2 H2O2(aq) → 2 H2O(l) + O2(g) ΔH° = −196 kJ/mol_rxn The decomposition of H2O2(aq) is represented by — Thermodynamics Chemistry Question

STEPS:

1. Relate thermodynamic favorability to Gibbs free energy ($\Delta G^\circ$): By definition, a physical or chemical process is thermodynamically favorable (spontaneous) under standard conditions if and only if the change in standard Gibbs free energy ($\Delta G^\circ$) is negative ($\Delta G^\circ < 0$). Because the problem explicitly states that this decomposition reaction is thermodynamically favorable, $\Delta G^\circ$ must be negative. This immediately eliminates Options A and C.
2. Understand the concept of entropy ($S$): Entropy is a thermodynamic measure of the molecular disorder, randomness, or number of microstates available to a system. A positive change in entropy ($\Delta S^\circ > 0$) indicates that the products of a reaction are more disordered than the reactants, whereas a negative change in entropy ($\Delta S^\circ < 0$) indicates the system has become more ordered.
3. Analyze the states of matter in the reaction equation: Look at the chemical equation provided for the decomposition of hydrogen peroxide:
$$2\text{ H}_2\text{O}_2(aq) \rightarrow 2\text{ H}_2\text{O}(l) + \text{O}_2(g)$$
* The reactant side consists of $2\text{ moles}$ of an aqueous solution ($\text{H}_2\text{O}_2(aq)$).
* The product side consists of $2\text{ moles}$ of liquid water ($\text{H}_2\text{O}(l)$) and $1\text{ mole}$ of highly disordered oxygen gas ($\text{O}_2(g)$).
4. Determine the sign of the entropy change ($\Delta S^\circ$): Gas molecules have substantially more freedom of motion, kinetic energy, and spatial distribution than molecules in the liquid or dissolved aqueous phases. The generation of a gaseous product ($\text{O}_2(g)$) from a starting aqueous solution represents a dramatic increase in the disorder of the system. Therefore, the system becomes more disordered as the reaction goes to completion, meaning $\Delta S^\circ$ must be positive.
5. Select the correct pairing: Combining our findings, $\Delta G^\circ$ is negative and $\Delta S^\circ$ is positive, which identifies Option B as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This option incorrectly claims that $\Delta G^\circ$ is positive. A positive $\Delta G^\circ$ value corresponds to a reaction that is thermodynamically *unfavorable* under standard conditions, which contradicts the given premise.
• Option C is incorrect: This option incorrectly identifies $\Delta G^\circ$ as positive and further claims that $\Delta S^\circ$ is negative. A negative $\Delta S^\circ$ would imply that the system is becoming more ordered, which is incorrect because the reaction produces highly disordered gas particles from an aqueous solution.
• Option D is incorrect: While this option correctly identifies $\Delta G^\circ$ as negative, it incorrectly claims that $\Delta S^\circ$ is negative. As established, the formation of oxygen gas increases the system's molecular randomness, making $\Delta S^\circ$ positive rather than negative.