1. John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend http://academic.cengage.com/kotz Chapter 19 Principles of Reactivity: Entropy and Free Energy

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3. 33 © 2009 Brooks/Cole - Cengage 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 PLAY MOVIE

4. 44 © 2009 Brooks/Cole - Cengage 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

5. 55 © 2009 Brooks/Cole - Cengage 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) e 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.

6. 66 © 2009 Brooks/Cole - Cengage 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. PLAY MOVIE

7. 77 © 2009 Brooks/Cole - Cengage Spontaneous Reactions In general, spontaneous reactions are exothermic. Fe 2 O 3 (s) + 2 Al(s) f 2 Fe(s) + Al 2 O 3 (s) ∆ r H = - 848 kJ

8. 88 © 2009 Brooks/Cole - Cengage Spontaneous Reactions But many spontaneous reactions or processes are endothermic or even have ∆H = 0. NH 4 NO 3 (s) + heat f NH 4 NO 3 (aq) ∆H = 0 PLAY MOVIE

9. 99 © 2009 Brooks/Cole - Cengage Entropy, S One property common to spontaneous processes is that the energy of the final state is more dispersed. In a spontaneous process energy goes from being more concentrated to being more dispersed. The thermodynamic property related to energy dispersal is ENTROPY, S. 2nd Law of Thermo — a spontaneous process results in an increase in the entropy of the universe. Reaction of K with water

10. 1010 © 2009 Brooks/Cole - Cengage Probability suggests that a spontaneous reaction will result in the dispersal of energy. Energy Dispersal Directionality of Reactions PLAY MOVIE

11. 1111 © 2009 Brooks/Cole - Cengage 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.

12. 1212 © 2009 Brooks/Cole - Cengage Energy Dispersal To begin, particle 1 has 2 packets of energy and 2-4 have none (upper left). With time it is more probable energy is dispersed over two particles. Each of these ways to distribute energy is called a microstate. BeginBegin See Figure 19.4 Energy over 2 particles

13. 1313 © 2009 Brooks/Cole - Cengage Matter & energy dispersal Directionality of Reactions As the size of the container increases, the number of microstates accessible to the system increases, and the density of states increases. Entropy increases. PLAY MOVIE

14. 1414 © 2009 Brooks/Cole - Cengage The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C. Energy is more dispersed in liquid water than in solid water.

15. 1515 © 2009 Brooks/Cole - Cengage S (solids) < S (liquids) < S (gases) 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 Energy dispersal PLAY MOVIE

16. 1616 © 2009 Brooks/Cole - Cengage Entropy and States of Matter S˚(Br 2 liq) < S˚(Br 2 gas) S˚(H 2 O sol) < S˚(H 2 O liq)

17. 1717 © 2009 Brooks/Cole - Cengage Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy, S PLAY MOVIE

18. 1818 © 2009 Brooks/Cole - Cengage Increase in molecular complexity generally leads to increase in S. Entropy, S PLAY MOVIE

19. 1919 © 2009 Brooks/Cole - Cengage 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 - PLAY MOVIE

20. 2020 © 2009 Brooks/Cole - Cengage Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase in entropy. Matter (and energy) are more dispersed. Entropy, S

21. 2121 © 2009 Brooks/Cole - Cengage Standard Molar Entropies

22. 2222 © 2009 Brooks/Cole - Cengage Entropy Changes for Phase Changes For a phase change, ∆S = q/T where q = heat transferred in phase change For H 2 O (liq) f H 2 O(g) ∆H = q = +40,700 J/mol PLAY MOVIE

23. 2323 © 2009 Brooks/Cole - Cengage Entropy and Temperature S increases slightly with T S increases a large amount with phase changes

24. 2424 © 2009 Brooks/Cole - Cengage Consider 2 H 2 (g) + O 2 (g) f 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/K·mol) - [2 mol (130.7 J/K·mol) + 1 mol (205.3 J/K·mol)] ∆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)

25. 2525 © 2009 Brooks/Cole - Cengage 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 Calculate the entropy created by energy dispersal in the system and surroundings.

26. 2626 © 2009 Brooks/Cole - Cengage Dissolving NH 4 NO 3 in water—an entropy driven process. 2nd Law of Thermodynamics ∆S universe = ∆S system + ∆S surroundings PLAY MOVIE

27. 2727 © 2009 Brooks/Cole - Cengage 2 H 2 (g) + O 2 (g) f 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 ∆ r H o = ∆H o system = -571.7 kJ

28. 2828 © 2009 Brooks/Cole - Cengage 2 H 2 (g) + O 2 (g) f 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. 2nd Law of Thermodynamics

29. 2929 © 2009 Brooks/Cole - Cengage 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.

30. 3030 © 2009 Brooks/Cole - Cengage 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

31. 3131 © 2009 Brooks/Cole - Cengage ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system - energy lost in energy dispersal - energy lost in energy dispersal If reaction is exothermic (negative ∆H o ) exothermic (negative ∆H o ) and entropy increases (positive ∆S o )and entropy increases (positive ∆S o ) then ∆G o must be NEGATIVEthen ∆G o must be NEGATIVE reaction is spontaneous (and product-favored).

