The Born-Haber cycle

Result

The Born-Haber cycle

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Statement

Let \(M_pX_q(s)\) be a binary ionic solid built from cations \(M^{z_+}\) and anions \(X^{z_-}\), with the stoichiometry fixed by electroneutrality, \(p\,z_+ = q\,z_-\). Suppose every species is taken in its standard state at one common temperature \(T\) and pressure \(p^{\circ}\), that a single convention for the enthalpy of the free electron is used throughout, and that the standard enthalpy of formation of the solid, the atomisation enthalpies of both elements, the \(z_+\) successive ionisation enthalpies of \(M\) and the \(z_-\) successive electron-gain enthalpies of \(X\) are all known at that same \(T\). Then, because enthalpy is a function of state, the standard lattice enthalpy \(\Delta H^{\circ}_{\mathrm{L}}\) — the enthalpy change of \(p\,M^{z_+}(g) + q\,X^{z_-}(g) \to M_pX_q(s)\), which has no direct experimental route — is determined exactly by the closure of a thermodynamic cycle: \(\Delta H^{\circ}_{\mathrm{L}} = \Delta H^{\circ}_{\mathrm{f}} - p\big[\Delta H^{\circ}_{\mathrm{at}}(M) + \sum_i I_i\big] - q\big[\Delta H^{\circ}_{\mathrm{at}}(X) + \sum_j E_j\big]\).

Why it matters

Two of the quantities this unit spends its time on — ionisation energy and electron affinity — are properties of isolated gas-phase atoms, measured by spectroscopy and by photodetachment, far from any beaker. The Born-Haber cycle is the bridge that carries them into bulk thermochemistry: it is the only route by which a number extracted from an atomic spectrum can be checked against the heat evolved when a metal burns in chlorine. That is why the cycle belongs beside the periodic trends rather than after them; it is the audit that makes those trends thermochemically consequential.

It runs in both directions, and the reverse direction is the historically important one. Second and third electron affinities cannot be measured at all — a doubly charged anion such as \(\mathrm{O^{2-}}(g)\) is not even a bound species in isolation, so there is no beam of them to photodetach — and the tabulated values for them exist only because a lattice enthalpy from an independent electrostatic model was fed into a cycle like this one and the remaining unknown solved for. The same manoeuvre prices hypothetical compounds (\(\mathrm{CaCl}\), \(\mathrm{NaCl_2}\)) that have never been made, and so explains why the oxidation states we observe are the ones we observe.

Hypotheses

Enthalpy is a function of state.\(H = U + pV\) is built from state functions alone, so \(\Delta H\) between two states is path-independent and the sum of \(\Delta H\) around any closed cycle is exactly zero. Drop this and the six-legged path has no computable relation to the one-legged direct path; the cycle becomes an arbitrary list of reactions whose sum means nothing. Everything else in the derivation is bookkeeping.

Every leg is referred to the same temperature and standard pressure.Standard tables are almost always \(T = 298.15\,\mathrm{K}\), \(p^{\circ} = 1\,\mathrm{bar}\); spectroscopic ionisation energies and electron affinities are \(0\,\mathrm{K}\) energies. Mixing the two silently inserts the thermal correction of each leg as an error — small per leg but systematic, and it grows with the ionic charge because the number of electron-carrying legs grows with it.

Electroneutrality of the formula unit, \(p\,z_+ = q\,z_-\).This is what makes the free electrons cancel between the ionisation legs and the electron-attachment legs. Without it the cycle does not close on a neutral solid at all: an unbalanced count of electrons is left over, and the residue depends on an arbitrary convention rather than on chemistry.

One convention for the enthalpy of the free electron, used in every leg.Under the electron convention the electron is an ideal monatomic gas, so \(H(\mathrm{e^-},T) - H(\mathrm{e^-},0) = \tfrac{5}{2}RT\); under the ion convention its enthalpy is defined to be zero at all \(T\). Individual ionisation and electron-gain enthalpies differ by \(\tfrac{5}{2}RT = 6.20\,\mathrm{kJ\,mol^{-1}}\) at \(298.15\,\mathrm{K}\) between the two, so tabulated values must never be mixed — although, by Step 7 below, \(\Delta H^{\circ}_{\mathrm{L}}\) itself is invariant once one convention is applied consistently.

The tabulated \(\Delta H^{\circ}_{\mathrm{f}}\) refers to the same solid phase the lattice enthalpy is being asked about.Polymorphs differ: the cycle returns the lattice enthalpy of whichever phase the formation enthalpy was measured for. For a compound with a phase transition below \(298\,\mathrm{K}\), or for one tabulated in a different structure type, the extracted \(\Delta H^{\circ}_{\mathrm{L}}\) is the wrong crystal's.

Interpreting \(\Delta H^{\circ}_{\mathrm{L}}\) as an electrostatic lattice energy needs the ions to be real, integrally charged species.The cycle itself is exact thermochemistry whatever the bonding is, because it only ever adds measured enthalpies. But if the solid has substantial covalent character the number returned is still correct and still meaningful — it simply no longer equals what a point-charge model would predict, and the gap is the diagnostic (see the Corollaries).

