AQA A-Level Chemistry Paper 1, 2022: Question 3

17 marks · Medium difficulty · State/Explain/Numerical

Perform redox titration calculations, ideal gas law calculations, and bond enthalpy determinations involving hydrogen peroxide.

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Question text

03 This question is about hydrogen peroxide, H2O2

The half-equation for the oxidation of hydrogen peroxide is

H O O + 2 H+ + 2e–

22 2

Hair bleach solution contains hydrogen peroxide.

A sample of hair bleach solution is diluted with water.

The concentration of hydrogen peroxide in the diluted solution is 5.00% of that in the

original solution.

A 25.0 cm3 sample of the diluted hair bleach solution is acidified with dilute sulfuric

acid.

This acidified sample is titrated with 0.0200 mol dm−3 potassium manganate(VII)

solution.

The reaction is complete when 35.85 cm3 of the potassium manganate(VII) solution

are added.

03.1 Give an ionic equation for the reaction between potassium manganate(VII) and

acidified hydrogen peroxide.

Calculate the concentration, in mol dm–3, of hydrogen peroxide in the original hair

bleach solution.

(If you were unable to write an equation for the reaction you may assume that the

mole ratio of potassium manganate(VII) to hydrogen peroxide is 3:4

This is not the correct mole ratio.)

[5 marks]

11 –3

Concentration mol dm

03.2 State why an indicator is not added in this titration.

[1 mark]

03.3 Give the oxidation state of oxygen in hydrogen peroxide.

[1 mark]

03.4 Hydrogen peroxide decomposes to form water and oxygen.

Give an equation for this reaction.

Calculate the amount, in moles, of hydrogen peroxide that would be needed to

produce 185 cm3 of oxygen gas at 100 kPa and 298 K

The gas constant, R = 8.31 J K−1 mol−1

[5 marks]

Equation

Amount mol

03.5 Hydrazine (N2H4) is used as a rocket fuel that is oxidised by hydrogen peroxide.

The equation for this reaction in the gas phase is

The enthalpy change for this reaction, ∆H = –789 kJ mol−1

Table 3 shows some mean bond enthalpy values.

Table 3

N–H N–N N≡N O–H

Mean bond

−1 388 163 944 463

enthalpy / kJ mol

Define the term mean bond enthalpy.

Use the equation and the data in Table 3 to calculate a value for the O–O bond

enthalpy in hydrogen peroxide.

[5 marks]

Definition

Bond enthalpy kJ mol–1

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

M1 2 MnO – + 6 H+ + 5 H O 2 Mn2+ + 8 H O + 5 O ignore state symbols 1

42 2 2 2

M2 n(MnO –) = 0.020 × 35.85 = 7.17 × 10–4 (mol) 1

1000

M3 n(H O ) = 7.17 × 10–4 × 5/2 = 1.793 × 10–3 (mol) M3 = M2 × 5/2 1

M4 conc (H O in sample) = 1.793 × 10–3 = 0.0717 (mol dm–3) M4 = M3 × 1000 1

25 × 10–3 25

03.1

M5 original conc of H O (= 0.0717 × 20 ) = 1.43 (mol dm–3) M5 = M4 × 100 1

5 AO2

allow 1.43–1.44

alternative answer using 3:4 ratio given on question

paper

M3 = 7.17 × 10–4 × 4/3 = 9.56 × 10–4

M4 = 0.0382 (mol dm–3)

M5 = 0.765 (mol dm–3)

– A-LEVEL CHEMISTRY – –

KMnO4 is self-indicating 1

or AO1

03.2 KMnO4 is no longer decolourised at end point

(solution) changes (from colourless) to (pale) pink/purple at end point

03.3 –1

AO2

– A-LEVEL CHEMISTRY – –

M1 2H2O2 2H2O + O2 allow multiples 1

ignore state symbols

M2 V = 185 × 10–6 (m3) and P = 100 000(Pa) unit conversions 1

M3 n = PV = 100 000 × 185 × 10–6 rearrangement of ideal gas equation 1

RT 8.31 × 298

2003.4

M4 n(O ) = 7.47 × 10–3 (mol ) calculation 1

M5 n(H O ) = (7.47 × 10–3 × 2) = 0.0149 (mol) allow M4 × 2 to 2 sig fig or more 1

