AQA A-Level Chemistry Paper 3, June 2018: Question 4

16 marks · Medium difficulty · Practical Techniques & Data Analysis

Plot a cooling curve to find the temperature rise for a neutralisation reaction, evaluate uncertainties and heat loss, calculate the enthalpy of neutralisation between ethanedioic acid and potassium hydroxide, and explain the difference in enthalpy compared to a strong acid.

Practise this question

Question

Question text

04 A student carried out an experiment to find the temperature rise for a reaction

between hydrochloric acid and sodium hydroxide solution.

• The student used a measuring cylinder to place 50 cm3 of 0.400 mol dm–3

hydrochloric acid into a glass beaker.

• The student recorded the temperature at one-minute intervals for

three minutes.

• At the fourth minute the student added 50 cm3 of 0.400 mol dm–3

sodium hydroxide solution and stirred to mix the solutions, but did not record

the temperature.

• The student recorded the temperature at one-minute intervals for a further

eight minutes.

The results are shown in Table 3.

Table 3

Time/min 0 1 2 3 4 5 6 7 8 9 10 11 12

Temperature

19.8 19.8 19.8 19.8 21.4 21.7 21.6 21.5 21.4 21.3 21.2 21.1

/ °C

04.1 Plot a graph of temperature against time on the grid opposite.

Use your graph to find the temperature rise, ∆T, at the fourth minute.

Show your working on the graph by drawing suitable lines of best fit.

[5 marks]

∆T °C

04.2 The uncertainty in each of the temperature readings from the thermometer used in

this experiment was ±0.1°C

Calculate the percentage uncertainty in the value for the temperature rise.

[1 mark]

Percentage uncertainty

04.3 Suggest a change to the experiment that would minimise heat loss.

[1 mark]

04.4 Suggest and explain another change to the experiment that would decrease the

percentage uncertainty in the use of the same thermometer.

[2 marks]

04.5 A second student completed an experiment to determine the enthalpy of neutralisation

for the reaction between ethanedioic acid solution (HOOCCOOH) and potassium

hydroxide solution.

The student added 25 cm3 of 0.80 mol dm–3 ethanedioic acid solution to 75 cm3 of

0.60 mol dm–3 potassium hydroxide solution.

The temperature increased by 3.2°C

Give an equation for the reaction between ethanedioic acid solution and potassium

hydroxide solution.

Calculate the enthalpy change (∆H) per mole of water formed in this reaction.

Assume that the specific heat capacity of the reaction mixture is 4.2 J K–1 g–1

Assume that the density of the reaction mixture is 1.00 g cm–3

[5 marks]

Equation

∆H kJ mol–1

04.6 In a similar experiment to that in Question 04.5, the enthalpy of neutralisation for the

reaction between sulfuric acid and potassium hydroxide solution was found to be

–57.0 kJ mol–1 per mole of water formed.

Suggest an explanation for the difference between this value and your answer to

Question 04.5.

(If you were unable to obtain an answer to Question 04.5 you should assume a value

of –28.5 kJ mol–1. This is not the correct answer.)

[2 marks]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

M1: Temperature on the y-axis and uses sensible scales (i.e. Lose mark if temperature scale starts at 0°C

minimum 20 little squares for each °C on y-axis) This mark scores if all points fit on the grid. Do not

penalise M1 if extrapolation to 4 mins goes off the

grid – this is penalised in M3. 1

M2: Plots all the points correctly (± half a small square) Lose mark if the points go off the grid

Ignore a plotted point at 4 mins used to work out ∆T

M3: Draws two best fit lines (0-3 mins) and (6-12 mins) 1

Both lines must be straight and through all points

except 5th minute; lose mark if the lines are

04.1 kinked/doubled. Any line through 5th minute loses

mark

“S-shaped curve” through points loses M3 and M4

M4: Extrapolates both lines to at least the 4th minute

Lose mark if the extrapolation goes off the grid.