Proof

The argument is one physical idea (path independence) followed by careful accounting. We construct a closed loop whose vertices are: the elements in their standard states, the free gaseous atoms, the free gaseous ions, and the solid. Five legs of the loop are measurable; the sixth is the target.

\[ H \equiv U + pV \quad\Longrightarrow\quad \oint \mathrm{d}H = 0 \]

\(U\), \(p\) and \(V\) are functions of state, so \(H\) is one; the integral of an exact differential around any closed path vanishes. Equivalently, for any two routes between the same initial and final states, \(\sum \Delta H\) is the same — Hess's law, stated as a property of \(H\) rather than as an empirical rule. B

\[ p\,M(s) + \tfrac{q}{a}\,X_a \;\xrightarrow{\ \Delta H^{\circ}_{\mathrm{f}}\ }\; M_pX_q(s) \]

The direct leg: formation of one mole of the solid from its elements in their standard states, with \(X_a\) the standard-state form of the non-metal — \(a=2\) for the diatomic gases \(\mathrm{Cl_2}\) and \(\mathrm{O_2}\), \(a=8\) for orthorhombic \(\mathrm{S_8}\), \(a=2\) for solid \(\mathrm{I_2}\), the last two being condensed rather than gaseous in their standard states. The value of \(a\) never survives into the Result, because Step 3 defines atomisation per mole of gaseous atoms produced and so absorbs it. This leg is calorimetric and tabulated. A

\[ p\,M(s) \xrightarrow{\ p\,\Delta H^{\circ}_{\mathrm{at}}(M)\ } p\,M(g), \qquad \tfrac{q}{a}X_a \xrightarrow{\ q\,\Delta H^{\circ}_{\mathrm{at}}(X)\ } q\,X(g) \]

Atomisation of both elements. Defining each by its enthalpy per mole of gaseous atoms produced absorbs the stoichiometric factors once and for all: \(\Delta H^{\circ}_{\mathrm{at}}(X) = \tfrac12 D(X_2)\) for a diatomic gas, and equals the sublimation enthalpy for a metal. Both are necessarily endothermic, since separating atoms against any attractive interaction costs energy. A

\[ p\,M(g) \;\longrightarrow\; p\,M^{z_+}(g) + p\,z_+\,\mathrm{e^-}(g), \qquad \Delta H = p\sum_{i=1}^{z_+} I_i \]

Successive ionisation, one electron at a time, \(I_i\) being the enthalpy of \(M^{(i-1)+}(g) \to M^{i+}(g) + \mathrm{e^-}\). Each \(I_i\) is positive and \(I_{i+1} \gt I_i\), because the electron is being removed from a species of higher positive charge and (at a shell boundary) from a more tightly bound shell — the same staircase this unit's ionisation result reads off the periodic table. A

\[ q\,X(g) + q\,z_-\,\mathrm{e^-}(g) \;\longrightarrow\; q\,X^{z_-}(g), \qquad \Delta H = q\sum_{j=1}^{z_-} E_j \]

Successive electron attachment, \(E_j\) the enthalpy of \(X^{(j-1)-}(g) + \mathrm{e^-} \to X^{j-}(g)\). \(E_1\) is usually negative (a neutral atom binds one extra electron); every \(E_j\) with \(j \ge 2\) is strongly positive, since the incoming electron is pushed against a net negative charge. Note the sign convention: these are electron-gain enthalpies, not electron affinities, which are quoted with the opposite sign. A

\[ p\,M^{z_+}(g) + q\,X^{z_-}(g) \;\xrightarrow{\ \Delta H^{\circ}_{\mathrm{L}}\ }\; M_pX_q(s) \]

The target leg, which closes the loop back onto the solid reached in Step 2. It is not directly measurable: a mole of separated gaseous ions at \(1\,\mathrm{bar}\) is not a preparable state, so no calorimeter can be built around this reaction. Everything else in the cycle exists to isolate it. A

\[ \text{electrons released} = p\,z_+ \;=\; q\,z_- = \text{electrons consumed} \]

This is the electroneutrality hypothesis, and it is what makes the cycle well posed. If the free electron is assigned molar enthalpy \(\varepsilon\) (whatever the convention), Step 4 carries \(+p\,z_+\varepsilon\) and Step 5 carries \(-q\,z_-\varepsilon\); the two are equal and opposite, so \(\varepsilon\) cancels identically from the sum and \(\Delta H^{\circ}_{\mathrm{L}}\) is convention-independent. What is not permitted is using \(I_i\) from one convention and \(E_j\) from the other, which leaves an uncancelled residue of \(p\,z_+\,\tfrac52 RT\). C

\[ \begin{aligned} \Delta H^{\circ}_{\mathrm{f}} &= p\,\Delta H^{\circ}_{\mathrm{at}}(M) + p\sum_{i=1}^{z_+} I_i \\ &\quad + q\,\Delta H^{\circ}_{\mathrm{at}}(X) + q\sum_{j=1}^{z_-} E_j + \Delta H^{\circ}_{\mathrm{L}} \end{aligned} \]

