1. 11 © 2006 Brooks/Cole - Thomson Chemistry and Chemical Reactivity 6th Edition John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 19 Principles of Reactivity: Entropy and Free Energy © 2006 Brooks/Cole Thomson Lectures written by John Kotz

2. 22 © 2006 Brooks/Cole - Thomson Entropy and Free Energy How to predict if a reaction can occur, given enough time? THERMODYNAMICS How to predict if a reaction can occur at a reasonable rate? KINETICS

3. 33 © 2006 Brooks/Cole - Thomson Thermodynamics If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system---If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system--- AND the K is greater than 1,AND the K is greater than 1, then this is a product-favored system.then this is a product-favored system. Most product-favored reactions are exothermic —but this is not the only criterionMost product-favored reactions are exothermic —but this is not the only criterion

4. 44 © 2006 Brooks/Cole - Thomson Thermodynamics Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneous process.Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneous process. AgCl(s) Æ Ag + (aq) + Cl – (aq) K = 1.8 x 10 -10 Reaction is not product-favored, but it moves spontaneously toward equilibrium. Spontaneous does not imply anything about time for reaction to occur.Spontaneous does not imply anything about time for reaction to occur.

5. 55 © 2006 Brooks/Cole - Thomson Thermodynamics and Kinetics Diamond is thermodynamically favored to convert to graphite, but not kinetically favored. Paper burns — a product-favored reaction. Also kinetically favored once reaction is begun.

6. 66 © 2006 Brooks/Cole - Thomson Spontaneous Reactions In general, spontaneous reactions are exothermic. Fe 2 O 3 (s) + 2 Al(s) ---> 2 Fe(s) + Al 2 O 3 (s) ∆H = - 848 kJ

7. 77 © 2006 Brooks/Cole - Thomson Spontaneous Reactions But many spontaneous reactions or processes are endothermic or even have ∆H = 0. NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

8. 88 © 2006 Brooks/Cole - Thomson Entropy, S 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. Reaction of K with water

9. 99 © 2006 Brooks/Cole - Thomson The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C.

10. 1010 © 2006 Brooks/Cole - Thomson How probable is it that reactant molecules will react? PROBABILITY suggests that a spontaneous reaction will result in the dispersal *of energy *or of matter * or of energy & matter. Directionality of Reactions

11. 1111 © 2006 Brooks/Cole - Thomson Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both. Matter Dispersal Directionality of Reactions

12. 1212 © 2006 Brooks/Cole - Thomson Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both. Energy Dispersal Directionality of Reactions

13. 1313 © 2006 Brooks/Cole - Thomson Directionality of Reactions Energy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings. The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. The final state—with energy dispersed—is more probable and makes a reaction spontaneous.

14. 1414 © 2006 Brooks/Cole - Thomson 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

15. 1515 © 2006 Brooks/Cole - Thomson Entropy and States of Matter S˚(Br 2 liq) < S˚(Br 2 gas) S˚(H 2 O sol) < S˚(H 2 O liq)

16. 1616 © 2006 Brooks/Cole - Thomson Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy, S

17. 1717 © 2006 Brooks/Cole - Thomson Increase in molecular complexity generally leads to increase in S. Entropy, S

18. 1818 © 2006 Brooks/Cole - Thomson 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 -

19. 1919 © 2006 Brooks/Cole - Thomson Entropy usually increases when a pure liquid or solid dissolves in a solvent. Entropy, S

20. 2020 © 2006 Brooks/Cole - Thomson Standard Molar Entropies

21. 2121 © 2006 Brooks/Cole - Thomson 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

22. 2222 © 2006 Brooks/Cole - Thomson Entropy and Temperature S increases slightly with T S increases a large amount with phase changes

23. 2323 © 2006 Brooks/Cole - Thomson 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)

24. 2424 © 2006 Brooks/Cole - Thomson 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 )

25. 2525 © 2006 Brooks/Cole - Thomson Dissolving NH 4 NO 3 in water—an entropy driven process. 2nd Law of Thermodynamics ∆S universe = ∆S system + ∆S surroundings

26. 2626 © 2006 Brooks/Cole - Thomson 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

27. 2727 © 2006 Brooks/Cole - Thomson 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

28. 2828 © 2006 Brooks/Cole - Thomson Spontaneous or Not? Remember that –∆H˚ sys is proportional to ∆S˚ surr An exothermic process has ∆S˚ surr > 0. Remember that –∆H˚ sys is proportional to ∆S˚ surr An exothermic process has ∆S˚ surr > 0.

29. 2929 © 2006 Brooks/Cole - Thomson Gibbs Free Energy, G Multiply through by -T -T∆S univ = ∆H sys - T∆S sys -T∆S univ = change in Gibbs free energy for the system = ∆G system Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys ∆S univ = ∆S surr + ∆S sys J. Willard Gibbs 1839-1903

30. 3030 © 2006 Brooks/Cole - Thomson ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is exothermic (negative ∆ H o ) (energy dispersed) exothermic (negative ∆ H o ) (energy dispersed) and entropy increases (positive ∆S o ) (matter dispersed)and entropy increases (positive ∆S o ) (matter dispersed) then ∆G o must be NEGATIVEthen ∆G o must be NEGATIVE reaction is spontaneous (and product- favored).

