry Paper 2 (Higher), 2023: Question 9

13 marks · Standard Demand difficulty · Short Answer

Analyze reversible reactions and dynamic equilibrium, including calculating mass changes, energy transfers, and the effects of temperature, pressure, and catalysts on equilibrium position.

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Question

The image shows an exam question about reversible reactions. It includes calculations of mass and energy changes for the hydration of copper sulfate, multiple-choice questions on the effects of temperature and pressure on the equilibrium of nitrogen dioxide and dinitrogen tetroxide, an explanation question on the effect of pressure on the decomposition of hydrogen iodide, and questions on catalysts and dynamic equilibrium.
Question text

09 This question is about reversible reactions.

When 4.68 g of hydrated copper sulfate changes into anhydrous copper sulfate:

• 2.99 g of anhydrous copper sulfate is produced

• 1.47 kJ of energy is taken in from the surroundings.

The equation for the reversible reaction is:

hydrated copper sulfate ⇌ anhydrous copper sulfate + water

09.1 Calculate the maximum mass of water that can be produced from

11.7 g of hydrated copper sulfate.

[3 marks]

Mass = g

09.2 15.0 g of anhydrous copper sulfate completely changes into hydrated copper sulfate

when water is added.

Calculate the amount of energy transferred to the surroundings.

[2 marks]

Energy = kJ

The gases nitrogen dioxide and dinitrogen tetroxide reach dynamic equilibrium in a

sealed container.

The equation for the reaction is:

*29* 2 NO2(g) ⇌ N2O4(g)

nitrogen dioxide (brown) dinitrogen tetroxide (colourless)

The forward reaction is exothermic.

09.3 What happens to the position of the equilibrium in this reaction if the temperature

is increased?

[1 mark]

Tick ( ) one box.

Shifts to the left

Stays the same

Shifts to the right 31

09.4 A teacher seals a brown-coloured mixture of nitrogen dioxide and dinitrogen tetroxide

in a gas syringe.

Figure 8 shows the sealed gas syringe.

Figure 8

The teacher pushes the syringe piston in.

This increases the pressure in the gas syringe.

*30* What is the colour of the mixture when a new equilibrium position is reached?

[1 mark]

Tick ( ) one box.

The mixture is a darker shade of brown.

The mixture is the same shade of brown.

The mixture is a lighter shade of brown.32

Hydrogen iodide gas decomposes into hydrogen gas and iodine gas at

high temperatures.

The equation for the reaction is:

2 HI(g) ⇌ H2(g) + I2(g)

09.5 Explain the effect of increasing the pressure on the equilibrium position of

this reaction.

[2 marks]

09.6 Suggest the effect of adding a catalyst on the equilibrium position of this reaction.

[1 mark]

Copper forms coloured compounds.

Hydrochloric acid is added to an aqueous solution of copper compound A.

The word equation for the reaction is:

copper compound A + hydrochloric acid ⇌ copper compound B + water

(blue) (yellow)

09.7 The reaction mixture is green when both copper compounds are present in a solution

at equilibrium.

How can the equilibrium position be shifted to make the reaction mixture more yellow?

[1 mark]

Tick ( ) one box.

Add more hydrochloric acid

*32* Add more water

Leave the reaction mixture for 30 minutes

09.8 The concentrations of the substances in this reaction do not change at

dynamic equilibrium.

Explain why.

[2 marks]

Mark scheme

Show the mark scheme The mark scheme shows the answers and marking criteria for Question 9. For 9.1, the mass of water is calculated as 4.23 g. For 9.2, the energy transferred is calculated as 7.37 kJ. For 9.3, the equilibrium shifts to the left. For 9.4, the mixture becomes a lighter shade of brown. For 9.5, there is no effect because there are equal moles of gas on both sides. For 9.6, there is no effect. For 9.7, add more hydrochloric acid. For 9.8, concentrations do not change because the forward and reverse reactions occur at the same rate.

Question 9

AO /

Question Answers Extra information Mark

Spec. Ref.

