OCR A-Level Chemistry Unified chemistry (03), June 2019: Question 5
13 marks · Hard difficulty · Structured Questions
Calculate the enthalpy change of hydration of anhydrous copper(II) sulfate using experimental thermochemical data and Hess's law, determine experimental temperature change from uncertainty, calculate standard entropy from Gibbs free energy data, and predict the bond angle, shape, and structure of sulfur-containing oxyanions.
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Question text
5 This question is about copper(II) sulfate, CuSO4, and sodium thiosulfate, Na2S2O3.
(a) The enthalpy change of reaction, ∆rH, for converting anhydrous copper(II) sulfate to hydrated
copper(II) sulfate is difficult to measure directly by experiment.
CuSO4(s) + 5H2O(l) → CuSO4•5H2O(s) ∆rH reaction 5.1
The enthalpy changes of solution of anhydrous and hydrated copper(II) sulfate can be
measured by experiment. The reactions are shown below.
In the equations, ‘aq’ represents an excess of water.
CuSO (s) + aq → Cu2+(aq) + SO 2−(aq) ∆ H(CuSO (s)) reaction 5.2
44 sol 4
CuSO •5H O(s) + aq → Cu2+(aq) + SO 2−(aq) ∆ H(CuSO •5H O(s)) reaction 5.3
42 4 sol 4 2
Experiment 1
A student carries out an experiment to find ∆solH(CuSO4(s)) for reaction 5.2.
Student’s method
• Weigh a bottle containing CuSO4(s) and weigh a polystyrene cup.
• Add about 50 cm3 of water to the polystyrene cup and measure its temperature.
• Add the CuSO4(s), stir the mixture, and measure the final temperature.
• Weigh the empty bottle and weigh the polystyrene cup with final solution.
Mass readings
Mass of bottle + CuSO4(s) / g 28.04
Mass of empty bottle / g 20.06
Mass of polystyrene cup / g 23.43
Mass of polystyrene cup + final solution / g 74.13
Temperature readings
Initial temperature of water / °C 20.5
Temperature of final solution / °C 34.0
Experiment 2
The student carries out a second experiment with CuSO4•5H2O (reaction 5.3). The student
uses the same method as in Experiment 1.
The student calculates ∆ H(CuSO •5H O(s)) as +8.43 kJ mol−1.
sol 4 2
(i)* Calculate ∆solH(CuSO4(s)) for reaction 5.2 and determine the enthalpy change of
reaction 5.1, ∆rH.
Assume that the specific heat capacity, c, of the solution is the same as for water.
Show your working, including an energy cycle linking the enthalpy changes. [6]
Additional answer space if required
(ii) The thermometer had an uncertainty in each temperature reading of ± 0.1 °C.
The student calculates a 20% uncertainty in the temperature change in Experiment 2.
Calculate the temperature change in Experiment 2.
temperature change = … °C [1]
(b) The standard enthalpy change of reaction, ∆ H o, and the standard free energy change, ∆G o,
r
for converting anhydrous sodium thiosulfate to hydrated sodium thiosulfate are shown below.
Na S O (s) + 5H O(l) → Na S O •5H O(s) ∆ H o = –55.8 kJ mol−1
22 3 2 2 2 3 2 r
∆G o = –16.1 kJ mol−1
Standard entropies are given in the table.
Compound S o / J K−1 mol−1
Na2S2O3•5H2O(s) 372.4
H2O(l) 69.9
Determine the standard entropy, S o, of anhydrous sodium thiosulfate, Na S O (s).
22 3
Give your answer to 3 significant figures.
S o = … J K−1 mol−1 [4]
(c) Sodium thiosulfate contains the thiosulfate ion, S O 2−.
The displayed formula of S O 2− can be shown as below.
O
–O S S–
O
thiosulfate ion
(i) Predict the O–S–S bond angle and name of the shape of the thiosulfate ion.
Bond angle …
Name of shape …
[1]
(ii) In some of its reactions, the thiosulfate ion forms the tetrathionate ion, S O 2−.
The S O 2− ion is a ‘dimer’ of S O 2−.
46 2 3
Draw a displayed formula for the S O 2− ion.
