Chapter 19 Principles of Reactivity: Entropy and Free Energy

1. Chapter 19Principles of Reactivity: Entropy and Free Energy

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3. Entropy and Free Energy How to predict if a reaction can occur, given enough time? THERMODYNAMICS PLAY MOVIE How to predict if a reaction can occur at a reasonable rate? KINETICS PLAY MOVIE

4. 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--- • AND the K is greater than 1, • then this is a product-favoredsystem. • Most product-favored reactions are exothermic—but this is not the only criterion

5. Thermodynamics • Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneousprocess. 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.

6. 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. Spontaneous Reactions In general, spontaneous reactions are exothermic. Fe2O3(s) + 2 Al(s) f2 Fe(s) + Al2O3(s) ∆rH = - 848 kJ

8. Spontaneous Reactions But many spontaneous reactions or processes are endothermic or even have ∆H = 0. PLAY MOVIE ∆H = 0 NH4NO3(s) + heat f NH4NO3(aq)

9. 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. Directionality of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy. Energy Dispersal PLAY MOVIE

11. Directionality of ReactionsEnergy 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. Begin Energy over 2 particles 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. See Figure 19.4

13. Directionality of Reactions Matter & energy dispersal 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. 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. Energy dispersal Entropy, S So (J/K•mol) H2O(liq) 69.95 H2O(gas) 188.8 S (solids) < S (liquids) < S (gases) PLAY MOVIE

16. Entropy and States of Matter S˚(Br2 liq) < S˚(Br2 gas) S˚(H2O sol) < S˚(H2O liq)

17. Entropy, S Entropy of a substance increases with temperature. Molecular motions of heptane at different temps. Molecular motions of heptane, C7H16 PLAY MOVIE PLAY MOVIE

18. Entropy, S Increase in molecular complexity generally leads to increase in S. PLAY MOVIE

19. Entropy, S Entropies of ionic solids depend on coulombic attractions. PLAY MOVIE So (J/K•mol) MgO 26.9 NaF 51.5 Mg2+ & O2- Na+ & F-

20. Entropy, S Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase in entropy. Matter (and energy) are more dispersed.

21. Standard Molar Entropies

22. Entropy Changes for Phase Changes For a phase change, ∆S = q/T where q = heat transferred in phase change For H2O (liq) f H2O(g) ∆H = q = +40,700 J/mol PLAY MOVIE

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

24. Calculating ∆S for a Reaction ∆So =  So (products) -  So (reactants) Consider 2 H2(g) + O2(g) f 2 H2O(liq) ∆So = 2 So (H2O) - [2 So (H2) + So (O2)] ∆So = 2 mol (69.9 J/K·mol) - [2 mol (130.7 J/K·mol) + 1 mol (205.3 J/K·mol)] ∆So = -326.9 J/K Note that there is a decrease in Sbecause 3 mol of gas give 2 mol of liquid.

25. 2nd Law of Thermodynamics A reaction is spontaneous if ∆S for the universe is positive. ∆Suniverse = ∆Ssystem + ∆Ssurroundings ∆Suniverse > 0 for spontaneous process Calculate the entropy created by energy dispersal in the system and surroundings.

26. 2nd Law of Thermodynamics Dissolving NH4NO3 in water—an entropy driven process. PLAY MOVIE ∆Suniverse = ∆Ssystem + ∆Ssurroundings PLAY MOVIE

27. 2nd Law of Thermodynamics 2 H2(g) + O2(g) f 2 H2O(liq) ∆Sosystem = -326.9 J/K Can calc. that ∆rHo = ∆Hosystem = -571.7 kJ ∆Sosurroundings = +1917 J/K

28. 2 H2(g) + O2(g) f 2 H2O(liq) ∆Sosystem = -326.9 J/K ∆Sosurroundings = +1917 J/K ∆Souniverse = +1590. J/K The entropy of the universe is increasing, so the reaction is product-favored. 2nd Law of Thermodynamics

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

30. Gibbs Free Energy, G ∆Suniv = ∆Ssurr + ∆Ssys Multiply through by -T -T∆Suniv = ∆Hsys - T∆Ssys -T∆Suniv = change in Gibbs free energy for the system = ∆Gsystem Under standard conditions — ∆Gosys = ∆Hosys - T∆Sosys J. Willard Gibbs1839-1903

31. ∆Go = ∆Ho - T∆So Gibbs free energy change = total energy change for system - energy lost in energy dispersal If reaction is • exothermic (negative ∆Ho) • and entropy increases (positive ∆So) • then ∆Go must be NEGATIVE • reaction is spontaneous (and product-favored).

