Match the transformations in column I with appropriate options in column II
Column-I | Column-II |
(A) CO2(s) → CO2 (g) |
(p) phase transition |
(B) CaCO3(s) → CaO(s) + CO2(g) |
(q) allotropic change |
(C) 2H → H2(g) | (r) ΔH is positive |
(D) P(white, solid) → P(red, solid) | (s) ΔS is positive |
(t) ΔS is negative |
[A] CO2(s) CO2(g)
p, r, s
[B] CaCO3 (s) CaO (s) + CO2(g)
r, s
[C] 2H H2 (g)
t
[D] Pwhite Pred
p, q, t
This question involves matching chemical transformations (Column I) with their correct descriptions (Column II). We need to analyze each transformation for type of change, enthalpy change (ΔH), and entropy change (ΔS).
Step 1: Analyze Transformation (A) CO2(s) → CO2(g)
This is sublimation (solid to gas). It is a phase transition (p). Energy is absorbed (endothermic), so ΔH > 0 (r). Disorder increases (gas > solid), so ΔS > 0 (s).
Matches: (p), (r), (s)
Step 2: Analyze Transformation (B) CaCO3(s) → CaO(s) + CO2(g)
This is decomposition. It is not a phase transition or allotropic change. It is endothermic (requires heat), so ΔH > 0 (r). A solid produces a solid and a gas, so disorder increases, ΔS > 0 (s).
Matches: (r), (s)
Step 3: Analyze Transformation (C) 2H → H2(g)
This represents formation of H2 molecule from atoms. Bond formation releases energy, so exothermic: ΔH < 0 (not positive). Two free atoms form one molecule, freedom decreases, so ΔS < 0 (t).
Matches: (t)
Step 4: Analyze Transformation (D) P(white, solid) → P(red, solid)
This is a change between allotropes of phosphorus (different solid forms). It is an allotropic change (q). White phosphorus is less stable than red; conversion is exothermic (ΔH < 0, not positive). White P has higher disorder than red P, so ΔS < 0 (t).
Matches: (q), (t)
(A) → (p), (r), (s); (B) → (r), (s); (C) → (t); (D) → (q), (t)
Gibbs Free Energy:
Spontaneity depends on ΔG < 0. For phase changes like sublimation (A), ΔG = 0 at equilibrium.
Entropy Sign: ΔS > 0 for processes increasing molecular freedom (e.g., solid→gas, dissolution, number of gas molecules increase).
Enthalpy Sign: ΔH > 0 for bond breaking (endothermic), ΔH < 0 for bond formation (exothermic).