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+(g) + Cl-(g) -> NaCl(s) [delta H = -788 kJ mol-1]
Example: NaCl(s) -> Na+(g) + Cl-(g) [delta H = +788 kJ mol-1]
Example: Na(s) -> Na(g) [delta H = +107 kJ mol-1]
Example: 1/2 Cl2(g) -> Cl(g) [delta H = +122 kJ mol-1]
Example: Cl(g) + e- -> Cl-(g) [delta H = -348 kJ mol-1] (exothermic)
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) [delta H = +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:
delta Hf = delta Hat(metal) + IE1(metal) + delta Hat(non-metal) + delta HEA1(non-metal) + delta H_latt(formation)
Rearranging for the standard lattice enthalpy of formation gives:
delta H_latt(formation) = delta Hf - [delta Hat(metal) + IE1(metal) + delta Hat(non-metal) + delta HEA1(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):
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) [delta H_sol = +3.9 kJ mol-1]
Example: Na+(g) + aq -> Na+(aq) [delta H_hyd = -406 kJ mol-1]
Example: Cl-(g) + aq -> Cl-(aq) [delta H_hyd = -364 kJ mol-1]
The relationship between these terms is given by the Hess cycle:
delta H_sol = delta H_latt(dissociation) + sum[delta H_hyd(ions)]
Or equivalently: delta H_sol = -delta H_latt(formation) + delta H_hyd(cations) + delta H_hyd(anions)
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 modeled as perfect, unpolarisable spheres.
- Electrostatic charges are treated as point charges located at ion centers.
- 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 | -2526 | Moderate difference: small, highly charged Mg2+ ion polarises Cl- electron clouds. |
| Al2O3 | -13800 | -15100 | Substantial difference: high charge density Al3+ polarises O2- giving significant covalent character. |
| AgI | -736 | -858 | Large discrepancy (16.6% difference): large iodide ion is easily polarised by silver cation, introducing substantial covalent bonding. |
6. Entropy (delta S)
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 (delta S > 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:
delta S = sum[S(products)] - sum[S(reactants)]
7. Gibbs Free Energy (delta G) & Feasibility
Whether a chemical reaction is thermodynamically feasible depends on both the enthalpy change (delta H) and the entropy change (delta S), linked through temperature (T in Kelvin) via the Gibbs free energy equation:
delta G = delta H - T * delta S
delta H is usually given in kJ mol-1, while delta S is given in J K-1 mol-1. To use the Gibbs equation correctly, you MUST convert delta S into kJ K-1 mol-1 by dividing by 1000 before evaluating delta G!
| Enthalpy Change (delta H) | Entropy Change (delta S) | Gibbs Equation Term (-T delta S) | Feasibility Outcome |
|---|---|---|---|
| Exothermic (delta H < 0) | Positive (delta S > 0) | Always negative | Feasible at all temperatures (delta G < 0 always). |
| Endothermic (delta H > 0) | Negative (delta S < 0) | Always positive | Never feasible at any temperature (delta G > 0 always). |
| Endothermic (delta H > 0) | Positive (delta S > 0) | Negative at high T | Feasible only above threshold temperature: T >= delta H / delta S. |
| Exothermic (delta H < 0) | Negative (delta S < 0) | Positive at high T | Feasible only below threshold temperature: T <= delta H / delta S. |
Calculating the Threshold Feasibility Temperature
At the exact boundary where a reaction transitions from non-feasible to feasible, delta G = 0:
0 = delta H - T * delta S => T = delta H / delta S
Linear Graphical Analysis (delta G vs T)
Plotting delta G on the y-axis against absolute temperature T on the x-axis gives a straight line following y = mx + c:
- y-intercept: delta H
- Gradient: -delta S (negative of entropy change)
- x-intercept: Feasibility transition temperature (T = delta H / delta S)
8. Worked Calculations
delta Hf[NaCl(s)] = -411 kJ mol-1
delta Hat[Na(s)] = +107 kJ mol-1
First ionisation energy of Na = +496 kJ mol-1
delta Hat[Cl2(g)] = +122 kJ mol-1
First electron affinity of Cl = -348 kJ mol-1
Step 1: Write out the Hess's Law relationship:
delta Hf = delta Hat(Na) + IE1(Na) + delta Hat(Cl) + delta HEA1(Cl) + delta H_latt
Step 2: Substitute the known values:
-411 = (+107) + (+496) + (+122) + (-348) + delta H_latt
-411 = +377 + delta H_latt
Step 3: Solve for delta H_latt:
delta H_latt = -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)
delta H = +178 kJ mol-1 and delta S = +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:
delta H = +178 kJ mol-1
delta S = +161 / 1000 = +0.161 kJ K-1 mol-1
Step 2: Set delta G = 0 for the feasibility threshold:
delta G = delta H - T * delta S = 0
T = delta H / delta S
Step 3: Calculate the threshold temperature:
T = 178 / 0.161 = 1105.6 K (or 832.6 degrees C)
At temperatures above 1106 K, T * delta S exceeds delta H, causing delta G to become negative and making decomposition feasible.
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 delta H = -85 kJ mol-1 and delta S = -120 J K-1 mol-1. Under what temperature conditions is this reaction thermodynamically feasible?
Show Answer & Explanation
Correct Answer: B
When delta H < 0 and delta S < 0, the -T delta S term is positive. For delta G to remain negative, the temperature must be low enough so that -T delta S does not outweigh the exothermic delta H.
Question 4: On a graph of delta G versus absolute temperature T, what chemical quantities are represented by the gradient and y-intercept?
Show Answer & Explanation
Correct Answer: B
Comparing delta G = -delta S * T + delta H to y = mx + c shows that the gradient m = -delta S and the y-intercept c = delta H.