AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2024: Question 5
14 marks · Standard Demand difficulty · Extended Answer
Complete a dot-and-cross diagram for ammonia, calculate the mass of hydrogen needed to make 25 g of ammonia, calculate the H–H bond energy from bond energies, and complete the exothermic reaction profile.
Practise this questionQuestion
Question text
05 Nitrogen reacts with hydrogen to produce ammonia (NH3).
05.1 Complete the dot and cross diagram for an ammonia molecule.
[2 marks]
05.2 The equation for the reaction between nitrogen and hydrogen to produce ammonia is:
N2 + 3H2 → 2NH3
Calculate the mass of hydrogen that is needed to produce 25 g of ammonia.
Relative atomic masses (Ar): H = 1 N = 14
[4 marks]
Mass of hydrogen = g
Figure 7 shows the displayed formulae equation for the reaction of nitrogen
with hydrogen.
Figure 7
In the reaction the energy released forming new bonds is 93 kJ/mol greater than the
energy needed to break existing bonds.
Table 3 shows bond energies.
Table 3
Bond H — H N — H
Bond energy in kJ/mol 945 X 391
05.3 Calculate the bond energy X for the H — H bond.
*17* Use Figure 7 and Table 3.
[5 marks]
19 X = kJ/mol
05.4 Energy is released from the reaction to produce ammonia.
Figure 8 shows part of the reaction profile for the reaction between nitrogen and
*18* hydrogen to produce ammonia.
Complete Figure 8.
You should:
• complete the profile line
• label the energy level of the reactants and the product
• label the overall energy change.
[3 marks]
Figure 8
Mark scheme
Show the mark scheme
Question 5
AO /
Question Answers Extra information Mark
Spec. Ref.
05.1 allow any combination of circles, AO1
dots, crosses or e(-) 5.2.1.4
ignore any inner shell electrons
on nitrogen
one shared pair of electrons in 1
each overlap
2 non-bonding electrons on the do not accept non-bonding 1
outer shell of nitrogen electrons on hydrogen
an answer of
– scores 2 marks – –
AO /
Spec. Ref.
05.2 (Mr NH3 = 14 + (3 × 1) =) 17 1 AO2
5.3.1.1
5.3.2.2
25 allow correct use of incorrectly 1
(moles NH3= ) = 1.47 (mols) determined M of NH
17 r 3
3 allow correct use of an 1
(moles H2 = 1.47 × ) incorrectly calculated value of
= 2.205 (mols) moles of NH3
(mass H2 = 2.205 × 2) allow 4.4117647 (g) correctly 1
= 4.41 (g) rounded to at least 2 significant
figures
allow correct use of an
incorrectly calculated value of
moles of H2
alternative approach
(2 × Mr NH3 = 17 × 2 =) 34 (1)
6 g H2 → 34 g NH3 (1) allow correct use of an
incorrectly calculated value of
2 × Mr NH3
× 25 (g) H2 → 25 g NH3 (1)
4.41 (g H2 → 25 g NH3)–(1) – –
AO /
Spec. Ref.
05.3 (bonds broken =) 1 AO2
945 + 3X 5.5.1.1
5.5.1.3
(bonds made = 6 × 391 =) 1
2346
(energy released = bonds made allow correct use of incorrectly 1
– bonds broken =) determined value(s) of bonds
93 = 2346 – [945 + 3X] broken and/or bonds formed
(3X =) 1308 (kJ/mol) 1 15
(X =) 436 (kJ/mol) allow correct use of an 1
incorrectly determined value
– for 3X – –
AO /
Spec. Ref.
05.4 correct shape for exothermic 1 AO2
reaction with the product line
below the level of the reactants
line
labelled horizontal lines for labelled horizontal lines for 1 AO1
reactants and product nitrogen and hydrogen and
ammonia
labelled overall energy change 1 AO1
5.5.1.2
an answer of
16 scores 3 marks
Total Question 5 14
How to answer it
Ammonia: bonding, masses and energy changes
What this question tests
This question checks your understanding of covalent bonding in ammonia, using balanced equations and relative formula mass in calculations, bond energy calculations, and reaction profiles for exothermic reactions. To score well, you need to use ratios from equations carefully, show working clearly, and label diagrams precisely.
💡 Big ideas in this question
- Ammonia is NH₃ with covalent bonds.
- Nitrogen forms 3 shared pairs and has 1 lone pair.
- Use the balanced equation: N₂ + 3H₂ → 2NH₃.
- Exothermic reactions have products at a lower energy than reactants.
- Bond energies use: energy change = bonds broken − bonds made, or rearranged carefully from the information given.
🧠 Examiner insight
- Students often lost marks by not showing enough calculation steps.
- In the dot-and-cross diagram, some put lone pairs on hydrogen, which is wrong.
- For the reaction profile, labels matter as much as the curve shape.
- Top answers linked each number directly to the balanced equation and the bonds shown.
Part (a) – 05.1 Dot-and-cross diagram for ammonia
2 marks
✅ Correct answer
- One shared pair of electrons in each N—H overlap.
- Two non-bonding electrons on the outer shell of nitrogen (one lone pair).
The mark scheme allows any sensible combination of dots/crosses/circles/e⁻ symbols.
💡 Key knowledge
- Nitrogen has 5 electrons in its outer shell.
- Hydrogen has 1 electron in its outer shell.
- Nitrogen shares 3 electrons with 3 hydrogen atoms, making 3 covalent bonds.
- That leaves 2 electrons as a lone pair on nitrogen.
🧠 Exam technique
- Count the overlaps: there must be 3 shared pairs.
