OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2017: Question 18
12 marks · Medium difficulty · Structured Questions
Analyze free energy changes, Gibbs equation graphical interpretation, heterogeneous equilibria, Kp expressions, feasibility temperature calculations, and enthalpy of formation calculations.
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Question text
18 This question is about free energy changes, ∆G, enthalpy changes, ∆H, and temperature, T.
(a) The Gibbs’ equation is shown below.
∆G = ∆H – T∆S
A chemist investigates a reaction to determine how ∆G varies with T.
The results are shown in Fig. 18.1.
+
ΔG Q
/ kJ mol–1 0,0
T/K
–
P
Fig. 18.1
What is significant about the gradient of the line and the values P and Q shown in Fig. 18.1?
Explain your reasoning.
… [4]
(b) Iron can be extracted from its ore Fe3O4 using carbon.
Several equilibria are involved including equilibrium 18.1, shown below.
equilibrium 18.1 Fe O (s) + 4C(s) 3Fe(s) + 4CO(g) ∆H = +676.4 kJ mol–1
∆S = +703.1 J K–1 mol–1
(i) Why is equilibrium 18.1 a heterogeneous equilibrium?
… [1]
(ii) Write the expression for Kp for equilibrium 18.1.
[1]
(iii) The forward reaction in equilibrium 18.1 is only feasible at high temperatures.
• Show that the forward reaction is not feasible at 25 °C.
• Calculate the minimum temperature, in K, for the forward reaction to be feasible.
minimum temperature = … K [3]
(iv) Another equilibrium involved in the extraction of iron from Fe3O4 is shown below.
Fe O (s) + 4CO(g) 3Fe(s) + 4CO (g) ∆H = –13.5 kJ mol–1
34 2
Enthalpy changes of formation, ∆fH, for Fe3O4(s) and CO2(g) are shown in the table.
Compound ∆ H / kJ mol–1
f
Fe3O4(s) –1118.5
CO2(g) –393.5
Calculate the enthalpy change of formation, ∆fH, for CO(g).
∆ H, for CO(g) = … kJ mol–1 [3]
f
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
18 (a) ∆G = ∆H – T∆S 4 Could be:
linked to y = mx + c (somewhere) ∆G = –∆S T + ∆H
gradient = –∆S – sign required
ALLOW ∆S = –gradient
P: ∆H / enthalpy change
Q: (temperature) for reaction to be feasible/unfeasible ALLOW ‘point of feasibility’
OR For Feasibility:
(temperature) at which feasibility changes ALLOW can take place/happen OR is spontaneous
IGNORE ‘minimum/maximum temperature’
(b) (i) (Species have) different states/phases 1
(ii) (K =) p(CO(g))4 1 Allow species without state symbols and without
p
brackets, e.g. p 4 , ppCO4, PCO4, p(CO4) etc.
CO
DO NOT ALLOW square brackets
(iii) ∆G at 25 C 3
∆G = ∆H – T∆S = 676.4 – (298 0.7031) IGNORE units
= (+) 467 (kJ mol–1) OR (+) 466876 (J mol–1) ALLOW (+) 467 up to calculator value of 466.8762
correctly rounded
Non-feasibility statement
Non-feasible when ∆G > 0 ECF for any positive value determined in M1
OR ∆G > 0 OR ∆H > T∆S
Minimum temperature
∆H 676.4
minimum temperature = = ALLOW 962 up to calculator value of 962.0253165
∆S 0.7031
correctly rounded
= 962(.0) K
(iv) FIRST, CHECK THE ANSWER ON ANSWER LINE 11 3 For answer,
IF answer = –110.5, Award 3 marks. ALLOW –111 (kJ mol–1)
-------------------------------------------------------------------- --------------------------------------------------------------------
Correct expression NOTE: IF any values are omitted, DO NOT
–13.5 = (4 –393.5) – (–1118.5 + 4 ∆fH(CO)) AWARD any marks. e.g. –393.5 OR –13.5 may be
missing
Correct subtraction using ∆H and ∆fH(Fe3O4)
4 ∆fH(CO) = (4 –393.5) – (–1118.5) + 13.5 --------------------------------------------------------------------
= –442(.0) (kJ mol–1) Common errors
Calculation of ∆fH(CO) formation (+)110.5 wrong/omitted sign 2 marks
442 –1 (+)184.625 / 184.63 / 184.6 / 185 2 marks
∆fH(CO) = – 4 = –110.5 (kJ mol ) No 4CO2
(+)738.5 / 739 No 4CO2 and no CO/4 1 mark
–117.25 / –117.3 / –117 Wrong cycle 2 marks
–469 Wrong cycle, no CO/4 1 mark
(+)177.875 / 177.88 / 177.9 / 178 1 mark
Wrong cycle, no 4CO2
–360.5 Used 118.5 2 marks
Any other number: CHECK for ECF from 1st
marking point for expressions using ALL values
with ONE error only e.g. one transcription error:,
e.g.395.3 for 393.5
Total 12
How to answer it
Free Energy, Equilibria and Enthalpy Changes
What this question tests
This question assesses your understanding of chemical energetics and equilibria. Key skills tested include interpreting the graphical form of the Gibbs equation (y = mx + c), defining heterogeneous equilibria, writing Kp expressions, applying Gibbs free energy calculations to determine reaction feasibility and threshold temperatures, and using enthalpy cycles/data to calculate enthalpy of formation.
