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.

No, enthalpy is one of the contributing factors in deciding spontaneity but it is not the only factor. Another contributory factor, entropy factor has also to be taken into consideration.

The given diagram represent that the process is carried out in infinite steps, hence it is isothermal reversible expansion of the ideal gas from pressure 2.0 atm to 1.0 atm 298 K .

$$\begin{aligned} & W=-2.303 n R T \log \frac{p_1}{p_2} \\ & W=-2.303 \times 1 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{Klog} 2 \quad \left(\because \frac{p_1}{p_2}=\frac{2}{1}\right)\\ & W=-2.303 \times 1 \times 8.314 \times 298 \times 0.3010 \mathrm{~J} \\ & W=-1717.46 \mathrm{~J} \end{aligned}$$

In the first case, as the expansion is against constant external pressure

$$\begin{aligned} W & =-p_{\text {ext }}\left(V_2-V_1\right)=-2 \text { bar } \times(50-10) \mathrm{L} \\ & =-80 \mathrm{~L} \text { bar } \quad \text{(1L bar = 100 J)}\\ & =-80 \times 100 \mathrm{~J} \\ & =-8 \mathrm{~kJ} \end{aligned}$$

If the given expansion was carried out reversibly, the internal pressure of the gas should be greater than the external pressure at every stage. Hence, the work done will be more.

$$\begin{aligned} &\begin{aligned} & \text { I. } \rightarrow(8) \quad \text { J. } \rightarrow(9) \quad \text { K. } \rightarrow(1,12,13) \quad \text { L. } \rightarrow(7,11) \end{aligned}\\ &\text { I. } \rightarrow(8) \quad \text { J. } \rightarrow(9) \quad \text { K. } \rightarrow(1,12,13) \quad \text { L. } \rightarrow(7,11) \end{aligned}$$

Correct matching can be done as