Energetics Problem Solving

This final lesson brings together all the topics from the course into exam-style problem solving. The key to success in energetics questions is a systematic approach: identify what data you have, decide which method to use, set up the calculation clearly, and check your answer makes sense. This lesson works through multi-step problems covering every area of the syllabus, with detailed solutions and exam technique advice.

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Strategy for Multi-Step Hess's Law Problems

Some exam questions give you several thermochemical equations and ask you to calculate ΔH for a target reaction. The approach is:

1. Write the target equation clearly.
2. Examine each given equation — decide if it needs to be reversed, multiplied, or used as-is to build the target.
3. Reverse equations as needed (and change the sign of ΔH).
4. Multiply equations as needed (and multiply ΔH by the same factor).
5. Add the manipulated equations — intermediate species should cancel.
6. Add the ΔH values to get the answer.

Worked Example 1: Multi-Step Hess's Law

Calculate ΔH for: C₂H₂(g) + 2H₂(g) → C₂H₆(g)

Given:

• (1) C₂H₂(g) + 5/2 O₂(g) → 2CO₂(g) + H₂O(l), ΔH₁ = −1300 kJ mol⁻¹
• (2) C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH₂ = −1560 kJ mol⁻¹
• (3) H₂(g) + ½O₂(g) → H₂O(l), ΔH₃ = −286 kJ mol⁻¹

Solution:

Use (1) as-is: C₂H₂ on the left. ΔH = −1300 Reverse (2): C₂H₆ on the right. ΔH = +1560 Multiply (3) by 2: 2H₂ on the left. ΔH = 2 × (−286) = −572

Cancel CO₂, H₂O, and O₂ terms:

C₂H₂(g) + 2H₂(g) → C₂H₆(g)

ΔH = −1300 + 1560 + (−572) = −312 kJ mol⁻¹

Sign check: Hydrogenation adds hydrogen across a triple bond, forming more stable single bonds. This should be exothermic. ΔH = −312. Correct.

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Born-Haber Cycle Calculations with Missing Values

These problems give you all but one value and ask you to calculate the unknown. The method is always the same: write the Hess's law equation, substitute known values, and solve for the unknown.

Worked Example 2: Calculate the Electron Affinity of Bromine

Construct a Born-Haber cycle for KBr and calculate EA₁(Br).

Quantity	Value / kJ mol⁻¹
ΔH°f(KBr)	−394
ΔH°at(K)	+89
ΔH°at(Br)	+112
IE₁(K)	+419
ΔH°latt(KBr)	−679
EA₁(Br)	?

ΔH°f = ΔH°at(K) + ΔH°at(Br) + IE₁(K) + EA₁(Br) + ΔH°latt

−394 = (+89) + (+112) + (+419) + EA₁ + (−679)

−394 = +620 + (−679) + EA₁

−394 = −59 + EA₁

EA₁(Br) = −394 + 59 = −335 kJ mol⁻¹

The data book value is −342 kJ mol⁻¹, close to our answer.

Worked Example 3: Born-Haber Cycle for MgO (Two Ionisation Energies and Two Electron Affinities)

Quantity	Value / kJ mol⁻¹
ΔH°f(MgO)	−602
ΔH°at(Mg)	+148
ΔH°at(O)	+249
IE₁(Mg)	+738
IE₂(Mg)	+1451
EA₁(O)	−141
EA₂(O)	+798
ΔH°latt(MgO)	?

ΔH°f = ΔH°at(Mg) + ΔH°at(O) + IE₁ + IE₂ + EA₁ + EA₂ + ΔH°latt

−602 = (+148) + (+249) + (+738) + (+1451) + (−141) + (+798) + ΔH°latt

−602 = +3243 + ΔH°latt

ΔH°latt = −602 − 3243 = −3845 kJ mol⁻¹

The literature value is −3850 kJ mol⁻¹. The enormous magnitude reflects the 2+ and 2− charges and small ionic radii.

graph TD
    A["Mg(s) + ½O₂(g)"] -->|"ΔH°f = −602"| H["MgO(s)"]
    A -->|"+148"| B["Mg(g) + ½O₂(g)"]
    B -->|"+249"| C["Mg(g) + O(g)"]
    C -->|"+738"| D["Mg⁺(g) + O(g)"]
    D -->|"+1451"| E["Mg²⁺(g) + O(g)"]
    E -->|"−141"| F["Mg²⁺(g) + O⁻(g)"]
    F -->|"+798"| G["Mg²⁺(g) + O²⁻(g)"]
    G -->|"ΔH°latt = −3845"| H

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Combined Calorimetry and Hess's Law

Worked Example 4

A student burns 0.92 g of ethanol (Mᵣ = 46.0) and heats 200 g of water from 20.0 °C to 31.5 °C. Calculate the experimental ΔH°c and compare it with the value calculated from enthalpies of formation.

