OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 16
17 marks · Hard difficulty · Structured Questions
Calculate enthalpy changes, Gibbs free energy, activation energy on profiles, and lattice enthalpies for hydrogen peroxide decomposition and manganese oxide formation.
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Question text
16 This question is about energy changes.
Hydrogen peroxide decomposes as shown in Reaction 16.1.
H O (l) H O(l) + 1O (g) Reaction 16.1
22 2 2 2
(a) The table shows enthalpy changes of formation and entropies.
∆∆H ө / kJ mol–1 Sө / J K–1 mol–1
f
H2O2(l) –188 110
H2O(l) –286 70.0
O2(g) 0 205
(i) Calculate the free-energy change, ∆G, in kJ mol–1, of Reaction 16.1 at 25 °C.
Give your answer to 3 significant figures.
∆G = … kJ mol–1 [4]
(ii) The decomposition of hydrogen peroxide shown in Reaction 16.1 is feasible.
Suggest why Reaction 16.1 does not take place at 25°C despite being feasible.
… [1]
(b) The rate of decomposition of hydrogen peroxide shown in Reaction 16.1 can be increased by
adding a small amount of powdered manganese(IV) oxide, MnO2.
The MnO2 acts as a catalyst.
(i) Complete the enthalpy profile diagram for Reaction 16.1 using formulae for the reactants and
products.
• Use Ea to label the activation energy without MnO2.
• Use Ec to label the activation energy with MnO2.
• Use ∆H to label the enthalpy change of reaction.
enthalpy
progress of reaction
[3]
(ii) Explain why MnO2 is described as a heterogeneous catalyst for this reaction.
… [1]
(iii) Mn3O4 is a compound in which Mn has two different oxidation states. The two oxidation states
are different from the Mn in MnO2.
Suggest the two oxidation states of manganese in Mn3O4.
… [1]
(c) Manganese(II) oxide, MnO, has a giant ionic lattice structure.
The table shows the enthalpy changes that are needed to determine the lattice enthalpy of MnO.
enthalpy change /
kJ mol–1
atomisation of manganese +281
atomisation of oxygen +249
first ionisation energy of manganese +717
second ionisation energy of manganese +1509
first electron affinity of oxygen –141
second electron affinity of oxygen +798
formation of manganese(II) oxide −385
(i) Define the term lattice enthalpy.
… [2]
(ii) The diagram shows an incomplete Born-Haber cycle that can be used to determine the lattice
enthalpy of MnO.
Mn2+(g) + O(g) + 2e– Mn2+(g) + O2–(g)
Mn2+(g) + 1 O (g) + 2e–
Mn+(g) + 1O (g) + e–
Mn(s) + 1 O (g)
Complete the diagram by adding the species present on the dotted lines, include state symbols.
[3]
(iii) Calculate the lattice enthalpy of MnO.
lattice enthalpy = … kJ mol–1 [2]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
16 (a) (i) FIRST CHECK ANSWER ON ANSWER LINE 4 ALLOW ECF throughout
If answer = –117 kJ mol–1, award 4 marks.
-----------------------------------------------------------------------------
∆H = –286– (–188)
= –98 kJ mol–1
∆S = 70 + ½(205) – 110 = 62.5 (J K–1 mol–1)
or 0.0625 (kJ K–1 mol–1)
∆G = ∆H – T∆S
= –98 – (298 0.0625) ALLOW –98000 – (298 62.5)
∆G = –117 kJ mol–1 (3SF)
Common Errors for ∆G
3 marks
–18700 ( S not converted to kJ)
–493 ( H = –286 + (–188) = –474)
–147 ( S = 165: not halving 205)
– 99.6 (T not converted to K)
–18.7 ( H not converted J but S J K–1 mol–1 )
(+)79.4 (–188 – (–286) = +98)
2 marks
(+) 117 (incorrect signs for H and S)
Final Answer MUST BE 3 SF
Question Answer 10 Marks Guidance
(ii) (Rate of reaction) slow 1 ALLOW ∆G takes no account of rate of reaction
OR Activation energy high
ALLOW molecules do not have sufficient energy to
equal or exceed the activation energy.
IGNORE molecules do not have sufficient energy to
react.
DO NOT ALLOW there is not enough activation
energy
(b) (i) 3
Care enthalpy profile must match ΔH sign in16 a) i)
– check calculation
ALLOW endothermic profile as ECF from + ΔH
calculated in 16 a) i) for all three marks
H2O2 on LHS AND H2O + ½ O2 on RHS
AND
ΔH labelled with product line below
reactant line
State symbols not required
AND
Arrow downwards ΔH DO NOT ALLOW –ΔH
DO NOT ALLOW double headed arrow on ΔH
ALLOW ΔH arrow even with small gap at the top and
bottom,
i.e. line does not quite reach reactant or product line.
