Ncert Exercises & Solutions

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCl(s).

The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. For the reaction

{Na}^{+} {Cl}^{-} (s) \to {Na}^{+} (g) + {Cl}^{-} (g); ∆_{lattice} H^{⊝} = + 788 k J {mol}^{- 1}

Since, it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber cycle.

Let us now calculate the lattice enthalpy of

{Na}^{+} {Cl}^{-} (s)

by following steps given below

(i)

{Na}^{+} (s) \to Na (g);

Sublimation of sodium metal,

∆_{sub} H^{⊝} = 108.4 k J {mol}^{- 1}

(ii)

Na (g) \to {Na}^{+} (g) + e^{-} (g);

The ionisation of sodium atoms, ionisation enthalpy

∆_{i} H^{⊝} = 496 kJ {mol}^{- 1}

(iii)

\frac{1}{2} {Cl}_{2} (g) \to Cl (g);

The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy

\frac{1}{2} ∆_{bond} H^{⊝} = 121 kJ {mol}^{- 1}

(iv)

Cl (g) + e^{-} (g) \to {Cl}^{-} (g);

electron gained by chlorine atoms. The electron gain enthalpy,

∆_{eg} H^{⊝} = - 348.6 kJ {mol}^{- 1}

(v)

{Na}^{+} (g) + {Cl}^{-} (g) \to {Na}^{+} {Cl}^{-} (s)

The sequence of steps is shown in given figure and is known as Born-Haber cycle. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.

Applying Hess's law, we get

∆_{lattice} H^{⊝} = 411.2 + 108.4 + 121 + 496 - 348.6

∆_{lattice} H^{⊝} = 788

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