1. Entropy By 120570119127

2. Introduction One property common to spontaneous processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S.

3. The increase in entropy from solid to liquid to gas

4. The entropy change accompanying the dissolution of a salt pure solid pure liquid solution MIX

5. Ethanol WaterSolution of ethanol and water The small increase in entropy when ethanol dissolves in water

6. The large decrease in entropy when a gas dissolves in a liquid O 2 gas Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7. S (gases) > S (liquids) > S (solids) S o (J/Kmol) H 2 O(liq)69.95 H 2 O(gas)188.8 S o (J/Kmol) H 2 O(liq)69.95 H 2 O(gas)188.8 Entropy, S

8. Entropy and States of Matter S˚(Br 2 liq) < S˚(Br 2 gas) S˚(H 2 O sol) < S˚(H 2 O liq)

9. Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy, S

10. Increase in molecular complexity generally leads to increase in S. Entropy, S

11. Entropies of ionic solids depend on coulombic attractions. S o (J/Kmol) MgO26.9 NaF51.5 S o (J/Kmol) MgO26.9 NaF51.5 Entropy, S Mg 2+ & O 2- Na + & F -

12. Entropy usually increases when a pure liquid or solid dissolves in a solvent. Entropy, S

13. Standard Molar Entropies

14. Entropy Changes for Phase Changes For a phase change, ∆S = q/T where q = heat transferred in phase change For H 2 O (liq) ---> H 2 O(g) ∆H = q = +40,700 J/mol

15. Entropy and Temperature S increases slightly with T S increases a large amount with phase changes

16. Consider 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) ∆S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )] ∆S o = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)] ∆S o = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating ∆S for a Reaction ∆S o =  S o (products) -  S o (reactants)

17. 2nd Law of Thermodynamics A reaction is spontaneous if ∆S for the universe is positive. ∆S universe = ∆S system + ∆S surroundings ∆S universe > 0 for spontaneous process First calc. entropy created by matter dispersal (∆S system ) Next, calc. entropy created by energy dispersal (∆S surround )

18. Dissolving NH 4 NO 3 in water—an entropy driven process. 2nd Law of Thermodynamics ∆S universe = ∆S system + ∆S surroundings

19. 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) ∆S o system = -326.9 J/K 2nd Law of Thermodynamics ∆S o surroundings = +1917 J/K Can calc. that ∆H o rxn = ∆H o system = -571.7 kJ

20. 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) ∆S o system = -326.9 J/K ∆S o surroundings = +1917 J/K ∆S o universe = +1590. J/K The entropy of the universe is increasing, so the reaction is product-favored.The entropy of the universe is increasing, so the reaction is product-favored. 2nd Law of Thermodynamics

21. Spontaneous or Not? Remember that –∆H˚ sys is proportional to ∆S˚ surr An exothermic process has ∆S˚ surr > 0.