Entropy and Energetics

Q: Do I need to use Gibbs free energy (ΔG) to calculate feasibility?

No, Edexcel IAL Unit 4 requires you use ΔS<subtotal</sub = ΔS<subsystem</sub + ΔS<subsurroundings</sub 0 to assess feasibility. ΔG is an optional alternative but will not be required for exam questions.

Edexcel International A-Level Chemistry· 12.1–12.19 (12A–12B)· 45 min read

1. Entropy and Reaction Feasibility★★★☆☆⏱ 15 min

✓ Calculator OK

📘 Definition

Entropy (S)

A measure of the disorder or dispersal of energy and molecules in a system, with units J K⁻¹ mol⁻¹. Higher entropy = greater disorder.

Enthalpy change alone cannot predict if a reaction will be spontaneous, as many endothermic reactions (e.g. ammonium nitrate dissolving in water) occur naturally. Entropy accounts for this disorder change: entropy increases with temperature, for state changes solid → liquid → gas, when ionic solids dissolve, and when the number of moles of gas increases in a reaction. A perfect crystal at 0 K has 0 entropy (third law of thermodynamics).

\Delta S_{system} = \sum S(products) - \sum S(reactants)

\Delta S_{surroundings} = \frac{-\Delta H}{T}

\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}

A reaction is feasible if ΔS_total > 0, even if one component is negative. You can calculate the minimum temperature at which a reaction becomes feasible by setting ΔS_total ≥ 0 and solving for T.

📐 Worked Example

Calculate the minimum temperature at which the thermal decomposition of sodium hydrogencarbonate becomes feasible: $2 N a H C O_{3} (s) \to N a_{2} C O_{3} (s) + C O_{2} (g) + H_{2} O (g)$ . Given: ΔS_system = +335 J K⁻¹ mol⁻¹, ΔH = +129 kJ mol⁻¹.

1
Step 1: Convert ΔH to J mol⁻¹ to match entropy units: $129 \times 1000 = 129000$ J mol⁻¹
2
Step 2: Set ΔS_total ≥ 0 for feasibility, substitute the entropy formula:
3
$335 + \frac{-129000}{T} \geq 0$
4
Step 3: Rearrange to solve for T:
5
$T \geq \frac{129000}{335} \approx 385 K$
6
Conclusion: The reaction is feasible at temperatures above 385 K (112 °C).

Distinguish between thermodynamic and kinetic stability: if ΔS_total < 0, the reaction is thermodynamically impossible. If ΔS_total > 0 but the reaction has very high activation energy, it is kinetically stable and will not proceed at a measurable rate (e.g. diamond turning to graphite).

Exam tip:

Always convert ΔH from kJ mol⁻¹ to J mol⁻¹ and use temperature in Kelvin for all entropy calculations to avoid unit errors.

2. Born-Haber Cycles and Lattice Energy★★★★☆⏱ 15 min

✓ Calculator OK

📘 Definition

Lattice Energy

The enthalpy change when 1 mole of solid ionic compound forms from its constituent gaseous ions. Edexcel defines this as an exothermic process (negative ΔH value).

Born-Haber cycles are applications of Hess's law used to calculate lattice energy for ionic compounds, using measurable enthalpy values including enthalpy of atomisation, ionisation energy, electron affinity, and enthalpy of formation. Note: 1st electron affinity is exothermic (adding an electron to a neutral gaseous atom), 2nd electron affinity is endothermic (adding an electron to a negative ion requires energy to overcome repulsion).

📐 Worked Example

Calculate the lattice energy of sodium chloride (NaCl) using the following values: Δ_fH(NaCl) = -411 kJ mol⁻¹, Δ_atH(Na) = +107 kJ mol⁻¹, Δ_atH(Cl) = +122 kJ mol⁻¹, 1st IE(Na) = +496 kJ mol⁻¹, 1st EA(Cl) = -349 kJ mol⁻¹.