Calorimetry: q = mcΔT = 200 × 4.18 × 11.5 = 9614 J = 9.614 kJ Moles = 0.92 / 46.0 = 0.020 mol ΔH°c(experimental) = −9.614 / 0.020 = −481 kJ mol⁻¹

Hess's law (formation data): C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

ΔH°c = [2(−394) + 3(−286)] − [(−277) + 0]

ΔH°c = [−788 − 858] − [−277]

ΔH°c = −1646 + 277 = −1369 kJ mol⁻¹

The experimental value (−481) is much less exothermic than the accepted value (−1369) due to significant heat losses in the simple calorimetry setup.

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Entropy and Gibbs Free Energy Combined Problems

Worked Example 5: Full ΔG Calculation

For the reaction: 2SO₂(g) + O₂(g) → 2SO₃(g)

Given:

• ΔH = −198 kJ mol⁻¹
• S°[SO₂(g)] = 248.1 J K⁻¹ mol⁻¹
• S°[O₂(g)] = 205.0 J K⁻¹ mol⁻¹
• S°[SO₃(g)] = 256.6 J K⁻¹ mol⁻¹

Step 1: Calculate ΔS°.

ΔS° = [2(256.6)] − [2(248.1) + 205.0]

ΔS° = 513.2 − 701.2 = −188.0 J K⁻¹ mol⁻¹

Convert: ΔS° = −0.1880 kJ K⁻¹ mol⁻¹

Step 2: Calculate ΔG at 298 K.

ΔG = ΔH − TΔS = (−198) − (298)(−0.1880) = −198 + 56.0 = −142.0 kJ mol⁻¹

Feasible at 298 K.

Step 3: Find the temperature above which the reaction becomes infeasible.

T = ΔH / ΔS = −198 / −0.1880 = 1053 K (≈ 780 °C)

This is Case 3 (ΔH negative, ΔS negative): feasible at low temperatures, infeasible above 1053 K. The Contact process operates at 400–450 °C (673–723 K), well within the feasible range but high enough to give a reasonable rate with a V₂O₅ catalyst.

Worked Example 6: Finding ΔS from ΔG and ΔH

A reaction has ΔG = −100 kJ mol⁻¹ at 400 K and ΔH = −160 kJ mol⁻¹. Calculate ΔS.

ΔG = ΔH − TΔS

−100 = −160 − (400)ΔS

(400)ΔS = −160 + 100 = −60

ΔS = −60 / 400 = −0.150 kJ K⁻¹ mol⁻¹ = −150 J K⁻¹ mol⁻¹

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Bond Enthalpy Problem with Multiple Bond Types

Worked Example 7: Combustion of Propene

CH₂=CHCH₃(g) + 9/2 O₂(g) → 3CO₂(g) + 3H₂O(g)

Using bond enthalpies: C=C +614, C–C +347, C–H +413, O=O +498, C=O +805, O–H +463.

Bonds broken in propene and O₂:

• 1 × C=C = 614
• 1 × C–C = 347
• 6 × C–H = 6 × 413 = 2478
• 4.5 × O=O = 4.5 × 498 = 2241
• Total = 5680 kJ

Bonds formed in CO₂ and H₂O:

• 6 × C=O = 6 × 805 = 4830
• 6 × O–H = 6 × 463 = 2778
• Total = 7608 kJ

ΔH = 5680 − 7608 = −1928 kJ mol⁻¹

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Dissolving and Hydration Cycle Problem

Worked Example 8: Calculate ΔH°sol for LiF

Given:

• ΔH°latt(LiF, formation) = −1037 kJ mol⁻¹
• ΔH°hyd(Li⁺) = −520 kJ mol⁻¹
• ΔH°hyd(F⁻) = −506 kJ mol⁻¹

ΔH°sol = −(−1037) + (−520) + (−506) = 1037 − 520 − 506 = +11 kJ mol⁻¹

Slightly endothermic. LiF is sparingly soluble — the enthalpy of solution is positive and the entropy increase on dissolving is relatively small (both ions are small and strongly hydrated, restricting the water molecules' freedom).

