Question

List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy \((\Delta H)\) and entropy \((\Delta S)\). Match each entry in List-I to the appropriate entry in List-II and choose the correct option.

• A. \(P \rightarrow 2;\ Q \rightarrow 3;\ R \rightarrow 5;\ S \rightarrow 4\)
• B. \(P \rightarrow 4;\ Q \rightarrow 3;\ R \rightarrow 5;\ S \rightarrow 1\)
• C. \(P \rightarrow 2;\ Q \rightarrow 5;\ R \rightarrow 1;\ S \rightarrow 4\)
• D. \(P \rightarrow 2;\ Q \rightarrow 5;\ R \rightarrow 1;\ S \rightarrow 3\)

Hint:

Important thermodynamic trends:
• Adsorption: \[ \Delta H < 0,\quad \Delta S < 0 \]
• Greater disorder: \[ \Delta S > 0 \]
• Cyclization generally decreases entropy: \[ \Delta S < 0 \] Always think physically about whether molecular freedom increases or decreases.

Answer 1

The Correct Option is C

Concept: To solve this matching problem, we analyze whether each process is:
• exothermic or endothermic
• associated with increase or decrease in randomness Recall: \[ \Delta H < 0 \] means exothermic process. \[ \Delta H > 0 \] means endothermic process. Also, \[ \Delta S > 0 \] means entropy increases. \[ \Delta S < 0 \] means entropy decreases.

Step 1: Analyzing physisorption. Physisorption involves adsorption due to weak van der Waals forces. During adsorption:
• gas molecules lose freedom of movement
• randomness decreases Hence: \[ \Delta S < 0 \] Also adsorption releases heat. Thus: \[ \Delta H < 0 \] Therefore: \[ P \rightarrow (2) \]

Step 2: Analyzing conversion of diamond into graphite. Graphite is thermodynamically more stable than diamond. Hence: \[ \Delta H < 0 \] Also graphite has greater disorder because of layered structure. Therefore: \[ \Delta S > 0 \] Thus: \[ Q \rightarrow (5) \]

Step 3: Analyzing denaturation of protein. Denaturation unfolds the highly ordered protein structure. Thus randomness increases. Therefore: \[ \Delta S > 0 \] The process generally requires absorption of heat. Hence: \[ \Delta H > 0 \] Thus: \[ R \rightarrow (1) \]

Step 4: Analyzing conversion of propene into cyclopropane. Cyclization decreases randomness because open chain becomes cyclic. Thus: \[ \Delta S < 0 \] The enthalpy change is nearly zero in this approximation. Hence: \[ \Delta H = 0 \] Therefore: \[ S \rightarrow (4) \]

Step 5: Final matching. Thus the correct matching is: \[ P \rightarrow 2 \] \[ Q \rightarrow 5 \] \[ R \rightarrow 1 \] \[ S \rightarrow 4 \] This corresponds to option (C). Final Answer: \[ \boxed{(C)} \]