Question

How is the lattice enthalpy of NaCl calculated explain by Born - Haber cycle.

Answer 1

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The heat of formation of sodium chloride (ΔHf0) from the sodium metal and chlorine gas can be experimentally measured.

Na (s) + 1/2 Cl2 (g) → NaCl(s) ΔHf0 = -411kJ/mol

The formation of ionic solid sodium chloride form solid sodium metal and gaseous chlorine is not a single step process but goes through several processes. Heat changes of all the processes except the lattice energy can be experimentally measured.

The processes or steps in the formation of sodium chloride are-

are-

1. Solid sodium atom sublimes to gaseous atom by absorbing heat energy (∆Hsub).

Na (s) → Na (g),

Sublimation energy ΔHsub = + 107kJ/mol

2. Gaseous sodium atom absorbs the ionization energy to release one electron and forms gaseous sodium ion.

Na(g) → Na+(g) + 1e– ,

Ionization energy ∆HIE = +502kJ/mol

3. Diatomic gaseous chlorine breaks into two individual atoms by absorbing bond energy, such that each chlorine atom absorbs half of the bond energy of chlorine molecule.

Cl2(g) → 2Cl(g) 1/2

Bond dissociation energy of chlorine = 1/2

∆Hdiss= 1/2 242 = +121kJ/mol

4. Chlorine atom accepts an electron to form chloride ion and releases energy equivalent to electron affinity.

Cl(g) + 1e– → Cl– (g)

Electron affinity = ∆HEA = -355kJ/mol

5. Gaseous sodium ion and gaseous chloride ion combine to form solid sodium chloride molecule and releases energy equivalent to lattice energy.

Na+(g) + Cl– (g) → Na+ Cl– (s)

Lattice energy = ∆HLE = U = ?

Summation of enthalpy of all the processes from step 1 to step 5) give the net enthalpy of formation of solid crystalline sodium chloride from sodium and chlorine in their standard conditions of solid and gas respectively. This should be equal to the experimentally measured enthalpy of formation of solid sodium chloride.

The enthalpies are represented as a cycle in the figure.

(see the above figure)

So, ΔHf0 = ΔHsub + ∆HIE + 1/2 ∆Hdis + ∆HEA + U or ΔHf0 – (ΔHsub + ∆HIE + 1/2 ∆Hdis + ∆HEA + U)=0

411 + 107 + 502 +121 -355 +U = 0

Here, except lattice energy, all other enthalpies can be experimentally measured.

Lattice energy of the sodium chloride solid = U = ΔHf0 – (ΔHsub + ∆HIE + ½ ∆Hdis + ∆HEA).

= -411 -107 -502 -121 +355

= – 786kJ/mol