AO1

if incorrect rearrangement in M3 can score M1, M2 AO2

and M5

– A-LEVEL CHEMISTRY – –

M1 enthalpy (change) to break 1 mol bonds (in gaseous state) allow heat energy (change) to break 1 mol bonds 1

allow the enthalpy needed to break 1 mol bonds

do not accept enthalpy released

M2 averaged over a range of compounds / molecules 1

M3 –789 = 4(388) + 163 + 4(463) + 2(O–O) – 944 – 8(463) 1

or –789 = 4(388) + 163 + 2(O–O) – 944 – 4(463)

or –789 = 3567 + 2(O–O) – 4648

03.5

or –789 = 1715 + 2(O–O) – 2796

M4 2(O–O) = 292 (kJ mol–1) 1

M5 O–O = 146 (kJ mol–1) M5 = M4 ÷ 2 1

AO1

AO2

How to answer it

Hydrogen Peroxide: Redox, Titrations, Gas Laws & Energetics

AQA A-Level Chemistry • Comprehensive Multi-Topic Question Breakdown (17 Marks Total)

WHAT THIS QUESTION TESTS

Redox Chemistry & Balancing Equations: Combining half-equations to form full ionic redox equations and determining oxidation states in peroxides.
Redox Titrations & Dilutions: Multi-step stoichiometric titration calculations involving potassium manganate(VII) and handling percentage dilution factors.
Ideal Gas Equation (PV = nRT): Converting standard units (cm³ to m³, kPa to Pa) and relating molar gas quantities back to stoichiometric equations.
Energetics & Mean Bond Enthalpies: Defining mean bond enthalpy precisely and applying Hess's Law / bond breaking minus bond making to deduce unknown bond values.

Question 03.1 • 5 Marks

Redox Equation & Back-Calculation of Diluted Bleach

Ionic equation for MnO₄⁻ with H₂O₂ and calculating the original concentration

💡 Key Knowledge

Reduction half-equation:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (×2)
Oxidation half-equation:
H₂O₂ → O₂ + 2H⁺ + 2e⁻ (×5)
Overall ionic equation:
2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
Reacting mole ratio: 2 mol MnO₄⁻ : 5 mol H₂O₂

📐 Step-by-Step Calculation

Moles of MnO₄⁻:
n = c × V = 0.0200 × (35.85 / 1000) = 7.17 × 10⁻⁴ mol [M2]
Moles of H₂O₂ in 25.0 cm³ diluted sample:
Mole ratio is 2 : 5
n(H₂O₂) = (7.17 × 10⁻⁴) × (5 / 2) = 1.7925 × 10⁻³ mol [M3]
Concentration of diluted H₂O₂:
c = n / V = 1.7925 × 10⁻³ / 0.0250 = 0.0717 mol dm⁻³ [M4]
Concentration of original hair bleach:
The sample was diluted to 5.00% (dilution factor = 100 / 5 = 20):
Original conc = 0.0717 × 20 = 1.43 mol dm⁻³ [M5]

❌ Common Calculation Traps

Cancelling H⁺ incorrectly: Remember 16H⁺ on the left and 10H⁺ on the right simplifies to 6H⁺ on the left.
Inverting the dilution: Students often multiply by 0.05 instead of dividing by 0.05 (or multiplying by 20). The original bleach must be more concentrated than the diluted sample!
Safety Net: The question provided a fallback ratio (3:4) in case you could not balance the equation. If used, M5 gave 0.765 mol dm⁻³.

✅ Final Answers & Mark Scheme

M1: 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
M2: n(MnO₄⁻) = 7.17 × 10⁻⁴ mol
M3: n(H₂O₂ in 25 cm³) = 1.793 × 10⁻³ mol
M4: conc(dilute H₂O₂) = 0.0717 mol dm⁻³
M5: 1.43 mol dm⁻³ (allow 1.43–1.44)

Question 03.2 • 1 Mark

Choice of Indicator in Manganate(VII) Titrations

Why is an indicator not added in this titration?

💡 Key Knowledge

Potassium manganate(VII) acts as its own indicator (it is self-indicating).

During the titration, deep purple MnO₄⁻ is reduced to almost colourless Mn²⁺.
At the end point, the very first excess drop of MnO₄⁻ remains unreacted, giving a permanent pale pink / pale purple colour.

✅ Acceptable Answers (Any One)

KMnO₄ is self-indicating.
KMnO₄ is no longer decolourised at the end point.
Solution changes colour (from colourless) to (pale) pink / purple at the end point.