09.1 (mass of water in 4.68 g = AO2

4.68 - 2.99) 4.6.2.2

= 1.69 (g) 1

(mass of water in 11.7 g =)

11.7 allow correct use of an 1

× 1.69

4.68 incorrectly determined mass of

water in 4.68 g

= 4.23 (g) allow 4.2 / 4.225 (g) 1

alternative approach:

�mass of anhydrous compound =

11.7

× 2.99�

4.68

= 7.475 (g) (1)

(mass of water =)

11.7 – 7.475 (1) allow correct use of an

incorrectly determined mass of

anhydrous compound

= 4.23 (g) (1) allow 4.2 / 4.225 (g)

AO /

Spec. Ref.

09.2 (energy =) AO2

15.0 1

× 1.47 4.6.2.2

2.99

= 7.37 (kJ) allow 7.37458194 correctly 1

rounded to at least 2 significant

figures

AO /

Spec. Ref.

09.3 shifts to the left 1 AO2

4.6.2.1

4.6.2.4

22 4.6.2.6

AO /

Spec. Ref.

09.4 the mixture is a lighter shade of 1 AO2

brown

4.6.2.4

4.6.2.7

AO /

Spec. Ref.

09.5 no effect (on equilibrium allow (equilibrium position) stays 1 AO2

position) the same

4.6.2.4

4.6.2.7

(because) there are equal 1

numbers of (gas) moles /

molecules on each side (of the

equation)

AO /

Spec. Ref.

09.6 no effect (on equilibrium allow (equilibrium position) stays 1 AO2

position) the same 4.6.2.3

4.6.2.4

AO /

Spec. Ref.

09.7 add more hydrochloric acid 1 AO2

4.6.2.4

4.6.2.5

AO /

Spec. Ref.

09.8 ignore references to closed AO1

systems 4.6.2.3

(because the) forward and 2

reverse reactions are taking

place at (exactly) the same rate

allow for 1 mark 23

(because) the reactions are

taking place at (exactly) the

same rate

Total Question 9 13

How to answer it

Reversible Reactions and Dynamic Equilibrium

What this question tests

This exam question assesses your understanding of reversible reactions and dynamic equilibrium. You will need to apply the conservation of mass to calculate reacting masses, determine energy changes in reversible processes, and use Le Chatelier's Principle to predict how changes in temperature, pressure, concentration, and catalysts affect the position of equilibrium.

Part (a) — Mass Calculations in Reversible Reactions

Question 09.1

3 Marks

Calculate the maximum mass of water that can be produced from 11.7 g of hydrated copper sulfate, given that 4.68 g of hydrated copper sulfate produces 2.99 g of anhydrous copper sulfate.

📐 Step-by-Step Calculation

1
Find the mass of water in the initial sample:
Mass of water = Hydrated mass - Anhydrous mass
4.68 g - 2.99 g = 1.69 g of water
2
Find the scaling factor for the new mass:
Scaling factor = 11.7 g / 4.68 g = 2.5
3
Calculate the new mass of water:
New mass of water = 1.69 g × 2.5 = 4.23 g

✅ Correct Answer

4.23 g

Note: The mark scheme also accepts 4.2 g or 4.225 g .

Mark Breakdown:
• 1 Mark: Calculating 1.69 g of water.
• 1 Mark: Setting up the correct ratio calculation.
• 1 Mark: Final correct answer.

🧠 Exam Technique: Error Carried Forward (ECF)

If you make an arithmetic error in Step 1 (e.g., writing 4.68 - 2.99 = 1.79 g), you can still get the remaining 2 marks if you use your incorrect number correctly in the next steps. Always show your working clearly!

❌ Common Errors

Many students lose marks by trying to use complex molar mass calculations (Ar and Mr) from the periodic table. Because the formula of hydrated copper sulfate isn't fully specified here, you must use the simple experimental mass ratios provided in the question.

Part (b) — Energy Transfer in Reversible Reactions

Question 09.2

2 Marks

15.0 g of anhydrous copper sulfate completely changes into hydrated copper sulfate when water is added. Calculate the amount of energy transferred to the surroundings. (Given: 1.47 kJ is taken in when 4.68 g of hydrated copper sulfate decomposes to form 2.99 g of anhydrous copper sulfate).

📐 Step-by-Step Calculation

1
Identify the key relationship:
The formation of 2.99 g of anhydrous copper sulfate is associated with 1.47 kJ of energy.
2
Scale up to 15.0 g:
Energy = (15.0 / 2.99) × 1.47 kJ
3
Calculate and round:
Energy = 7.37 kJ (rounded to 3 significant figures)

✅ Correct Answer

7.37 kJ

Note: The mark scheme accepts any correctly rounded value from 7.37458... to at least 2 significant figures (e.g. 7.4 or 7.37 ).

