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Lecture 8: The Second and Third Laws of Thermodynamics

Lecture 8: The Second and Third Laws of Thermodynamics. Reading: Zumdahl 10.5, 10.6 Outline Definition of the Second Law Determining D S Definition of the Third Law. The Second Law.

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Lecture 8: The Second and Third Laws of Thermodynamics

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  1. Lecture 8: The Second and Third Laws of Thermodynamics Reading: Zumdahl 10.5, 10.6 Outline Definition of the Second Law Determining DS Definition of the Third Law

  2. The Second Law • The Second Law: In any spontaneous process, there is always an increase in the entropy of the universe. • From our definitions of system and surroundings: DSuniverse = DSsystem + DSsurroundings

  3. There are only three possibilities: If DSuniv > 0, then the process is spontaneous. If DSuniv < 0, then the process is spontaneous in the opposite direction. If DSuniv = 0, the system is in equilibrium. • Here’s the catch: We need to know DS for both • the system and surroundings to predict whether a • reaction will be spontaneous.

  4. Consider a reaction driven by heat flow from the surroundings at constant pressure: Exothermic Process: DSsurr = heat/T Endothermic Process: DSsurr = -heat/T (Note: sign of DS is from the surrounding’s point of view!) Heat transferred = qP,surr = - qP,system= -DHsys

  5. Example: calculating DSsur What is DSsurr for the following reaction at 298 K? Sb4O6(s) + 6C(s) 4Sb(s) + 6CO2(g) DH = 778 kJ DSsurr= -DH/T = -778 kJ/298K = -2.6 kJ/K

  6. The Third Law of Thermodynamics Recall: in determining enthalpy changes (∆H) we had standard state values to use for reference. Q: Do standard reference states exist for entropy S? The Third Law: The entropy S of a perfect crystal at absolute zero (0 degrees K ) is zero. The third law provides the reference state for use in calculating absolute entropies.

  7. What is a Perfect Crystal? Perfect crystal at 0 K, so S =0 Crystal deforms for T > 0 K, slight random motions (wiggling in place), S>0

  8. Standard Entropies With reference to this state, standard entropies have been tabulated (Appendix 4). Recall, entropy S is a state function; therefore, the entropy change for a chemical reaction can be calculated as follows:

  9. Example: calculating DS°rxn Balance the following reaction and determine DS°rxn Fe(s) + H2O(g) Fe2O3(s) + H2(g) 2Fe(s) + 3H2O(g) Fe2O3(s) + 3H2(g) DS°rxn = (S°(Fe2O3(s)) + 3S°H2(g)) - (2S°Fe(s) + 3S°H2O(g)) DS°rxn = -141.5 J/K

  10. Example: determining rxn spontaneity Is the following reaction spontaneous at 298 K? (That is to say, is DS°univ > 0?) 2Fe(s) + 3H2O(g) Fe2O3(s) + 3H2(g) DS°rxn = DS°system= -141.5 J/K DS°surr = -DH°sys/T = -DH°rxn/T DH°rxn= DH°f(Fe2O3(s)) + 3DH°f(H2(g)) - 2DH°f(Fe(s)) - 3 DH°f(H2O(g))

  11. DH°rxn= -100 kJ DS°surr = -DH°sys/T = 348 J/K DS°univ = DS°sys + DS°surr = -141.5 J/K + 348 J/K = 207.5 J/K DS°univ > 0 ; therefore, yes, the reaction is spontaneous

  12. Entropy and Phase Changes Phase Change: Reaction in which a substance goes from one phase of state to another. Example: H2O(l) H2O(g) @ 373 K Key point: phase changes are equilibrium processes such that: DSuniv = 0

  13. H2O(l) H2O(g) @ 373 K So, DS°rxn = S°(H2O(g)) - S°(H2O(l)) = 195.9 J/K - 86.6 J/K = 109.1 J/K and, DS°surr = -DHsys/T = -40.7 kJ/373 K = -109.1 J/K Therefore, DSuniv = DSsys + DSsurr = 0

  14. Another example Determine the temperature at which liquid bromine boils: Br2(l) Br2(g) DS°rxn = S°(Br2 (g)) - S°(Br2(l)) = 245.38 J/K - 152.23 J/K = 93.2 J/K

  15. DS°surr = - DS°sys = -93.2 J/K = -DHsys/T Therefore, calculate DHsys and solve for T! 0 Now, DH°rxn = DH°f(Br2(g)) - DH°f(Br2(l)) = 30.91 kJ - 0 = 30.91 kJ (standard state) Finally, -93.2 J/K = -30.91 kJ/T Tboiling = 331.6 K

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