Under which of the following conditions can an endothermic reaction be thermodynamically favorable? — Thermodynamics Chemistry Question

STEPS:

1. Define Thermodynamic Favorability: In chemistry, a process is considered thermodynamically favorable (historically called spontaneous) if the change in Gibbs Free Energy ($\Delta G$) is negative ($\Delta G < 0$).
2. Define Endothermic: An endothermic reaction is one that absorbs heat from the surroundings, meaning the change in enthalpy ($\Delta H$) is positive ($\Delta H > 0$).
3. Apply the Gibbs Free Energy Equation: The relationship between these variables is given by the equation:
$$\Delta G = \Delta H - T\Delta S$$
where $T$ is the absolute temperature and $\Delta S$ is the change in entropy.
4. Analyze the Requirements for Favorability: For an endothermic reaction to be favorable, the result of the calculation ($\Delta H - T\Delta S$) must be a negative number. Since $\Delta H$ is positive, the only way to arrive at a negative $\Delta G$ is to subtract a value ($T\Delta S$) that is larger than $\Delta H$.
5. Conclusion: Therefore, for $\Delta G$ to be less than zero when $\Delta H$ is positive, the term $T\Delta S$ must be greater than $\Delta H$ ($T\Delta S > \Delta H$). This usually occurs at high temperatures when there is a significant increase in entropy.

WHY_OTHERS_WRONG:

• A) $\Delta G$ is positive: By definition, if $\Delta G$ is positive, the reaction is thermodynamically unfavorable.
• B) $\Delta S$ is negative: If $\Delta H$ is positive and $\Delta S$ is negative, the term $-T\Delta S$ becomes positive. In this case, $\Delta G$ would be the sum of two positive numbers, making it always positive and unfavorable regardless of temperature.
• D) $T\Delta S = 0$: If this term is zero, then $\Delta G = \Delta H$. Since the reaction is endothermic ($\Delta H > 0$), $\Delta G$ would remain positive and unfavorable.
• E) There are no conditions: This is incorrect because many endothermic processes, such as the melting of ice at temperatures above $0^\circ\text{C}$ or the evaporation of water, are thermodynamically favorable because the increase in entropy ($T\Delta S$) outweighs the heat absorbed ($\Delta H$).