[1]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
5 (a) (i)* Please refer to the marking instructions on page 4 of this mark 6 AO3.1 Indicative scientific points may include:
scheme for guidance on how to mark this question. ×4 1. Processing experimental data
Energy change from mcΔT
Level 3 (5–6 marks) AO3.2 • Energy in J OR kJ
Calculates CORRECT enthalpy change with correct – signs ×2 Using 50.70 g, 50.0 g
for = 50.70 × 4.18 × 13.5 = 2861 (J) OR 2.861 (kJ)
∆solH (CuSO4(s)) for reaction 5.2 3SF or more (2.861001 unrounded)
AND OR 50.0 × 4.18 × 13.5 = 2821.5 (J) OR 2.8215 (kJ)
∆rH, for reaction 5.1. Amount in mol of CuSO4
7.98
There is a well-developed line of reasoning which is clear and • n(CuSO4) = = 0.0500 (mol)
159.6
logically structured.
The information presented is relevant and substantiated. ----------------------------------------------
2. ± value of ∆solH(CuSO4(s)) for reaction 5.2
Level 2 (3–4 marks) 2.861 –1
From m = 50.70 g = ± = ±57.22 (kJ mol )
Calculates a value of ∆solH (CuSO4(s)) for reaction 5.2 from 0.0500
the: (–57.22002 unrounded)
Energy change 2.8215 –1
From m = 50.0 g = ± = ±56.43 (kJ mol )
AND 0.0500
Amount in mol of CuSO4. ----------------------------------------------
3. CORRECT enthalpy changes for reactions
There is a line of reasoning presented with some structure. 5.2 and 5.1 with signs (using 50.70 g ONLY)
The information presented is relevant and supported by some Reaction 5.2 = –57.22 (kJ mol–1)
evidence. 3SF or more with correct – sign
Reaction 5.1
Level 1 (1–2 marks) ∆rH = ∆solH(CuSO4(s)) – ∆solH(CuSO4•5H2O(s))
Processes experimental data to obtain the: = –57.22 – 8.43 = –65.65 (kJ mol–1)
Energy change from mc∆T 3SF or more with correct – sign
OR NOTE: A clear and logically structured response
Amount in mol of CuSO4. would include an energy cycle
ALLOW omission of trailing zeroes
There is an attempt at a logical structure with a line of reasoning. ALLOW minor slips
The information is in the most part relevant.
AO
element
0 marks – No response or no response worthy of credit. 16
(a) (ii) 100 1 AO2.8 IGNORE direction of temperature change
Temperature change = 0.2 × 20 = 1(.0)ºC Working NOT required
(b) FIRST CHECK THE ANSWER IN ON ANSWER LINE 4 AO2.4
If answer = (+)156 (J K–1 mol–1) award 4 marks
×4
--------------------------------------------------------------------------
Part 1: Calc of ∆rS
Use of 298 K (seen anywhere) 1 mark
• e.g. –16.1 = –55.8 – 298 × ∆S –55.8 – (–16.1) –39.7
Using 298 K, ∆S = 298 = 298
= –0.133…(kJ K–1mol–1)
CORRECT use of Gibbs’ equation 1 mark
OR –133… (J K–1mol–1)
• using candidate’s temperature (e.g. 298)
• with –16.1 AND –55.8 Sign required IGNORE units
Calculator:
• to calculate ∆S in kJ OR J –0.133221 (kJ K–1 mol–1)
–133.221 (J K–1 mol–1)
ALLOW ECF from incorrect temperature.
-----------------------------------------------------------------
Part 2: Calc of S(Na2S2O3) 1 mark
CORRECT use of standard S data in question
Seen anywhere (could be within an expression) e.g.
• 372.4 – [ S(Na2S2O3) + (5 × 69.9) ]
• OR 372.4 – (5 × 69.9)
• OR 372.4 – 349.5
• OR 22.9
IGNORE sign, i.e. ALLOW –22.9, etc Using –133:
S(Na2S2O3) = 372.4– 349.5 – (–133)
CORRECT calculation of S(Na2S2O3) using candidate’s = 22.9 + 133
= (+)156 (J K–1 mol–1)
calculated ∆S in Part 1 to 3 SF 1 mark
3 SF required
ALLOW ECF from incorrect ∆rS (Part 1)
AO
element
(c) (i) 109.5(º) AND tetrahedral 17 1 AO1.2 ALLOW 109–110(º)
(ii) 1 AO3.1 IGNORE charges
ALLOW cyclic structures.
Three 6-ring structures possible, e.g.
