Born-Haber cycles

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BOND ENTHALPIES

ΔH for a chemical reaction=  Σ (breaking bonds of reactants) − Σ (forming bonds of products)

Mean bond dissociation enthalpy ΔH⦵B.E.
Standard enthalpy change that accompanies the separation of 1 mole of a covalently bound molecules (element or compound) in the gaseous state to its constituent atoms in the gaseous state.

A mean bond enthalpy indicates the average enthalpy value for a particular covalent bond given its occurrence in different compounds and molecular environments.
H−H(g) → 2H(g)            ΔH⦵B.E.   = +436 kJ mol−1
H−Cl(g) → H(g)+ Cl(g)   ΔH⦵B.E.   = +432 kJ mol−1
Cl−Cl(g) → 2Cl(g)           ΔH⦵B.E.   = +244 kJ mol−1
Br−Br(g) → 2Br(g)          ΔH⦵B.E.   = +193 kJ mol−1

The bond enthalpy for Cl−Cl bond is greater than Br−Br bond because Cl−Cl bond length is shorter (0.199nm vs. 0.228nm), so attraction between nucleus and covalent bond pair of electrons is stronger for the Cl−Cl bond .

Relationship between atomisation and bond enthalpies for gaseous molecules
For atomisation, only 1 mol of gaseous atoms are formed from half a mole of a diatomic gaseous element.
For bond dissociation enthalpy, breaking 1 mol of a diatomic gaseous molecules yields two moles of gaseous atoms.
Hence, half bond dissociation enthalpy equates to the atomisation enthalpy for a diatomic gaseous element.

½Cl2(g) → Cl(g)                ΔH⦵at   = +122 kJ mol−1 

Cl−Cl(g) → 2Cl(g)              ΔH⦵B.E.   = +244 kJ mol−1

½ Cl−Cl(g) → Cl(g)            ½ × ΔH⦵B.E. 

So,  ½ × ΔH⦵B.E. (Cl−Cl) = ΔH⦵at  Cl = +122 kJ mol−1

Examples of bond enthalpy calculations

Q1. The bond dissociation enthalpy for chlorine is +242 kJ mol−1 and that for fluorine is +158 kJ mol−1.
The standard enthalpy of formation of ClF(g) is −56 kJ mol−1.
i) Write an equation, including state symbols, for the reaction that has an enthalpy change equal to the standard enthalpy of formation of gaseous ClF.
ii) Calculate a value for the bond enthalpy of the Cl – F bond.
iii) Calculate the enthalpy of formation of gaseous chlorine trifluoride, ClF3(g).

½F2(g) + ½Cl2(g)  →  ClF(g)
ΔH=  Σ (breaking bonds) − Σ (forming bonds)
−56 = [(½ × 158) + (½ × 242)] − (bond Cl-F)
Bond enthalpy Cl-F=79+121+56=256 kJ mol−1

1½F2(g) + ½Cl2(g)  →  ClF3(g)
ΔH=  Σ (breaking bonds) − Σ (forming bonds) = [(1½ × 158) + (½ × 242)] − [(256 × 3)]= −410 kJ mol−1

NB. The bond enthalpy of Cl-F bond in ClF differs from that found in ClF3 .

Q2. Use data below to calculate a value for the enthalpy of formation for one mole of ammonia.
Mean bond enthalpy / kJ mol–1 :  NΞN  944     H−H  436       N−H  388

 ½N2 (g) + 1½H2 (g)   →  NH3 (g)

ΔH⦵f  NH3 (g)  = Σ (breaking bonds) − Σ (forming bonds) = [(½ × 944) + (1½ × 436] − [(388× 3)]= −38 kJ mol−1

Of significance, the bond enthalpy derived ΔH⦵f  NH3 (g) is less accurate than the accepted value of −46.1 kJ mol−1 . 

This is because mean bond enthalpies are determined for N−H bonds over from a range of compounds and molecular environments. The N-H bond strength in ammonia will be unique to the ammonia molecule and not necessarily closely correlate to the value of the mean bond enthalpy for the N-H bond.

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