Allow calculation of ΔT from S-shaped curve as:

M5: 21.9 – 19.8 = 2.1 (°C) Value at 4th minute – 19.8 but not if 0 (°C)

ΔT value ecf from incorrect lines/extrapolation

ΔT must be to at least 1dp

If value of ∆T = 2.1, then award M5

0.2 / 2.1 x 100 = 9.5 % Conseq on 04.1

04.2 1

Ignore no of sfs.

Replace the glass beaker with a polystyrene cup / insulate the glass Ignore use more dilute solutions

04.3 beaker / use a lid Ignore suggested materials for insulation 1

Do not allow copper calorimeter / bomb calorimeter

Increase magnitude of temperature change Ignore references to volume changes

04.4 by increasing the concentration of the acid/alkali

Mark independently

– – –

HOOCCOOH + 2KOH K2(OOCCOO) + 2H2O M1 – equation (allow ionic KOH / K2C2O4)

H C O + 2KOH K C O + 2H O 21

22 4 2 2 4 2

ignore state symbols

allow multiples

Mark independently

q ( = mc∆T) = 100 x 4.2 x 3.2 = 1344 J M2 – process

(ignore sign here)

(allow calculations involving 4.18 which leads to

1338 J)

n HOOCCOOH = 25 x 0.800 / 1000 = 0.020 M3 – calculations of amounts, in moles, of both the

n KOH = 75 x 0.6 / 1000 = 0.045 ethanedioic acid and potassium hydroxide 1

04.5 (both calculations needed)

Moles of water = 0.040 moles M4 – answer (stated or used in calculation of ∆H)

∆H = –1.344/0.04 M5 – ecf on M2 and M4

Answer must be negative and to at least 2sf

= – 33.6 ( kJ mol–1) ∆H = – M2 (in kJ) / M4

–32.5 - –34 scores 4/4 (M2-M5 + equation)

+32.5 - +34 scores 3/4 (M2, M3, M4 + equation)

–65 - –68 scores 3/4 (+ equation)

+65 - +68 scores 2/4 (+ equation)– – –

–52 - –54 scores 3/4 (+ equation)

+52 - +54 scores 2/4 (+ equation)

HOOCCOOH is a weak acid / not fully dissociated H2SO4 is a strong(er)acid / fully dissociated /

dissociates more

(more) energy needed to break bonds/complete dissociation / 1

So less energy is needed for dissociation of sulfuric

04.6

dissociation is endothermic acid

Ignore references to heat loss

Total 16

How to answer it

Calorimetry, Cooling Curves & Enthalpy of Neutralisation

What this question tests

This multi-part practical question examines core physical chemistry skills: plotting temperature-time graphs and extrapolating cooling curves to determine an accurate temperature change (ΔT); calculating percentage apparatus uncertainty for two readings; proposing experimental improvements to minimise heat transfer; executing multi-step calorimetry calculations (q = mcΔT) involving limiting reagents and diprotic acids; and theoretically accounting for variations in enthalpies of neutralisation using acid dissociation equilibria.

Question 04.1: Cooling Curve Graph & Temperature Rise

Plot temperature vs time, draw best-fit lines, extrapolate to the 4th minute, and find ΔT [5 Marks]

✅ Correct Values & Plotting Standards

Axes & Scale: Temperature plotted on the vertical y-axis; sensible linear scale where each 1 °C occupies at least 20 small squares. The y-axis scale must not start at 0 °C (false origin starting around 18 °C or 19 °C is required).
Points: All points from Table 3 plotted correctly to within ± half a small square.
Lines of best fit:
• Initial line (0–3 min): Horizontal straight line through 19.8 °C.
• Cooling line (6–12 min): Straight line through points at 6, 7, 8, 9, 10, 11, and 12 min. The point at 5 min (21.4 °C) must not be included as mixing/heating was incomplete.
Extrapolation: Both lines extrapolated cleanly to the 4th minute.
Temperature rise: Extrapolated peak = 21.9 °C.
ΔT = 21.9 − 19.8 = 2.1 °C (given to at least 1 d.p.).