Steps 3–6 form a route from the same initial state as Step 2 (elements in their standard states) to the same final state (the solid). By Step 1 the two routes have equal enthalpy change, so the five-term sum equals \(\Delta H^{\circ}_{\mathrm{f}}\). No new physics enters here; this single equation is the whole content of the cycle. B

\[ \Delta H^{\circ}_{\mathrm{L}} = \Delta H^{\circ}_{\mathrm{f}} - p\left[\Delta H^{\circ}_{\mathrm{at}}(M) + \sum_{i=1}^{z_+} I_i\right] - q\left[\Delta H^{\circ}_{\mathrm{at}}(X) + \sum_{j=1}^{z_-} E_j\right] \]

Rearranged for the one unknown. The equation is linear in every input, which is worth noting twice over: any single term may be made the subject instead (the inverse use), and the uncertainties combine in quadrature with unit sensitivity coefficients, so no term's error is amplified. A

\[ \Delta U^{\circ}_{\mathrm{L}} = \Delta H^{\circ}_{\mathrm{L}} - \Delta n_{\mathrm{gas}} RT = \Delta H^{\circ}_{\mathrm{L}} + (p+q)\,RT \]

The cycle delivers an enthalpy; an electrostatic model such as the Born-Landé calculation delivers an internal energy. Converting between them uses \(\Delta H = \Delta U + \Delta(pV)\) with \(\Delta(pV) = \Delta n_{\mathrm{gas}} RT\) for ideal gases and negligible \(pV\) for the solid; here \(p+q\) moles of gas are consumed, so \(\Delta n_{\mathrm{gas}} = -(p+q)\) and the lattice energy is the less negative of the two by \((p+q)RT\), which is \(4.96\,\mathrm{kJ\,mol^{-1}}\) for a \(1{:}1\) salt at \(298.15\,\mathrm{K}\). B

Result

Reading. The lattice enthalpy of \(M_pX_q\) is whatever is needed to make the atomising-then-ionising route agree with the direct formation route. Each bracket is one element's full cost of being converted from its standard state into the gaseous ion the crystal actually contains, and each is multiplied by how many of those ions the formula unit holds.

Scope. Any binary ionic solid whose five input quantities are known at a common \(T\) and under a common electron convention. It reduces to the familiar five-term form for a \(1{:}1\) salt (\(p=q=z_+=z_-=1\)), and generalises unchanged to ternary compounds if the extra element's atomisation and ionisation terms are appended. Convert to a lattice energy with \(\Delta U^{\circ}_{\mathrm{L}} = \Delta H^{\circ}_{\mathrm{L}} + (p+q)RT\) before comparing with an electrostatic calculation.

Corollaries & converses

Inverse use. Because the relation is linear with unit coefficients, any one term can be made the subject given the other five. This is the only route to second and third electron-gain enthalpies, since \(\mathrm{O^{2-}}(g)\) and \(\mathrm{N^{3-}}(g)\) are unbound in isolation and cannot be prepared for measurement.
Ionic-model test. Comparing the cycle's \(\Delta U^{\circ}_{\mathrm{L}}\) against a point-charge electrostatic value tests the ionic model itself. Agreement within a few per cent (alkali halides) supports it; a thermochemical value substantially more negative than the electrostatic one (silver and thallium halides, zinc and lead chalcogenides) is the standard evidence for covalent contributions, and correlates with cation polarisability exactly as the soft-acid/soft-base picture predicts.
Pricing compounds that do not exist. Estimating \(\Delta H^{\circ}_{\mathrm{L}}\) for a hypothetical stoichiometry (from ionic radii and a Kapustinskii- or Born-Landé-type formula) and running the cycle forwards gives \(\Delta H^{\circ}_{\mathrm{f}}\) for a compound never made. Comparing it against the observed stoichiometry decides which oxidation state wins; see the fourth problem.
Uncertainty budget. With uncorrelated inputs, \(u(\Delta H^{\circ}_{\mathrm{L}})^2 = u(\Delta H^{\circ}_{\mathrm{f}})^2 + p^2\big[u_{\mathrm{at}}^2 + \sum_i u_{I_i}^2\big] + q^2\big[u_{\mathrm{at}}^2 + \sum_j u_{E_j}^2\big]\). The stoichiometric coefficients enter squared, so a \(2{:}3\) compound inherits its inputs' errors amplified, not averaged.
Converse. Two independent lattice enthalpies for the same solid (thermochemical and electrostatic) over-determine the cycle; the residual is a consistency check on the input tables rather than a new physical quantity, and historically flagged several bad electron-affinity values.