31. 3131 © 2006 Brooks/Cole - Thomson ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is endothermic (positive ∆H o ) endothermic (positive ∆H o ) and entropy decreases (negative ∆S o )and entropy decreases (negative ∆S o ) then ∆G o must be POSITIVEthen ∆G o must be POSITIVE reaction is not spontaneous (and is reactant- favored). reaction is not spontaneous (and is reactant- favored).

32. 3232 © 2006 Brooks/Cole - Thomson Gibbs Free Energy, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o ∆H o ∆S o ∆G o Reaction exo(–)increase(+)–Prod-favored endo(+)decrease(-)+React-favored exo(–)decrease(-)?T dependent endo(+)increase(+)?T dependent

33. 3333 © 2006 Brooks/Cole - Thomson Gibbs Free Energy, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o Two methods of calculating ∆G o a)Determine ∆H o rxn and ∆S o rxn and use GIbbs equation. b)Use tabulated values of free energies of formation, ∆G f o. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

34. 3434 © 2006 Brooks/Cole - Thomson Free Energies of Formation Note that ∆G˚ f for an element = 0

35. 3535 © 2006 Brooks/Cole - Thomson Calculating ∆G o rxn Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate ∆H o rxn = -1238 kJ ∆H o rxn = -1238 kJ Use standard molar entropies to calculate ∆S o rxn = -97.4 J/K or -0.0974 kJ/K ∆S o rxn = -97.4 J/K or -0.0974 kJ/K ∆G o rxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ = -1209 kJ Reaction is product-favored in spite of negative ∆S o rxn. Reaction is “enthalpy driven”

36. 3636 © 2006 Brooks/Cole - Thomson Calculating ∆G o rxn Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

37. 3737 © 2006 Brooks/Cole - Thomson Calculating ∆G o rxn From tables of thermodynamic data we find ∆H o rxn = +25.7 kJ ∆H o rxn = +25.7 kJ ∆S o rxn = +108.7 J/K or +0.1087 kJ/K ∆G o rxn = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ = -6.7 kJ Reaction is product-favored in spite of negative ∆H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

38. 3838 © 2006 Brooks/Cole - Thomson Gibbs Free Energy, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o Two methods of calculating ∆G o a)Determine ∆H o rxn and ∆S o rxn and use GIbbs equation. b)Use tabulated values of free energies of formation, ∆G f o. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

39. 3939 © 2006 Brooks/Cole - Thomson Calculating ∆G o rxn Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) ∆G o rxn = ∆G f o (CO 2 ) - [∆G f o (graph) + ∆G f o (O 2 )] ∆G o rxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. ∆G o rxn = -394.4 kJ Reaction is product-favored as expected. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

40. 4040 © 2006 Brooks/Cole - Thomson Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g) ∆H o rxn = +467.9 kJ ∆S o rxn = +560.3 J/K ∆G o rxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does ∆G o rxn just change from being (+) to being (-)? When ∆G o rxn = 0 = ∆H o rxn - T∆S o rxn

41. 4141 © 2006 Brooks/Cole - Thomson  FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1.  Therefore, both ∆G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq

42. 4242 © 2006 Brooks/Cole - Thomson K eq is related to reaction favorability and so to ∆G o rxn. The larger the value of K the more negative the value of ∆G o rxn ∆G o rxn = - RT lnK ∆G o rxn = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

43. 4343 © 2006 Brooks/Cole - Thomson Calculate K for the reaction N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ ∆G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K ∆G o rxn = - RT lnK Thermodynamics and K eq K = 0.14 When ∆G o rxn > 0, then K 0, then K < 1

44. 4444 © 2006 Brooks/Cole - Thomson ∆G, ∆G˚, and K eq ∆G is change in free energy at non- standard conditions.∆G is change in free energy at non- standard conditions. ∆G is related to ∆G˚∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q K, reaction is spontaneous.When Q K, reaction is spontaneous. When Q = K reaction is at equilibriumWhen Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibriumWhen ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln KTherefore, ∆G˚ = - RT ln K

45. 4545 © 2006 Brooks/Cole - Thomson But systems can reach equilibrium when reactants have NOT converted completely to products. In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. ∆G, ∆G˚, and K eq Product Favored, ∆G˚ negative, K > 1 Active Figure 19.13

46. 4646 © 2006 Brooks/Cole - Thomson Product-favoredProduct-favored 2 NO 2 ---> N 2 O 42 NO 2 ---> N 2 O 4 ∆G o rxn = – 4.8 kJ∆G o rxn = – 4.8 kJ State with both reactants and products present is more stable than complete conversion.State with both reactants and products present is more stable than complete conversion. K > 1, more products than reactants.K > 1, more products than reactants. ∆G, ∆G˚, and K eq

47. 4747 © 2006 Brooks/Cole - Thomson Reactant-favoredReactant-favored N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJN 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ State with both reactants and products present is more stable than complete conversion.State with both reactants and products present is more stable than complete conversion. K < 1, more reactants than productsK < 1, more reactants than products ∆G, ∆G˚, and K eq

48. 4848 © 2006 Brooks/Cole - Thomson  K eq is related to reaction favorability.  When ∆G o rxn < 0, reaction moves energetically “downhill”  ∆G o rxn is the change in free energy when reactants convert COMPLETELY to products. Thermodynamics and K eq