1. Standard Enthalpy Definitions
Thermodynamics examines the energy changes and driving forces that dictate whether chemical reactions occur spontaneously. To construct energy cycles for ionic lattices, several standard enthalpy terms are required:
Example: Na(s) + ½Cl₂(g) → NaCl(s) [ΔH = −411 kJ mol⁻¹]
Example: Na(g) → Na⁺(g) + e⁻ [ΔH = +496 kJ mol⁻¹]
The second ionisation energy removes an electron from each 1+ ion: Mg⁺(g) → Mg²⁺(g) + e⁻ [ΔH = +1451 kJ mol⁻¹]
Example: Cl₂(g) → 2Cl(g) [ΔH = +244 kJ mol⁻¹]
For chlorine, the bond enthalpy (+244) is twice the enthalpy of atomisation (+122), because atomisation makes only one mole of Cl atoms.
Example: Na⁺(g) + Cl⁻(g) → NaCl(s) [ΔH = −788 kJ mol⁻¹]
Example: NaCl(s) → Na⁺(g) + Cl⁻(g) [ΔH = +788 kJ mol⁻¹]
Example: Na(s) → Na(g) [ΔH = +107 kJ mol⁻¹]
Example: ½Cl₂(g) → Cl(g) [ΔH = +122 kJ mol⁻¹]
Example: Cl(g) + e⁻ → Cl⁻(g) [ΔH = −348 kJ mol⁻¹] (exothermic)
The enthalpy of hydration is defined in Section 4. Data books differ slightly, so the same quantity can have a slightly different value in different questions (for example, EA(Cl) is given as −348 or −349 kJ mol⁻¹). Always use the values in the question.
The first electron affinity is exothermic because the incoming electron is attracted to the positive nuclear charge of the neutral atom. However, the second electron affinity (e.g. O⁻(g) + e⁻ → O²⁻(g) [ΔH = +798 kJ mol⁻¹]) is always endothermic because energy must be supplied to overcome the strong electrostatic repulsion between the negative ion and the incoming electron.
2. Born-Haber Energy Cycles
Lattice enthalpies cannot be measured directly through experimental calorimetry because it is practically impossible to assemble a solid crystal directly from isolated gaseous ions. Instead, we apply Hess's Law using a thermodynamic cycle known as a Born-Haber cycle.
A Born-Haber cycle connects the standard enthalpy of formation of an ionic solid with the series of steps required to vaporise, atomise, and ionise the constituent elements into gaseous ions:
ΔHf = ΔHat(metal) + IE₁(metal) + ΔHat(non-metal) + ΔHEA1(non-metal) + ΔHlatt(formation)
Rearranging for the standard lattice enthalpy of formation gives:
ΔHlatt(formation) = ΔHf − [ΔHat(metal) + IE₁(metal) + ΔHat(non-metal) + ΔHEA1(non-metal)]
3. Born-Haber Cycle Diagram
The diagram below displays the complete energy level progression for sodium chloride. Endothermic stages are represented by upward arrows (+ values) and exothermic stages by downward arrows (- values):
Born-Haber Cycle for MgCl₂
For a compound such as MgCl₂, the cycle has more steps:
- Mg forms Mg²⁺, so you need the first and second ionisation energies.
- There are two chloride ions, so you need 2 × the enthalpy of atomisation of chlorine and 2 × the electron affinity of chlorine.
ΔfH(MgCl₂) = −642
ΔatH(Mg) = +148
First IE(Mg) = +738; second IE(Mg) = +1451
ΔatH(Cl) = +122 (½Cl₂(g) → Cl(g))
EA(Cl) = −349
Step 1: Add up the steps from the elements to the gaseous ions:
148 + 738 + 1451 + (2 × 122) + (2 × −349)
= 148 + 738 + 1451 + 244 − 698 = +1883 kJ mol⁻¹
Step 2: Use Hess's law:
Lattice formation = ΔfH − (sum of the other steps)
= −642 − 1883 = −2525 kJ mol⁻¹
The most common mistake is to forget the ×2 for chlorine. Each mole of MgCl₂ contains two moles of Cl⁻ ions.
4. Enthalpies of Solution and Hydration
When an ionic solid dissolves in water, the crystal lattice breaks apart and the separated ions become surrounded by polar water molecules. The overall energy change is known as the standard enthalpy of solution.
