OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 18
11 marks · Hard difficulty · Calculations
Calculate enthalpy changes of combustion using calorimetry and Hess's law, and determine entropy change, minimum temperature of feasibility, and enthalpy change from a Gibbs free energy graph.
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Question text
18 This question is about enthalpy changes of reactions involving hydrocarbons.
(a) A student determines the enthalpy change of combustion, ∆cH, of heptane, C7H16, using the
method outlined below.
• Add 150 g of water to a beaker and measure its temperature.
• Weigh a spirit burner containing heptane and use it to heat the water.
• Extinguish the flame and record the maximum temperature reached by the water.
• Reweigh the spirit burner.
The temperature of the water increased by 10.5 °C.
The spirit burner decreased in mass by 0.133 g.
Use the student’s results to determine the enthalpy change of combustion of heptane,
∆ H (C H ), in kJ mol–1.
c 7 16
∆ H (C H ) = … kJ mol–1 [3]
c 7 16
(b) Nonane, C9H20, can be broken down by heat to form pentane, C5H12, and ethene, C2H4.
C H (g) C H (g) + 2C H (g) ∆H = +186 kJ mol–1 Reaction 1
9 20 5 12 2 4
The enthalpy changes of combustion of C9H20(g) and C2H4(g) are shown in the table below.
Hydrocarbon ∆ H / kJ mol–1
c
C9H20(g) –6171
C2H4(g) –1411
Use ∆H in Reaction 1 and the enthalpy changes of combustion in the table to determine the
enthalpy change of combustion of C5H12(g).
∆ H (C H (g)) = … kJ mol–1 [2]
c 5 12
(c) Ethene can be produced from ethanol, as shown in Reaction 2 below.
C2H5OH(g) C2H4(g) + H2O(g) Reaction 2
(i) Predict the sign of the entropy change, ∆S, for Reaction 2.
Explain your reasoning.
… [1]
(ii) Reaction 2 is repeated:
C2H5OH(g) C2H4(g) + H2O(g) Reaction 2
The Gibbs equation is shown below.
∆G = ∆H – T∆S
The enthalpy change, ∆H, and the entropy change, ∆S, can be assumed to be constant
at different temperatures.
Fig. 18.1 shows values of the free energy change, ∆G, in kJ mol–1, at different
temperatures, T, in K, for Reaction 2.
Fig. 18.1
∆G
–1 0
/kJ mol
–10
–20
–30
–40
–50
–60
0 100 200 300 400 500 600 700 800 900
T/K
Use the graph in Fig. 18.1 to answer the following:
• Draw the best-fit line on the graph in Fig. 18.1.
• Determine ∆S, in J K–1 mol–1, for Reaction 2.
• Determine the minimum temperature, T, at which the reaction is feasible.
• Determine ∆H for Reaction 2.
∆S = … J K–1 mol–1
minimum T = … K
∆H = … kJ mol–1
[5]
Mark scheme
Show the mark scheme
Answer AO
Question Marks Guidance
element
18 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 3 ALLOW 3 SF up to the calculated value
If answer = –4950 award 3 marks Ignore RE after 3SF
-----------------------------------------------
• q = mc T
= 150 x 4.18 x 10.5
= 6583.5 (J) OR 6.5835 (kJ) ✓ AO2.4 IGNORE sign
• n(C7H16)
0.133 –3
= = 1.33 × 10 ✓ AO2.8 ALLOW ECF from incorrect q and/or n
• cH = q n Common errors for 2 marks
6.5835 +4950 kJ mol–1 (wrong sign)
=
1.33x10⁻³ -5077 (use of 0.0013 and 6.6 2SF)
= –4950 kJ mol–1 -5064 (use of 0.0013 2SF)
– sign required ✓ -4962 (use of 6.6kJ use of 2SF)
(b) FIRST CHECK THE ANSWER ON ANSWER LINE 2
If answer = –3535 award 2 marks
---------------------------------------------------
186 = H(C H ) – H(C H ) – 2 H(C H ) AO2.2 IGNORE any incorrect combustion products on bottom
c 9 20 c 5 12 c 2 4
line.
