Enthalpy is an extensive property. In general, if enthalpy of an overall reaction $A \rightarrow B$ along one route is $\Delta_r H$ and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3 \ldots$ represent enthalpies of intermediate reactions leading to product $B$. What will be the relation between $\Delta_r H$ overall reaction and $\Delta_r H_1, \Delta_r H_2 \ldots$ etc for intermediate reactions.
In general, if enthalpy of an overall reaction $A \rightarrow B$ along one route is $\Delta_r H$ and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3 \ldots$ representing enthalpies of reaction leading to same product $B$ along another route, then we have
$$\Delta_r H=\Delta_r H_1+\Delta_r H_2+\Delta_r H_3+\ldots$$
The enthalpy of atomisation for the reaction $\mathrm{CH}_4(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})$ is $1665 \mathrm{~kJ} \mathrm{~mol}^{-1}$. What is the bond energy of $\mathrm{C}-\mathrm{H}$ bond?
In $\mathrm{CH}_4$, there are four $\mathrm{C}-\mathrm{H}$ bonds. The enthalpy of atomisation of 1 mole of $\mathrm{CH}_4$ means dissociation of four moles of $\mathrm{C}-\mathrm{H}$ bond.
$$\begin{aligned} \therefore \quad \mathrm{C}-\mathrm{H} \text { bond energy per } \mathrm{mol} & =\frac{1665 \mathrm{~kJ}}{4 \mathrm{~mol}} \\ & =416.25 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$
Use the following data to calculate $\Delta_{\text {lattice }} H^{\mathrm{s}}$ for $\mathrm{NaBr} . \Delta_{\text {sub }} H^{\mathrm{s}}$ for sodium metal $=108.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$, ionisation enthalpy of sodium $=496 \mathrm{~kJ} \mathrm{~mol}^{-1}$, electron gain enthalpy of bromine $=-325 \mathrm{~kJ} \mathrm{~mol}^{-1}$, bond dissociation enthalpy of bromine $=192 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_f H^{\mathrm{s}}$ for $\mathrm{NaBr}(s)=-360.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Given that, $\Delta_{\text {sub }} H^5$ for Na metal $=108.4 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$
IE of $\mathrm{Na}=496 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\text {eg }} H^{\mathrm{s}}$ of $\mathrm{Br}=-325 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\text {diss }} H^{\mathrm{s}}$ of $\mathrm{Br}=192 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_f H^{\mathrm{s}}$ for $\mathrm{NaBr}=-360.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Born-Haber cycle for the formation of NaBr is as
By applying Hess's law,
$$\begin{aligned} \Delta_f H^{\mathrm{s}} & =\Delta_{\text {sub }} H^{\mathrm{s}}+I E+\Delta_{\text {diss }} H^{\mathrm{s}}+\Delta_{\text {eg }} H^{\mathrm{s}}+U \\ -360.1 & =108.4+496+96+(-325)-U \\ U & =+735.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$
Given that $\Delta H=0$ for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
The mixing of two gases have $\Delta H$ equal to zero. Therefore, it is spontaneous process because energy factor has no role to play but randomness increases i.e., randomness factor favours the process.
Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is
$$\Delta S=\frac{q_{\mathrm{rev}}}{T}$$
Here, $\quad \Delta S=$ change in entropy
$$\begin{aligned} \mathrm{q}_{\mathrm{rev}} & =\text { heat of reversible reaction } \\ T & =\text { temperature } \end{aligned}$$