Question 03.3 • 1 Mark

Oxidation State in Peroxides

Give the oxidation state of oxygen in hydrogen peroxide (H₂O₂)

💡 Key Knowledge

Hydrogen is bonded to a non-metal, so its oxidation state is +1.

2(+1) + 2(Oxidation State of O) = 0 ⇒ 2(O) = -2 ⇒ O = -1

Peroxides are a fundamental syllabus exception to the rule that oxygen is always -2 (the other exception being bonded to fluorine in OF₂ where it is +2).

✅ Mark Scheme

-1 (or -I)

❌ Do not write just "1" or "-2". The negative sign is mandatory.

Question 03.4 • 5 Marks

Decomposition of H₂O₂ & The Ideal Gas Equation

Calculate the moles of H₂O₂ required to produce 185 cm³ of O₂ at 100 kPa and 298 K

📐 Step-by-Step Calculation

Write the balanced decomposition equation [M1]:
2H₂O₂ → 2H₂O + O₂
Convert units to SI base units [M2]:
Volume, V = 185 cm³ = 185 × 10⁻⁶ m³
Pressure, P = 100 kPa = 100 000 Pa = 1.00 × 10⁵ Pa
Temperature, T = 298 K
Rearrange ideal gas equation [M3]:
n = PV / RT
n(O₂) = (100 000 × 185 × 10⁻⁶) / (8.31 × 298)
Calculate amount of O₂ gas [M4]:
n(O₂) = 18.5 / 2476.38 = 7.4706 × 10⁻³ mol
Apply mole ratio to find H₂O₂ needed [M5]:
From equation, 1 mol O₂ requires 2 mol H₂O₂:
n(H₂O₂) = 7.4706 × 10⁻³ × 2 = 0.0149 mol (or 1.49 × 10⁻² mol)

🧠 Exam Technique & Pitfalls

Unit Conversions: The most common place students drop M2 is converting cm³ to m³. Remember: 1 m³ = 10⁶ cm³ , so multiply by 10⁻⁶ .
Final Stoichiometry: A very frequent error is stopping at M4 after finding the moles of O₂. Always re-read what the question asks for: amount of hydrogen peroxide!
Significant Figures: Give your final answer to at least 2 or 3 sig figs (0.0149 mol).

Question 03.5 • 5 Marks

Mean Bond Enthalpies & Rocket Fuel Oxidation

Definition and calculation of the O-O single bond enthalpy

🧠 Part 1: Precise Definition (2 Marks)

Examiners are extremely strict on mean bond enthalpy definitions:

M1: Enthalpy / heat energy change to break 1 mol of bonds in gaseous state / molecules.
M2: Averaged over a range of compounds / molecules.

❌ Do not say: "Enthalpy released" (bond breaking is always endothermic!). Never forget the word gaseous or the quantity 1 mol.

📐 Part 2: Calculation of O-O Bond Enthalpy (3 Marks)

Equation: N₂H₄(g) + 2H₂O₂(g) → N₂ + 4H₂O(g) ΔH = -789 kJ mol⁻¹

Bonds Broken (Reactants):

4 × (N–H) = 4 × 388 = 1552
1 × (N–N) = 163
4 × (O–H) [from 2 H-O-O-H] = 4 × 463 = 1852
2 × (O–O) [from 2 H-O-O-H] = 2(O–O)
Total Broken = 3567 + 2(O–O)

Bonds Formed (Products):

1 × (N≡N) = 944
8 × (O–H) [from 4 H-O-H] = 8 × 463 = 3704
Total Formed = 944 + 3704 = 4648

Setting up ΔH = Σ(broken) - Σ(formed) [M3]:

-789 = [3567 + 2(O–O)] - 4648

-789 = 2(O–O) - 1081

2(O–O) = 1081 - 789 = +292 kJ mol⁻¹ [M4]

O–O = +292 / 2 = +146 kJ mol⁻¹ [M5]

Examiner Insight: Notice that there are 4 O-H bonds broken in the reactants and 8 O-H bonds formed in products. You can cancel them out directly as net forming 4 O-H bonds: -789 = 4(388) + 163 + 2(O-O) - 944 - 4(463) . Don't forget that 2 H₂O₂ molecules contain 2 O-O single bonds, so you must divide by 2 at the end!

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.2 Amount of Substance · 3.1.4 Energetics · 3.1.7 Oxidation, Reduction and Redox Equations · 3.2.5 Transition Metals

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.