Fails without

Common temperature dropped: feeding \(0\,\mathrm{K}\) spectroscopic ionisation energies into a \(298.15\,\mathrm{K}\) cycle. Under the electron convention each ionisation leg is short by \(\tfrac52 RT = 6.20\,\mathrm{kJ\,mol^{-1}}\), so \(\mathrm{MgCl_2}\) picks up a systematic \(2 \times 6.20 = 12.4\,\mathrm{kJ\,mol^{-1}}\) and \(\mathrm{Al_2O_3}\) a \(6 \times 6.20 = 37.2\,\mathrm{kJ\,mol^{-1}}\) bias — invisible in any single calculation and fatal when the residual of the cycle is the quantity of interest.
Electroneutrality dropped: if the ion charges assumed do not satisfy \(p\,z_+ = q\,z_-\) — writing the \(\mathrm{MgCl_2}\) cycle with one chloride, say — the electron counts no longer cancel in Step 7, the loop never returns to the neutral solid, and the extracted number is wrong by the whole of the omitted leg plus an arbitrary electron-convention residue.
Integral ionic charges dropped: for \(\mathrm{AgI}\) or \(\mathrm{ZnS}\) the arithmetic still returns a perfectly good enthalpy, but calling it “the electrostatic lattice energy” is meaningless because the crystal does not contain \(\mathrm{Ag^+}\) and \(\mathrm{I^-}\) as point charges. Any radius-based prediction of that number will then be badly out, and the discrepancy is chemistry, not experimental error.
Phase identity dropped: using \(\Delta H^{\circ}_{\mathrm{f}}\) for one polymorph and radii for another returns the lattice enthalpy of the phase the calorimetry was done on, not the phase being modelled. The two differ by the polymorph transition enthalpy, typically a few \(\mathrm{kJ\,mol^{-1}}\) — small, but the same size as the electron-convention residue it might be confused with.

Common errors

“Use the bond enthalpy of \(\mathrm{Cl_2}\).” The cycle needs enthalpy per mole of atoms: \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Cl}) = \tfrac12 D(\mathrm{Cl_2}) = 122\,\mathrm{kJ\,mol^{-1}}\), not \(244\). Defining atomisation per mole of atoms from the start, as in Step 3, removes this trap entirely.
“Electron affinity is \(+349\,\mathrm{kJ\,mol^{-1}}\), so add it.” The electron affinity of chlorine is \(+349\,\mathrm{kJ\,mol^{-1}}\) (energy released); the electron-gain enthalpy that the cycle needs is \(-349\,\mathrm{kJ\,mol^{-1}}\). The two names differ by a sign, so the error is \(2\times349 = 698\,\mathrm{kJ\,mol^{-1}}\) per mole of chloride — which in \(\mathrm{MgCl_2}\), where \(q=2\), is a \(1396\,\mathrm{kJ\,mol^{-1}}\) error in the extracted lattice enthalpy.
“Lattice enthalpy is positive.” Both conventions are in print: lattice formation enthalpy (gaseous ions \(\to\) solid) is large and negative; lattice dissociation enthalpy is the same magnitude, positive. Neither is wrong, but a cycle that mixes them changes the sign of its largest term.
Forgetting the multiplicities \(p\) and \(q\). Only the ionisation sum is commonly remembered as needing repetition; the anion's atomisation term needs the same factor \(q\). In \(\mathrm{MgCl_2}\) both the \(122\) and the \(-349\) must be doubled.
Reporting \(\Delta H^{\circ}_{\mathrm{L}}\) and \(\Delta U^{\circ}_{\mathrm{L}}\) as the same number. They differ by \((p+q)RT\): about \(5\,\mathrm{kJ\,mol^{-1}}\) for \(\mathrm{NaCl}\), \(7\) for \(\mathrm{MgCl_2}\). Negligible against \(-787\), but not against the \(20\,\mathrm{kJ\,mol^{-1}}\) discrepancies used to argue for covalency.
Over-precision. Quoting a cycle result to \(0.1\,\mathrm{kJ\,mol^{-1}}\) when the electron-gain enthalpy that went in is known to \(\pm 3\). The quadrature rule in the Corollaries sets the honest number of figures.

Discussion

Max Born and Fritz Haber published the construction independently in 1919, and the timing is not accidental: it arrived with the first quantitative electrostatic lattice-energy calculations, and the two were built to be compared. That comparison was the decisive early test of the ionic model of solids. Neither number could be measured directly, but they were computed from disjoint bodies of evidence — one from calorimetry and atomic spectra, the other from X-ray-determined interionic distances and compressibilities — so their agreement for the alkali halides was not something either construction could have arranged.

The cycle's real methodological interest is that it is an instrument for measuring the unmeasurable, and its accuracy is entirely inherited. It introduces no approximation of its own; Step 8 is an identity, not a model. So every kilojoule of error in the answer is a kilojoule of error carried in from a table, which is why the uncertainty budget in the Corollaries is not pedantry but the whole quality statement of the result. It also means the cycle cannot be “improved”: one improves the inputs.

The subtlety that trips up careful workers is the electron. There is no experiment that measures the enthalpy of a free electron, so a convention must be adopted, and two are in circulation: the electron convention treats \(\mathrm{e^-}\) as an ideal monatomic gas with \(H(T)-H(0) = \tfrac52 RT\), while the ion convention sets its enthalpy to zero at every temperature. Tabulated ionisation and electron-gain enthalpies therefore differ between sources by \(6.20\,\mathrm{kJ\,mol^{-1}}\) per electron at \(298.15\,\mathrm{K}\). Step 7 shows why this rarely bites: electroneutrality makes the electron terms cancel exactly, so the cycle's output is convention-free. The danger is only in mixing sources — and it is a systematic, sign-definite error, exactly the kind that survives averaging over many compounds and can masquerade as a physical trend in a plot of “covalent contribution” against cation charge.

