Which of the following equations represents a reaction for which the standard entropy change is posi — Thermodynamics Chemistry Question

STEPS:

1. Understand the concept of entropy ($S$): Entropy is a thermodynamic measure of the molecular disorder or randomness of a system. At the particulate level, entropy is determined by the number of possible microstates (arrangements) available to the system.
2. Relate physical states of matter to entropy: Because of the differences in particle freedom and spacing:
* Solids have highly ordered, locked crystal structures and the lowest entropy.
* Liquids have more freedom of movement and moderate entropy.
* Gases have particles that are widely separated and free to move rapidly in all directions, giving them the highest entropy by a wide margin ($S_{\text{gas}} \gg S_{\text{liquid}} > S_{\text{solid}}$).
3. Establish the rule for chemical reactions: The sign of the standard entropy change ($\Delta S^\circ$) of a reaction is primarily governed by the change in the number of moles of gas ($\Delta n_{\text{gas}} = \text{moles of product gas} - \text{moles of reactant gas}$):
* If a reaction produces more gas molecules than it consumes ($\Delta n_{\text{gas}} > 0$), or converts a solid/liquid into a gas, the system becomes significantly more disordered, and $\Delta S^\circ$ is positive ($\Delta S^\circ > 0$).
* If a reaction consumes gas molecules ($\Delta n_{\text{gas}} < 0$), the system becomes more ordered, and $\Delta S^\circ$ is negative ($\Delta S^\circ < 0$).
4. Evaluate Option C ($\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$):
* Reactants: 1 mole of solid ($\text{CaCO}_3$), 0 moles of gas.
* Products: 1 mole of solid ($\text{CaO}$) and 1 mole of gas ($\text{CO}_2$).
* Since this reaction starts with a highly ordered solid and produces a gas, there is a net increase of one mole of gas ($\Delta n_{\text{gas}} = +1$). The gas molecules have a vast number of positional microstates compared to the rigid solid reactant, leading to a highly positive standard entropy change ($\Delta S^\circ > 0$). This confirms Option C is the correct answer.

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WHY_OTHERS_WRONG:

• A is incorrect: In the reaction $3\text{ O}_2(g) \rightarrow 2\text{ O}_3(g)$, 3 moles of reactant gas are converted into 2 moles of product gas ($\Delta n_{\text{gas}} = -1$). Reducing the total number of gas molecules restricts the ways the particles can be distributed in space, which decreases entropy ($\Delta S^\circ < 0$).
• B is incorrect: In the reaction $2\text{ H}_2(g) + \text{O}_2(g) \rightarrow 2\text{ H}_2\text{O}(l)$, 3 moles of highly disordered gases are converted into 2 moles of a much more ordered liquid ($\Delta n_{\text{gas}} = -3$). This major reduction in particle movement and spatial distribution results in a highly negative entropy change ($\Delta S^\circ < 0$).
• D is incorrect: In the reaction $\text{I}_2(g) + 2\text{ K}(s) \rightarrow 2\text{ KI}(s)$, 1 mole of gas reacts with 2 moles of solid to form 2 moles of solid. The gaseous reactant is entirely consumed to form a solid crystal lattice ($\Delta n_{\text{gas}} = -1$), leading to a significant increase in molecular order and a negative entropy change ($\Delta S^\circ < 0$).