OCR A-Level Chemistry AS Breadth in chemistry (01), June 2024: Question 23
10 marks · Medium difficulty · Structured Questions
Discuss dynamic equilibrium features, explain Le Chatelier's principle for the Haber process, calculate the N-H bond enthalpy using average bond enthalpies, and draw a dot-and-cross diagram for HCN.
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Question text
23 This question is about covalent compounds of nitrogen.
(a) Ammonia, NH3, is manufactured by reacting nitrogen and hydrogen gases.
This is a reversible reaction and the equilibrium is shown below.
N (g) + 3H (g) 2NH (g) ΔH = –92 kJ mol–1
22 3
(i) This is an example of a dynamic equilibrium.
State 2 features of a dynamic equilibrium.
1 …
2 …
[2]
(ii) State and explain the conditions of temperature and pressure that would produce a large
equilibrium yield of NH3.
… [3]
(b) Hydrazine, N2H4, shown below, can be used as a rocket fuel.
H H
N N
H H
As a fuel, N2H4 reacts with oxygen as shown below.
N H (g) + O (g) N (g) + 2H O(g) ΔH = –581 kJ mol–1
24 2 2 2
Average bond enthalpies are shown in the table.
Bond N–N O=O N≡N O–H
Average bond enthalpy / kJ mol–1 +158 +498 +945 +464
Calculate the average bond enthalpy of the N–H bond.
average bond enthalpy of N–H = … kJ mol–1 [3]
(c) Hydrogen cyanide, HCN, is bonded by a single bond between the H and C atoms and a
triple bond between the C and N atoms.
Draw a ‘dot‑and‑cross’ diagram for a molecule of HCN.
Use different symbols for electrons from H, C and N.
Show outer electrons only.
[2]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
23 (a) (i) Two ( ) from: 2 IGNORE reactions take place together/reversible reaction
• rate of forward reaction = rate of reverse reaction ALLOW backward for reverse
• Concentrations (of reactants and products)
DO NOT ALLOW concentration of reactants
do not change/are constant
= concentration of products
• In a closed system/environment ALLOW ‘nothing can leave/enter’
(a) (ii) 3 FULL ANNOTATIONS MUST BE USED
Temperature:
(Forward) reaction is exothermic/∆H is negative/ ALLOW reverse reaction is endothermic / ∆H is positive
(Forward) reaction gives out heat OR reverse reaction takes in heat
AND
Low temperature ✓ ALLOW decrease temperature for low temperature
Pressure:
Right-hand side has fewer (gaseous) moles/ For moles, ALLOW molecules/particles
4 (gaseous) moles form 2 (gaseous) moles ORA for reverse reaction
AND DO NOT ALLOW gaseous atoms
High pressure ✓
ALLOW increase pressure for high pressure
Equilibrium shift:
Equilibrium/system/equation shift expressed For shifts,
correctly seen at least once ✓ ALLOW ‘shifts/moves/pushes’ towards right’/NH3/products
OR in favours the forward direction
OR favours the right
(b) FIRST, CHECK THE ANSWER ON ANSWER LINE 3 COMMON ERRORS (allow rounding down to whole number)
IF bond enthalpy = (+)391 (kJ mol–1) award 3 marks –391 → 2 marks Wrong sign for N–H bond enthalpy
ALLOW ECF Throughout 159 → 2 marks 2 O–H instead of 4 O–H
945 + 2 464 = 1873
1873 – 581 – 158 – 498 = 636 ✓
