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.
Practise this questionQuestion
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
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
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
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
Mass of water = Hydrated mass - Anhydrous mass
4.68 g - 2.99 g = 1.69 g of water
Scaling factor = 11.7 g / 4.68 g = 2.5
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 .
• 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
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
The formation of 2.99 g of anhydrous copper sulfate is associated with 1.47 kJ of energy.
Energy = (15.0 / 2.99) × 1.47 kJ
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 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
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
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
- Count gas moles: Left side has 2 moles of gas ( 2 NO₂ ). Right side has 1 mole of gas ( 1 N₂O₄ ).
- Apply pressure rule: Increasing pressure shifts equilibrium to the side with fewer moles of gas (the right side).
- 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
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
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
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
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.
• 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.