OR From Hess cycle with all numerical values used 2
and correct multiples used/labelled Common errors for 1 mark
+3535 (wrong sign for final answer)
+8807 (use of +6171)
-9179 (use of +1411)
-4946 (use of 1 x -1411)
-3163 (use of +186)
✓
For other answers, check for a single transcription
error or calculation error using all values which could
cH(C5H12) = (–6171) – 2(–1411) -186 merit 1 mark
= –3535 kJ mol–1 ✓
(c) (i) (∆S) is positive 1 AO2.5 ALLOW reaction produces more (gaseous)
AND 14 1 molecules /moles of products than reactants
more molecules / moles of (gaseous) product
/produced✓ IGNORE explanations based on ∆G
(ii) Best fit line drawn ✓ 5 Allow ECF throughout
Place tick for line of best fit on the graph
AO3.1
2 Allow lines that will extrapolate to y axis at 43-47
and x axis at 820-840
AO3.2 DO NOT ALLOW outside ranges
ALLOW rounding to 2SF
(–)∆S = correct gradient from graph e.g. (–)∆S = 105 / 830 = (–)0.127
OR (–)0.127 (kJ K–1 mol–1) ✓ ALLOW (–)0.122 to (–)0.131
∆S = gradient –1000 = (+)127 (J K–1 mol–1) ✓ ALLOW 122 to 131
This mark subsumes gradient mark
Minimum T (∆G = 0) = 370 (K) ✓ ALLOW 340 to 370
ALLOW 67 to 97 AND oC
DO NOT ALLOW -ve T in K
H (y-intercept) = (+)46 (kJ mol–1) ✓ ALLOW 43 to 47
Candidates can receive full credit for calculating
∆S, T and/or H from previously determined
values
e.g
∆H 46
T = ∆S = 0.127 = 362K
How to answer it
Enthalpy Changes and Thermodynamics Study Guide
What this question tests
This multi-part question assesses core physical chemistry concepts including calorimetry calculations, Hess's Law cycles using enthalpy of combustion data, entropy change predictions based on moles of gas, and graphical interpretation of the Gibbs free energy equation (delta-G = delta-H - T delta-S).
Enthalpy of Combustion Calorimetry
✅ Correct Answer
delta_c H (C₇H₁₆) = -4950 kJ mol⁻¹ (3 marks)
📐 Step-by-Step Calculation
- Calculate heat energy released (q):
q = m × c × delta_T
q = 150 × 4.18 × 10.5 = 6583.5 J (or 6.5835 kJ ) - Calculate moles of heptane (n):
Molar mass of C₇H₁₆ = (7 × 12.0) + (16 × 1.0) = 100.0 g mol⁻¹
n = mass / Mr = 0.133 / 100 = 1.33 × 10⁻³ mol - Calculate enthalpy change per mole:
delta_c H = q / n = 6.5835 / (0.133 / 100) = -4950 kJ mol⁻¹
❌ Common Errors
- Omitting the negative sign (must explicitly show -4950 since combustion is exothermic).
- Using the mass of the heptane (0.133 g) instead of the mass of water (150 g) in the q = mc delta_T calculation.
- Rounding intermediate values too early, leading to off-target final numbers.
Hess's Law and Enthalpy of Combustion Cycles
✅ Correct Answer
delta_c H (C₅H₁₂) = -3535 kJ mol⁻¹ (2 marks)
💡 Key Knowledge (Hess's Law)
For combustion data cycles, arrows point downwards from reactants and products to their common combustion products (CO₂ and H₂O).
Equation: delta_H = sum(delta_c H(reactants)) - sum(delta_c H(products))
🧠 Exam Technique
Set up a clear algebraic expression based on the reaction given:
186 = delta_c H(C₉H₂₀) - [delta_c H(C₅H₁₂) + 2 × delta_c H(C₂H₄)]
Rearrange to solve for the unknown:
delta_c H(C₅H₁₂) = (-6171) - 2(-1411) - 186 = -3535 kJ mol⁻¹
Predicting Entropy Changes
✅ Correct Answer
Positive (AND explanation: more molecules / moles of gaseous products produced)
💡 Key Knowledge
Look at the stoichiometry of gaseous species in C₂H₅OH(g) → C₂H₄(g) + H₂O(g) .
1 mole of gas forms 2 moles of gas. An increase in gaseous moles increases disorder, resulting in a positive delta_S.
Gibbs Free Energy Graphical Analysis
✅ Correct Answer
- delta_S: 127 J K⁻¹ mol⁻¹ (Acceptable range: 122 to 131)
- minimum T: 370 K (Acceptable range: 340 to 370 K)
- delta_H: +46 kJ mol⁻¹ (Acceptable range: 43 to 47)
🧠 Exam Technique & Calculations
- Best-fit line: Draw a straight line of best fit through the crosses provided on Fig. 18.1.
- Finding delta_S (Gradient): Calculate gradient of the line ( delta y / delta x ). The equation is delta_G = -delta_S(T) + delta_H (matching y = mx + c). Gradient = -delta_S . Multiply the negative gradient value by 1000 to convert from kJ to J K⁻¹ mol⁻¹.
- Finding minimum T (Feasibility): A reaction becomes feasible when delta_G ≤ 0 . Find the temperature where your line crosses the x-axis ( delta_G = 0 ).
- Finding delta_H (Y-intercept): Read the value where the line intercepts the y-axis at T = 0.
❌ Common Errors
- Forgetting to multiply the gradient by 1000, missing the unit conversion from kJ to J for delta_S.
- Not recognizing that the gradient of a delta_G vs T graph represents -delta_S .
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Practical Activity Groups · 3.2 Physical chemistry · 5.2 Energy · PAG 3: Enthalpy determination
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.