Example: NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)
Example: Na⁺(g) + aq → Na⁺(aq) [ΔHhyd = −406 kJ mol⁻¹]
Example: Cl⁻(g) + aq → Cl⁻(aq) [ΔHhyd = −364 kJ mol⁻¹]
The relationship between these terms is given by the Hess cycle:
ΔHsol = ΔHlatt(dissociation) + Σ[ΔHhyd(ions)]
Or equivalently: ΔHsol = −ΔHlatt(formation) + ΔHhyd(cations) + ΔHhyd(anions)
Example (NaCl): ΔHsol = +788 + (−406) + (−364) = +18 kJ mol⁻¹.
The measured value is +3.9 kJ mol⁻¹; the cycle value differs because the data come from different sources and are rounded.
Lattice dissociation enthalpy of KCl = +711
Enthalpy of hydration of K⁺ = −322
Enthalpy of hydration of Cl⁻ = −364
Step 1: Write the cycle: KCl(s) → K⁺(g) + Cl⁻(g) → K⁺(aq) + Cl⁻(aq)
Step 2: Add the steps:
ΔsolH = +711 + (−322) + (−364) = +25 kJ mol⁻¹
The positive value means KCl dissolves endothermically, so the solution cools. The measured value is about +17 kJ mol⁻¹. Values vary slightly between data sources, so a cycle value does not always match the measured value exactly.
5. The Ionic Model & Polarisation
Theoretical lattice enthalpies can be calculated using electrostatics and Coulomb's law. This calculation assumes the purely ionic model:
- Ions are modelled as perfect, unpolarisable spheres.
- Electrostatic charges are treated as point charges located at ion centres.
- The only bonding forces acting between ions are purely electrostatic attractions and repulsions (zero covalent character).
| Compound | Theoretical Lattice Enthalpy / kJ mol⁻¹ | Experimental Born-Haber Value / kJ mol⁻¹ | Discrepancy & Nature of Bonding |
|---|---|---|---|
| NaCl | -770 | -788 | Close agreement (2.3% difference): almost perfectly ionic. |
| MgCl₂ | -2326 | -2525 | Moderate difference: small, highly charged Mg²⁺ ion polarises Cl⁻ electron clouds. |
| AgCl | -833 | -905 | Large difference (8.6%): Ag⁺ polarises the Cl⁻ ion, so the bonding has some covalent character. |
| AgI | -736 | -858 | Large discrepancy (16.6% difference): large iodide ion is easily polarised by silver cation, introducing substantial covalent bonding. |
- If the two values are close, the compound is almost purely ionic (NaCl).
- If the Born-Haber value is larger (more exothermic) than the perfect ionic model value, the bonding has some covalent character.
- Example: AgCl. Perfect ionic model −833 kJ mol⁻¹; Born-Haber −905 kJ mol⁻¹. The Ag⁺ ion distorts (polarises) the electron cloud of Cl⁻. Some electron density is shared, so the lattice is stronger than the ionic model predicts.
6. Entropy (ΔS)
Entropy (S) is a quantitative measure of the disorder or randomness of a chemical system, measured in units of J K⁻¹ mol⁻¹. A system with higher randomness and more available energy distributions possesses higher entropy.
Factors Controlling Entropy
- Physical State: Solids have the lowest entropy (ordered crystal lattice); liquids have moderate entropy; gases have the highest entropy (rapid, random molecular movement): S(solid) < S(liquid) << S(gas).
- Number of Particles: Reactions that produce an increase in the number of moles of gas exhibit a large positive entropy change (ΔS > 0).
- Temperature: Increasing temperature increases molecular kinetic energy and rotational/vibrational microstates, increasing entropy.
The standard entropy change of a chemical reaction is calculated as:
ΔS = Σ[S(products)] − Σ[S(reactants)]
Predicting the Sign of ΔS
- Melting and boiling increase entropy (ΔS positive). Particles become more disordered: solid → liquid → gas.
- Dissolving a solid usually increases entropy. The ions or molecules spread out through the solution.
- A reaction that makes more moles of gas has a positive ΔS. A reaction that makes fewer moles of gas has a negative ΔS.
Standard entropies (J K⁻¹ mol⁻¹): N₂ = 192, H₂ = 131, NH₃ = 193
(a) Calculate ΔS. (b) Find the temperature at which ΔG = 0.
(a) Step 1: Products minus reactants:
ΔS = 2(193) − [192 + 3(131)] = 386 − 585 = −199 J K⁻¹ mol⁻¹
ΔS is negative because 4 moles of gas become 2 moles of gas.
(b) Step 2: Convert units. ΔS is in J K⁻¹ mol⁻¹ but ΔH is in kJ mol⁻¹. Either divide ΔS by 1000 (−0.199 kJ K⁻¹ mol⁻¹) or multiply ΔH by 1000 (−92 000 J mol⁻¹).
