# Born-Haber Cycle

Born-Haber Cycle

According to Born-Haber Cycle, the formation of an ionic compound from its constituent elements proceeds in a number of steps. For example, the formation of one mole of sodium chloride form sodium and chlorine atoms can be represented as follows: $Na (s) \overset{\text{Sublimation}}{\underset{+S} \longrightarrow} Na (g) \overset{\text{Ionization}}{\underset{+I} \longrightarrow} Na^+ (g) \\[3mm] \dfrac{1}{2}Cl_2(g) \overset{\text{Dissociation}}{\underset{+\frac{1}{2}D} \longrightarrow} Cl (g) \overset{\text{Electron}}{\underset{\text{Affinnity - E}} \rightarrow} Cl^-(g) \\[3mm] Na^+(g) + Cl^-(g) \overset{\text{Crystal formation}}{\underset{-U} \rightarrow} NaCl(s)$

Where,

I = Ionization energy of sodium

S = Heat of sublimation of sodium metal

D = Heat of dissociation of molecular chlorine

E = Electron affinity of chlorine

U = Lattice energy of sodium chloride.

The cycle can be represented as : The amount of heat liberated $( = - \Delta H_f)$ in overall reaction is heat of formation of sodium chloride.

Then, $-\Delta H_f = S + \dfrac{1}{2}D + I- E- U \\ \text{or,} \hspace{20 mm} U = S + \dfrac{1}{2}D + I- E + \delta H_f$

then higher the negative value of heat of formation the greater will be the stability of ionic compound produced. By putting the values of S, D, I, E and $( = \Delta H_f)$ for NaCl we get, $U = 108.8 + \dfrac{1}{2} (242.7) + 498.3- 380 + 410.9 \\ = 758.65 \text{Kj} \text{mol}^-1$

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