Calculating from Born–Haber Cycles

Spec Mapping — OCR H432 Module 5.2.1 — Born–Haber cycle calculations, covering the algebraic rearrangement of the Hess-law equation derived in Lesson 2 to solve for any unknown step in the cycle, numerical calculations of lattice enthalpies for Group 1 halides (NaCl, KBr), Group 2 halides (CaCl$_2$2​, MgF$_2$2​) and Group 2 oxides (MgO, CaO) using tabulated standard enthalpies of formation, atomisation, ionisation, and electron affinity; identification and correction of sign errors; comparison of theoretical (purely electrostatic) and experimental (Born–Haber) lattice enthalpies as a probe of covalent character; and application of Born–Haber analysis to explain why certain hypothetical ionic compounds (NaCl$_2$2​, MgCl) do not form (refer to the official OCR H432 specification document for exact wording).

Lesson 2 set up the Born–Haber cycle qualitatively — definitions, signs, and the layout of the cycle for NaCl, MgCl$_2$2​, and MgO. This lesson turns the cycle into a numerical machine: given a table of measurable thermodynamic data, we rearrange the Hess-law equation to solve for the lattice enthalpy (or for any other step, if the lattice enthalpy is known). The single hardest part of any Born–Haber calculation is sign bookkeeping — a stray sign error converts $-788$−788 kJ mol$^{-1}$−1 into $-1486$−1486 kJ mol$^{-1}$−1, and the answer suddenly looks wrong by a factor of two. The structured five-step procedure below — write the Hess equation, rearrange to isolate the unknown, substitute with explicit brackets around every value, simplify term by term, sanity-check against the expected order of magnitude — is designed to eliminate sign errors. We work through six numbered worked examples (NaCl, MgCl$_2$2​, MgO, CaO, KBr-for-EA, hypothetical NaCl$_2$2​), an AO3 comparison of theoretical and experimental lattice enthalpies for NaCl versus AgCl (Fajans's-rules covalent-character argument), and a Year-13 specimen question with tiered model answers.

Key Equation: for any Born–Haber cycle, Hess's law gives a single equation linking the enthalpy of formation to every other step in the indirect route: $\Delta_f H^\ominus = \sum_{\text{indirect steps}} \nu_i \,\Delta H_i$Δf​H⊖=∑indirect steps​νi​ΔHi​ where $\nu_i$νi​ is the stoichiometric coefficient for step $i$i. Rearrange to isolate the unknown step, substitute with explicit brackets around every signed value, and simplify term by term. Sanity-check the final answer against the expected order of magnitude for the charge product of that compound (Lesson 1 table).

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Learning Objectives (OCR Spec 5.2.1)

By the end of this lesson you should be able to:

• Rearrange a Hess-law expression from a Born–Haber cycle to solve for any unknown step (lattice enthalpy, electron affinity, enthalpy of formation, etc.)
• Carry out calculations for Group 1 halides (NaCl, KBr), Group 2 halides (MgCl$_2$2​, CaCl$_2$2​, MgF$_2$2​) and Group 2 oxides (MgO, CaO)
• Apply Born–Haber cycles to determine lattice enthalpies that cannot be measured directly by calorimetry
• Identify and correct sign errors during calculations, particularly around electron affinities and the lattice-enthalpy step itself
• Compare theoretical (purely electrostatic) and experimental (Born–Haber) lattice enthalpies, and attribute discrepancies to covalent character (Fajans's rules)
• Use Born–Haber analysis to explain why hypothetical compounds (NaCl$_2$2​, MgCl) do not form: an AO3 application
• Present numerical answers with correct units (kJ mol$^{-1}$−1) and an appropriate number of significant figures

The Five-Step Strategy

Every Born–Haber calculation in OCR follows the same disciplined procedure. Internalise these five steps and the sign-error problem essentially disappears:

1. Write the Hess-law equation for the cycle, including every indirect-route step. Include a coefficient ($\times 2$×2, $\times 3$×3, ...) on any step that appears more than once in the cycle (e.g. atomisation of Cl for MgCl$_2$2​).
2. Rearrange the equation to isolate the single unknown on the left-hand side. Always subtract; never multiply through by $-1$−1 unnecessarily, as that introduces an extra sign-flip risk.
3. Substitute every value with its sign in explicit brackets — for example, $-(+108)$−(+108) or $-(-349)$−(−349). Brackets force you to think about each sign twice.
4. Simplify term-by-term, collapsing the bracket-and-sign pairs one at a time. Keep an audit trail on the page so you can spot where any error crept in.
5. Sanity-check the answer: is the sign right? Is the order of magnitude right (1+/1- compounds $\sim$∼ $-700$−700 to $-1000$−1000; 2+/2- compounds $\sim$∼ $-3000$−3000 to $-4000$−4000; 3+/2- compounds $\sim$∼ $-15{,}000$−15,000)?

