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Chapter 4: Problem 20

Discuss Born-Haber cycle with the help of a suitable example.

Short Answer

Expert verified
The Born-Haber Cycle is a specific form of Hess’ Law cycle that uses the enthalpies of sublimation, ionization, dissociation, affinity and lattice formation to calculate the overall enthalpy change for the formation of an ionic compound from its constituent elements. The example used here is the formation of sodium chloride from sodium and chlorine.

Step by step solution

01

Sublimation of Sodium

The first step is to convert sodium from its solid phase to gaseous phase for which the energy required is the atomization enthalpy of sodium. So, \[ \text{Na(s)} \rightarrow \text{Na(g)} ; ΔH_1 = +108 Kcal \]
02

Ionization of Sodium

The gaseous sodium atom then loses an electron to form a sodium ion. The energy change associated with this step is the ionization enthalpy of sodium. So, \[ \text{Na(g)} \rightarrow \text{Na}^{+}(g) + e^- ; ΔH_2 = +496 Kcal \]
03

Dissociation of Chlorine

Chlorine is a diatomic molecule and hence one of its molecule will be dissociated into chlorine atoms. It's a bond dissociation hence the energy will be the bond dissociation enthalpy of chlorine. \[ \text{Cl2(g)} \rightarrow 2 \text{Cl(g)} ; ΔH_3 = +120 Kcal/mol \]
04

Electron Affinity of Chlorine

One of the atomic chlorine gains an electron to form chloride ion. This occurs with the release of energy equal to the electron affinity of chlorine. \[ \text{Cl(g)} + e^- \rightarrow \text{Cl}^- (g) ; ΔH_4 = -349 Kcal \]
05

Formation of Sodium Chloride

Finally, sodium ion and chloride ion combine to form one mole of sodium chloride. The energy released in this process is the lattice enthalpy of sodium chloride. \[ \text{Na}^{+}(g) + \text{Cl}^- (g) \rightarrow \text{NaCl(s)} ; ΔH_5 = -788 Kcal \] The overall enthalpy change of a reaction (ΔH_r) is the sum of the enthalpies of the individual steps.

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