Edexcel A-Level Chemistry Paper 3, November 2020: Question 4
10 marks · Medium difficulty · Calculations
Calculate enthalpy changes and construct an enthalpy level diagram for the hydration of copper(II) sulfate using experimental calorimetry data and Hess's law.
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Question text
4 Hess’s law can be used to determine enthalpy changes for reactions which cannot be
obtained directly.
An example is the reaction of anhydrous copper(II) sulfate with water to form
hydrated copper(II) sulfate, CuSO4.5H2O.
The following outline procedure was carried out.
Step 1 42.75g of deionised water was weighed out in a polystyrene cup and the
temperature measured.
Step 2 0.0250mol of hydrated copper(II) sulfate was added to the water in the
polystyrene cup with stirring, making a total of 45.00g of water.
Step 3 The temperature change was recorded.
Step 4 Steps 1 to 3 were repeated using 45.00g of deionised water and
0.0250mol of anhydrous copper(II) sulfate.
(a) Calculate the mass of 0.0250mol of hydrated copper(II) sulfate, CuSO4.5H2O.
(2)
(b) The reaction of hydrated copper(II) sulfate with water is shown.
CuSO .5H O(s) + aq → CuSO (aq) ∆H = +18.2 kJ mol−1
42 4 1
Calculate the temperature change that would have given this enthalpy change for
the stated experimental procedure.
Give your answer to a measurable number of significant figures and state whether
the temperature increases or decreases.
−1 oC−1]
[Specific heat capacity of the solution = 4.18J g
(3)
(c) The reaction of anhydrous copper(II) sulfate with water is shown.
CuSO (s) + aq → CuSO (aq) ∆H = −84.5 kJ mol−1
44 2
(i) Draw to scale, on the graph paper, a labelled enthalpy level diagram which
shows the enthalpy changes for the reactions of water with hydrated
copper(II) sulfate (∆H1)and anhydrous copper(II) sulfate(∆H2).
(3)
*P62670A01132*
Arbitrary CuSO4(aq)
zero
(ii) Use your enthalpy level diagram in (c)(i) to determine the enthalpy change, ∆rH,
for the reaction
CuSO4(s) + 5H2O(l) → CuSO4.5H2O(s)
You must show your working on the diagram.
12 (1)
*P62670A01232*
∆rH …
(d) State why the enthalpy change for the reaction of one mole of
anhydrous copper(II) sulfate with five moles of water to form
hydrated copper(II) sulfate, CuSO4.5H2O, cannot be measured directly.
(1)
(Total for Question 4 = 10 marks)
Mark scheme
Show the mark scheme
How to answer it
Hess's Law and Enthalpy Changes Study Guide
This question assesses core thermochemistry skills: calculating molar masses, applying calorimetry equations ( Q = mcΔT ), constructing accurate enthalpy level (Hess's Law) diagrams, and explaining experimental limitations regarding direct enthalpy measurement.
Calculating Mass from Moles
Calculate the mass of 0.0250 mol of hydrated copper(II) sulfate, CuSO₄·5H₂O. (2 marks)
✅ Correct Answer
Mᵣ(CuSO₄·5H₂O) = 63.5 + 32.1 + (16.0 × 4) + 5 × (1.0 × 2 + 16.0) = 249.6 g mol⁻¹
Mass = moles × Mᵣ = 0.0250 × 249.6 = 6.24 g (Accept 2 or 3 significant figures)
📐 Calculation Steps
- Find the molar mass (Mᵣ) of the hydrated salt by carefully including the 5 moles of water of crystallization.
- Multiply the number of moles (0.0250) by the calculated Mᵣ.
- Ensure correct units ( g ) are understood.
❌ Common Errors
- Omitting the mass of the 5 water molecules ( 5H₂O ) in the Mᵣ calculation.
- Rounding intermediate values too early, leading to slight inaccuracies in final mass.
