Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below
$$2 \mathrm{Zn}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{ZnO}(s) ; \Delta H=-693.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
18.0 g of water completely vaporises at $100^{\circ} \mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is $40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?
Given that, quantity of water $=18.0 \mathrm{~g}$, pressure $=1$ bar
As we know that, $18.0 \mathrm{~g~H}_2 \mathrm{O}=1 \mathrm{~mole~H}_2 \mathrm{O}$
Enthalpy change for vaporising $1 \mathrm{~mole}^{-1} \mathrm{H}_2 \mathrm{O}=40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore$ Enthalpy change for vaporising 2 moles of $\mathrm{H}_2 \mathrm{O}=2 \times 40.79 \mathrm{~kJ}=81.358 \mathrm{~kJ}$
Standard enthalpy of vaporisation at $100^{\circ} \mathrm{C}$ and 1 bar pressure, $\Delta_{\text {vap }} H^{\circ}=+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$
One mole of acetone requires less heat to vaporise than 1 mole of water. Which of the two liquids has higher enthalpy of vaporisation?
One mole of acetone requires less heat to vaporise than 1 mole of water. Hence, acetone has less enthalpy of vaporisation and water has higher enthalpy of vaporisation. It can be represented as $\left(\Delta H_V\right)$ water $>\left(\Delta H_V\right)$ acetone.
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 $+183.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$