Questions 44-46 relate to the following information. XY2 → X + Y2 The equation above represents the — Thermodynamics Chemistry Question

STEPS:

1. Identify the thermodynamic condition for favorability: For a chemical reaction to be thermodynamically favorable (spontaneous) under standard conditions, the change in standard Gibbs free energy ($\Delta G^\circ$) must be negative ($\Delta G^\circ < 0$).
2. Recall the Gibbs free energy relationship: The standard Gibbs free energy change of a reaction is mathematically related to its standard enthalpy change ($\Delta H^\circ$), standard entropy change ($\Delta S^\circ$), and absolute temperature ($T$) by the equation:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
3. Set up the inequality for favorability: Combine the thermodynamic favorability condition with the Gibbs free energy equation:
$$\Delta H^\circ - T\Delta S^\circ < 0$$
4. Solve the inequality for $\Delta S^\circ$: Rearrange the terms to isolate the entropy variable:
$$\Delta H^\circ < T\Delta S^\circ$$
Because absolute temperature in Kelvin ($T$) is always positive, you can divide both sides of the inequality by $T$ without flipping the direction of the inequality sign:
$$\frac{\Delta H^\circ}{T} < \Delta S^\circ \implies \Delta S^\circ > \frac{\Delta H^\circ}{T}$$
5. Substitute the given temperature: The question specifies that the reaction is thermodynamically favorable under standard conditions at a temperature of $298\text{ K}$. Substituting $T = 298\text{ K}$ yields:
$$\Delta S^\circ > \frac{\Delta H^\circ}{298\text{ K}}$$
This mathematically proves that the value of $\Delta S^\circ$ must be greater than $\Delta H^\circ / 298\text{ K}$, which directly corresponds to Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect: If $\Delta S^\circ$ were equal to zero, then the free energy equation would simplify to $\Delta G^\circ = \Delta H^\circ$. Because this decomposition reaction is endothermic ($\Delta H^\circ = +50\text{ kJ/mol}_{rxn} > 0$), as shown in the potential energy diagram, $\Delta G^\circ$ would be positive ($+50\text{ kJ/mol}_{rxn}$). A positive $\Delta G^\circ$ means the reaction is thermodynamically unfavorable, contradicting the given premise.
• Option B is incorrect: If $\Delta S^\circ$ were exactly equal to $\Delta H^\circ / 298\text{ K}$, then $298\text{ K} \times \Delta S^\circ = \Delta H^\circ$. Substituting this back into the free energy equation yields $\Delta G^\circ = \Delta H^\circ - \Delta H^\circ = 0$. A standard free energy change of zero means the system is at equilibrium under standard conditions, rather than being thermodynamically favorable.
• Option D is incorrect: If $\Delta S^\circ < \Delta H^\circ / 298\text{ K}$, then $T\Delta S^\circ < \Delta H^\circ$. Subtracting the entropy term from enthalpy would yield a positive value ($\Delta H^\circ - T\Delta S^\circ > 0$), meaning $\Delta G^\circ > 0$. This would make the reaction thermodynamically unfavorable under standard conditions.