Interpreting Gibbs Free Energy Graphs
✅ Correct Answer
- Link to linear equation: ΔG = ΔH − TΔS compared to y = mx + c .
- Gradient = −ΔS (negative entropy change).
- Intercept P = ΔH (enthalpy change).
- Point Q = Temperature where feasibility changes / point of feasibility ( ΔG = 0 ).
💡 Key Knowledge
- The Gibbs equation acts like a straight-line equation where y = ΔG , x = T , m = −ΔS , and c = ΔH .
- A reaction is feasible when ΔG ≤ 0 .
🧠 Exam Technique
- Always explicitly state the matching between ΔG = ΔH − TΔS and y = mx + c to secure the first mark.
- Do not forget the negative sign on the gradient: gradient = −ΔS .
❌ Common Errors
- Stating that the gradient is just ΔS without the minus sign.
- Vague descriptions of point Q, such as just calling it "zero" rather than linking it to the temperature where the reaction becomes feasible.
Heterogeneous Equilibria and Kp Expressions
✅ Correct Answers
- (b)(i): The species are in different states / phases (solid and gas).
- (b)(ii): Kp = p(CO)⁴ (or equivalent notation without square brackets).
💡 Key Knowledge
- Heterogeneous systems contain species in more than one physical state.
- In expression writing for Kp , solids and liquids are omitted because their concentrations/activities remain constant, leaving only gaseous species.
❌ Common Errors
- Using square brackets [ ] instead of round pressure terms or partial pressure notation for Kp expressions.
- Including Fe₃O₄ , C , or Fe in the Kp expression.
Feasibility at 25°C and Minimum Temperature Calculations
📐 Step-by-Step Calculation
- Convert units: Ensure ΔS is in kJ K⁻¹ mol⁻¹ by dividing by 1000.
ΔS = +703.1 J K⁻¹ mol⁻¹ = +0.7031 kJ K⁻¹ mol⁻¹ . - Calculate ΔG at 25°C (298 K):
ΔG = ΔH − TΔS
ΔG = 676.4 − (298 × 0.7031)
ΔG = +467 kJ mol⁻¹ (or 466876 J mol⁻¹ ). - Non-feasibility statement: Since ΔG > 0 (or ΔH > TΔS ), the reaction is not feasible at 25°C.
- Calculate minimum temperature for feasibility ( ΔG = 0 ):
T = ΔH / ΔS
T = 676.4 / 0.7031 = 962 K (allow 962 to 962.03 K).
🧠 Exam Technique & Traps
- Unit Trap: Always watch out for ΔH given in kJ mol⁻¹ while ΔS is given in J K⁻¹ mol⁻¹ . You must convert one so their units match before subtracting!
- Temperature Conversion: 25°C must be converted to Kelvin by adding 273 (giving 298 K).
Enthalpy of Formation Calculation
📐 Step-by-Step Calculation
- Set up the enthalpy change equation using ΔfH values:
ΔrH = ΣΔfH(products) − ΣΔfH(reactants)
−13.5 = [3(ΔfH of Fe) + 4(ΔfH of CO₂)] − [ΔfH of Fe₃O₄ + 4(ΔfH of C)] - Apply standard states (elements = 0):
ΔfH(Fe) = 0 and ΔfH(C) = 0 .
−13.5 = [4 × (−393.5)] − [−1118.5 + 4 × ΔfH(CO)] - Rearrange and solve for ΔfH(CO):
−13.5 = −1574 − (−1118.5 + 4ΔfH(CO))
−13.5 = −1574 + 1118.5 − 4ΔfH(CO)
−13.5 = −455.5 − 4ΔfH(CO)
4ΔfH(CO) = −442.0
ΔfH(CO) = −442.0 / 4 = −110.5 kJ mol⁻¹
❌ Common Calculation Errors & Sign Traps
- Sign errors: Forgetting to account for subtracting a negative reactant enthalpy ( −(−1118.5) becomes +1118.5 ).
- Stoichiometry multipliers: Forgetting to multiply the enthalpy of formation of CO₂ and CO by their stoichiometric coefficients (4).
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 5.2 Energy
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.