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

$$\mathrm{Na}^{+} \mathrm{Cl}^{-}(\mathrm{s}) \rightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(g) ; \Delta_{\text {lattice }} H^{\mathrm{s}}=+788 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

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 $\mathrm{Na}^{+} \mathrm{Cl}^{-}$(s) by following steps given below

(i) $\mathrm{Na}^{+}$(s) $\rightarrow \mathrm{Na}(\mathrm{g})$; Sublimation of sodium metal, $\Delta_{\text {sub }} H^{\mathrm{S}}=108.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $\mathrm{Na}(g) \rightarrow \mathrm{Na}^{+}(g)+e^{-}(g)$; The ionisation of sodium atoms, ionisation enthalpy $\Delta_i H^{\mathrm{s}}=496 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{Cl}$ ( g ); The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy $\frac{1}{2} \Delta_{\text {bond }} H^{\mathrm{S}}=121 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iv) $\mathrm{Cl}(g)+e^{-}(g) \rightarrow \mathrm{Cl}^{-}(g)$; electron gained by chlorine atoms. The electron gain enthalpy, $\Delta_{\mathrm{eg}} H^{\mathrm{s}}=-348.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Enthalply diagram for lattice enthalpy of $\mathrm{NaCl}$

(v) $\mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(g) \longrightarrow \mathrm{Na}^{+} \mathrm{Cl}^{-}(\mathrm{s})$

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

$$\begin{aligned} & \Delta_{\text {lattice }} H^{\mathrm{S}}=411.2+108.4+121+496-348.6 \\ & \Delta_{\text {lattice }} H^{\mathbb{s}}=+788 \mathrm{~kJ} \end{aligned}$$

Gibbs free energy is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.

Mathematically, this results may be derived as follows

The relationship between heat absorbed by a system $q$, the change in its internal energy, $\Delta U$ and the work done by the system is given by the equation of the first law of thermodynamics, therefore,

$q=\Delta U+W_{\text {expansion }}+W_{\text {non-expansion }}$ .... (i)

Under constant pressure condition, the expansion work is given by $p \Delta V$. $$ \begin{aligned} \therefore \quad q & =\Delta U+p \Delta V+W_{\text {non expansion }} \quad (\because \Delta U+p \Delta V=\Delta H) \\ & =\Delta H+W_{\text {non expansion }} \quad \ldots \text { (ii) } \end{aligned}$$

For a reversible change taking place at constant temperature,

$$\Delta S=\frac{q_{\mathrm{rev}}}{T} \text { or } q_{\mathrm{rev}}=T \Delta \mathrm{S}\quad \text{.... (iii)}$$

Substituting the value of $q$ from Eq. (iii) in Eq. (ii), we get

$$\begin{aligned} T \Delta S & =\Delta H+W_{\text {non expansion }} \\ \text{or}\quad \Delta H-T \Delta S & =-W_{\text {non-expansion }}\quad \text{... (iv)} \end{aligned}$$

For a change taking place under conditions of constant temperature and pressure,

$$\Delta G=\Delta H-T \Delta S$$

Substituting this value in equation (iv), we get

$$\Delta G=-W_{\text {non expansion }}\quad \text{... (v)}$$

Thus, free energy change can be taken as a measure of work other than the work of expansion. For most changes, the work of expansion can not be converted to other useful work, whereas the non-expansion work is convertible to useful work.

Rearranging equation (v), it may write as

$\Delta G=\Delta H-T \Delta S$

If $\Delta H=$ positive and $\Delta S=$ positive, then

$\Delta G$ will be negative i.e., process will be spontaneous only when $T \Delta S>\Delta H$ in magnitude, which will be so when temperature is high.

(i) Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from $\left(p_i V_i\right)$ to $\left(p_f, V_f\right)$. Reversible work is represented by the combined areas $A B C$ and $B C V_i V_f$.

(ii) Work against constant pressure, $p_f$ is represented by the area $B C V_i V_t$. Work (i) $>$ work (ii)