🧠 Exam Technique: Why Extrapolate?

Reaction is not instantaneous; heat is lost to the surroundings concurrently as the solutions mix.

The 4th minute: Reactants were mixed here, but no reading was taken. Extrapolating both lines back/forward to 4 min simulates what the peak temperature would have been with instantaneous mixing and zero heat loss.
Ignoring anomalous points: At minute 5, reaction and thermal equilibration were still occurring. Top students recognise this and omit minute 5 when ruling the cooling line.

❌ Common Student Errors

Drawing an "S-shaped" continuous curve through all the data points (automatically loses marks M3 and M4).
Starting the temperature scale at 0 °C, which compresses the plot so it occupies less than half the page.
Forcing the cooling line through the point at 5 min, giving an incorrectly shallow slope and wrong ΔT.

Mark Breakdown: 1 mark for correct axes & sensible scale (scale must not start at 0 °C) | 1 mark for accurate plotting | 1 mark for two straight best-fit lines (excluding min 5) | 1 mark for extrapolating both lines to 4 min | 1 mark for ΔT = 2.1 °C (consequential on student's lines).

Question 04.2: Thermometer Percentage Uncertainty

Calculate percentage uncertainty in the value of ΔT (uncertainty = ±0.1 °C) [1 Mark]

📐 Calculation

Calculating a temperature rise (ΔT) requires two temperature readings (an initial and a final reading), so the absolute uncertainty is doubled:

Total Uncertainty = 2 × (±0.1 °C) = ±0.2 °C

% Uncertainty = (0.2 / 2.1) × 100 = 9.5%

Note: Consequential on your ΔT from 04.1 (e.g., if ΔT = 2.0 °C, % = 10.0%).

❌ Common Error

Forgetting there are two readings: Many candidates calculated (0.1 / 2.1) × 100 = 4.8% . Temperature difference always involves measuring T_initial and T_final, meaning two reading uncertainties must be summed!

Mark Breakdown: 1 mark for (0.2 / 2.1) × 100 = 9.5% (any significant figures accepted; error carried forward from 04.1).

Question 04.3: Minimising Heat Loss

Suggest one modification to the experimental apparatus to minimise heat loss [1 Mark]

✅ Acceptable Answers (Any One)

Replace the glass beaker with an expanded polystyrene cup.
Put a lid on the beaker/cup.
Wrap the glass beaker in mineral wool / lagging / insulating material.

🧠 Examiner Commentary

Glass is a relatively good conductor of heat compared to expanded polystyrene. Do not suggest using a "copper calorimeter" or "bomb calorimeter" (copper increases heat loss; bomb calorimeters are for combustion of solids/gases, not aqueous solutions).

Mark Breakdown: 1 mark for identifying insulation (polystyrene cup, lagging) or a lid.

Question 04.4: Reducing Percentage Uncertainty

Suggest and explain a change (using the same thermometer) to reduce percentage uncertainty [2 Marks]

✅ Correct Answer

Change: Increase the concentration of the acid and/or alkali solution. [1 mark]
Explanation: This increases the magnitude of the temperature change (ΔT), thereby making the fixed thermometer uncertainty a smaller fraction of the reading. [1 mark]

❌ Common Error: Changing Volume

Students frequently suggest "increase the volume of both solutions". This earns 0 marks because doubling volumes doubles the heat released ( q ), but also doubles the mass of solution ( m ). Since ΔT = q / (m × c) , the temperature rise remains completely unchanged!

Mark Breakdown: Mark 1 for increasing magnitude of ΔT | Mark 2 for increasing the concentration of reactants (marks are independent).

Question 04.5: Enthalpy of Neutralisation Calculation

Balanced equation and calculation of ΔH per mole of water formed [5 Marks]

📐 Step-by-Step Calculation

1 Balanced Chemical Equation
Ethanedioic acid is dicarboxylic ( HOOCCOOH or H₂C₂O₄ ), requiring 2 moles of KOH :
HOOCCOOH + 2KOH → K₂(OOCCOO) + 2H₂O
Also accepted: H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O (state symbols not required).