The critical step is sign handling — a single sign error can flip your final lattice enthalpy by $\sim$∼ 700 kJ mol$^{-1}$−1, or from negative to positive. Slow down, track every sign explicitly, and never try to do "minus minus = plus" in your head while also doing the addition. Split the operations.

Key Definitions (for sign-tracking discipline):

• Endothermic step (positive $\Delta H$ΔH): atomisation, ionisation energies, second electron affinity. Goes upward in the energy-level diagram.
• Exothermic step (negative $\Delta H$ΔH): lattice enthalpy (under OCR), first electron affinity (usually), enthalpy of formation (usually). Goes downward.
• Stoichiometric coefficient ($\nu_i$νi​): multiplies the molar enthalpy of each step. For MgCl$_2$2​, atomisation and EA$_1$1​ of Cl each carry a $\nu = 2$ν=2.

Worked Example 1: Lattice Enthalpy of NaCl

Use the following Data-Booklet values to calculate the lattice enthalpy of sodium chloride.

Quantity	Value / kJ mol$^{-1}$−1
$\Delta_f H^\ominus(\text{NaCl})$Δf​H⊖(NaCl)	$-411$−411
$\Delta_{at} H^\ominus(\text{Na})$Δat​H⊖(Na)	$+108$+108
$\Delta_{at} H^\ominus(\text{Cl})$Δat​H⊖(Cl)	$+122$+122
$\text{IE}_1(\text{Na})$IE1​(Na)	$+496$+496
$\text{EA}_1(\text{Cl})$EA1​(Cl)	$-349$−349

Step 1 — Write the Hess-law equation.

$\Delta_f H^\ominus(\text{NaCl}) = \Delta_{at} H^\ominus(\text{Na}) + \Delta_{at} H^\ominus(\text{Cl}) + \text{IE}_1(\text{Na}) + \text{EA}_1(\text{Cl}) + \Delta_{LE} H^\ominus$Δf​H⊖(NaCl)=Δat​H⊖(Na)+Δat​H⊖(Cl)+IE1​(Na)+EA1​(Cl)+ΔLE​H⊖

Step 2 — Rearrange to isolate $\Delta_{LE} H^\ominus$ΔLE​H⊖.

$\Delta_{LE} H^\ominus = \Delta_f H^\ominus - \Delta_{at} H^\ominus(\text{Na}) - \Delta_{at} H^\ominus(\text{Cl}) - \text{IE}_1(\text{Na}) - \text{EA}_1(\text{Cl})$ΔLE​H⊖=Δf​H⊖−Δat​H⊖(Na)−Δat​H⊖(Cl)−IE1​(Na)−EA1​(Cl)

Step 3 — Substitute with explicit brackets.

$\Delta_{LE} H^\ominus = (-411) - (+108) - (+122) - (+496) - (-349)$ΔLE​H⊖=(−411)−(+108)−(+122)−(+496)−(−349)

Step 4 — Simplify term by term.

$\Delta_{LE} H^\ominus = -411 - 108 - 122 - 496 + 349$ΔLE​H⊖=−411−108−122−496+349 $\Delta_{LE} H^\ominus = -788\;\text{kJ mol}^{-1}$ΔLE​H⊖=−788kJ mol−1

Step 5 — Sanity check. The answer is negative (as it must be for any lattice enthalpy under the OCR convention) and its magnitude $\sim 800$∼800 kJ mol$^{-1}$−1 is in the expected band for a 1+/1- salt. The literature value is $-787$−787 kJ mol$^{-1}$−1; the 1 kJ discrepancy is from data-table rounding.

Worked Example 2: Lattice Enthalpy of MgCl$_2$2​

The Group 2 halide cycle introduces three new sign-tracking traps: a second ionisation energy (IE$_2$2​), and a $\times 2$×2 multiplier on both the atomisation of Cl and the electron affinity of Cl.