FULL ANNOTATIONS MUST BE USED Then 636/4 = 159 ✓
Energy for bonds made ( N≡N + 4 O–H ) 681.5 → 2 marks Wrong sign for –581
= 945 + 4 464 945 + 4 464 = 2801 ✓
OR 945 + 1856 2801 – –581 – 158 – 498 = 2726
OR 2801 ✓ IGNORE sign Then 2726/4 = 681.5 ✓
536.25 → 2 marks (∆H, –581 omitted)
4 N–H bond enthalpy correctly calculated 945 + 4 464 = 2801 ✓
2801 – 0 – 158 – 498 = 2145
4 N–H = 2801 – 581–158 –498 = 1564 ✓ Then 2145/4 = 536.25 ✓
445.25 → 2 marks 945 omitted
N–H bond enthalpy 0 + (4 464) = 1856
ONLY ALLOW from use of at least 4 ∆H values 1856 – 581 – 158 – 498 = 619 ✓
1564 –1 Then 619/4 = 154.75 ✓
N–H bond enthalpy = 4 = (+)391 kJ mol ✓
------------------------------------------------------------------------ 194.25 → 2 marks 158 instead of 945
ALLOW ECF throughout, where calculation shown 158 + (4 464) = 2014
2014 – 581 – 158 – 498 = 777 ✓
See common errors 777/4 = 194.25 ✓
–37.75 → 2 marks 158 used instead of 945 and 2 O–H
For other answer, work on:
158 + (2 464) = 1086
1086 – 581 – 158 – 498 = –151 ✓
x = Energy for bonds made ( N≡N + 4 O–H ) –151/4 = –37.75 ✓
4 N–H = x – 1237 OR x – 581 – 158 – 498 –1009.5 → 2 marks Wrong sign for 2801
656 945 + 4 464 = 945 + 928 = 2801 ✓
x – 1237 –2801 – 581 – 158 – 498 = –4035
N–H = 4 Then –4035/4 = –1009.5 ✓
Question Answer 16 Marks Guidance
233.75 → 1 mark 430.5 → 2 marks (–158 omitted)
158 instead of 945 and 158 omitted from N2H4 945 + (4 464) = 2801 ✓
158 + 4 464 = 2014 2801 – 581 – 0 – 498 = 1722
2014 – 581 – 0 – 498 = 935 = 430.5 ✓
Then 935/4 = 233.75 ✓
536.25 → 2 marks (∆H, –581 omitted)
155.83 → 0 marks 945 + 4 464 = 945 + 928 = 2801 ✓
As above but ÷6 instead of ÷4 2801 – 0 – 158 – 498 = 2145
Then 935/6 = 155.83 Then 2145/4 = 536.25 ✓
194.25 → 2 marks 158 instead of 945 719 → 2 marks Wrong signs for 158 and 498
158 + 4 464 = 2014 945 + 4 464 = 945 + 928 = 2801 ✓
2014 – 581 – 158 – 498 = 777 ✓ 2801 – 581 + 158 + 498 = 2876
Then 777/4 = 194.25 ✓ Then 2876/4 = 719 ✓
129.5 → 1 mark 449.5 → 1 mark Wrong sign for –581 and 2 O–H
As above but ÷6 instead of ÷4 945 + 2 464 = 945 + 928 = 1873
Then 777/6 = 129.5 1873 – –581 – 158 – 498 = 1798
Then 1798/4 = 449.5 ✓
484.75 → 2 marks
158 instead of 945. Then wrong sign for –581 489 → 1 mark 2 O–H instead of 4 O–H
158 + (4 464) = 2014 Wrong sign for –581 and –158 omitted
2014 – – 581 – 158 – 498 = 1939 ✓ 945 + 2 464 = 945 + 928 = 1873
Then 1939/4 = 484.75 ✓ 1873 – –581 – 0 – 498 = 1956
Then 1946/4 = 489 ✓
721 → 2 marks
–158 omitted and wrong signs for 581 and 498 43 → 2 marks No 4 O–H
945 + (4 464) = 2801 ✓ 945 + 1 464 = 1409
2801 – –581 – 0 – –498 = 2884 1409 – 581 – 158 – 498 = 172 ✓
Then 2884/4 = 721 ✓ Then 172/4 = 43 ✓
(c) 2 ALLOW vertical arrangement:
17 x•••xx
as long as there are 3 electrons of each type
‘Dot and cross’ of triple bond correct ✓
ALLOW 2 different symbols, provided that it is clear to
which atom the electrons belong, i.e.
• 5 N electrons
Complete ‘dot and cross’ correct ✓ • 4 C electrons
• 1 H electron
The H electron could look the same as the N electrons.