Calorimetry and Temperature Change
Calculate the temperature change for the dissolution of hydrated copper(II) sulfate. State significant figures and whether it increases or decreases. (3 marks)
✅ Correct Answer
Q = ΔH × n = 18.2 kJ mol⁻¹ × 0.0250 mol = 0.455 kJ = 455 J
ΔT = Q / (m × c) = 455 / (45.00 × 4.18) = 2.4189... °C
Final Answer: 2.4 °C (or 2 °C), and state that the temperature decreases.
💡 Key Knowledge
- ΔH₁ = +18.2 kJ mol⁻¹ is endothermic, meaning heat is absorbed from the solution, causing the temperature to drop.
- Watch out for unit conversions: kJ must be converted to J by multiplying by 1000 before using Q = mcΔT .
🧠 Exam Technique
- Significant figures: The data in the prompt uses 3 s.f. ( 18.2 , 45.00 , 4.18 ), so give your answer to 1 or 2 significant figures ( 2.4 °C or 2 °C ) as requested by the mark scheme.
- Don't forget to explicitly state "decreases" to secure the final marking point.
Enthalpy Level Diagram Construction
Draw to scale a labelled enthalpy level diagram for the reactions of water with hydrated and anhydrous copper(II) sulfate. (3 marks)
✅ Correct Answer Requirements
- y-axis: Labelled clearly as "Enthalpy" or "H" with units ( kJ mol⁻¹ ) and an appropriate linear scale.
- Entities & Levels: Place CuSO₄(s) + 5H₂O(l) at the top, CuSO₄(aq) in the middle, and CuSO₄·5H₂O(s) at the bottom.
- Arrows & Values: Draw directed arrows showing ΔH₂ = -84.5 kJ mol⁻¹ downwards from solid anhydrous to aqueous, and ΔH₁ = +18.2 kJ mol⁻¹ upwards from hydrated solid to aqueous.
🧠 Exam Technique
- Always use a ruler for energy levels and arrow lines.
- Make sure arrowheads point strictly in the correct vertical direction (down for exothermic, up for endothermic).
- Include state symbols ( s , l , aq ) for all chemical species.
❌ Common Errors
- Omitting units on the enthalpy axis.
- Placing the hydrated level above the anhydrous level (failing to reflect the relative exothermic/endothermic magnitudes).
- Using double-headed arrows instead of single-headed reaction pathway arrows.
Applying Hess's Law
Use your enthalpy level diagram to determine the enthalpy change, ΔᵣH, for: CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s). (1 mark)
✅ Correct Answer
ΔᵣH = ΔH₂ - ΔH₁ = -84.5 - (+18.2) = -102.7 kJ mol⁻¹ (Also accept -103 kJ mol⁻¹ )
💡 Hess's Law Route Analysis
According to Hess's Law, the total enthalpy change of a reaction is independent of the route taken. Going from CuSO₄(s) + 5H₂O(l) to CuSO₄(aq) releases 84.5 kJ mol⁻¹ ( ΔH₂ ), and reversing the hydration step removes 18.2 kJ mol⁻¹ ( ΔH₁ ).
Experimental Limitations
State why the enthalpy change for the direct reaction cannot be measured directly. (1 mark)
✅ Correct Answer
It is impossible to react exactly 5 moles of water with 1 mole of anhydrous copper(II) sulfate and measure the temperature change of a solid reaction mixture accurately.
❌ Common Errors & Rejections
- Simply stating "heat loss to surroundings" (this applies to all calorimetry, but isn't the primary reason the direct hydration of a solid is impossible).
- Stating "heat is needed to start the reaction" (incorrect, the reaction is spontaneous).
- Vague statements about incomplete mixing without mentioning the specific stoichiometric water constraint (5:1 ratio).
Topics
Physical Chemistry · Core Practicals · Core Practical 8: Determine the enthalpy change of a reaction using Hess’s law · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics I
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.