32. ∆Go = ∆Ho - T∆So Gibbs free energy change = total energy change for system - energy lost in energy dispersal If reaction is • endothermic (positive ∆Ho) • and entropy decreases (negative ∆So) • then ∆Go must be POSITIVE • reaction is not spontaneous (and is reactant-favored).

33. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So ∆Ho ∆So ∆Go Reaction exo(–) increase(+) – Prod-favored endo(+) decrease(-) + React-favored exo(–) decrease(-) ? T dependent endo(+) increase(+) ? T dependent

34. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So Two methods of calculating ∆Go a) Determine ∆rHo and ∆rSo and use Gibbs equation. b) Use tabulated values of free energies of formation, ∆fGo. ∆rGo =  ∆fGo (products) -  ∆fGo (reactants)

35. Free Energies of Formation Note that ∆fG˚ for an element = 0

36. Calculating ∆rGo Combustion of acetylene C2H2(g) + 5/2 O2(g) f 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆rHo = -1238 kJ Use standard molar entropies to calculate ∆rSo = -97.4 J/K or -0.0974 kJ/K ∆rGo = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1209 kJ Reaction is product-favored in spite of negative ∆rSo. Reaction is “enthalpy driven”

37. Calculating ∆rGo Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven? PLAY MOVIE NH4NO3(s) + heat f NH4NO3(aq)

38. Calculating ∆rGo NH4NO3(s) + heat f NH4NO3(aq) From tables of thermodynamic data we find ∆rHo = +25.7 kJ ∆rSo = +108.7 J/K or +0.1087 kJ/K ∆rGo = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ Reaction is product-favored in spite of negative ∆rHo. Reaction is “entropy driven”

39. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So Two methods of calculating ∆Go a) Determine ∆rHo and ∆rSo and use Gibbs equation. b) Use tabulated values of free energies of formation, ∆fGo. ∆rGo =  ∆fGo (products) -  ∆fGo (reactants)

40. Calculating ∆Gorxn ∆rGo =  ∆Gfo (products) -  ∆Gfo (reactants) Combustion of carbon C(graphite) + O2(g) f CO2(g) ∆rGo = ∆fGo(CO2) - [∆fGo(graph) + ∆fGo(O2)] ∆rGo = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. ∆rGo = -394.4 kJ Reaction is product-favored as expected.

41. Free Energy and Temperature 2 Fe2O3(s) + 3 C(s) f 4 Fe(s) + 3 CO2(g) ∆rHo = +467.9 kJ∆rSo = +560.3 J/K ∆rGo = +300.8 kJ Reaction is reactant-favored at 298 K At what T does ∆rGo just change from being (+) to being (-)? When ∆rGo = 0 = ∆rHo - T∆rSo

42. Thermodynamics and Keq • FACT: ∆rGo is the change in free energy when pure reactants convert COMPLETELY to pure products. • FACT: Product-favored systems have Keq > 1. • Therefore, both ∆rG˚ and Keq are related to reaction favorability.

43. Thermodynamics and Keq Keq is related to reaction favorability and so to ∆rGo. The larger the value of K the more negative the value of ∆rGo ∆rGo = - RT lnK where R = 8.31 J/K•mol

44. Thermodynamics and Keq ∆rGo = - RT lnK Calculate K for the reaction N2O4f 2 NO2 ∆rGo = +4.8 kJ ∆rGo = +4800 J = - (8.31 J/K)(298 K) ln K K = 0.14 When ∆rGo > 0, then K < 1

45. ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K ∆G, ∆G˚, and Keq

46. ∆G, ∆G˚, and Keq Product Favored, ∆G˚ negative, K > 1 See Active Figure 19.13

47. ∆G, ∆G˚, and Keq • Product-favored • 2 NO2e N2O4 • ∆rGo = – 4.8 kJ • State with both reactants and products present is more stable than complete conversion. • K > 1, more products than reactants. PLAY MOVIE

48. ∆G, ∆G˚, and Keq Reactant Favored, ∆G˚ positive, K < 1 See Active Figure 19.13

49. ∆G, ∆G˚, and Keq • Reactant-favored • N2O4e 2 NO2 ∆rGo = +4.8 kJ • State with both reactants and products present is more stable than complete conversion. • K < 1, more reactants than products PLAY MOVIE

50. Thermodynamics and Keq • Keq is related to reaction favorability. • When ∆rGo < 0, reaction moves energetically “downhill” • ∆rGo is the change in free energy when reactants convert COMPLETELY to products.