• 1 Mark: Correct ratio setup.
• 1 Mark: Correct final value.

💡 Key Knowledge: Reversible Energy Changes

If a reversible reaction is endothermic in the forward direction (takes in energy), it must be exothermic in the reverse direction (releases energy to the surroundings) by the exact same amount.

❌ Common Calculation Trap

Do not use the hydrated mass ( 4.68 g ) in your ratio calculation here. The question asks about 15.0 g of anhydrous copper sulfate, so you must pair it with the anhydrous mass from the prompt ( 2.99 g ).

Part (c) — Temperature and Equilibrium

Question 09.3

1 Mark

What happens to the position of the equilibrium in this reaction if the temperature is increased?
2 NO₂(g) (brown) ⇌ N₂O₄(g) (colourless) [Forward reaction is exothermic]

✅ Correct Answer

☑ Shifts to the left

💡 Le Chatelier's Principle: Temperature

  • An increase in temperature always shifts the equilibrium in the endothermic direction (to absorb the extra heat).
  • Since the forward reaction is exothermic, the reverse reaction (left) is endothermic.

Part (d) — Pressure and Gas Equilibrium

Question 09.4

1 Mark

A teacher pushes the syringe piston in, which increases the pressure. What is the colour of the mixture when a new equilibrium position is reached?

✅ Correct Answer

☑ The mixture is a lighter shade of brown.

🧠 Exam Technique: Step-by-Step Logic

  1. Count gas moles: Left side has 2 moles of gas ( 2 NO₂ ). Right side has 1 mole of gas ( 1 N₂O₄ ).
  2. Apply pressure rule: Increasing pressure shifts equilibrium to the side with fewer moles of gas (the right side).
  3. Determine colour: Shifting right produces more colourless N₂O₄ , making the brown colour of NO₂ fade to a lighter shade.

Part (e) — Pressure with Equal Gas Moles

Question 09.5

2 Marks

Explain the effect of increasing the pressure on the equilibrium position of this reaction:
2 HI(g) ⇌ H₂(g) + I₂(g)

✅ Correct Answer

No effect on the equilibrium position 1 Mark

Because there are equal numbers of gas moles/molecules on each side of the equation 1 Mark

❌ Common Errors

Students often write "no effect" but forget to state the reason. To get the second mark, you must explicitly mention that the number of moles of gas is the same on both sides (2 moles on the left vs 2 moles on the right).

Part (f) — Effect of a Catalyst

Question 09.6

1 Mark

Suggest the effect of adding a catalyst on the equilibrium position of this reaction.

✅ Correct Answer

No effect (equilibrium position stays the same)

💡 Key Knowledge: Catalysts & Equilibrium

A catalyst increases the rate of both the forward and reverse reactions equally. It helps the system reach equilibrium faster, but it does not change the final position of equilibrium.

Part (g) — Concentration and Equilibrium

Question 09.7

1 Mark

The reaction mixture is green when both copper compounds are present in a solution at equilibrium. How can the equilibrium position be shifted to make the reaction mixture more yellow?
copper compound A (blue) + hydrochloric acid ⇌ copper compound B (yellow) + water

✅ Correct Answer

☑ Add more hydrochloric acid

💡 Le Chatelier's Principle: Concentration

To make the mixture more yellow, we need to shift the equilibrium to the right (towards the yellow product). Adding more of a reactant (hydrochloric acid) forces the system to shift right to use it up.

Part (h) — Defining Dynamic Equilibrium

Question 09.8

2 Marks

The concentrations of the substances in this reaction do not change at dynamic equilibrium. Explain why.

✅ Correct Answer

Because the forward and reverse reactions are taking place at the exact same rate.

• 1 Mark: Mentioning reactions take place at the same rate.
• 2 Marks: Specifying that the forward and reverse reactions are at the same rate.

❌ Common Misconception

Do not say "the concentrations are equal". At equilibrium, the concentrations of reactants and products are constant (they don't change), but they are rarely equal to each other.

Topics

Chemistry · C3: Quantitative Chemistry · C5: Energy Changes · C6: The Rate and Extent of Chemical Change

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.