Step 3: Set ΔG = 0, so T = ΔH / ΔS:
T = −92 000 / −199 = 462 K
ΔH and ΔS are both negative, so the reaction is feasible below 462 K. Above this temperature, the positive −TΔS term is bigger in size than the negative ΔH, so ΔG is positive.
7. Gibbs Free Energy (ΔG) & Feasibility
Whether a chemical reaction is thermodynamically feasible depends on both the enthalpy change (ΔH) and the entropy change (ΔS), linked through temperature (T in Kelvin) via the Gibbs free energy equation:
ΔG = ΔH − TΔS
ΔH is usually given in kJ mol⁻¹, while ΔS is given in J K⁻¹ mol⁻¹. To use the Gibbs equation correctly, you MUST convert ΔS into kJ K⁻¹ mol⁻¹ by dividing by 1000 before evaluating ΔG!
| Enthalpy Change (ΔH) | Entropy Change (ΔS) | Gibbs Equation Term (−TΔS) | Feasibility Outcome |
|---|---|---|---|
| Exothermic (ΔH < 0) | Positive (ΔS > 0) | Always negative | Feasible at all temperatures (ΔG < 0 always). |
| Endothermic (ΔH > 0) | Negative (ΔS < 0) | Always positive | Never feasible at any temperature (ΔG > 0 always). |
| Endothermic (ΔH > 0) | Positive (ΔS > 0) | Negative at high T | Feasible only above threshold temperature: T ≥ ΔH / ΔS. |
| Exothermic (ΔH < 0) | Negative (ΔS < 0) | Positive at high T | Feasible only below threshold temperature: T ≤ ΔH / ΔS. |
Calculating the Threshold Feasibility Temperature
At the exact boundary where a reaction transitions from non-feasible to feasible, ΔG = 0:
0 = ΔH − TΔS ⇒ T = ΔH / ΔS
Linear Graphical Analysis (ΔG vs T)
Plotting ΔG on the y-axis against absolute temperature T on the x-axis gives a straight line following y = mx + c:
- y-intercept: ΔH
- Gradient: −ΔS (negative of entropy change)
- x-intercept: Feasibility transition temperature (T = ΔH / ΔS)
8. Worked Calculations
ΔHf[NaCl(s)] = −411 kJ mol⁻¹
ΔHat[Na(s)] = +107 kJ mol⁻¹
First ionisation energy of Na = +496 kJ mol⁻¹
ΔatH (½Cl₂(g) → Cl(g)) = +122 kJ mol⁻¹
First electron affinity of Cl = −348 kJ mol⁻¹
Step 1: Write out the Hess's Law relationship:
ΔHf = ΔHat(Na) + IE₁(Na) + ΔHat(Cl) + ΔHEA1(Cl) + ΔHlatt
Step 2: Substitute the known values:
−411 = (+107) + (+496) + (+122) + (−348) + ΔHlatt
−411 = +377 + ΔHlatt
Step 3: Solve for ΔHlatt:
ΔHlatt = −411 − 377 = −788 kJ mol⁻¹
The standard lattice enthalpy of formation of sodium chloride is −788 kJ mol⁻¹ (or lattice dissociation enthalpy = +788 kJ mol⁻¹).
CaCO₃(s) → CaO(s) + CO₂(g)
ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹.
Calculate the minimum temperature in Kelvin at which this decomposition becomes thermodynamically feasible.
Step 1: Reconcile units between enthalpy and entropy:
ΔH = +178 kJ mol⁻¹
ΔS = +161 / 1000 = +0.161 kJ K⁻¹ mol⁻¹
Step 2: Set ΔG = 0 for the feasibility threshold:
ΔG = ΔH − TΔS = 0
T = ΔH / ΔS
Step 3: Calculate the threshold temperature:
T = 178 / 0.161 = 1105.6 K (or 832.6 °C)
At temperatures above 1106 K, TΔS exceeds ΔH, causing ΔG to become negative and making decomposition feasible.
Lattice enthalpy of formation of NaCl = −788
ΔfH(NaCl) = −411
ΔatH(Na) = +107; ΔatH(Cl) = +122
First IE(Na) = +496
Step 1: Write the Hess's law relationship:
ΔfH = ΔatH(Na) + ΔatH(Cl) + IE₁(Na) + EA(Cl) + lattice formation
Step 2: Substitute:
−411 = 107 + 122 + 496 + EA − 788
−411 = −63 + EA
Step 3: Rearrange:
EA = −411 + 63 = −348 kJ mol⁻¹
The value is negative, as expected for a first electron affinity.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: Why is the second electron affinity of oxygen endothermic (+798 kJ mol⁻¹)?