1
Step 1: Apply Hess's law to the Born-Haber cycle:
2
$\Delta_f H = \Delta_{at}H(Na) + \Delta_{at}H(Cl) + 1st\ IE(Na) + 1st\ EA(Cl) + \Delta_{LE}H$
3
Step 2: Rearrange to solve for lattice energy Δ_LEH:
4
$\Delta_{LE}H = \Delta_f H - \Delta_{at}H(Na) - \Delta_{at}H(Cl) - 1st\ IE(Na) - 1st\ EA(Cl)$
5
Step 3: Substitute values:
6
$\Delta_{LE}H = -411 - 107 - 122 - 496 - (-349) = -787\ kJ\ mol^{-1}$
7
Conclusion: The lattice energy of NaCl is -787 kJ mol⁻¹, consistent with its exothermic definition.

If experimental lattice energy from Born-Haber cycles is significantly more exothermic than the theoretical value calculated using the perfect ionic model, the ionic compound has covalent character. This occurs when a small, highly charged cation polarises a large anion, distorting its electron cloud (Fajans' Rules).

Exam tip:

Draw Born-Haber cycles with upward arrows for endothermic steps and downward arrows for exothermic steps to avoid sign errors in calculations.

3. Enthalpy of Solution and Hydration★★★☆☆⏱ 10 min

✓ Calculator OK

📘 Definition

Enthalpy of Solution (Δ<sub>sol</sub>H)

Enthalpy change when 1 mole of ionic compound dissolves in excess water to form a dilute solution, can be exothermic or endothermic.

📘 Definition

Enthalpy of Hydration (Δ<sub>hyd</sub>H)

Enthalpy change when 1 mole of gaseous ions dissolves in excess water to form dilute aqueous ions, always exothermic due to ion-dipole attractions between ions and water molecules.

You can calculate enthalpy of solution using a Hess cycle combining lattice energy and hydration enthalpies:

\Delta_{sol}H = -\Delta_{LE}H + \sum \Delta_{hyd}H(cations) + \sum \Delta_{hyd}H(anions)

Lattice energy and hydration enthalpy both become more exothermic as ionic charge increases and ionic radius decreases, due to stronger electrostatic attractions.

📐 Worked Example

Calculate the enthalpy of solution of NaCl, given Δ_LEH(NaCl) = -787 kJ mol⁻¹, Δ_hydH(Na⁺) = -406 kJ mol⁻¹, Δ_hydH(Cl⁻) = -364 kJ mol⁻¹.

1
Step 1: Substitute values into the enthalpy of solution formula:
2
$\Delta_{sol}H = -(-787) + (-406) + (-364)$
3
$\Delta_{sol}H = 787 - 770 = +17\ kJ\ mol^{-1}$
4
Conclusion: Dissolving NaCl in water is a slightly endothermic process.

4. Predicting Solubility Trends★★★★☆⏱ 5 min

Solubility depends on both enthalpy of solution and entropy change of the system. Even if Δ_solH is endothermic, a large positive ΔS_system from dissolving can result in a positive ΔS_total, making the compound soluble.

Key Group 2 solubility trends you must recall and explain:

Sulfates: Solubility decreases down Group 2. The Δ_hydH of the Group 2 cation decreases more rapidly than the lattice energy of the sulfate as cation radius increases, so Δ_solH becomes more endothermic.
Hydroxides: Solubility increases down Group 2. The lattice energy of the hydroxide decreases more rapidly than Δ_hydH of the Group 2 cation as cation radius increases, so Δ_solH becomes less endothermic/more exothermic.

📐 Worked Example

Explain why MgSO₄ is soluble but BaSO₄ is insoluble at 298 K.

1
Step 1: Compare ionic radii: Mg²⁺ is much smaller than Ba²⁺.
2
Step 2: Δ_hydH of Mg²⁺ is very exothermic, so Δ_solH of MgSO₄ is only slightly endothermic. The positive ΔS_system from dissolving offsets this, giving ΔS_total > 0, so MgSO₄ is soluble.
3
Step 3: Δ_hydH of Ba²⁺ is much less exothermic, so Δ_solH of BaSO₄ is very endothermic. The ΔS_system is not large enough to offset this, giving ΔS_total < 0, so BaSO₄ is insoluble.

