Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIGibb's Free EnergyEasy2 minQB
ChemistryEasymatching list
Match Column I (Change at 1 atm) with Column II (Signs)
Column I | Column II |
|---|---|
P) Condensation of steam at 100°C | i) Δ G = 0, Δ S > 0 |
Q) Melting of ice at 0°C | ii) Δ G = 0, Δ S < 0 |
R) Vaporization of water at 110°C | iii) Δ G < 0, Δ S < 0 |
S) Freezing of water at −10°C | iv) Δ G < 0, Δ S > 0 |
Options:
Answer:
B
Solution:
At 100°C and 1 atm steam and water are in equilibrium, Δ G = 0, and gas → liquid loses entropy (P – ii). At 0°C ice and water are in equilibrium, Δ G = 0, and melting gains entropy (Q – i). At 110°C (above the normal boiling point) vaporization is spontaneous, Δ G < 0, with Δ S > 0 (R – iv). Supercooled water freezes spontaneously at −10°C, Δ G < 0, and freezing loses entropy (S – iii).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy
⏱ 2mℹ️ Source: QB
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