Standard molar enthalpy of formation, $\Delta_f H^{\mathrm{s}}$ is just a special case of enthalpy of reaction, $\Delta_r H^{\mathrm{s}}$. Is the $\Delta_r H^{\mathrm{s}}$ for the following reaction same as $\Delta_f H^s$ ? Give reason for your answer.
$$\mathrm{CaO}(s)+\mathrm{CO}_2(\mathrm{~g}) \rightarrow \mathrm{CaCO}_3(\mathrm{~s}) ; \Delta_f H^{\mathrm{s}}=-178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
No, the $\Delta_r H^5$ for the given reaction is not same as $\Delta_r H^5$. The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, $\Delta_t H^{\mathrm{s}}$.
$$\mathrm{Ca}(\mathrm{s})+\mathrm{C}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CaCO}_3(\mathrm{~s}) ; \Delta_{\mathrm{f}} \mathrm{H}^{\mathrm{s}}$$
This reaction is different from the given reaction. Hence,
$$\Delta_r H^{\circ} \neq \Delta_f H^{\circ}$$
The value of $\Delta_f H^{\mathrm{s}}$ for $\mathrm{NH}_3$ is $-91.8 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$. Calculate enthalpy change for the following reaction.
$$2 \mathrm{NH}_3(g) \rightarrow \mathrm{N}_2(g)+3 \mathrm{H}_2(g)$$
Given, $\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightarrow \mathrm{NH}_3(g) ; \Delta_t H^{\mathrm{s}}=-91.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
($\Delta_f H^5$ means enthalpy of formation of 1 mole of $\mathrm{NH}_3$ )
$\therefore$ Enthalpy change for the formation of 2 moles of $\mathrm{NH}_3$
$$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g) ; \Delta_t H^5=2 \times-91.8=-183.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
And for the reverse reaction,
$$2 \mathrm{NH}_3(g) \rightarrow \mathrm{N}_2(g)+3 \mathrm{H}_2(g) ; \Delta_f H^5=+183.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
Hence, the value of $\Delta_t H^5$ for $\mathrm{NH}_3$ is $+183.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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,
$$\begin{aligned} \Delta_f H^{\mathrm{s}} & =\Delta_{\text {sub }} H^{\mathrm{s}}+I E+\Delta_{\text {diss }} H^{\mathrm{s}}+\Delta_{\text {eg }} H^{\mathrm{s}}+U \\ -360.1 & =108.4+496+96+(-325)-U \\ U & =+735.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$