As a sample of KNO3(s) is stirred into water at 25°C, the compound dissolves endothermically. Which — Thermodynamics Chemistry Question

STEPS:

1. Identify the thermodynamic state functions of the system:
The question states that potassium nitrate ($\text{KNO}_3$) dissolves endothermically. This indicates that heat is absorbed from the surroundings, meaning the change in enthalpy is positive ($\Delta H > 0$).
Because the salt spontaneously dissolves at $25^\circ\text{C}$, the process is thermodynamically favorable. A thermodynamically favorable process at a constant temperature and pressure must have a negative change in Gibbs free energy ($\Delta G < 0$).

2. Relate favorability using the Gibbs free energy equation:
The relationship between free energy, enthalpy, temperature, and entropy is defined by the fundamental thermodynamic equation:
$$\Delta G = \Delta H - T\Delta S$$
where $T$ is the absolute temperature in Kelvin ($298\text{ K}$ at $25^\circ\text{C}$) and $\Delta S$ is the entropy change of the system.

3. Analyze the algebraic conditions for favorability:
- We know that $\Delta H$ is positive ($+$) because the process is endothermic.
- For $\Delta G$ to be negative ($-$), the subtraction term ($T\Delta S$) must be positive and have a greater magnitude than the positive $\Delta H$ term.
- Algebraically:
$$\text{Since } \Delta G = \Delta H - T\Delta S < 0 \implies T\Delta S > \Delta H$$
- Because $T$ is always positive in Kelvin, the entropy change ($\Delta S$) must be positive ($\Delta S > 0$), meaning the entropy of the system increases during dissolution.

4. Connect the math to particulate behavior:
- Solid potassium nitrate ($\text{KNO}_3(s)$) consists of $\text{K}^+$ and $\text{NO}_3^-$ ions arranged in a highly ordered, rigid crystalline lattice.
- When the salt dissolves, these ions break free from their fixed positions and disperse randomly throughout the aqueous solvent.
- This transition from a highly ordered solid phase to a highly mobile, dispersed aqueous phase represents a significant increase in the positional disorder of the particles, causing entropy to increase ($\Delta S > 0$).
- At $25^\circ\text{C}$, this entropy increase is large enough that $T\Delta S$ exceeds $\Delta H$, driving $\Delta G$ negative and making the dissolution favorable. This identifies Option D as the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Most endothermic processes are actually thermodynamically unfavorable at standard temperature. An endothermic process is only favorable if it is accompanied by a sufficiently large increase in entropy ($\Delta S > 0$) and occurs at a temperature high enough to make $T\Delta S > \Delta H$.
• Option B is incorrect: While stirring is a mechanical action that speeds up the kinetics (the rate) of dissolution, it does not alter the thermodynamic state functions ($\Delta H$, $\Delta S$, or $\Delta G$) of the chemical system. Stirring does not supply the thermodynamic driving force required to make an unfavorable process occur.
• Option C is incorrect: A decrease in entropy ($\Delta S < 0$) combined with an endothermic enthalpy ($\Delta H > 0$) would result in a positive free energy change ($\Delta G > 0$) at all temperatures, making the dissolution completely unfavorable. *(Note: In the original exam text, Option C states "Dissolving the salt decreases the enthalpy of the system". This is also incorrect because an endothermic process always increases the enthalpy of the system.)*