Ea and Ec
Ea correctly labelled ALLOW no arrowhead or arrowheads at both end of
Ea or Ec lines
Ea or Ec lines must reach maximum (or near to
maximum) on curve
ALLOW overlapping lines OR lines on side reaching
Ec correctly labelled with Ec < Ea maximum
For Ea, ALLOW AE OR AE OR Eact OR suitable
alternatives
ALLOW ECF marks for Ea and Ec for correctly
labelled endothermic diagram from a –ΔH value
(from16 a) i))
(ii) (MnO2) is in different phase/state (to the reactant / H2O2) 1 ASSUME ‘it’ is MnO2
OR ALLOW ‘species in the reaction’
catalyst is a solid AND reactant is liquid IGNORE references to products
(iii) Mn is +2 AND +3 1 + required
OR ALLOW 2+ and 3+
Mn is +1 AND +6 DO NOT ALLOW Mn2+ Mn3+
DO NOT ALLOW + 4 (this is the oxidation state in
MnO2)
(c) (i) (Enthalpy / heat energy change / released when) 1 mol of12 2 ALLOW 1 mol of (ionic)
(ionic lattice) compound/product/substance
IGNORE energy released/required
Is formed from its gaseous ions ALLOW M+(g) + X–(g) → MX(s)
DO NOT ALLOW one mole of gaseous ions
(ii) 3
Care: State symbols are required
(iii) FIRST CHECK ANSWER ON ANSWER LINE 2 Common errors for 1 mark
If answer = –3798 award 2 marks -4080 (use of -141)
----------------------------------------------------------------- -3674 (use of +249/2 and correctly rounded)
H lattice -3673.5 (use of +249/2)
= – 281 – 249 – 717 – 1509 – (–141) – 798 + (–385) ✓ -3236 (use of +281)
-3300 (use of +249)
H lattice = –3798 (kJ mol–1) ✓ -3028 (use of -385)
-2364 (use of +717)
-2202 (use of +798)
-780 (use of +1509)
+3798 (wrong sign on answer)
For other answers, check for a single transcription
error or calculation error which could merit 1 mark
How to answer it
Energy Changes, Catalysts & Born-Haber Cycles
What this question tests
This comprehensive question assesses your mastery of thermodynamics and kinetics. Key skills include calculating Gibbs free-energy change (ΔG) using enthalpy of formation and entropy data, understanding reaction feasibility versus rate, sketching detailed enthalpy profile diagrams with catalytic pathways, defining heterogeneous catalysis and oxidation states, recalling exact definitions of lattice enthalpy, and accurately constructing and calculating steps within Born-Haber cycles.
Free-Energy Change and Feasibility
Part (i): Calculating ΔG
✅ Correct Answer
ΔG = -117 kJ mol⁻¹ (Must be to 3 significant figures)
📐 Step-by-Step Calculation
- Calculate ΔH:
ΣΔH_f(products) - ΣΔH_f(reactants)
= [-286 + 0] - [-188] = -98 kJ mol⁻¹ - Calculate ΔS:
ΣS(products) - ΣS(reactants)
= [70.0 + (0.5 × 205)] - [110]
= [70.0 + 102.5] - 110 = 62.5 J K⁻¹ mol⁻¹ (or 0.0625 kJ K⁻¹ mol⁻¹) - Apply Gibbs Equation (ΔG = ΔH - TΔS):
T = 25 + 273 = 298 K
ΔG = -98 - (298 × 0.0625)
= -98 - 18.625 = -116.625 kJ mol⁻¹ - Round to 3 sig figs: -117 kJ mol⁻¹
❌ Common Errors
- Forgetting to convert entropy units from J to kJ (leads to -18700 or -493).
- Failing to convert temperature to Kelvin (using T = 25 instead of 298).
- Omitting the coefficient 0.5 for O₂ when calculating ΔS.
- Failing to round the final answer strictly to 3 significant figures.
💡 Key Knowledge: Feasibility
A reaction is thermodynamically feasible when ΔG ≤ 0. Even though ΔG is negative here, the decomposition does not happen rapidly at room temperature.
✅ Part (ii) Answer
The reaction has a high activation energy / slow rate of reaction. (Note: Thermodynamics tells us *if* a reaction can happen, but kinetics tells us *how fast* it goes).
Enthalpy Profiles and Catalysis
💡 Part (i): Enthalpy Profile Diagram
- Reactants & Products: Write H₂O₂ on the LHS reactant line and H₂O + 0.5O₂ on the RHS product line.
- ΔH Label: Product line must sit vertically lower than the reactant line with an arrow pointing downwards labelled ΔH.
- Activation Energies: Curve must feature a tall hump labelled E_a (without catalyst) and a lower hump passing underneath labelled E_c (with MnO₂ catalyst). Both must peak at or near the top of the curves.
🧠 Part (ii) & (iii): Catalyst & Oxidation States
Part (ii): MnO₂ is a heterogeneous catalyst because it is in a different phase/state (solid) compared to the reactant (liquid H₂O₂).
Part (iii): In Mn₃O₄, manganese exists in two different oxidation states: +2 and +3 (or +1 and +6).
Lattice Enthalpy and Born-Haber Cycles
💡 Part (i): Definition of Lattice Enthalpy
The enthalpy change that accompanies the formation of 1 mole of an ionic compound from its gaseous ions under standard conditions.
Key phrasing required: "1 mole", "ionic compound / lattice", and "gaseous ions".
🧠 Part (ii): Completing the Born-Haber Cycle
Working from bottom to top on the missing dotted lines:
- 1st dotted line (bottom): Mn⁺(g) + ½O₂(g) + e⁻
- 2nd dotted line (middle): Mn²⁺(g) + O⁻(g) + e⁻
- 3rd dotted line (top right): Mn²⁺(g) + O²⁻(g)
📐 Part (iii): Calculating Lattice Enthalpy of MnO
Using Hess's Law / Born-Haber energy cycle:
- Set up equation:
ΔH_lattice = ΔH_atom(Mn) + ΔH_atom(O) + IE₁ + IE₂ + EA₁ + EA₂ - ΔH_form - Substitute values:
= (+281) + (+249) + (+717) + (+1509) + (-141) + (+798) - (-385) - Evaluate:
ΔH_lattice = -3798 kJ mol⁻¹
❌ Calculation Traps
- Forgetting to subtract formation enthalpy (misinterpreting sign changes).
- Using incorrect electron affinity values (e.g., using second EA values incorrectly or missing signs).
- Transcription errors copying numbers directly from the table.
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry · 5.2 Energy · 5.3 Transition elements
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.