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Exam Technique Checklist

What to check	Why
Are all units consistent?	ΔH in kJ, ΔS in J — convert before combining
Temperature in kelvin?	T(K) = T(°C) + 273
Correct sign on every quantity?	One wrong sign can reverse your answer
Stoichiometric coefficients applied?	2 moles of Cl⁻ means 2 × ΔH°hyd(Cl⁻)
Sign check on the answer?	Does the sign make chemical sense? Combustion should be exothermic.
Correct formula for the method?	Formation: products − reactants. Combustion: reactants − products.
State symbols included?	Essential for full marks in equation-writing questions
Bond enthalpy: which bonds actually break?	Only break/make bonds that change in the reaction

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Reference Table: Key Formulae

Formula	Use
q = mcΔT	Calorimetry: energy transferred
ΔH = ΣΔH°f(products) − ΣΔH°f(reactants)	Hess's law with formation data
ΔH = ΣΔH°c(reactants) − ΣΔH°c(products)	Hess's law with combustion data
ΔH = Σ(bonds broken) − Σ(bonds formed)	Bond enthalpy method
ΔH°f = sum of Born-Haber cycle steps	Born-Haber cycle (rearrange for unknown)
ΔH°sol = −ΔH°latt(formation) + ΣΔH°hyd	Dissolving energy cycle
ΔS° = ΣS°(products) − ΣS°(reactants)	Entropy change
ΔG = ΔH − TΔS	Gibbs free energy
T = ΔH / ΔS	Transition temperature (when ΔG = 0)

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Summary

This lesson has worked through representative problems from every major area of the energetics course. The keys to exam success are:

1. Identify the data type (formation, combustion, bond enthalpies, Born-Haber, entropy) and choose the correct formula.
2. Show all working clearly, with substituted values before the final calculation.
3. Check signs at every stage and at the end.
4. Convert units (J to kJ, °C to K) before substituting.
5. Apply stoichiometric coefficients to all data values.
6. Perform a reasonableness check: does the sign and magnitude of your answer make chemical sense?

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A-Level Deep Dive: Energetics Problem Solving

Spec mapping

Edexcel 9CH0 specification Topics 8 + 13 are integrated synoptically in this lesson, with the highest density of cross-strand connections in the energetics chapter (refer to the official specification document for exact wording). Synoptic energetics questions appear primarily on Paper 3 (synoptic), where 12–15 mark items routinely combine: Hess cycles (Topic 8.2) with Born–Haber cycles (Topic 13); enthalpies of solution and hydration (Topic 13) with entropy and Gibbs free energy (Topic 13); bond enthalpies (Topic 8.3) with kinetics (Topic 9); and lattice-energy trends (Topic 13) with group 2 chemistry (Topic 4). Multi-step problems demand a procedure: identify what is being asked; identify the data given and missing; choose the appropriate cycle (Hess, Born–Haber, or thermodynamic); calculate; sanity-check sign and order of magnitude; relate to spontaneity and/or equilibrium. Mastery of this lesson is the discriminator between a high A and an A* on Paper 3.

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Worked example with full mark scheme

Question (12 marks):

A student investigates the thermal decomposition of MgCO₃(s) → MgO(s) + CO₂(g).

Data:

• ΔH°_f[MgCO₃(s)] = −1096 kJ mol⁻¹
• ΔH°_f[MgO(s)] = −602 kJ mol⁻¹
• ΔH°_f[CO₂(g)] = −394 kJ mol⁻¹
• S°[MgCO₃(s)] = +66 J K⁻¹ mol⁻¹
• S°[MgO(s)] = +27 J K⁻¹ mol⁻¹
• S°[CO₂(g)] = +214 J K⁻¹ mol⁻¹

(a) Use Hess's law to calculate ΔH° for the decomposition. (2) (b) Calculate ΔS° for the decomposition. (2) (c) Determine the temperature above which decomposition is thermodynamically spontaneous. (3) (d) The lattice energy of MgCO₃ is −3122 kJ mol⁻¹; that of MgO is −3791 kJ mol⁻¹. Comment on how the difference relates to the thermal decomposition temperature you calculated. (3) (e) Predict (with reasoning) how the decomposition temperature of BaCO₃ would compare. (2)

Solution with mark scheme:

(a) ΔH° = ΣΔH°_f(prod) − ΣΔH°_f(react) = (−602) + (−394) − (−1096) = +100 kJ mol⁻¹.

M1 — Hess equation. A1 — value with sign.

(b) ΔS° = ΣS°(prod) − ΣS°(react) = (27 + 214) − 66 = +175 J K⁻¹ mol⁻¹.

M1 — substitution. A1 — value with units.

(c) ΔG° = ΔH° − TΔS° = 0 ⟹ T = ΔH°/ΔS° = 100 / 0.175 = 571 K (≈ 298 °C).

M1 — recognise condition. M1 — convert ΔS° to kJ K⁻¹ (or ΔH° to J). A1 — answer with units.

(d) B1 — the lattice energy of MgO is more exothermic than MgCO₃ by 669 kJ mol⁻¹; this drives the reaction forward thermodynamically.

B1 — together with the entropy gain from CO₂ release, the difference in lattice energies sets a moderate threshold T (~571 K), accessible at ordinary furnace temperatures.

B1 — the calculated threshold (571 K = 298 °C) is consistent with the observed decomposition of MgCO₃ in the range 300–500 °C; the kinetic onset typically lags the thermodynamic threshold.