4.3.2Unit 4. Rates, Equilibria and Further Organic Chemistry
Born-Haber Cycles — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 24 of 56 Unit quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Part 1 of 6The Cycle
A Born–Haber cycle is an application of Hess's law to the formation of an ionic compound.
Principle
- Principle Hess's law states that the enthalpy change between two states is the same whatever route is taken.
Direct route
- Direct route This is the enthalpy of formation of the ionic solid from its elements.
Indirect route
- Indirect route The elements are atomised, ionised, then the gaseous ions form the solid, with lattice energy as the final step.
Equation
- Equation ΔHf = ΔHat(metal) + ΔHat(non-metal) + IE + EA + lattice energy.
Part 2 of 6Building a Cycle
Building a Cycle
Start
- Start Begin with the elements in their standard states at the base of the cycle.
Atomisation
- Atomisation Convert each element into gaseous atoms; for Cl₂ use ½Cl₂(g) → Cl(g).
Ions
- Ions Ionise the metal atom, then add electrons to the non-metal atom to give gaseous ions.
Lattice
- Lattice Form the solid lattice from the gaseous ions, which is the step to be found.
| Step | NaCl / kJ mol⁻¹ |
|---|---|
| ΔHf NaCl(s) | −411 |
| Na(s) → Na(g) | +107 |
| Na(g) → Na⁺(g) + e⁻ | +496 |
| ½Cl₂(g) → Cl(g) | +122 |
| Cl(g) + e⁻ → Cl⁻(g) | −349 |
Part 3 of 6Calculation Method
1
Draw the cycle with the elements at the base and the solid compound below, and put enthalpy values on each arrow.
2
Equate the enthalpy of formation to the sum of all the steps in the indirect route.
3
Rearrange to make lattice energy the subject: lattice energy = ΔHf − (all other steps).
4
Substitute the values with the correct signs and calculate the total carefully.
5
Check that the answer is large and negative, as expected for lattice energy of formation.
Part 4 of 6Theoretical Comparison
Theoretical value
- Theoretical value This is calculated assuming perfectly spherical ions with purely electrostatic attraction.
Experimental value
- Experimental value The Born–Haber value uses measured data and so reflects real bonding.
Differences
- Differences When the experimental value is more exothermic than the theoretical one, the bonding has some covalent character.
Examples
- Examples Sodium chloride matches theory closely, but silver iodide shows a larger difference.
Part 5 of 6Worked Example
Question Use the NaCl data in the table to calculate the lattice energy of formation of sodium chloride.
- ΔHf = ΔHat(Na) + ΔHat(Cl) + IE₁(Na) + EA₁(Cl) + lattice energy.
- −411 = 107 + 122 + 496 + (−349) + lattice energy, and 107 + 122 + 496 − 349 = 376.
- Lattice energy = −411 − 376 = −787 kJ mol⁻¹.
Answer Lattice energy of formation = −787 kJ mol⁻¹
Part 6 of 6Exam Checklist & Tips
4.3.2 – Key Points for the Exam
- Hess's law The enthalpy change of a reaction is the same regardless of the route taken.
- Born–Haber cycle A cycle linking the formation of an ionic solid to its gaseous ions via atomisation, ionisation and electron affinity.
- Lattice energy The enthalpy change when one mole of an ionic solid forms from its gaseous ions.
- Covalent character Partial sharing of electrons in a bond that is mainly ionic.
- Theoretical lattice energy A value calculated from the ionic model alone.
Common Mistakes
- Using the wrong sign for electron affinity or lattice energy in the final sum.
- Forgetting to halve the atomisation of chlorine, which is ½Cl₂ → Cl.
- For a 2+ ion, including only the first ionisation energy and omitting the second.
- Ignoring the number of moles, such as two chlorine atoms in MgCl₂.
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