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In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g) D. Cl₂(g) → 2Cl(g) Choose the correct answer from the options given below:
More gas / higher randomness = ΔS > 0! Liquid to gas (A: ΔS>0); Cooling to absolute zero (B: ΔS<0); Solid producing 2 gas moles (C: ΔS>0); Dissociation 1 mol -> 2 mol gas (D: ΔS>0) -> A, C, and D!
Entropy S is a state function that quantifies the number of accessible microstates (disorder/randomness) of a system. For any process: - Transition from ordered to disordered phase increases entropy (ΔS > 0). - An increase in the number of gaseous molecules increases entropy (Δng > 0 ⟹ ΔS > 0). - Cooling a substance decreases thermal energy and accessible states (ΔS < 0).
Analyzing each process individually: - A: Evaporation: Molecules in vapour move freely in 3D space with much higher entropy than in the condensed liquid: ΔS_vap > 0. - B: Cooling crystalline solid: As T drops from 130 K to 0 K, atomic vibrations cease. By the Third Law, S approaches 0 at absolute zero: ΔS < 0. - C: Decomposition: 2 NaHCO3(s) ⟶ Na2CO3(s) + CO2(g) + H2O(g). Formation of 2 moles of gaseous products from a solid creates enormous spatial disorder: ΔS > 0. - D: Homolytic dissociation: Cl2(g) ⟶ 2 Cl(g). A single bound diatomic molecule dissociates into two independent free particles: ΔS > 0.
Processes in which entropy increases are A, C, and D. This corresponds precisely to Option (1).
Diagnose every choice: Know exactly why the right option works and why distractors fail.
A (vaporization), C (gas evolution), and D (molecular dissociation) all increase disorder.
Omits vaporization of a liquid (A), which is the most classic entropy-increasing process.
Omits Cl2 dissociation (D), which doubles the number of gas particles.
Includes B, where cooling decreases entropy.
Students sometimes think zero Kelvin means "maximum freedom". In reality, the Third Law of Thermodynamics states that at absolute zero, a perfect crystal has ZERO entropy (minimum possible disorder). Cooling always decreases entropy!
"Entropy increases during vaporization, gas-producing decomposition, and dissociation of molecules into atoms; it decreases upon cooling."
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