Show Answer & Explanation
Correct Answer: B
The O⁻ ion already carries a net negative charge. Adding a second electron requires energy to overcome the repulsive force between like negative charges.
Question 2: The theoretical lattice enthalpy of silver iodide is −736 kJ mol⁻¹, whereas the experimental Born-Haber value is −858 kJ mol⁻¹. What accounts for this discrepancy?
Show Answer & Explanation
Correct Answer: B
Large iodide ions are easily distorted (polarised) by silver cations. This electron cloud sharing introduces partial covalent bonding, making the crystal lattice stronger than predicted by the pure ionic model.
Question 3: A reaction has ΔH = −85 kJ mol⁻¹ and ΔS = −120 J K⁻¹ mol⁻¹. Under what temperature conditions is this reaction thermodynamically feasible?
Show Answer & Explanation
Correct Answer: B
When ΔH < 0 and ΔS < 0, the −TΔS term is positive. For ΔG to remain negative, the temperature must be low enough so that −TΔS does not outweigh the exothermic ΔH.
Question 4: On a graph of ΔG versus absolute temperature T, what chemical quantities are represented by the gradient and y-intercept?
Show Answer & Explanation
Correct Answer: B
Comparing ΔG = −ΔS × T + ΔH to y = mx + c shows that the gradient m = −ΔS and the y-intercept c = ΔH.
Question 5 (3 marks): Use the data (kJ mol⁻¹) to calculate the lattice enthalpy of formation of magnesium oxide, MgO. ΔfH(MgO) = −602; ΔatH(Mg) = +148; first IE(Mg) = +738; second IE(Mg) = +1451; ΔatH(O) = +249; first EA(O) = −141; second EA(O) = +798.
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Answer
Sum of steps from the elements to the gaseous ions = 148 + 738 + 1451 + 249 − 141 + 798 = +3243 kJ mol⁻¹.
Lattice formation = −602 − 3243 = −3845 kJ mol⁻¹.
Mark points
- Both ionisation energies and both electron affinities included (second EA positive)
- Sum of steps = +3243
- Lattice enthalpy of formation = −3845 kJ mol⁻¹ (sign needed)
Question 6 (2 marks): Calculate the enthalpy of solution of potassium bromide, KBr. Lattice dissociation enthalpy of KBr = +679 kJ mol⁻¹; enthalpy of hydration of K⁺ = −322 kJ mol⁻¹; enthalpy of hydration of Br⁻ = −335 kJ mol⁻¹. State whether the solution warms or cools as KBr dissolves.
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Answer
ΔsolH = +679 + (−322) + (−335) = +22 kJ mol⁻¹. The value is positive (endothermic), so the solution cools.
Mark points
- ΔsolH = lattice dissociation + sum of hydration enthalpies = +22 kJ mol⁻¹
- Endothermic, so the temperature of the solution falls
Question 7 (4 marks): 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197 kJ mol⁻¹. Standard entropies (J K⁻¹ mol⁻¹): SO₂ = 248, O₂ = 205, SO₃ = 257. Calculate ΔS and the temperature above which the reaction is no longer feasible.
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Answer
ΔS = 2(257) − [2(248) + 205] = 514 − 701 = −187 J K⁻¹ mol⁻¹.
At ΔG = 0, T = ΔH / ΔS = −197 000 / −187 = 1053 K (1050 K to 3 s.f.).
Both ΔH and ΔS are negative, so the reaction is feasible below 1053 K and not feasible above it.
Mark points
- ΔS = −187 J K⁻¹ mol⁻¹
- Units converted (ΔH in J mol⁻¹, or ΔS in kJ K⁻¹ mol⁻¹)
- T = ΔH / ΔS = 1053 K
- Not feasible above this temperature, because the positive −TΔS term outweighs the negative ΔH, so ΔG > 0
Question 8 (2 marks): Predict the sign of ΔS for each change and give a reason: (a) H₂O(l) → H₂O(g); (b) CaCO₃(s) → CaO(s) + CO₂(g).
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Answer
(a) Positive: a liquid becomes a gas, so the particles are more disordered.
(b) Positive: a solid makes a gas, so the number of moles of gas increases from 0 to 1.
Mark points
- (a) Positive, liquid to gas (more disorder)
- (b) Positive, moles of gas increase