Quantity	Value / kJ mol$^{-1}$−1
$\Delta_f H^\ominus(\text{MgCl}_2)$Δf​H⊖(MgCl2​)	$-641$−641
$\Delta_{at} H^\ominus(\text{Mg})$Δat​H⊖(Mg)	$+148$+148
$\Delta_{at} H^\ominus(\text{Cl})$Δat​H⊖(Cl)	$+122$+122
$\text{IE}_1(\text{Mg})$IE1​(Mg)	$+738$+738
$\text{IE}_2(\text{Mg})$IE2​(Mg)	$+1451$+1451
$\text{EA}_1(\text{Cl})$EA1​(Cl)	$-349$−349

Step 1 — Hess-law equation (note the $\nu = 2$ν=2 on the chloride terms):

$\Delta_f H^\ominus = \Delta_{at} H^\ominus(\text{Mg}) + 2\,\Delta_{at} H^\ominus(\text{Cl}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + 2\,\text{EA}_1(\text{Cl}) + \Delta_{LE} H^\ominus$Δf​H⊖=Δat​H⊖(Mg)+2Δat​H⊖(Cl)+IE1​(Mg)+IE2​(Mg)+2EA1​(Cl)+ΔLE​H⊖

Step 2 — Rearrange.

$\Delta_{LE} H^\ominus = \Delta_f H^\ominus - \Delta_{at} H^\ominus(\text{Mg}) - 2\,\Delta_{at} H^\ominus(\text{Cl}) - \text{IE}_1(\text{Mg}) - \text{IE}_2(\text{Mg}) - 2\,\text{EA}_1(\text{Cl})$ΔLE​H⊖=Δf​H⊖−Δat​H⊖(Mg)−2Δat​H⊖(Cl)−IE1​(Mg)−IE2​(Mg)−2EA1​(Cl)

Step 3 — Substitute.

$\Delta_{LE} H^\ominus = (-641) - (+148) - 2(+122) - (+738) - (+1451) - 2(-349)$ΔLE​H⊖=(−641)−(+148)−2(+122)−(+738)−(+1451)−2(−349)

Step 4 — Simplify.

$\Delta_{LE} H^\ominus = -641 - 148 - 244 - 738 - 1451 + 698 = -2524\;\text{kJ mol}^{-1}$ΔLE​H⊖=−641−148−244−738−1451+698=−2524kJ mol−1

Step 5 — Sanity check. Charge product $|q_+ q_-| = 2$∣q+​q−​∣=2, so the answer should be $\sim 2$∼2–$3 \times$3× the magnitude of a 1+/1- salt — confirmed. The literature value is $-2526$−2526 kJ mol$^{-1}$−1.

Worked Example 3: Lattice Enthalpy of MgO (the Endothermic EA$_2$2​)

MgO introduces the positive second electron affinity EA$_2$2​(O) — the classic Year-13 discriminator. Both EA$_1$1​ and EA$_2$2​ for oxygen must appear in the cycle; EA$_1$1​ is exothermic ($-141$−141) but EA$_2$2​ is endothermic ($+798$+798) because the electron is being forced onto an already-negative O$^-$− ion.

Quantity	Value / kJ mol$^{-1}$−1
$\Delta_f H^\ominus(\text{MgO})$Δf​H⊖(MgO)	$-602$−602
$\Delta_{at} H^\ominus(\text{Mg})$Δat​H⊖(Mg)	$+148$+148
$\Delta_{at} H^\ominus(\text{O})$Δat​H⊖(O)	$+249$+249
$\text{IE}_1(\text{Mg})$IE1​(Mg)	$+738$+738
$\text{IE}_2(\text{Mg})$IE2​(Mg)	$+1451$+1451
$\text{EA}_1(\text{O})$EA1​(O)	$-141$−141
$\text{EA}_2(\text{O})$EA2​(O)	$+798$+798

Hess equation:

$\Delta_f H^\ominus = \Delta_{at} H^\ominus(\text{Mg}) + \Delta_{at} H^\ominus(\text{O}) + \text{IE}_1 + \text{IE}_2 + \text{EA}_1 + \text{EA}_2 + \Delta_{LE} H^\ominus$Δf​H⊖=Δat​H⊖(Mg)+Δat​H⊖(O)+IE1​+IE2​+EA1​+EA2​+ΔLE​H⊖

Rearrange and substitute:

$\Delta_{LE} H^\ominus = (-602) - (+148) - (+249) - (+738) - (+1451) - (-141) - (+798)$ΔLE​H⊖=(−602)−(+148)−(+249)−(+738)−(+1451)−(−141)−(+798)

$\Delta_{LE} H^\ominus = -602 - 148 - 249 - 738 - 1451 + 141 - 798 = -3845\;\text{kJ mol}^{-1}$ΔLE​H⊖=−602−148−249−738−1451+141−798=−3845kJ mol−1

The literature value is $-3791$−3791 kJ mol$^{-1}$−1; the 54 kJ discrepancy comes from data-table rounding (different sources quote MgO's $\Delta_f H^\ominus$Δf​H⊖ anywhere from $-601$−601 to $-602$−602, and EA$_2$2​(O) from $+798$+798 to $+844$+844). The enormous magnitude — over four times that of NaCl — reflects the doubly charged ions and the short Mg–O distance: charge product 4 versus 1.

