49. Which of the following particulate representations shows a process during which the entropy of t — Thermodynamics Chemistry Question

STEPS:

1. Understand the concept of entropy ($S$):
Entropy is a thermodynamic measure of the positional disorder and the number of microstates (possible arrangements) available to a system. Generally, entropy increases ($\Delta S > 0$) when:
* A solid transitions to a liquid, or a liquid/solid transitions to a gas (since gases have far greater freedom of motion and spatial distribution than condensed phases).
* A chemical reaction results in an increase in the total number of gaseous particles.
* Large molecules dissociate or decompose into a larger number of smaller individual particles.

Conversely, entropy decreases ($\Delta S < 0$) when a system becomes more ordered, such as when a gas condenses into a liquid or solid, or when the total number of free-moving particles decreases.

2. Analyze Option A:
* Before: Shows four diatomic molecules ($\text{X}_2$) clustered in a gas-like distribution.
* After: Shows eight individual, separated atoms ($\text{X}$) dispersed throughout the container.
* Process: Chemical bond cleavage/dissociation: $\text{X}_2(g) \rightarrow 2\ \text{X}(g)$. Since the number of independent gas particles doubles (from 4 to 8), the positional disorder and microstates increase. Thus, entropy increases ($\Delta S > 0$).

3. Analyze Option B:
* Before: Shows a highly ordered, tightly packed crystalline lattice of atoms at the bottom of the beaker, representing a solid phase.
* After: Shows those same atoms widely separated and distributed randomly throughout the entire volume of the vessel, representing a gas phase.
* Process: Sublimation/vaporization: $\text{X}(s) \rightarrow \text{X}(g)$. Transitioning from a highly restricted solid lattice to a highly dispersed gaseous state significantly increases positional freedom. Thus, entropy increases ($\Delta S > 0$).

4. Analyze Option C:
* Before: Shows four triatomic molecules of type $\text{ABA}$ (with a black central atom and two white terminal atoms).
* After: Shows four diatomic molecules of type $\text{A}_2$ (white-white pairs) and four separate individual black atoms.
* Process: Decomposition reaction: $4\ \text{ABA}(g) \rightarrow 4\ \text{A}_2(g) + 4\ \text{B}(g)$. Because the reaction starts with 4 gas molecules and produces 8 independent gas particles, spatial disorder increases. Thus, entropy increases ($\Delta S > 0$).

5. Analyze Option D:
* Before: Shows five triatomic bent molecules (such as $\text{H}_2\text{O}$) dispersed randomly and moving freely throughout the container, representing a gaseous or liquid state.
* After: Shows those same five molecules clustered together in a highly organized, tightly packed localized arrangement at the bottom of the vessel, representing a crystalline solid state.
* Process: Condensation or freezing: $\text{X}(g/l) \rightarrow \text{X}(s)$. Because the molecules lose their translational and rotational freedom of motion and are locked into a highly structured solid arrangement, the number of available microstates drops dramatically. Therefore, the entropy of the system decreases ($\Delta S < 0$). This identifies Option D as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is wrong because the dissociation of molecules into free atoms increases the number of independent particles, which increases entropy.
• Option B is wrong because the transition of a substance from a solid to a gas represents sublimation, which increases the freedom of movement of the particles and increases entropy.
• Option C is wrong because a decomposition reaction that increases the net number of gaseous molecules in the vessel results in more possible spatial arrangements, increasing entropy.