Match Transformations with Phase Change and Thermodynamic Properties

To match the transformations in Column I with the appropriate thermodynamic and physical properties in Column II, we analyze each process systematically:

---

(A) $\text{CO}_2(\text{s}) \rightarrow \text{CO}_2(\text{g})$

1. Phase Transition:
   This transformation represents the sublimation of solid carbon dioxide (dry ice) directly into gaseous carbon dioxide without passing through the liquid phase. Since there is a change of state from solid to gas for a single chemical substance, it is a phase transition.
   $\implies \textbf{(p)}$

2. Enthalpy Change ($\Delta H$):
   Sublimation requires the input of energy to overcome the intermolecular van der Waals forces holding the $\text{CO}_2$ molecules in the crystal lattice. Thus, the process is endothermic:
   $\Delta H_{\text{sub}} > 0$
   $\implies \textbf{(r)}$

3. Entropy Change ($\Delta S$):
   Gaseous molecules have vastly higher degrees of freedom and translational motion compared to the ordered, fixed arrangement in a crystal lattice. Therefore, the disorder of the system increases:
   $\Delta S > 0$
   $\implies \textbf{(s)}$

Hence, (A) matches with p, r, s.

---

(B) $\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$

1. Enthalpy Change ($\Delta H$):
   The thermal decomposition of limestone ($\text{CaCO}_3$) into quicklime ($\text{CaO}$) and carbon dioxide gas is a well-known endothermic chemical reaction requiring high temperatures:
   $\Delta H > 0$
   $\implies \textbf{(r)}$

2. Entropy Change ($\Delta S$):
   The reaction produces one mole of gas ($\text{CO}_2$) from a purely solid reactant ($\Delta n_g = 1 - 0 = +1$). Since gases possess much higher molar entropy than solids, the entropy of the system increases significantly:
   $\Delta S > 0$
   $\implies \textbf{(s)}$

Hence, (B) matches with r, s.

---

(C) $2\text{ H}^\bullet \rightarrow \text{H}_2(\text{g})$

1. Enthalpy Change ($\Delta H$):
   This process represents the combination of two free hydrogen radicals/atoms to form a stable covalent $\text{H}-\text{H}$ bond. Since bond formation releases energy, the process is exothermic:
   $\Delta H < 0$

2. Entropy Change ($\Delta S$):
   Two independent gaseous radicals combine to form a single diatomic molecule ($\Delta n_g = 1 - 2 = -1$). The reduction in the number of free particles reduces the total degrees of translational freedom and randomness in the system:
   $\Delta S < 0$
   $\implies \textbf{(t)}$

Hence, (C) matches with t.

---

(D) $\text{P}_{(\text{white, solid})} \rightarrow \text{P}_{(\text{red, solid})}$

1. Phase Transition and Allotropic Change:

   • White phosphorus and red phosphorus are allotropes of elemental phosphorus. Hence, the transformation is an allotropic change.
     $\implies \textbf{(q)}$
   • An allotropic change between different solid forms of an element is categorized as a solid-to-solid phase transition.
     $\implies \textbf{(p)}$

2. Enthalpy Change ($\Delta H$):
   White phosphorus consists of strained, discrete tetrahedral $\text{P}_4$ units, while red phosphorus exists as an interconnected, strain-relieved polymeric chain of $\text{P}_4$ units. Because red phosphorus is thermodynamically more stable than white phosphorus at standard conditions, the conversion is exothermic:
   $\Delta H < 0$

3. Entropy Change ($\Delta S$):
   Due to the rigid, interconnected polymeric network of red phosphorus compared to the discrete molecular units of white phosphorus, red phosphorus has a more ordered structure with lower entropy ($S^\circ_{\text{red}} < S^\circ_{\text{white}}$). Consequently:
   $\Delta S < 0$
   $\implies \textbf{(t)}$

Hence, (D) matches with p, q, t.

---

Final Matching:

• A $\rightarrow$ p, r, s
• B $\rightarrow$ r, s
• C $\rightarrow$ t
• D $\rightarrow$ p, q, t