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.
Show 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.
\(P \rightarrow 2;\ Q \rightarrow 3;\ R \rightarrow 5;\ S \rightarrow 4\)
\(P \rightarrow 4;\ Q \rightarrow 3;\ R \rightarrow 5;\ S \rightarrow 1\)
\(P \rightarrow 2;\ Q \rightarrow 5;\ R \rightarrow 1;\ S \rightarrow 4\)
\(P \rightarrow 2;\ Q \rightarrow 5;\ R \rightarrow 1;\ S \rightarrow 3\)
Show Solution
The Correct Option isC
Solution and Explanation
Step 1: Understanding the Concept:
This matching question explores the thermodynamic parameters \(\Delta H\) (enthalpy change) and \(\Delta S\) (entropy change) for various physical and chemical processes.
- \(\Delta H<0\) indicates an exothermic process; \(\Delta H>0\) indicates an endothermic process.
- \(\Delta S>0\) indicates an increase in randomness or disorder; \(\Delta S<0\) indicates a decrease. Step 3: Detailed Explanation: (P) Physisorption: Physical adsorption is always an exothermic process (\(\Delta H<0\)) because it involve the formation of weak Van der Waals forces. As gas molecules are localized on the surface, their randomness decreases, so \(\Delta S<0\).
Match: P \(\rightarrow\) 2. (Q) Diamond \(\longrightarrow\) Graphite: At standard conditions, Graphite is the more stable allotrope of carbon. Therefore, the conversion is exothermic (\(\Delta H<0\)). Graphite has a layered, more "loose" structure compared to the rigid 3D lattice of diamond, leading to higher entropy (\(\Delta S>0\)).
Match: Q \(\rightarrow\) 5. (R) Denaturation of protein: Denaturation involves the unfolding of the highly organized tertiary structure of a protein into a random coil. This increase in disorder means \(\Delta S>0\). Breaking the stabilizing hydrogen bonds and hydrophobic interactions requires energy, so the process is generally endothermic (\(\Delta H>0\)).
Match: R \(\rightarrow\) 1. (S) Propene \(\longrightarrow\) Cyclopropane: Cyclopropane is a highly strained three-membered ring (angle strain). Converting the stable open-chain propene to the strained cyclopropane is endothermic (\(\Delta H>0\)). Ring formation restricts the rotational freedom of the molecule, resulting in a decrease in entropy (\(\Delta S<0\)).
Match: S \(\rightarrow\) 4. Step 4: Final Answer:
Matching the thermodynamic signatures: Physisorption (Exo, \(\downarrow\)S), Diamond to Graphite (Exo, \(\uparrow\)S), Denaturation (Endo, \(\uparrow\)S), and Propene to Cyclopropane (Endo, \(\downarrow\)S). Option (C) is correct.
