# Entropy and Energetics

> Edexcel International A-Level Chemistry · Unit 4: Rates, Equilibria and Further Organic Chemistry
> Source: https://www.owlsprep.com/study/edexcel-ial-chemistry-u4-entropy-and-energetics/

This guide covers all Edexcel IAL Chemistry Unit 4 content for entropy, feasibility calculations, Born-Haber cycles, and solubility trends, aligned to specification points 12.1–12.19 for your WCH14 exam. All out-of-scope content is excluded for focused revision.

**Prerequisites:** Standard enthalpy change calculations (Hess's law); Ionic bonding and ion polarisation concepts; Mole and stoichiometry calculation fundamentals

## Learning objectives

- Explain entropy and use ΔS<sub>total</sub> to predict reaction feasibility
- Perform Born-Haber cycle calculations for lattice energy
- Relate lattice and hydration enthalpies to solubility of ionic compounds
- Distinguish between thermodynamic and kinetic stability

## Entropy and Reaction Feasibility

**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<sub>total</sub> > 0, even if one component is negative. You can calculate the minimum temperature at which a reaction becomes feasible by setting ΔS<sub>total</sub> ≥ 0 and solving for T.

**Worked example:** Calculate the minimum temperature at which the thermal decomposition of sodium hydrogencarbonate becomes feasible: $2NaHCO_3(s) \rightarrow Na_2CO_3(s) + CO_2(g) + H_2O(g)$. Given: ΔS<sub>system</sub> = +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<sub>total</sub> ≥ 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<sub>total</sub> < 0, the reaction is thermodynamically impossible. If ΔS<sub>total</sub> > 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.

*Calculator:* allowed

## Born-Haber Cycles and Lattice Energy

**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<sub>f</sub>H(NaCl) = -411 kJ mol⁻¹, Δ<sub>at</sub>H(Na) = +107 kJ mol⁻¹, Δ<sub>at</sub>H(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<sub>LE</sub>H:
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.

*Calculator:* allowed

## Enthalpy of Solution and Hydration

**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.

**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<sub>LE</sub>H(NaCl) = -787 kJ mol⁻¹, Δ<sub>hyd</sub>H(Na⁺) = -406 kJ mol⁻¹, Δ<sub>hyd</sub>H(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.

*Calculator:* allowed

## Predicting Solubility Trends

Solubility depends on both enthalpy of solution and entropy change of the system. Even if Δ<sub>sol</sub>H is endothermic, a large positive ΔS<sub>system</sub> from dissolving can result in a positive ΔS<sub>total</sub>, making the compound soluble.

Key Group 2 solubility trends you must recall and explain:

- **Sulfates**: Solubility decreases down Group 2. The Δ<sub>hyd</sub>H of the Group 2 cation decreases more rapidly than the lattice energy of the sulfate as cation radius increases, so Δ<sub>sol</sub>H becomes more endothermic.
- **Hydroxides**: Solubility increases down Group 2. The lattice energy of the hydroxide decreases more rapidly than Δ<sub>hyd</sub>H of the Group 2 cation as cation radius increases, so Δ<sub>sol</sub>H 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<sub>hyd</sub>H of Mg²⁺ is very exothermic, so Δ<sub>sol</sub>H of MgSO₄ is only slightly endothermic. The positive ΔS<sub>system</sub> from dissolving offsets this, giving ΔS<sub>total</sub> > 0, so MgSO₄ is soluble.
3. Step 3: Δ<sub>hyd</sub>H of Ba²⁺ is much less exothermic, so Δ<sub>sol</sub>H of BaSO₄ is very endothermic. The ΔS<sub>system</sub> is not large enough to offset this, giving ΔS<sub>total</sub> < 0, so BaSO₄ is insoluble.

## Common pitfalls

- **Wrong:** Using ΔH in kJ mol⁻¹ directly in ΔS<sub>surroundings</sub> calculations
  - Why it fails: Entropy units are J K⁻¹ mol⁻¹, so unit mismatch leads to a 1000x calculation error
  - Correct: Multiply ΔH by 1000 to convert to J mol⁻¹ before dividing by temperature in Kelvin
- **Wrong:** Using an endothermic lattice energy definition in Born-Haber cycles
  - Why it fails: Edexcel explicitly defines lattice energy as exothermic (formation of solid from gaseous ions), so incorrect sign leads to wrong cycle sums
  - Correct: Always use exothermic lattice energy, draw downward arrows for lattice formation steps in your cycle
- **Wrong:** Forgetting ΔS<sub>surroundings</sub> is negative for endothermic reactions
  - Why it fails: Endothermic reactions absorb heat from the surroundings, decreasing their disorder
  - Correct: Use the exact formula ΔS<sub>surroundings</sub> = -ΔH/T, so positive endothermic ΔH gives negative ΔS<sub>surroundings</sub>
- **Wrong:** Assuming endothermic dissolving means a compound is insoluble
  - Why it fails: A large positive ΔS<sub>system</sub> from dissolving can offset endothermic Δ<sub>sol</sub>H to give positive ΔS<sub>total</sub>
  - Correct: Calculate total entropy change, do not rely on Δ<sub>sol</sub>H alone for solubility predictions
- **Wrong:** Confusing thermodynamic and kinetic stability
  - Why it fails: A feasible reaction may not proceed if activation energy is very high
  - Correct: Thermodynamic stability = ΔS<sub>total</sub> < 0, kinetic stability = ΔS<sub>total</sub> > 0 but high Ea prevents measurable reaction rate

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| ΔS<sub>system</sub> | ΣS(products) - ΣS(reactants) | Units J K⁻¹ mol⁻¹; use given standard S values |
| ΔS<sub>surroundings</sub> | -ΔH / T | Convert ΔH to J, T in Kelvin |
| Feasibility condition | ΔS<sub>total</sub> > 0 | Feasible even if one entropy component is negative |
| Lattice energy (Edexcel) | Gaseous ions → 1 mol solid ionic compound | Exothermic (negative value) |
| Δ<sub>sol</sub>H | -Δ<sub>LE</sub>H + ΣΔ<sub>hyd</sub>H(ions) | Can be exothermic or endothermic |
| Group 2 solubility trends | Sulfates ↓ down group, Hydroxides ↑ down group | Explain using relative changes to Δ<sub>LE</sub>H and Δ<sub>hyd</sub>H |

## 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.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/edexcel-ial-chemistry-u4-entropy-and-energetics/