32. 3232 © 2009 Brooks/Cole - Cengage ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system - energy lost in energy dispersal - energy lost in energy dispersal 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).

33. 3333 © 2009 Brooks/Cole - Cengage 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

34. 3434 © 2009 Brooks/Cole - Cengage 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 ∆ r H o and ∆ r S o and use Gibbs equation. b) Use tabulated values of free energies of formation, ∆ f G o. ∆ r G o =  ∆ f G o (products) -  ∆ f G o (reactants)

35. 3535 © 2009 Brooks/Cole - Cengage Free Energies of Formation Note that ∆ f G˚ for an element = 0

36. 3636 © 2009 Brooks/Cole - Cengage Calculating ∆ r G o Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) f 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate ∆ r H o = -1238 kJ ∆ r H o = -1238 kJ Use standard molar entropies to calculate ∆ r S o = -97.4 J/K or -0.0974 kJ/K ∆ r S o = -97.4 J/K or -0.0974 kJ/K ∆ r G o = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1209 kJ = -1209 kJ Reaction is product-favored in spite of negative ∆ r S o. Reaction is “enthalpy driven”

37. 3737 © 2009 Brooks/Cole - Cengage Calculating ∆ r G o Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat f NH 4 NO 3 (aq) PLAY MOVIE

38. 3838 © 2009 Brooks/Cole - Cengage Calculating ∆ r G o From tables of thermodynamic data we find ∆ r H o = +25.7 kJ ∆ r H o = +25.7 kJ ∆ r S o = +108.7 J/K or +0.1087 kJ/K ∆ r G o = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ = -6.7 kJ Reaction is product-favored in spite of negative ∆ r H o. Reaction is “entropy driven” NH 4 NO 3 (s) + heat f NH 4 NO 3 (aq)

39. 3939 © 2009 Brooks/Cole - Cengage 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 ∆ r H o and ∆ r S o and use Gibbs equation. b) Use tabulated values of free energies of formation, ∆ f G o. ∆ r G o =  ∆ f G o (products) -  ∆ f G o (reactants)

40. 4040 © 2009 Brooks/Cole - Cengage Calculating ∆G o rxn Combustion of carbon C(graphite) + O 2 (g) f CO 2 (g) ∆ r G o = ∆ f G o (CO 2 ) - [∆ f G o (graph) + ∆ f G o (O 2 )] ∆ r G o = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. ∆ r G o = -394.4 kJ Reaction is product-favored as expected. ∆ r G o =  ∆G f o (products) -  ∆G f o (reactants)

41. 4141 © 2009 Brooks/Cole - Cengage Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) f 4 Fe(s) + 3 CO 2 (g) ∆ r H o = +467.9 kJ ∆ r S o = +560.3 J/K ∆ r G o = +300.8 kJ Reaction is reactant-favored at 298 K At what T does ∆ r G o just change from being (+) to being (-)? When ∆ r G o = 0 = ∆ r H o - T∆ r S o

42. 4242 © 2009 Brooks/Cole - Cengage  FACT: ∆ r G o is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1.  Therefore, both ∆ r G˚ and K eq are related to reaction favorability. Thermodynamics and K eq

43. 4343 © 2009 Brooks/Cole - Cengage K eq is related to reaction favorability and so to ∆ r G o. The larger the value of K the more negative the value of ∆ r G o ∆ r G o = - RT lnK ∆ r G o = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

44. 4444 © 2009 Brooks/Cole - Cengage Calculate K for the reaction N 2 O 4 f 2 NO 2 ∆ r G o = +4.8 kJ ∆ r G o = +4800 J = - (8.31 J/K)(298 K) ln K ∆ r G o = - RT lnK Thermodynamics and K eq K = 0.14 When ∆ r G o > 0, then K 0, then K < 1

45. 4545 © 2009 Brooks/Cole - Cengage ∆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

46. 4646 © 2009 Brooks/Cole - Cengage ∆G, ∆G˚, and K eq Product Favored, ∆G˚ negative, K > 1 See Active Figure 19.13

47. 4747 © 2009 Brooks/Cole - Cengage Product-favoredProduct-favored 2 NO 2 e N 2 O 42 NO 2 e N 2 O 4 ∆ r G o = – 4.8 kJ∆ r G o = – 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 PLAY MOVIE

48. 4848 © 2009 Brooks/Cole - Cengage ∆G, ∆G˚, and K eq Reactant Favored, ∆G˚ positive, K < 1 See Active Figure 19.13

49. 4949 © 2009 Brooks/Cole - Cengage Reactant-favoredReactant-favored N 2 O 4 e 2 NO 2 ∆ r G o = +4.8 kJN 2 O 4 e 2 NO 2 ∆ r G o = +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 PLAY MOVIE

50. 5050 © 2009 Brooks/Cole - Cengage  K eq is related to reaction favorability.  When ∆ r G o < 0, reaction moves energetically “downhill”  ∆ r G o is the change in free energy when reactants convert COMPLETELY to products. Thermodynamics and K eq

51. 5151 © 2009 Brooks/Cole - Cengage A Summary The relation of ∆ r G, ∆ r G˚, Q, K, reaction spontaneity, and product- or reactant favorability.