The enthalpy of vaporisation of $\mathrm{CCl}_4$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the heat required for the vaporisation of 284 g of $\mathrm{CCl}_4$ at constant pressure. (Molar mass of $\mathrm{CCl}_4=154 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$$\begin{aligned} & \text { Given that, } 1 \mathrm{~mol} \mathrm{of} \mathrm{CCl}_4=154 \mathrm{~g} \\ & \Delta_{\text {vap }} H \text { for } 154 \mathrm{~g} \mathrm{CCl}_4=30.5 \mathrm{~kJ} \\ & \therefore \quad \Delta_{\text {vap }} H \text { for } 284 \mathrm{~g} \mathrm{CCl}_4=\frac{30.5 \times 284}{154} \mathrm{~kJ}=56.25 \mathrm{~kJ} \end{aligned}$$
The enthalpy of reaction for the reaction
$$2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \text { is } \Delta_r H^{\mathrm{s}}=-572 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
What will be standard enthalpy of formation of $\mathrm{H}_2 \mathrm{O}(l)$ ?
Given that,
$2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l), \Delta_r H^{\circ}=-572 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements then
$$\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta_r H^{\circ}=?$$
This can be obtained by dividing the given equation by 2.
Therefore, $$\Delta_f H^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-\frac{572 \mathrm{kJmol}^{-1}}{2}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{\text {ext }}$ in a single step as shown in figure? Explain graphically.
Suppose total volume of the gas is $V_i$ and pressure of the gas inside cylinder is $p$. After compression by constant external pressure, $\left(p_{\text {ext }}\right)$ in a single step, final volume of the gas becomes $V_f$.
Then volume change, $\Delta v=\left(V_f-V_i\right)$
If $W$ is the work done on the system by movement of the piston, then
$$\begin{aligned} & W=p_{\text {ext }}(-\Delta V) \\ & W=-p_{\text {ext }}\left(V_f-V_i\right) \end{aligned}$$
This can be calculated from $p-V$ graph as shown in the figure. Work done is equal to the shaded area $A B V_f V_i$
The negative sign in this expression is required to obtain conventional sign for $W$ which will be positive. Because in case of compression work is done on the system, so $\Delta V$ will be negative.
How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
When compression is carried out in infinite steps with change in pressure, it is a reversible process. The work done can be calculated from $p-V$ plot as shown in the given figure. Shaded area under the curve represents the work done on the gas.
Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b) $\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HCl}(\mathrm{g}) \Delta_r H^{\mathrm{s}}=-92.32 \mathrm{~kJ} \mathrm{~mol}^{-1}$
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Representation of potential energy/enthalpy change in the following processes
(a) Throwing a stone from the ground to roof.
(b) $\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{Cl}_2(g) \rightleftharpoons \mathrm{HCl}(g) ; \Delta_r H^s=-92.32 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Energy increases in (a) and it decreases in (b) process. Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.