AQA A-Level Chemistry AS Paper 1, June 2025: Question 18
1 mark · Medium difficulty · Multiple Choice
Identify which combination of hydrochloric acid and sodium hydroxide solutions produces the greatest temperature increase when mixed.
Practise this questionQuestion
Question text
18 The reaction between hydrochloric acid and sodium hydroxide is exothermic.
Which row shows the solutions that will give the greatest temperature increase when
mixed?
[1 mark]
HCl(aq) NaOH(aq)
3 Concentration / 3 Concentration /
Volume / cm –3 Volume / cm –3
mol dm mol dm
A 200 1.0 200 1.0
B 50 1.0 50 1.0
C 50 1.5 100 1.0
D 50 1.5 50 1.5
Mark scheme
Show the mark scheme
18 D 1 (AO1)
D 50 1.5 50 1.5
How to answer it
Enthalpy of Neutralisation & Temperature Change
This question assesses your ability to evaluate the relationship between moles reacted, total solution volume (mass heated), and temperature rise (ΔT) in calorimetry experiments.
- Understanding calorimetry principles using q = mcΔT .
- Distinguishing between total heat energy released (q) and temperature change (ΔT).
- Identifying limiting reagents in neutralisation reactions.
Determining Greatest Temperature Increase
AQA A-Level Chemistry • Physical Chemistry • Energetics
✅ Correct Answer
Option D
Mark: 1 / 1 (AO1)
Mixing 50 cm³ of 1.5 mol dm⁻³ HCl with 50 cm³ of 1.5 mol dm⁻³ NaOH yields the greatest ratio of moles reacted per unit volume, giving the highest ΔT.
💡 Key Knowledge
- Neutralisation Equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) (1:1 ratio) - Heat released (q): Directly proportional to moles of water formed ( q = n × |ΔH| ).
- Mass of solution (m): Proportional to total volume ( m ≈ Vtotal × 1 g cm⁻³ ).
- Temperature rise (ΔT):
ΔT = q / (m × c) ∝ n(reacted) / Vtotal
📐 Step-by-Step Analysis of Options
Because ΔT ∝ n(reacted) / Vtotal , calculate the effective concentration of reaction for each row:
| Row | Moles HCl ( c × V ) | Moles NaOH ( c × V ) | Reacting Moles ( n ) | Total Volume ( Vtotal ) | Ratio ( n / Vtotal ) | Relative ΔT |
|---|---|---|---|---|---|---|
| A | 0.200 × 1.0 = 0.200 mol | 0.200 × 1.0 = 0.200 mol | 0.200 mol | 200 + 200 = 400 cm³ | 0.200 / 400 = 0.00050 | 1.0× |
| B | 0.050 × 1.0 = 0.050 mol | 0.050 × 1.0 = 0.050 mol | 0.050 mol | 50 + 50 = 100 cm³ | 0.050 / 100 = 0.00050 | 1.0× |
| C | 0.050 × 1.5 = 0.075 mol (limiting) | 0.100 × 1.0 = 0.100 mol (excess) | 0.075 mol | 50 + 100 = 150 cm³ | 0.075 / 150 = 0.00050 | 1.0× |
| D | 0.050 × 1.5 = 0.075 mol | 0.050 × 1.5 = 0.075 mol | 0.075 mol | 50 + 50 = 100 cm³ | 0.075 / 100 = 0.00075 | 1.5× (Greatest) |
Rows A, B, and C all produce the exact same temperature rise because their mole-to-volume ratio is identical (0.50 mol dm⁻³). Row D has a 50% higher ratio (0.75 mol dm⁻³), producing 1.5× the temperature increase.
❌ Common Errors
- Selecting Option A: The most common trap. Students see the largest moles (0.200 mol) and assume it gives the largest ΔT. While q (heat released) is 4× larger than in B, the volume heated is also 4× larger, so ΔT is identical.
- Ignoring limiting reagents in Option C: Assuming all 0.100 mol of NaOH reacts, ignoring that HCl limits the reaction to 0.075 mol.
- Forgetting to add volumes together: Calculating n / V using only the volume of acid rather than the total combined mixture volume.
🧠 Exam Technique & Shortcut
- The Equal Volumes Shortcut: When equal volumes of equimolar strong monoprotic acid and base are mixed (e.g. 50 cm³ of each), the mixture is simply diluted by a factor of 2.
- In A: 1.0 mol dm⁻³ diluted to 0.5 mol dm⁻³ reacted.
- In B: 1.0 mol dm⁻³ diluted to 0.5 mol dm⁻³ reacted.
- In D: 1.5 mol dm⁻³ diluted to 0.75 mol dm⁻³ reacted → Highest concentration of reaction, so highest ΔT! Takes under 15 seconds to solve.
Topics
Physical Chemistry · 3.1.4 Energetics · 3.1.2 Amount of Substance
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.