Common misconceptions. First, that the cycle predicts or explains lattice enthalpy. It does neither: it measures it, by difference. Why lattice enthalpies are large and negative, and why they scale as \(z_+z_-/r_0\), is the business of the electrostatic model, which the cycle only cross-checks. Second, that a large negative \(\Delta H^{\circ}_{\mathrm{L}}\) is by itself the reason an ionic solid forms — formation is governed by \(\Delta G\), and for dissolution or for high-temperature stability the \(T\Delta S\) term routinely overturns the enthalpic ordering. Third, that the electron-gain enthalpy of oxygen being positive overall (\(-141 + 748 = +607\,\mathrm{kJ\,mol^{-1}}\) to reach \(\mathrm{O^{2-}}\)) is a paradox: it simply means oxide formation is paid for entirely by the lattice term, which is why \(\mathrm{O^{2-}}\) exists in crystals and never in the gas phase.

Worked examples

Example 1. Find the standard lattice enthalpy of magnesium chloride at \(298.15\,\mathrm{K}\), and convert it to a lattice energy for comparison with an electrostatic model. Standard values, all in \(\mathrm{kJ\,mol^{-1}}\): \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Mg}) = +148\); \(I_1(\mathrm{Mg}) = +738\), \(I_2(\mathrm{Mg}) = +1451\); \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Cl}) = \tfrac12 D(\mathrm{Cl_2}) = +122\); \(E_1(\mathrm{Cl}) = -349\); \(\Delta H^{\circ}_{\mathrm{f}}(\mathrm{MgCl_2},s) = -641\).

\[ p = 1,\ q = 2,\ z_+ = 2,\ z_- = 1 \quad\Longrightarrow\quad p\,z_+ = 2 = q\,z_- \]

Fix the stoichiometry first and check electroneutrality, which is the hypothesis that licenses Step 7 of the Proof. Two electrons leave magnesium and two are taken up by the two chlorines. A

\[ p\left[\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Mg}) + I_1 + I_2\right] = 1\,(148 + 738 + 1451) = +2337\,\mathrm{kJ\,mol^{-1}} \]

The cation bracket: the full cost of turning one mole of magnesium metal into one mole of gaseous \(\mathrm{Mg^{2+}}\). Both ionisations are required because the crystal contains \(\mathrm{Mg^{2+}}\), not \(\mathrm{Mg^+}\). A

\[ q\left[\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Cl}) + E_1\right] = 2\,(122 - 349) = 2(-227) = -454\,\mathrm{kJ\,mol^{-1}} \]

The anion bracket, with \(q = 2\) applied to both terms inside it. The bracket is net exothermic because chlorine's electron-gain enthalpy outweighs half the \(\mathrm{Cl}\)–\(\mathrm{Cl}\) bond enthalpy. A

\[ \Delta H^{\circ}_{\mathrm{L}} = -641 - (+2337) - (-454) = -641 - 1883 = -2524\,\mathrm{kJ\,mol^{-1}} \]

The Result, with symbols cleared before numbers were substituted. The non-lattice legs cost \(+1883\,\mathrm{kJ\,mol^{-1}}\) in total, yet formation is exothermic by \(641\) — the lattice term alone pays for both, which is the quantitative form of the statement that ionic solids exist because of their lattices. A

\[ \Delta U^{\circ}_{\mathrm{L}} = \Delta H^{\circ}_{\mathrm{L}} + (p+q)RT = -2524 + 3(8.314)(298.15)\times10^{-3} = -2524 + 7.4 = -2517\,\mathrm{kJ\,mol^{-1}} \]

Step 10 of the Proof, with \(RT = 2.479\,\mathrm{kJ\,mol^{-1}}\) at \(298.15\,\mathrm{K}\). Three moles of gas are consumed, so the correction is \(3RT\). Only the internal energy may be set beside a Born-Landé number. B

\[ \Delta H^{\circ}_{\mathrm{L}}(\mathrm{MgCl_2}) = -2524\,\mathrm{kJ\,mol^{-1}}, \qquad \Delta U^{\circ}_{\mathrm{L}}(\mathrm{MgCl_2}) = -2517\,\mathrm{kJ\,mol^{-1}} \]

Reading. Roughly \(3.2\) times the magnitude of \(\mathrm{NaCl}\)'s \(-787\,\mathrm{kJ\,mol^{-1}}\), which is what the doubled cation charge and the doubled ion count in the formula unit together demand.

Scope. The inputs are rounded to the nearest \(\mathrm{kJ\,mol^{-1}}\), so the output is trustworthy to a few \(\mathrm{kJ\,mol^{-1}}\); quoting \(-2524\) rather than \(-2500\) is justified only because every input carried at least that precision.