Sign-tracking spotlight: notice how EA$_1$1​ (negative) becomes a $+141$+141 contribution to the lattice enthalpy after the subtraction, while EA$_2$2​ (positive) becomes a $-798$−798 contribution. Getting either of these the wrong way round is the most-marked Year-13 sign error in this topic.

Worked Example 4: Lattice Enthalpy of CaO

CaO uses the same algebra as MgO but with calcium replacing magnesium throughout. This is a useful self-check exercise — the cycle structure is identical, only the numerical inputs change.

Quantity	Value / kJ mol$^{-1}$−1
$\Delta_f H^\ominus(\text{CaO})$Δf​H⊖(CaO)	$-635$−635
$\Delta_{at} H^\ominus(\text{Ca})$Δat​H⊖(Ca)	$+178$+178
$\Delta_{at} H^\ominus(\text{O})$Δat​H⊖(O)	$+249$+249
$\text{IE}_1(\text{Ca})$IE1​(Ca)	$+590$+590
$\text{IE}_2(\text{Ca})$IE2​(Ca)	$+1145$+1145
$\text{EA}_1(\text{O})$EA1​(O)	$-141$−141
$\text{EA}_2(\text{O})$EA2​(O)	$+798$+798

Substitute:

$\Delta_{LE} H^\ominus = -635 - 178 - 249 - 590 - 1145 + 141 - 798 = -3454\;\text{kJ mol}^{-1}$ΔLE​H⊖=−635−178−249−590−1145+141−798=−3454kJ mol−1

Literature value $\approx -3401$≈−3401 kJ mol$^{-1}$−1; the 53 kJ discrepancy is again rounding. Notice the comparison to MgO: CaO's lattice enthalpy is less exothermic ($-3454$−3454 vs $-3845$−3845) because Ca$^{2+}$2+ is larger than Mg$^{2+}$2+ (100 pm vs 72 pm) — longer inter-ionic distance, weaker Coulombic attraction. Charge product is identical (both 4), so size is the sole discriminator. This is exactly the kind of "compare and explain" reasoning that Lesson 4 develops in detail.

Worked Example 5: Finding an Unknown Electron Affinity (KBr)

A Born–Haber cycle can be rearranged to solve for any unknown step. Suppose we want EA$_1$1​(Br) — historically the cycle was used precisely this way before EA values became spectroscopically accessible.

Quantity	Value / kJ mol$^{-1}$−1
$\Delta_f H^\ominus(\text{KBr})$Δf​H⊖(KBr)	$-394$−394
$\Delta_{at} H^\ominus(\text{K})$Δat​H⊖(K)	$+90$+90
$\Delta_{at} H^\ominus(\text{Br})$Δat​H⊖(Br)	$+112$+112
$\text{IE}_1(\text{K})$IE1​(K)	$+419$+419
$\Delta_{LE} H^\ominus(\text{KBr})$ΔLE​H⊖(KBr)	$-670$−670

Hess equation:

$\Delta_f H^\ominus = \Delta_{at} H^\ominus(\text{K}) + \Delta_{at} H^\ominus(\text{Br}) + \text{IE}_1 + \text{EA}_1(\text{Br}) + \Delta_{LE} H^\ominus$Δf​H⊖=Δat​H⊖(K)+Δat​H⊖(Br)+IE1​+EA1​(Br)+ΔLE​H⊖

Rearrange for EA$_1$1​(Br):

$\text{EA}_1(\text{Br}) = \Delta_f H^\ominus - \Delta_{at} H^\ominus(\text{K}) - \Delta_{at} H^\ominus(\text{Br}) - \text{IE}_1 - \Delta_{LE} H^\ominus$EA1​(Br)=Δf​H⊖−Δat​H⊖(K)−Δat​H⊖(Br)−IE1​−ΔLE​H⊖

Substitute:

$\text{EA}_1(\text{Br}) = (-394) - (+90) - (+112) - (+419) - (-670) = -394 - 90 - 112 - 419 + 670 = -345\;\text{kJ mol}^{-1}$EA1​(Br)=(−394)−(+90)−(+112)−(+419)−(−670)=−394−90−112−419+670=−345kJ mol−1

This matches the accepted spectroscopic EA$_1$1​(Br) of $-325$−325 kJ mol$^{-1}$−1 to within the 20 kJ that data-table rounding allows. The negative sign confirms EA$_1$1​ is exothermic (electron attracted to nucleus of neutral Br atom).

