Edexcel A-Level Chemistry Paper 1, June 2017: Question 5
9 marks · Medium difficulty · Calculations
Calculate enthalpy changes using mean bond enthalpies, determine entropy changes of systems, and calculate free energy changes and reaction feasibility.
Practise this questionQuestion
Question text
5 This question is about enthalpy changes and entropy changes.
(a) Which is the equation for the standard enthalpy change of formation, ¨fH ,
of aluminium oxide?
(1)
A 4Al(s) + 3 O2(g) r 2Al2O3(s)
B 4Al(s) + 6 O(g) r 2Al2O3(s)
C 2Al(s) + 1½ O2(g) r Al2O3(s)
D 2Al(s) + 3 O(g) r Al2O3(s)
(b) Propan-1-ol is dehydrated to form propene.
H H
H H H
C C −1
H C C C O H + H2O ¨rH = +42 kJ mol
H C H
H H H H
H
The relevant mean bond enthalpies are given in the table.
Mean bond enthalpy
Bond –1
/ kJ mol
C C 347
C C 612
C H 413
O H 464
Calculate the C O mean bond enthalpy, using the mean bond enthalpies given
in the table and the enthalpy change of reaction.
(3)
(c) Which reaction has a negative value for ¨Ssystem?
(1)
A 2Cu(s) + O2(g) r 2CuO(s)
B 2H2O2(l) r 2H2O(l) + O2(g)
C MgCO3(s) + H2SO4(aq) r MgSO4(aq) + H2O(l) + CO2(g)
D Zn(s) + 2HCl(aq) r ZnCl2(aq) + H2(g)
(d) What is the expression for ¨Stotal?
(1)
10 ¨H
A ¨Ssurroundings + –
T*P48058A01028*
¨H
B ¨Ssurroundings – –
T
¨H
C ¨Ssystem + –
T
¨H
D ¨Ssystem – –
T
(e) Calcium carbonate decomposes on heating.
CaCO3(s) r CaO(s) + CO2(g)
¨ H = +178 kJ mol–1
r
¨S = +165 J mol–1 K–1
system
Show, by calculating the value for the free energy change, ¨G, that this
decomposition is not feasible at 298 K, and then calculate the minimum
temperature to which calcium carbonate must be heated to make it decompose.
(3)
*P48058A01128*
(Total for Question 5 = 9 marks)
Mark scheme
Show the mark scheme
Question
Answer Mark
Number
5(a) The only correct answer is C (1)
A is not correct because standard enthalpy of formation is for making 1 moles of a compound
B is not correct because standard enthalpy of formation is for making 1 moles of a compound
D is not correct because oxygen must be O2
Question
Answer Additional Guidance Mark
Number
5(b) Example of calculation (3)
calculation of energy needed to break bonds Energy to break bonds: (C―C) + (C―H) + (C―O)
(1) = 347 + 413 + (C―O)
= (C―O) + 760 (kJ)
calculation of energy released when bonds Energy released in forming bonds: (C=C) + (O―H)
are made (1) = 612 + 464 = (−)1076 (kJ)
calculation of mean bond enthalpy of C―O (C―O) + 760 – 1076 = 42
(1) (C―O) = (+)358 (kJ mol―1)
TE on M1 and M2
If all bonds broken:
Energy to break bonds = (C―O) + 4049 (kJ)
Energy released in forming bonds = (−)4365 (kJ)
Ignore units
Correct answer with no working scores (3)
Allow correct working in M1 and M2 if answers not
evaluated
Question
Answer Mark
Number
5(c) The only correct answer is A (1)
B is not correct because all increase in entropy as disorder increases when gases are formed
C is not correct because all increase in entropy as disorder increases when gases are formed
D is not correct because all increase in entropy as disorder increases when gases are formed
Question
Answer Mark
Number
5(d) The only correct answer is D (1)
A is not correct because ∆Ssurroundings is incorrect
B is not correct because ∆Ssurroundings is incorrect
C is not correct because sign of ∆H/T is incorrect
Question
Answer Additional Guidance Mark
Number
5(e) Penalise incorrect units in M1 or M3 once only (3)
Working is not required for the calculations
Example of calculation
calculation of ∆G (1) ∆G = 178 ― (298 x 165)
1000
= (+)128.83 / 129 (kJ mol―1)
or
∆G = 178000 ― (298 x 165)
= (+)128830 / 129000 (J mol―1)
∆G is positive / >0 so reaction is not Stand alone mark
feasible (1) Allow ∆G must be negative for a reaction to be feasible
Ignore ‘so reaction is not feasible’ without a reason
No TE on a calculated negative value
Example of calculation
calculation of T (1) ∆G = 0, so ∆H = T∆S(sys) or T = ∆H/∆S(sys)
or
∆Stotal = ∆Ssys − ∆H/T = 0, so T = ∆H/∆S(sys)
T = 178/0.165 = 1078.8 / 1079 / 1080 (K)
or
T = 178000/165 = 1078.8 / 1079 / 1080 (K)
or
T = 806 (oC)
Ignore SF except 1 SF
(Total for Question 5 = 9 marks)
How to answer it
Enthalpy Changes and Entropy Changes
This question assesses core energetics and thermodynamics concepts from Edexcel A-Level Chemistry. It covers definitions of standard enthalpy of formation, mean bond enthalpy calculations, predicting entropy changes of the system (ΔS_system), relating total entropy change to system entropy and enthalpy, and applying the Gibbs free energy equation (ΔG = ΔH - TΔS) to determine reaction feasibility and threshold temperatures.
