From the first law of thermodynamics, $\quad q=\Delta U+p \Delta V$

If the process carried out at constant volume, $\Delta V=0$

Hence, $$q_v=\Delta U$$

[Here, $q_v=$ Heat absorbed at constant volume, $\Delta U=$ change in internal energy]

Similarly, $q_p=\Delta H$

Here, $\quad q_p=$ heat absorbed at constant pressure

$\Delta H=$ enthalpy change of the system.

Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure.

As we know that at constant pressure, $\Delta H=\Delta U+p \Delta V$ where, $\Delta V$ is the change in volume.

This equation can be rewritten as $\Delta H=\Delta U+p\left(V_f-V_i\right)=\Delta U+\left(p V_f-p V_i\right)\quad \text{... (i)}$

where, $$V_i$$ = initial volume of the system $$V_f$$ = final volume of the system

But for the ideal gases, $$p V=n R T$$

So that $$p V_1=n_1 R T$$

and $$p V_2=n_2 R T$$

where, $n_1=$ number of moles of the gaseous reactants

$n_2=$ number of moles of the gaseous products.

Substituting these values in Eq. (i), we get

$$\begin{aligned} & \Delta H=\Delta U+\left(n_2 R T-n_1 R T\right) \\ \text{or}\quad & \Delta H=\Delta U+\left(n_2-n_1\right) R T \end{aligned}$$

where, $\Delta n_g=n_2-n_1$ is the difference between the number of moles of the gaseous products and gaseous reactants.

Putting the values of $\Delta H$ and $\Delta U$ we get

$$q_p=q_v+\Delta n_g R T$$

Extensive properties Those properties whose value depends on the quantity or size of matter present in the system is known as extensive properties.

e.g., mass, internal energy, heat capacity.

Intensive properties Those properties which do not depend on the quantity or size of matter present are known as intensive properties. e.g., pressure, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.

Mole fraction or molarity of a solution is same whether we take a small amount of the solution or large amount of the solution. Ratio of two extensive properties is always intensive.

$\frac{\text { Extensive }}{\text { Extensive }}=$ Intensive

So, mole fraction and molarity are intensive properties.

e.g., $\quad$ mole fraction $=\frac{\text { Moles of the component }}{\text { Total no. of moles }}=\frac{\text { Extensive }}{\text { Extensive }}$

and $\quad$ molarity $=\frac{\text { Mole }}{\text { Volume }}=\frac{\text { Extensive }}{\text { Extensive }}$

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)