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

3. 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:  S universe =  S system +  S surroundings

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

5. Consider a reaction driven by heat flow from the surroundings at constant pressure: Exothermic Process:  S surr = heat/T Endothermic Process:  S surr = -heat/T Heat transferred = q P,surr = - q P,system= -  H sys (Note: sign of  S is from the surrounding’s point of view!)

6. Example: calculating  S sur What is  S surr for the following reaction at 298 K? Sb 4 O 6 (s) + 6C(s) 4Sb(s) + 6CO 2 (g)  H = 778 kJ  S surr = -  H/T = -778 kJ/298K = -2.6 kJ/K

7. 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.

8. 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

9. 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:

10. Example: calculating  S° rxn Balance the following reaction and determine  S° rxn Fe(s) + H 2 O(g) Fe 2 O 3 (s) + H 2 (g) 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (s) + 3H 2 (g)  S° rxn = (S°(Fe 2 O 3 (s)) + 3S°H 2 (g)) - (2S°Fe(s) + 3S ° H 2 O(g))  S° rxn = -141.5 J/K

11. Example: determining rxn spontaneity Is the following reaction spontaneous at 298 K? 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (s) + 3H 2 (g)  S° rxn =  S° system = -141.5 J/K  S° surr = -  H° sys /T = -  H° rxn /T (That is to say, is  S° univ > 0?)  H° rxn =  H° f (Fe 2 O 3 (s)) + 3  H° f (H 2 (g)) - 2  H° f (Fe (s)) - 3  H° f (H 2 O(g))

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

13. Entropy and Phase Changes Phase Change: Reaction in which a substance goes from one phase of state to another. Example: H 2 O(l) H 2 O(g) @ 373 K Key point: phase changes are equilibrium processes such that:  S univ = 0

14. H 2 O(l) H 2 O(g) @ 373 K So,  S° rxn = S°(H 2 O(g)) - S°(H 2 O(l)) = 195.9 J/K - 86.6 J/K = 109.1 J/K and,  S° surr = -  H sys /T = -40.7 kJ/373 K = -109.1 J/K Therefore,  S univ =  S sys +  S surr = 0

15. Another example Determine the temperature at which liquid bromine boils: Br 2 (l) Br 2 (g)   S° rxn = S°(Br 2 (g)) - S°(Br 2 (l)) = 245.38 J/K - 152.23 J/K = 93.2 J/K

16.  S° surr = -  S° sys = -93.2 J/K = -  sys /T Therefore, calculate  H sys and solve for T! Now,  H° rxn =  H° f (Br 2 (g)) -  H° f (Br 2 (l)) = 30.91 kJ - 0 = 30.91 kJ Finally, -93.2 J/K = -  kJ/T T boiling = 331.6 K (standard state) 0