Example 2. The second electron-gain enthalpy of oxygen, \(\mathrm{O^-}(g) + \mathrm{e^-} \to \mathrm{O^{2-}}(g)\), cannot be measured: \(\mathrm{O^{2-}}(g)\) is unbound. Extract it from the magnesium oxide cycle, given an independent electrostatic lattice enthalpy \(\Delta H^{\circ}_{\mathrm{L}}(\mathrm{MgO}) = -3795\,\mathrm{kJ\,mol^{-1}}\), and state its uncertainty if that model value carries a \(1\%\) uncertainty while each thermochemical input carries \(\pm 2\,\mathrm{kJ\,mol^{-1}}\). Further data, in \(\mathrm{kJ\,mol^{-1}}\): \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Mg}) = +148\), \(I_1 = +738\), \(I_2 = +1451\), \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{O}) = \tfrac12 D(\mathrm{O_2}) = +249\), \(E_1(\mathrm{O}) = -141\), \(\Delta H^{\circ}_{\mathrm{f}}(\mathrm{MgO},s) = -602\).

\[ E_2 = \Delta H^{\circ}_{\mathrm{f}} - \left[\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Mg}) + I_1 + I_2\right] - \left[\Delta H^{\circ}_{\mathrm{at}}(\mathrm{O}) + E_1\right] - \Delta H^{\circ}_{\mathrm{L}} \]

Make the unknown the subject before any number is written. Here \(p = q = 1\) and \(z_+ = z_- = 2\), so the anion bracket contains two electron-gain terms and only \(E_2\) is unknown; the linearity noted in Step 9 of the Proof means this rearrangement costs nothing. B

\[ \left[148 + 738 + 1451\right] = +2337, \qquad \left[249 - 141\right] = +108 \]

The two known brackets, evaluated separately as a guard against sign slips. The cation bracket is identical to Example 1's because it is the same metal in the same oxidation state — a useful check that the tables are being read consistently. A

\[ E_2 = -602 - 2337 - 108 - (-3795) = -3047 + 3795 = +748\,\mathrm{kJ\,mol^{-1}} \]

Substitution. The sign is the expected one and its size is reasonable: forcing a second electron onto an already-anionic \(\mathrm{O^-}\) costs several times what the first one released, precisely because the incoming electron works against a net negative charge. A

\[ u(E_2) = \sqrt{(0.01 \times 3795)^2 + 6\,(2)^2} = \sqrt{1440 + 24} \approx 38\,\mathrm{kJ\,mol^{-1}} \]

Quadrature, with unit sensitivity coefficients from the Corollaries and six thermochemical inputs at \(\pm2\) each. The model lattice enthalpy dominates so completely that the thermochemistry contributes under \(2\%\) of the variance — the extracted electron-gain enthalpy is only as good as the electrostatic model behind it. C

\[ E_2(\mathrm{O}) = \mathrm{O^-}(g) + \mathrm{e^-} \to \mathrm{O^{2-}}(g): \quad +748 \pm 38\,\mathrm{kJ\,mol^{-1}} \]

Reading. Consistent with the commonly tabulated \(+744\,\mathrm{kJ\,mol^{-1}}\), and the agreement is not evidence of much, because the stated uncertainty is ten times the difference. What the calculation really shows is that the quantity is inaccessible except through a cycle, and that its published value is a model-dependent one.

Scope. The same inversion recovers any single missing term. It is only legitimate when exactly one term is unknown: with two unknowns the cycle is one equation short and no amount of rearrangement helps.