OR
OR
NOTE: There MUST be 2 atoms in centre
between 6-bonded S atoms.
e.g. DO NOT ALLOW
IGNORE absence of charges OR incorrect charges
For other structures, contact TL
Total 13
How to answer it
Thermodynamics & Sulfur Chemistry: CuSO₄ & Na₂S₂O₃
What this question tests
This question brings together core physical and inorganic chemistry concepts across practical enthalpy determination and chemical bonding:
- Calorimetry & Hess's Law: Calculating enthalpy change of solution ( q = mcΔT ) and applying an indirect Hess's Law cycle to determine the enthalpy of hydration for copper(II) sulfate.
- Apparatus Uncertainty: Calculating absolute and percentage uncertainties from two temperature readings.
- Gibbs Free Energy & Standard Entropy: Manipulating ΔG° = ΔH° - TΔS° with unit conversions (kJ to J) to deduce unknown standard entropy values ( S° ).
- Shapes of Molecules & Dimer Structures: Deducing VSEPR shapes/bond angles and drawing extended oxyanion structures (thiosulfate and tetrathionate).
Enthalpy of Hydration & Hess's Law Cycle
Calculate ΔsolH(CuSO₄) and determine ΔrH for Reaction 5.1
📐 Step-by-Step Calculation
- Mass of anhydrous CuSO₄:
28.04 g - 20.06 g = 7.98 g - Amount of CuSO₄ (moles):
M(CuSO₄) = 63.5 + 32.1 + (4 × 16.0) = 159.6 g mol⁻¹
n(CuSO₄) = 7.98 / 159.6 = 0.0500 mol - Mass of solution heated:
74.13 g (cup + solution) - 23.43 g (cup) = 50.70 g
(Note: The mark scheme also accepts using 50.0 g of pure water). - Temperature change:
ΔT = 34.0 - 20.5 = +13.5 °C - Heat energy released (q):
Using m = 50.70 g :
q = mcΔT = 50.70 × 4.18 × 13.5 = 2861 J = 2.861 kJ
(If m = 50.0 g is used: q = 50.0 × 4.18 × 13.5 = 2821.5 J = 2.822 kJ) - Enthalpy of solution of anhydrous CuSO₄ (Reaction 5.2):
Reaction is exothermic (temperature increased), so the sign must be negative:
ΔsolH(CuSO₄) = -q / n = -2.861 / 0.0500 = -57.22 kJ mol⁻¹
(If m = 50.0 g was used: -56.43 kJ mol⁻¹) - Indirect Determination of ΔrH (Reaction 5.1):
Using Hess's Law via aqueous ions:
ΔrH = ΔsolH(CuSO₄) - ΔsolH(CuSO₄·5H₂O)
ΔrH = -57.22 - (+8.43) = -65.65 kJ mol⁻¹ (or -65.7 kJ mol⁻¹)
(Using 50.0 g gives: -56.43 - 8.43 = -64.86 kJ mol⁻¹)
💡 The Hess's Law Cycle
To reach Level 3 (5–6 marks), examiners require an explicit, logically linked energy cycle:
🧠 Level of Response Breakdown
- Level 3 (5–6 marks): Both enthalpy values correctly calculated with unambiguous negative signs ( -57.22 and -65.65 kJ mol⁻¹ ), supported by a fully labelled energy cycle.
- Level 2 (3–4 marks): Successfully processes mcΔT , calculates moles, and finds the numeric value of ΔsolH , but might drop a negative sign or miss the cycle link.
- Level 1 (1–2 marks): Calculates either energy change ( q ) OR amount of moles ( n ) correctly.
❌ Common Traps
- Mass Confusion: Using 7.98 g (the solid) instead of 50.70 g (the liquid solution) in q = mcΔT . The thermometer measures the temperature of the liquid!
- Sign Errors: Forgetting the negative sign on ΔsolH despite the temperature rising from 20.5 °C to 34.0 °C.
- Hess Cycle Signs: Adding instead of subtracting +8.43 kJ mol⁻¹ . Follow the arrows carefully.
Thermometer Uncertainty
Calculate the temperature change in Experiment 2
📐 Calculation
Each temperature reading has an uncertainty of ±0.1 °C .
