Matching Physical Chemical Processes with Enthalpy and Entropy Changes

To determine the correct matching between List-I and List-II, we analyze the thermodynamic parameters, namely the enthalpy change ($\Delta H$) and entropy change ($\Delta S$), for each process:

1. (P) Physisorption:

   • Adsorption involves attractive van der Waals forces binding gas molecules onto a solid surface, releasing heat. Thus, the process is exothermic: $\Delta H < 0$
   • During adsorption, gas particles lose their translational degrees of freedom as they are restricted to the surface, resulting in a decrease in randomness: $\Delta S < 0$
   • Therefore, $\text{P} \rightarrow 2$ ($\Delta H < 0$ and $\Delta S < 0$).

2. (Q) Diamond $\rightarrow$ Graphite:

   • Graphite is the thermodynamically more stable allotrope of carbon at standard conditions ($298\text{ K}$, $1\text{ atm}$). The conversion of diamond to graphite releases energy: $\Delta H < 0$
   • Diamond has a highly rigid 3D covalent network structure, whereas graphite has a layered hexagonal structure with weaker inter-layer forces, allowing higher vibrational freedom and structural randomness. Thus, the entropy increases ($S_{\text{graphite}}^\circ > S_{\text{diamond}}^\circ$): $\Delta S > 0$
   • Therefore, $\text{Q} \rightarrow 5$ ($\Delta H < 0$ and $\Delta S > 0$).

3. (R) Denaturation of protein:

   • Denaturation breaks non-covalent interactions (such as hydrogen bonds, hydrophobic interactions, and ionic bonds) stabilizing the native protein structure. Energy is required to disrupt these bonds: $\Delta H > 0$
   • The highly ordered helical and folded native conformation of the protein unfolds into a disordered, random coil structure, leading to a significant increase in entropy: $\Delta S > 0$
   • Therefore, $\text{R} \rightarrow 1$ ($\Delta H > 0$ and $\Delta S > 0$).

4. (S) Propene $\rightarrow$ Cyclopropane:

   • Cyclopropane possesses high ring strain (angle strain and eclipsing interactions) compared to open-chain propene, making cyclopropane higher in enthalpy. Thus, conversion from propene to cyclopropane is endothermic: $\Delta H > 0$
   • The formation of a cyclic system from an open-chain isomer restricts internal rotations, reducing molecular randomness: $\Delta S < 0$
   • Therefore, $\text{S} \rightarrow 4$ ($\Delta H > 0$ and $\Delta S < 0$).

Matching the results: $\text{P} \rightarrow 2, \quad \text{Q} \rightarrow 5, \quad \text{R} \rightarrow 1, \quad \text{S} \rightarrow 4$

This corresponds to Option C.