Problems

Find the standard lattice enthalpy of potassium chloride, given (in \(\mathrm{kJ\,mol^{-1}}\)) \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{K}) = +89\), \(I_1(\mathrm{K}) = +419\), \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Cl}) = +122\), \(E_1(\mathrm{Cl}) = -349\), \(\Delta H^{\circ}_{\mathrm{f}}(\mathrm{KCl},s) = -437\). Compare with \(\mathrm{NaCl}\)'s \(-787\,\mathrm{kJ\,mol^{-1}}\) and account for the difference.
Solution
Here \(p=q=z_+=z_-=1\), so the Result reduces to the five-term form. Cation bracket: \(89 + 419 = +508\). Anion bracket: \(122 - 349 = -227\). Sum of the non-lattice legs: \(508 - 227 = +281\,\mathrm{kJ\,mol^{-1}}\).
\(\Delta H^{\circ}_{\mathrm{L}} = -437 - 281 = -718\,\mathrm{kJ\,mol^{-1}}\), in line with tabulated values near \(-715\,\mathrm{kJ\,mol^{-1}}\).
It is \(69\,\mathrm{kJ\,mol^{-1}}\) less negative than \(\mathrm{NaCl}\)'s. Charges are identical (\(z_+z_- = 1\) for both) and both adopt the rock-salt structure, so the Madelung constant is the same; the only variable left is the interionic separation, and \(\mathrm{K^+}\) is the larger cation. Since the electrostatic term scales as \(1/r_0\), a larger \(r_0\) gives a smaller magnitude — the ionic-radius trend down a group, read backwards out of a calorimeter.
Sodium chloride's lattice enthalpy is \(-787\,\mathrm{kJ\,mol^{-1}}\) at \(298.15\,\mathrm{K}\). Convert it to a lattice energy, and say whether the correction matters when testing an electrostatic model that predicts \(-756\,\mathrm{kJ\,mol^{-1}}\).
Solution
The association reaction \(\mathrm{Na^+}(g) + \mathrm{Cl^-}(g) \to \mathrm{NaCl}(s)\) consumes \(p + q = 2\) moles of gas, so \(\Delta n_{\mathrm{gas}} = -2\) and, by Step 10, \(\Delta U^{\circ}_{\mathrm{L}} = \Delta H^{\circ}_{\mathrm{L}} + 2RT\).
\(2RT = 2(8.314)(298.15) = 4958\,\mathrm{J\,mol^{-1}} = 4.96\,\mathrm{kJ\,mol^{-1}}\), so \(\Delta U^{\circ}_{\mathrm{L}} = -787 + 4.96 = -782\,\mathrm{kJ\,mol^{-1}}\).
The model’s shortfall is \(782 - 756 = 26\,\mathrm{kJ\,mol^{-1}}\) against the correctly converted value, versus \(31\) against the unconverted one. The correction is under \(1\%\) of the lattice term but nearly \(20\%\) of the discrepancy being interpreted — so it does not matter for the lattice enthalpy and does matter for the argument being built on top of it.
Show that \(\Delta H^{\circ}_{\mathrm{L}}\) is unchanged if the free electron is assigned an arbitrary molar enthalpy \(\varepsilon\), and hence that the choice between the electron convention and the ion convention cannot affect the answer. Then quantify the error made by a student who takes \(I_1, I_2\) for magnesium from an electron-convention table and \(E_1(\mathrm{Cl})\) from an ion-convention table.
Solution
Let \(I_i(\varepsilon) = I_i(0) + \varepsilon\), since each ionisation leg produces one mole of free electrons, and \(E_j(\varepsilon) = E_j(0) - \varepsilon\), since each attachment leg consumes one. The cation bracket of the Result contributes \(p\sum_i I_i = p\sum_i I_i(0) + p\,z_+\varepsilon\); the anion bracket contributes \(q\sum_j E_j = q\sum_j E_j(0) - q\,z_-\varepsilon\).
Both brackets are subtracted in the Result, so the total \(\varepsilon\)-dependence of \(\Delta H^{\circ}_{\mathrm{L}}\) is \(-(p\,z_+ - q\,z_-)\varepsilon\), which vanishes identically by electroneutrality, \(p\,z_+ = q\,z_-\). Hence \(\Delta H^{\circ}_{\mathrm{L}}\) is convention-independent, as Step 7 asserts.
Mixing sources breaks the cancellation. The two conventions differ by \(\varepsilon = \tfrac52 RT = 2.5(8.314)(298.15) = 6197\,\mathrm{J\,mol^{-1}} = 6.20\,\mathrm{kJ\,mol^{-1}}\). Taking both magnesium ionisations on the electron convention while the two chlorine attachment legs use the ion convention leaves \(p\,z_+\varepsilon = 2(6.20) = 12.4\,\mathrm{kJ\,mol^{-1}}\) uncancelled, which is subtracted in the Result: the extracted lattice enthalpy comes out \(12.4\,\mathrm{kJ\,mol^{-1}}\) too negative, \(-2536\) rather than \(-2524\,\mathrm{kJ\,mol^{-1}}\). It is a systematic bias proportional to the cation charge, so it will not average away over a series of compounds.
Calcium chloride is \(\mathrm{CaCl_2}\), never \(\mathrm{CaCl}\), yet the cycle shows \(\mathrm{CaCl}(s)\) would itself be exothermic with respect to its elements. Using (in \(\mathrm{kJ\,mol^{-1}}\)) \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Ca}) = +178\), \(I_1(\mathrm{Ca}) = +590\), \(I_2(\mathrm{Ca}) = +1145\), \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Cl}) = +122\), \(E_1(\mathrm{Cl}) = -349\), \(\Delta H^{\circ}_{\mathrm{L}}(\mathrm{CaCl_2}) = -2258\), and an estimate \(\Delta H^{\circ}_{\mathrm{L}}(\mathrm{CaCl}) \approx -717\) (from the near-equality of the \(\mathrm{Ca^+}\) and \(\mathrm{K^+}\) radii), compute both formation enthalpies and decide the question.
Solution
Run the cycle forwards, using the Result rearranged for \(\Delta H^{\circ}_{\mathrm{f}}\) as in Step 8.