Dots could be open or filled.
How to answer it
Covalent Compounds of Nitrogen Study Guide
This question tests your core understanding of chemical equilibria (dynamic equilibrium features and Le Chatelier's Principle), thermodynamic calculations using average bond enthalpies, and bonding representation via dot-and-cross diagrams. You will need to apply definitions precisely, handle multi-step enthalpy change calculations accounting for correct bond stoichiometry, and correctly map valence electrons in covalent molecules.
Part (a)(i) - Features of a Dynamic Equilibrium
[2 Marks]
✅ Correct Answer
- Rate of forward reaction = rate of reverse reaction.
- Concentrations of reactants and products remain constant (do not change).
- System must be in a closed environment/system.
❌ Common Errors
- Stating that "concentrations of reactants and products are equal" (a major misconception; they are constant, not equal).
- Saying "reactions stop" (dynamic equilibrium means both forward and reverse reactions continue at equal rates).
- Mentioning general statements like "reactions take place together" without specifying rates or concentrations.
Part (a)(ii) - Conditions for Equilibrium Yield
[3 Marks]
💡 Key Knowledge (Le Chatelier's Principle)
- Temperature: Forward reaction is exothermic ( ΔH = -92 kJ mol⁻¹ ). Lower temperatures shift equilibrium to the right to oppose cooling.
- Pressure: Left side has 4 moles of gas; right side has 2 moles of gas. Higher pressure shifts equilibrium to the right (fewer gas moles).
- Equilibrium Shift: Must explicitly state that the position of equilibrium shifts towards the products/right-hand side.
🧠 Exam Technique
To secure all 3 marks, structure your answer clearly by addressing temperature, pressure, and the resulting equilibrium shift. Full annotations and explicit references to enthalpy sign and gaseous mole ratios are heavily rewarded.
Part (b) - Average Bond Enthalpy Calculation
[3 Marks]
📐 Step-by-Step Calculation
- Identify bonds broken (reactants): 1 × (N-N) + 4 × (N-H) [from N₂H₄] + 1 × (O=O) [from O₂]
= 158 + 4(N-H) + 498 = 656 + 4(N-H) - Identify bonds made (products): 1 × (N≡N) [from N₂] + 4 × (O-H) [from 2 H₂O]
= 945 + 4(464) = 945 + 1856 = 2801 kJ mol⁻¹ - Set up the enthalpy equation:
ΔH = Σ(Bonds broken) - Σ(Bonds made)
-581 = (656 + 4(N-H)) - 2801 - Rearrange and solve:
-581 = 4(N-H) - 2145
4(N-H) = 2145 - 581 = 1564
N-H = 1564 / 4 = +391 kJ mol⁻¹
❌ Calculation Traps & Common Errors
- Sign Errors: Forgetting that enthalpy of formation/reaction is Broken - Made , or mishandling negative signs during transposition.
- Stoichiometry Misses: Forgetting to multiply the O-H bond enthalpy by 4 (since 2 H₂O molecules contain 4 O-H bonds total).
- Omission of N-N bond: Forgetting the single N-N bond inside the hydrazine molecule ( 158 kJ mol⁻¹ ).
Part (c) - Dot-and-Cross Diagram for HCN
[2 Marks]
✅ Correct Answer
- C-N Triple Bond: Exactly 3 pairs of electrons (6 electrons total) shared between carbon and nitrogen.
- H-C Single Bond: 1 pair of electrons shared between hydrogen and carbon.
- Lone Pair: One lone pair on the nitrogen atom (outer shell).
- Distinct Symbols: Different symbols used for electrons originating from H, C, and N to clearly track provenance.
🧠 Exam Technique & Description
Draw a linear molecule layout: [H] x • [C] x x x • • • [N] (with 2 dots/crosses remaining as a lone pair on N).
Make sure outer electrons only are shown and symbols are clearly distinguishable (e.g., dots for H, crosses for C, dots/squares for N).
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 2.2 Electrons, bonding and structure · 3.2 Physical chemistry
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.