1. 1 20-4-2011

2. 2 In general, the more atoms in its molecules, the greater is the entropy of a substance Entropy is a function of temperature

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4. 4  The second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases.  For a reversible process: S univ =  S sys +  S surr = 0  For a spontaneous process (irreversible): S univ =  S sys +  S surr > 0 The Second Law of Thermodynamics

5. 5  In any spontaneous process, the entropy of the universe increases. S univ = S sys + S surr  Entropy is not conserved: S univ is increasing.

6. 6 Third Law of Thermodynamics The entropy of a pure perfect crystalline substance at absolute zero is 0.

7. 7 Standard Molar Entropies The standard molar entropy, S o, is the entropy of one mole of a substance in its standard state.  S o rxn =  n S o (products) –  mS o (reactants) You should be able to recognize the similarity between this equation and that used for the calculation of the enthalpy change: According to the Third Law of Thermodynamics, the entropy of a pure, perfect crystalline material is zero at 0 K

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9. 9 Which substances in each of the following pairs would you expect to have the higher standard molar entropy? Why? a.C 2 H 2 (g) or C 2 H 6 (g) b. CO 2 (g) or CO (g) c. I 2 (s) or I 2 (g) d. CH 3 OH (g) or CH 3 OH (l)

10. 10 Entropy Changes in Chemical Reactions Can Calculate  S° for chemical reactions Calculate  S  for the dissolution of ammonium nitrate. NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq) 151.04 J/mol*K112.8 146.4  S o = {1*112.8 + 1*146.4) – (151.04) = 108.2

11. 11 Calculate the standard entropy change for the following reaction: Al 2 O 3 (s) + 3H 2 (g)  2Al (s) + 3 H 2 O (g)  S o = {2*28.3 + 3*188.7) – {1*50.99 + 3*131.0) = 178.7 Al 2 O 3 (s) + 3H 2 (g)  2Al (s) + 3 H 2 O (g) 188.728.3131.050.99

12. 12 Calculate the standard entropy change for the reaction: 2 NaHCO 3 (s)  Na 2 CO 3 (s) + CO 2 (g) + H 2 O (g) Provided that S 0 (NaHCO 3 (s)) = 155 J/mol K, S 0 (Na 2 CO 3 (s)) = 136 J/mol K, S 0 (CO 2 (g)) = 213.6 J/mol K, and S 0 (H 2 O (g)) = 188.7 J/mol K. Solution  S 0 =  nS 0 (products) -  mS 0 (reactants)  S 0 = {S 0 (Na 2 CO 3 (s)) + S 0 (CO 2 (g)) + S 0 (H 2 O (g))} – {2 S 0 (NaHCO 3 (s))}  S 0 = {136 + 213.6 + 188.7} – {2*155} = 228 J/K

13. 13 Predict whether the entropy of each of the following reactions is positive or negative. a.2H 2 (g) + O 2 (g)  2H 2 O(l) The entropy will have a negative sign as gaseous molecules are converted to liquid, thus decreasing entropy. b. NH 4 Cl(s)  NH 3 (g) + HCl(g) Here the entropy will have a positive sign since a lower entropy solid is converted to high entropy gases. c. N 2 (g) + O 2 (g)  2NO(g) Same number of gaseous molecules on both sides so we can not define the sign of entropy, however the entropy change will be very small.

14. 14 Calculate the change in entropy for the production of ammonia from nitrogen and hydrogen gas. At 298K as a standard temperature, the standard entropies are: S 0 (NH 3 ) = 192.5 J/mol K; S 0 (H 2 ) = 130.6 J/mol K; S 0 (N 2 ) = 191.5 J/mol K N 2 (g) + 3H 2 (g)  2NH 3 (g) Solution From the balanced equation we can write the equation for  S 0 (the change in the standard molar entropy for the reaction):  S 0 = [2* S 0 (NH 3 (g))] - [S 0 (N 2 (g) ) + (3* S 0 (H 2 (g)))]  S 0 = [2*192.5] - [191.5 + (3*130.6)]  S 0 = -198.3 J/mol K The negative sign means that entropy decreases due to production of less number of moles than reactants.

15. 15 Entropy Changes in the Surroundings Heat that flows into or out of the system also changes the entropy of the surroundings. For an isothermal process and at constant pressure,  S surr  –  H o sys The –ve sign is because for exothermic reactions  H o sys is negative and thus –  H o sys will be possitive, suggesting an increase in  S surr, therefore we can write:

16. 16 In the reaction: N 2 (g) + 3H 2 (g)  2 NH 3, If  H o = -92.6 kJ/mol, find  S sys,  S surr and  S univ ?  S sys o = {2*193.0) – {191.5+ 3*131.0) = -199 J/K  S surr = -(-92.6*1000)/298 = 311J/K  S univ =  S sys +  S surr = -199 + 311 = 112 J/K N 2 (g) + 3H 2 (g)  2 NH 3 193.0 131.0191.5

17. 17 Phase changes A phase change is isothermal (no change in T). Entropy system For water:  H fusion = 6 kJ/mol  H vap = 41 kJ/mol If we do this reversibly:  S surr = –  S sys