For \(\mathrm{CaCl_2}\) (\(p=1, q=2, z_+=2, z_-=1\)): cation bracket \(178 + 590 + 1145 = +1913\); anion bracket \(2(122 - 349) = -454\); so \(\Delta H^{\circ}_{\mathrm{f}} = 1913 - 454 - 2258 = -799\,\mathrm{kJ\,mol^{-1}}\), against the tabulated \(-796\,\mathrm{kJ\,mol^{-1}}\) — agreement within the rounding of the inputs, which validates the estimate procedure.
For hypothetical \(\mathrm{CaCl}\) (\(p=q=z_+=z_-=1\)): cation bracket \(178 + 590 = +768\); anion bracket \(122 - 349 = -227\); so \(\Delta H^{\circ}_{\mathrm{f}} = 768 - 227 - 717 = -176\,\mathrm{kJ\,mol^{-1}}\). Exothermic: \(\mathrm{CaCl}(s)\) is stable with respect to \(\mathrm{Ca}(s)\) and \(\tfrac12\mathrm{Cl_2}(g)\).
So stability against the elements is the wrong test. Test it against disproportionation, \(2\,\mathrm{CaCl}(s) \to \mathrm{Ca}(s) + \mathrm{CaCl_2}(s)\), whose enthalpy follows from the same state-function argument: \(\Delta H = \Delta H^{\circ}_{\mathrm{f}}(\mathrm{CaCl_2}) - 2\Delta H^{\circ}_{\mathrm{f}}(\mathrm{CaCl}) = -799 - 2(-176) = -447\,\mathrm{kJ\,mol^{-1}}\). Strongly exothermic, with a negligible entropy change (all condensed phases), so \(\mathrm{CaCl}\) is thermodynamically doomed.
The physical reading: \(I_2(\mathrm{Ca}) = +1145\,\mathrm{kJ\,mol^{-1}}\) is a large price, but it is a \(4s\) electron, not a core electron, and doubling the cation charge roughly triples the lattice term (\(-717 \to -2258\)). That is an extra \(1541\,\mathrm{kJ\,mol^{-1}}\) of lattice stabilisation against a \(1145\,\mathrm{kJ\,mol^{-1}}\) price — a margin of only about a third, but a margin, and it is the whole reason the \(2+\) state wins. The reason \(\mathrm{CaCl_3}\) does not exist in turn is that \(I_3(\mathrm{Ca})\) removes a \(3p\) core electron and is several times larger, more than the extra lattice stabilisation can cover — which is exactly how the building-up principle's shell structure becomes visible in a table of formulae.
For silver chloride, \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Ag}) = +284\), \(I_1(\mathrm{Ag}) = +731\), \(\Delta H^{\circ}_{\mathrm{at}}(\mathrm{Cl}) = +122\), \(E_1(\mathrm{Cl}) = -349\), \(\Delta H^{\circ}_{\mathrm{f}}(\mathrm{AgCl},s) = -127\), all in \(\mathrm{kJ\,mol^{-1}}\). A point-charge electrostatic calculation for the same crystal gives \(-770\,\mathrm{kJ\,mol^{-1}}\). Compute the thermochemical lattice enthalpy, convert it for a fair comparison, and interpret the residual. Compare the size of the residual with the electron-convention error of the third problem and say why one is chemistry and the other is a mistake.
Solution
Cation bracket: \(284 + 731 = +1015\). Anion bracket: \(122 - 349 = -227\). Non-lattice total \(+788\,\mathrm{kJ\,mol^{-1}}\).
\(\Delta H^{\circ}_{\mathrm{L}} = -127 - 788 = -915\,\mathrm{kJ\,mol^{-1}}\). Converting for comparison with a model internal energy: \(\Delta U^{\circ}_{\mathrm{L}} = -915 + 2RT = -915 + 4.96 = -910\,\mathrm{kJ\,mol^{-1}}\).
Residual: the crystal is bound by \(910 - 770 = 140\,\mathrm{kJ\,mol^{-1}}\) more than a purely ionic point-charge model can account for, about \(15\%\) of the total. Contrast \(\mathrm{NaCl}\), where the same comparison leaves roughly \(26\,\mathrm{kJ\,mol^{-1}}\), about \(3\%\). The excess is the standard signature of covalent character: \(\mathrm{Ag^+}\) is large, highly polarisable and carries a filled \(4d\) shell that shields the nuclear charge poorly, so it polarises \(\mathrm{Cl^-}\) and shares electron density with it rather than sitting as an inert sphere of charge. The effect grows down the halide series (\(\mathrm{AgCl} \lt \mathrm{AgBr} \lt \mathrm{AgI}\)) as the anion becomes more polarisable, and it is why the silver halides are insoluble and photosensitive while \(\mathrm{NaCl}\) is neither.
Why the two discrepancies are different in kind: the \(12.4\,\mathrm{kJ\,mol^{-1}}\) of the third problem arises entirely inside the arithmetic, from mixing two conventions for a quantity that provably cancels — it is an error of size \(p\,z_+\times 6.20\,\mathrm{kJ\,mol^{-1}}\), with a sign fixed by which table was misread and no physical content whatever, and it can be removed by reading one table instead of two. The \(140\,\mathrm{kJ\,mol^{-1}}\) here arises from the cycle being exactly right and the model being incomplete; it cannot be removed by better arithmetic, only by a better description of the bonding. Distinguishing the two is precisely why the uncertainty budget of the Corollaries has to be carried explicitly: for a \(1{:}1\) salt the convention residue is \(6.20\) (the third problem’s \(12.4\) is doubled only because \(\mathrm{Mg^{2+}}\) carries two charges), and a systematic \(6.20\) hidden inside \(\mathrm{AgCl}\)’s residual of \(140\) leaves the covalency conclusion untouched, whereas the same \(6.20\) inside \(\mathrm{NaCl}\)’s \(26\) is already a quarter of the effect being interpreted.