Worked Example 6: Why NaCl$_2$2​ Does Not Exist (AO3)

The Born–Haber cycle can be run predictively to test the existence of hypothetical compounds. If $\Delta_f H^\ominus$Δf​H⊖ comes out hugely positive, the compound is thermodynamically forbidden. Consider hypothetical "NaCl$_2$2​" containing Na$^{2+}$2+:

Data:

• $\Delta_{at} H^\ominus(\text{Na}) = +108$Δat​H⊖(Na)=+108, $\Delta_{at} H^\ominus(\text{Cl}) = +122$Δat​H⊖(Cl)=+122 kJ mol$^{-1}$−1
• $\text{IE}_1(\text{Na}) = +496$IE1​(Na)=+496, $\text{IE}_2(\text{Na}) = +4562$IE2​(Na)=+4562 kJ mol$^{-1}$−1 (enormous — second electron from inner 2p shell)
• $\text{EA}_1(\text{Cl}) = -349$EA1​(Cl)=−349 kJ mol$^{-1}$−1
• Estimated $\Delta_{LE} H^\ominus(\text{NaCl}_2) \approx -2300$ΔLE​H⊖(NaCl2​)≈−2300 kJ mol$^{-1}$−1 (rough estimate by Kapustinskii)

Predicted $\Delta_f H^\ominus$Δf​H⊖:

$\Delta_f H^\ominus = (+108) + 2(+122) + (+496) + (+4562) + 2(-349) + (-2300)$Δf​H⊖=(+108)+2(+122)+(+496)+(+4562)+2(−349)+(−2300) $\Delta_f H^\ominus = 108 + 244 + 496 + 4562 - 698 - 2300 = +2412\;\text{kJ mol}^{-1}$Δf​H⊖=108+244+496+4562−698−2300=+2412kJ mol−1

Hugely positive — NaCl$_2$2​ would be wildly unstable with respect to its elements. The reason is plain from the Hess sum: IE$_2$2​(Na) of +4562 kJ mol$^{-1}$−1 dominates, because removing a second electron from sodium means stripping an electron from the inner 2p shell (next noble-gas core), at a cost of more than ten times IE$_1$1​. No realistic lattice enthalpy can compensate for that. This is precisely why sodium is observed only as Na$^+$+ in nature: thermodynamics, via the Born–Haber cycle, forbids Na$^{2+}$2+ ionic compounds. The same logic explains why MgO forms readily (IE$_2$2​(Mg) only 1451 — the second electron is still from the valence shell) but MgF$_3$3​ does not (IE$_3$3​(Mg) is $\sim 7733$∼7733 kJ mol$^{-1}$−1 — stripping from the Ne core).

Theoretical vs Experimental Lattice Enthalpies (AO3)

Lattice enthalpies obtained from a Born–Haber cycle are sometimes called experimental lattice enthalpies — every other step in the cycle is measurable, so the cycle effectively "transfers" experimental data into the lattice term. An alternative is the theoretical lattice enthalpy, calculated from a purely electrostatic model (point charges interacting via Coulomb's law, summed over the crystal geometry):

$\Delta_{LE} H^\ominus_{\text{theoretical}} \propto -\frac{|q_+ q_-|}{r_+ + r_-} \cdot M$ΔLE​Htheoretical⊖​∝−r+​+r−​∣q+​q−​∣​⋅M

where $M$M is the Madelung constant for the crystal structure. The theoretical value assumes the bonding is purely ionic: ions are perfect spheres with their full formal charge, no electron-density overlap between cation and anion.

| Compound | Experimental $\Delta_{LE} H$ΔLE​H / kJ mol$^{-1}$−1 | Theoretical $\Delta_{LE} H$ΔLE​H / kJ mol$^{-1}$−1 | $|\text{diff}|$∣diff∣ | |----------|-----------------------------------------:|---------------------------------------:|:---------------:| | NaF | $-918$−918 | $-895$−895 | 23 | | NaCl | $-787$−787 | $-766$−766 | 21 | | NaBr | $-751$−751 | $-732$−732 | 19 | | AgCl | $-905$−905 | $-770$−770 | 135 | | AgBr | $-878$−878 | $-758$−758 | 120 | | AgI | $-889$−889 | $-736$−736 | 153 | | LiI | $-759$−759 | $-728$−728 | 31 |