Physical chemistryLesson 2 of 3 in the course Thermodynamics

Entropy Criteria for Spontaneity and Absolute Entropy

5 flashcards · 4 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

Entropy can help predict whether a process occurs spontaneously, but the entropy change of the system alone does not always tell the whole story. This lesson introduces the second law and shows why we consider both a system and its surroundings.

ΔSuniv=ΔSsys+ΔSsurr\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}

The entropy change of the system alone may be positive or negative. To assess spontaneity, the relevant quantity is the combined entropy change of the system and surroundings.

All flashcards

What is the entropy change of the universe?
It is the sum of the entropy changes of the system and surroundings: ΔSuniv=ΔSsys+ΔSsurr\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}.
State the second law of thermodynamics in terms of entropy.
Every spontaneous change increases the entropy of the universe, so a spontaneous direction has ΔSuniv>0\Delta S_{\text{univ}} > 0.
What does a negative ΔSuniv\Delta S_{\text{univ}} mean?
The process is nonspontaneous in the direction considered; the opposite direction is spontaneous.
Why does heat flow spontaneously from hot to cold?
The entropy gained by the cooler surroundings exceeds the entropy lost by the hotter system, making ΔSuniv\Delta S_{\text{univ}} positive.
What is the entropy criterion for equilibrium?
At equilibrium, ΔSuniv=0\Delta S_{\text{univ}} = 0; for objects at the same temperature, their entropy changes from heat exchange balance.

Quiz questions

  1. What entropy value does the third law assign to a pure, perfect crystal at 0 K0\ \text{K}?Answer: Zero, because the crystal has one possible microstate.
  2. How is standard reaction entropy calculated from tabulated molar entropies?Answer: ∑νS∘(products)−∑νS∘(reactants)\sum \nu S^\circ(\text{products}) - \sum \nu S^\circ(\text{reactants})
  3. Melting has ΔSsys=+22.1 J/K\Delta S_{\text{sys}} = +22.1\ \text{J/K} and qsurr=−6000 Jq_{\text{surr}} = -6000\ \text{J}. At 263.15 K263.15\ \text{K}, what is the approximate result for melting?Answer: ΔSuniv≈−0.7 J/K\Delta S_{\text{univ}} \approx -0.7\ \text{J/K}; melting is nonspontaneous.
  4. For condensation, use S∘(H2O(g))=188.8 J mol−1K−1S^\circ(\mathrm{H_2O(g)}) = 188.8\ \text{J mol}^{-1}\text{K}^{-1} and S∘(H2O(l))=70.0 J mol−1K−1S^\circ(\mathrm{H_2O(l)}) = 70.0\ \text{J mol}^{-1}\text{K}^{-1}. What is ΔS∘\Delta S^\circ?Answer: −118.8 J mol−1K−1-118.8\ \text{J mol}^{-1}\text{K}^{-1}

Based on Chemistry 2e by Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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