A temperature change involves two readings (initial and final):
Total uncertainty = 2 × 0.1 = ±0.2 °C
Percentage uncertainty is given as 20% :
% uncertainty = (Total uncertainty / ΔT) × 100
20 = (0.2 / ΔT) × 100
ΔT = 0.2 × (100 / 20) = 1.0 °C
✅ Correct Answer
Temperature change = 1.0 °C (or 1 °C )
❌ Common Error
Using only 0.1 °C instead of doubling the uncertainty to 0.2 °C for a difference calculation. This would incorrectly yield 0.5 °C.
Gibbs Free Energy & Standard Entropy
Calculate S° of anhydrous sodium thiosulfate, Na₂S₂O₃
📐 Step-by-Step Calculation
- Standard Temperature:
Standard conditions imply T = 298 K . - Find ΔrS° using Gibbs equation:
ΔG° = ΔH° - TΔS°
-16.1 kJ mol⁻¹ = -55.8 kJ mol⁻¹ - (298 × ΔS°)
298 × ΔS° = -55.8 - (-16.1) = -39.7 kJ mol⁻¹
ΔS° = -39.7 / 298 = -0.13322 kJ K⁻¹ mol⁻¹
Convert to J K⁻¹ mol⁻¹: ΔS° = -133.22 J K⁻¹ mol⁻¹ - Set up entropy equation:
Reaction: Na₂S₂O₃(s) + 5H₂O(l) → Na₂S₂O₃·5H₂O(s)
ΔrS° = ΣS°(products) - ΣS°(reactants)
-133.22 = S°(Na₂S₂O₃·5H₂O) - [S°(Na₂S₂O₃) + 5 × S°(H₂O)]
-133.22 = 372.4 - [S°(Na₂S₂O₃) + (5 × 69.9)]
-133.22 = 372.4 - S°(Na₂S₂O₃) - 349.5
-133.22 = 22.9 - S°(Na₂S₂O₃) - Solve for S°(Na₂S₂O₃):
S°(Na₂S₂O₃) = 22.9 - (-133.22) = +156.12 J K⁻¹ mol⁻¹
To 3 significant figures: +156 J K⁻¹ mol⁻¹
✅ Mark Breakdown (4 Marks)
- Mark 1: Stating or using T = 298 K .
- Mark 2: Correctly rearranging Gibbs to find ΔS = -133 J K⁻¹ mol⁻¹ (or -0.133 kJ K⁻¹ mol⁻¹ ).
- Mark 3: Correct use of standard entropy expression involving stoichiometry: 372.4 - [S° + (5 × 69.9)] .
- Mark 4: Final answer of +156 (must be 3 significant figures).
❌ Major Calculation Pitfall
Unit Mismatch: ΔH° and ΔG° are given in kJ mol⁻¹, whereas S° values are in J K⁻¹ mol⁻¹.
Students who forget to multiply ΔS° by 1000 before combining with S° data lose the final two marks.
Structure & Bonding of Sulfur Oxyanions
Shapes of thiosulfate (S₂O₃²⁻) and the tetrathionate ion dimer (S₄O₆²⁻)
Part (c)(i): O—S—S Bond Angle and Shape (1 Mark)
✅ Correct Answer
Bond angle: 109.5° (accept 109°–110°)
Name of shape: Tetrahedral
💡 Explanation
The central sulfur atom has 4 regions of electron density (bonding regions to 3 oxygen atoms and 1 sulfur atom) and 0 lone pairs. By VSEPR theory, 4 electron pairs repel equally to adopt a tetrahedral arrangement with ideal bond angles of 109.5°.
Part (c)(ii): Displayed Formula of Tetrathionate Ion, S₄O₆²⁻ (1 Mark)
✅ How to Draw the Structure
The question states that S₄O₆²⁻ is a "dimer of S₂O₃²⁻". Two thiosulfate units link via a central disulfide bond:
(Examiners allow resonance/delocalised forms, cyclic isomers, or unlabelled formal charges, but there must be two sulfur atoms bridging between two 6-bonded terminal sulfurs).
❌ Examiner Trap
Do NOT link through oxygen: Structures featuring —S—O—O—S— bridges or putting all 4 sulfurs in a terminal row with central oxygens are heavily penalised unless the connectivity maintains the correct dimer linkage.
Topics
Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Practical Activity Groups · 1.1 Practical skills assessed in a written examination · 2.2 Electrons, bonding and structure · 3.2 Physical chemistry · 5.2 Energy · PAG 3: Enthalpy determination
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.