5. Common Pitfalls

Wrong move:

Using ΔH in kJ mol⁻¹ directly in ΔS_surroundings calculations

Why:

Entropy units are J K⁻¹ mol⁻¹, so unit mismatch leads to a 1000x calculation error

Correct move:

Multiply ΔH by 1000 to convert to J mol⁻¹ before dividing by temperature in Kelvin

Wrong move:

Using an endothermic lattice energy definition in Born-Haber cycles

Why:

Edexcel explicitly defines lattice energy as exothermic (formation of solid from gaseous ions), so incorrect sign leads to wrong cycle sums

Correct move:

Always use exothermic lattice energy, draw downward arrows for lattice formation steps in your cycle

Wrong move:

Forgetting ΔS_surroundings is negative for endothermic reactions

Why:

Endothermic reactions absorb heat from the surroundings, decreasing their disorder

Correct move:

Use the exact formula ΔS_surroundings = -ΔH/T, so positive endothermic ΔH gives negative ΔS_surroundings

Wrong move:

Assuming endothermic dissolving means a compound is insoluble

Why:

A large positive ΔS_system from dissolving can offset endothermic Δ_solH to give positive ΔS_total

Correct move:

Calculate total entropy change, do not rely on Δ_solH alone for solubility predictions

Wrong move:

Confusing thermodynamic and kinetic stability

Why:

A feasible reaction may not proceed if activation energy is very high

Correct move:

Thermodynamic stability = ΔS_total < 0, kinetic stability = ΔS_total > 0 but high Ea prevents measurable reaction rate

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
ΔS_system	ΣS(products) - ΣS(reactants)	Units J K⁻¹ mol⁻¹; use given standard S values
ΔS_surroundings	-ΔH / T	Convert ΔH to J, T in Kelvin
Feasibility condition	ΔS_total > 0	Feasible even if one entropy component is negative
Lattice energy (Edexcel)	Gaseous ions → 1 mol solid ionic compound	Exothermic (negative value)
Δ_solH	-Δ_LEH + ΣΔ_hydH(ions)	Can be exothermic or endothermic
Group 2 solubility trends	Sulfates ↓ down group, Hydroxides ↑ down group	Explain using relative changes to Δ_LEH and Δ_hydH

7. Frequently Asked

Do I need to use Gibbs free energy (ΔG) to calculate feasibility?

No, Edexcel IAL Unit 4 requires you use ΔS_total = ΔS_system + ΔS_surroundings > 0 to assess feasibility. ΔG is an optional alternative but will not be required for exam questions.

Is lattice energy exothermic or endothermic for Edexcel exams?

Edexcel explicitly defines lattice energy as the enthalpy change when 1 mole of solid ionic compound forms from its gaseous ions, so it is always exothermic (negative value) for stable ionic compounds.

Why do I have to convert ΔH to joules for entropy calculations?

Entropy is measured in J K⁻¹ mol⁻¹ while enthalpy values are almost always given in kJ mol⁻¹. Converting ΔH to J ensures units match, avoiding a 1000x calculation error.

Going deeper

official_documentEdexcel IAL Chemistry SpecificationRefer to Unit 4 Section 12 for full topic guidance
resourceEdexcel IAL Chemistry Data BookletAll standard enthalpy and entropy values are provided in exams

What's Next

Now that you have mastered entropy and energetics for Edexcel IAL Chemistry Unit 4, you are ready to build on this knowledge for the rest of the Rates, Equilibria and Further Organic Chemistry unit. Your understanding of entropy will directly support your study of chemical equilibrium, as entropy changes drive the position of equilibrium for reversible reactions. Your Born-Haber cycle calculation skills are also foundational for advanced ionic structure and bonding questions. This topic appears in almost every WCH14 exam, typically as an 8-12 mark extended question, so practice topic-specific past papers to reinforce your accuracy and recall of definition terms.