2 Heat Released (q)
Total volume = 25 cm³ + 75 cm³ = 100 cm³
Mass of solution, m = 100 cm³ × 1.00 g cm⁻³ = 100 g
q = m × c × ΔT = 100 × 4.2 × 3.2 = 1344 J = 1.344 kJ

3 Moles of Reactants & Identifying the Limiting Reagent
• Moles of HOOCCOOH = (25 / 1000) × 0.80 = 0.020 mol
• Moles of KOH = (75 / 1000) × 0.60 = 0.045 mol
Stoichiometric demand: 0.020 mol acid needs 2 × 0.020 = 0.040 mol KOH.
Since 0.045 mol KOH is available, KOH is in excess and HOOCCOOH is the limiting reactant.

4 Moles of Water Formed
From the equation, 1 mol HOOCCOOH produces 2 mol H₂O (or 2 mol KOH reacted produces 2 mol H₂O):
Moles of H₂O = 0.020 × 2 = 0.040 mol (using the 0.040 mol of KOH that reacted).

5 Enthalpy Change of Neutralisation (ΔH)
Reaction is exothermic (temperature increased), so ΔH must be negative:
ΔH = − (q / n(H₂O)) = − 1.344 kJ / 0.040 mol = −33.6 kJ mol⁻¹

❌ Common Pitfalls

Forgetting the 1:2 ratio: Assuming a 1:1 reaction gives 0.020 mol H₂O , leading to −67.2 kJ mol⁻¹ (loses 2 marks).
Omitting the negative sign: Neutralisation is exothermic; omitting the negative sign loses the final mark.
Using only one volume for m: Using 25 g or 75 g instead of the total mixed volume of 100 g.

💡 Definition Checklist

Enthalpy of Neutralisation: The enthalpy change when solutions of an acid and alkali react together under standard conditions to produce one mole of water.

Always divide heat energy ( kJ ) by the moles of water formed, not merely the moles of acid added!

Mark Breakdown: M1: Balanced equation | M2: q = 1344 J (or 1.344 kJ) | M3: Both initial mole values calculated (0.020 and 0.045) | M4: Moles of water = 0.040 mol | M5: ΔH = −33.6 kJ mol⁻¹ (requires negative sign and at least 2 sig figs).

Question 04.6: Weak vs Strong Acid Enthalpy Difference

Explain why ΔH for ethanedioic acid (−33.6 kJ mol⁻¹) is less exothermic than sulfuric acid (−57.0 kJ mol⁻¹) [2 Marks]

✅ Model Answer

Point 1 (Acid strength): Ethanedioic acid is a weak acid / is only partially dissociated in aqueous solution (whereas sulfuric acid is a strong acid / fully dissociated). [1 mark]
Point 2 (Energetics of dissociation): Energy is required to break bonds to fully dissociate ethanedioic acid into H⁺ ions (dissociation is an endothermic process), making the overall enthalpy change less exothermic. [1 mark]

🧠 Hess's Law Perspective

Enthalpy of neutralisation can be broken into two steps:

1. HOOCCOOH(aq) → ⁻OOCCOO⁻(aq) + 2H⁺(aq) [ΔH > 0, Endothermic]
2. 2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l) [ΔH ≈ −57 kJ mol⁻¹ per H₂O]

Since step 1 absorbs energy, the net overall heat released is significantly less exothermic (−33.6 kJ mol⁻¹ vs −57.0 kJ mol⁻¹).

Mark Breakdown: 1 mark for stating ethanedioic acid is weak / not fully dissociated (or H₂SO₄ is strong / fully dissociated) | 1 mark for stating that dissociation is endothermic / requires energy to break bonds.

Topics

Physical Chemistry · Required Practicals · 3.1.4 Energetics · 3.1.2 Amount of Substance · 3.1.12 Acids and Bases · Required Practical 2: Measurement of an enthalpy change

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.