- Then check nitrogen still has 1 lone pair.
- You do not need to show inner shell electrons on nitrogen.
❌ Common errors
- Putting lone pairs on hydrogen.
- Missing one of the shared pairs.
- Drawing too many non-bonding electrons on nitrogen.
Part (b) – 05.2 Calculate the mass of hydrogen needed to produce 25 g of ammonia
4 marks
📐 Step-by-step calculation
- Work out Mᵣ of NH₃: 14 + (3 × 1) = 17
- Calculate moles of NH₃ in 25 g: moles NH₃ = 25 ÷ 17 = 1.47 mol
- Use the reacting ratio from N₂ + 3H₂ → 2NH₃
3 mol H₂ make 2 mol NH₃, so:moles H₂ = 1.47 × 3 ÷ 2 = 2.205 mol - Convert moles of H₂ to mass: mass H₂ = 2.205 × 2 = 4.41 g
Mass of hydrogen = 4.41 g
✅ Correct answer
4.41 g
The mark scheme accepts 4.4117647 g rounded to at least 2 significant figures.
💡 Key knowledge
- Mᵣ of NH₃ = 17.
- Mᵣ of H₂ = 2.
- Always use the mole ratio from the balanced equation, not the big numbers in grams.
🧠 Exam technique
- Write the balanced equation first.
- Find moles before using the ratio.
- Only convert back to grams at the end.
- Show every stage — this question has method marks.
❌ Common calculation traps
- Using 3:1 instead of 3:2 from the equation.
- Forgetting H₂ has Mᵣ = 2, not 1.
- Mixing up mass and moles.
- Rounding too early and getting a less accurate final answer.
💡 Alternative method from the mark scheme
- 2 × Mᵣ NH₃ = 17 × 2 = 34
- So 34 g NH₃ is made from 6 g H₂
- For 25 g NH₃: (6 ÷ 34) × 25 = 4.41 g
Part (c) – 05.3 Calculate bond energy X for H—H
5 marks
📐 Step-by-step calculation
- Identify bonds broken in the reactants:
- 1 × N≡N = 945
- 3 × H—H = 3X
bonds broken = 945 + 3X - Identify bonds made in the products:
2NH₃ contains 6 N—H bonds altogether.bonds made = 6 × 391 = 2346 - Use the information:
energy released forming new bonds is 93 kJ/mol greater than energy needed to break existing bonds
So:93 = 2346 − (945 + 3X) - Solve: 93 = 1401 − 3X
3X = 1308 - Divide by 3: X = 436 kJ/mol
✅ Correct answer
X = 436 kJ/mol
💡 Key knowledge
- Breaking bonds needs energy.
- Making bonds releases energy.
- In 2NH₃ there are 6 N—H bonds.
- Exothermic means more energy is released than taken in.
🧠 Exam technique
- Count bonds from the displayed formula carefully.
- Write separate lines for bonds broken and bonds made.
- Translate the wording carefully: “93 kJ/mol greater than” means the energy released is 93 more than the energy absorbed.
- If unsure, a sensible setup can still earn method marks.
❌ Common errors
- Counting only 3 N—H bonds instead of 6.
- Using the wrong sign in the equation.
- Forgetting there are 3 H—H bonds broken.
- Using 93 as the final answer instead of using it in the equation.
Part (d) – 05.4 Complete the reaction profile
3 marks
✅ Correct answer
- Draw an exothermic profile curve that rises to a peak and then falls.
- The product line must be below the reactant line.
- Label the reactants line as nitrogen and hydrogen.
- Label the product line as ammonia.
- Label the overall energy change as a downward arrow from reactants level to products level.
💡 Key knowledge
- Exothermic reactions release energy.
- That means products are at a lower energy than reactants.
- The hump shows activation energy, even though this question mainly asks for the overall energy change.
🧠 Exam technique
- Do not just draw the curve — the labels are worth marks too.
- Use horizontal lines for both reactants and products.
- Make the overall energy change arrow point down.
❌ Common errors
- Drawing an endothermic profile with products higher than reactants.
- Missing labels for reactants or products.
- Drawing the energy change arrow upwards.
📐 What should be drawn
Start from the given reactant energy line. Draw a curve up to a peak, then down to a lower horizontal line. Label the top-left horizontal line nitrogen and hydrogen . Label the lower horizontal line ammonia . Add a vertical downward arrow from the reactant level to the product level labelled overall energy change .
Quick full-mark checklist
✅ To score highly
- Show 3 bonding pairs and 1 lone pair in NH₃.
- Use Mᵣ and moles before using the equation ratio.
- Count bonds carefully in the bond energy question.
- Make sure the reaction profile is clearly exothermic and fully labelled.
🧠 What separates top answers
- Clear structure and all working shown.
- Correct use of the 3:2 mole ratio.
- Correct interpretation of “93 kJ/mol greater than”.
- Precise labels on diagrams, not just roughly correct shapes.
Final answers summary
| Part | Answer | Marks |
|---|---|---|
| 05.1 | 3 shared pairs between N and H; 1 lone pair on nitrogen | 2 |
| 05.2 | Mass of hydrogen = 4.41 g | 4 |
| 05.3 | Bond energy X = 436 kJ/mol | 5 |
| 05.4 | Exothermic reaction profile with products lower than reactants, labelled lines, downward overall energy change | 3 |
Total = 14 marks
Topics
Chemistry · C2: Bonding, Structure and the Properties of Matter · C3: Quantitative Chemistry · C5: Energy Changes
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Chemistry Paper 1 (Higher), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.