Part (a) — Standard Enthalpy of Formation
✅ Correct Answer
C: 2Al(s) + 1.5O₂(g) → Al₂O₃(s)
💡 Key Knowledge
- The standard enthalpy of formation ( Δ_fH ) is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
- Options A and B form 2 moles of Al₂O₃ , violating the "1 mole of product" rule.
- Option D uses atomic oxygen ( O ) instead of the standard state element form ( O₂ ).
Part (b) — Mean Bond Enthalpy Calculation
✅ Correct Answer
+358 kJ mol⁻¹
📐 Step-by-Step Calculation
- Energy to break bonds (Reactants):
Bonds broken: 1×(C-C), 7×(C-H), 1×(O-H), 1×(C-O)
= 347 + (7 × 413) + 464 + (C-O) = 347 + 2891 + 464 + (C-O) = 3702 + (C-O)
*(Alternative accepted grouping: breaking all bonds gives 4049 kJ)* - Energy released making bonds (Products):
Bonds formed: 1×(C=C), 6×(C-H), 2×(O-H)
= 612 + (6 × 413) + (2 × 464) = 612 + 2478 + 928 = -4018 kJ
*(When including water O-H separately: 612 + 2478 + 464 + 464 = -4018)* - Set up equation:
ΔH = Energy in (bonds broken) - Energy out (bonds formed)
+42 = [3702 + (C-O)] - 4018
+42 = (C-O) - 316
(C-O) = 42 + 316 = +358 kJ mol⁻¹
❌ Common Errors
Students often miscount the number of C-H bonds in propan-1-ol (there are 7 in total: 3 on C1, 2 on C2, 2 on C3). Mixing up signs for bonds broken (+) and bonds formed (-) is also a frequent source of lost marks.
Part (c) — Predicting Entropy Change of System
✅ Correct Answer
A: 2Cu(s) + O₂(g) → 2CuO(s)
💡 Key Knowledge
- Entropy measures disorder. Gases have much higher entropy than solids or liquids.
- In reaction A, 1 mole of gas ( O₂ ) is consumed to form a solid product, decreasing total gaseous moles from 1 to 0. This results in a negative ΔS_system .
- Options B, C, and D all produce gases or increase the number of moles of particles, leading to a positive ΔS_system .
Part (d) — Total Entropy Expression
✅ Correct Answer
D: ΔS_system - (ΔH / T)
🧠 Exam Technique
Recall the master equation connecting total entropy to enthalpy:
ΔS_total = ΔS_system + ΔS_surroundings
Since ΔS_surroundings = -ΔH / T , substituting this in gives:
ΔS_total = ΔS_system - (ΔH / T)
Part (e) — Gibbs Free Energy and Feasibility Temperature
✅ Correct Answer
ΔG = +129 kJ mol⁻¹ (Not feasible at 298 K)
Minimum temperature: 1079 K (or 806 °C )
📐 Step-by-Step Calculation
- Calculate ΔG at 298 K:
Convert ΔS from J mol⁻¹ K⁻¹ to kJ mol⁻¹ K⁻¹ by dividing by 1000 ( 165 / 1000 = 0.165 ).
ΔG = ΔH - TΔS
ΔG = 178 - (298 × 0.165)
ΔG = 178 - 49.17 = +128.83 kJ mol⁻¹ - Feasibility statement:
Since ΔG is positive ( > 0 ), the reaction is not feasible at 298 K. - Calculate minimum temperature for decomposition:
For a reaction to become feasible, ΔG ≤ 0 . Set ΔG = 0 :
0 = ΔH - TΔS_system ⇒ ΔH = TΔS_system
T = ΔH / ΔS_system
Using Joules consistently: T = 178000 / 165
T = 1078.78... = 1079 K (or 806 °C )
❌ Common Errors
Forgetting to convert ΔS from Joules to kiloJoules (or vice-versa) before combining with ΔH is the #1 reason students lose calculation marks here. Watch your units carefully!
Topics
Physical Chemistry · Topic 8: Energetics I · Topic 13: Energetics II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.