Thermodynamics Of Mixing And Excess Properties
For an ideal solution formed by mixing two liquids at constant temperature, which of the following sets of thermodynamic changes is correct?
Select the correct option:
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- thermodynamics of mixing and excess properties
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
ΔmixH=0,ΔmixS>0,ΔmixG<0
For an ideal solution, two defining conditions hold: (1) there is no enthalpy change upon mixing, ΔH_mix = 0, because the intermolecular interactions between unlike molecules (A–B) are identical to those between like molecules (A–A and B–B); (2) there is no volume change upon mixing, ΔV_mix = 0. However, mixing two distinct substances always increases the randomness or disorder of the system — molecules of each type are now spread throughout the mixture. Therefore, ΔS_mix > 0 (positive entropy change). Since ΔG_mix = ΔH_mix − TΔS_mix = 0 − T(positive) < 0, ideal solutions form spontaneously at all temperatures. Option B (ΔH > 0) describes a non-ideal solution with positive deviation from Raoult's law — mixing is endothermic for systems like ethanol-water. Option C (all zero) would imply no mixing ever occurred, contradicting physical reality. Option D (ΔH < 0) describes an exothermic mixing process typical of solutions with negative deviation (e.g., acetone-chloroform). The thermodynamic treatment of ideal solutions is from NCERT Chemical Thermodynamics and Solutions chapters and appears in JEE Advanced in integrated form. Plausibility check: ΔG < 0 confirms spontaneous mixing, consistent with the everyday observation that miscible liquids mix freely without external driving forces.
This hard difficulty chemistry question is from the chapter solutions, covering the topic of thermodynamics of mixing and excess properties. It appeared in the 2025 exam.
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