1. Chapter 17 Free Energy and Thermodynamics

2. Goals Entropy (S,  S) and spontaneity Free energy;  G,  G o  G, K, product- or reactant-favored Review: H (Enthalpy) and the 1st Law of Thermodynamics Chemical Equilibria (ch. 14, etc)

3. 3 First Law of Thermodynamics First Law of Thermodynamics: Energy cannot be created or destroyed the total energy of the universe cannot change it can be transfered from one place to another   E universe = 0 =  E system +  surroundings system = reactants & products surroundings = everything else (the transfer of energy from one to the other does not change the energy of the universe)

4. 4 First Law of Thermodynamics For an exothermic reaction, heat from the system goes into the surroundings two ways energy can be “lost” from a system, converted to heat, q used to do work, w Energy conservation requires that the energy change in the system = heat exchanged + work done on the system.  E = q + w (  E = internal energy change)  E =  H – P  V (at const. P, q p =  H, enthalpy change) State functions (H, P, V). q and w are not. internal energy change (  E) independent of how done

5. 5 Enthalpy, H related to (includes) the internal energy  H generally kJ/mol stronger bonds = more stable molecules if products more stable than reactants, energy released; exothermic  H = negative if reactants more stable than products, energy absorbed; endothermic  H = positive The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law:  H° rxn =  (  H f ° prod ) -  (  H f ° react )

6. 6 Thermodynamics and Spontaneity thermodynamics predicts whether a process will proceed (occur) under the given conditions spontaneous process  nonspontaneous process does not occur under specific conditions. spontaneity is determined by comparing the free energy (G) of the system before the reaction with the free energy of the system after reaction. if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable. spontaneity ≠ fast or slow (rate); this is kinetics

7. 7 Spontaneous Nonspontaneous ice melts @ 25 o C water freezes @ 25 o C 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq)  H 2 (g) + 2NaOH(aq) 2Na(s) + 2H 2 O(l) ball rolls downhill ball rolls uphill water freezes @ -10 o C ice melts @ -10 o C

8. 8 Diamond → Graphite Graphite is thermodynamically more stable than diamond, so the conversion of diamond into graphite is spontaneous – but it’s kinetically too slow (inert) it will never happen in many, many generations. kinetics Spontaneity: direction & extent kinetics: how fast

9. 9 Factors Affecting Whether a Reaction Is Spontaneous The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond.  H The entropy factors relate to the randomness/orderliness of a system  S The enthalpy factor is generally more important than the entropy factor

11. 11 Entropy, S Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases.  S generally in J/K (joules/K) S = k lnW k = Boltzmann Constant (R/N A ) = 1.38  10 -23 J/K W is the number of energetically equivalent ways, (microstates). It is unitless. Entropy is usually described as a measure of the randomness or disorder; the greater the disorder of a system, the greater its S. The greater the order  the smaller its S.

12. 12 Entropy & Microstates, W Energetically Equivalent States for the Expansion of a Gas (4 gas molecules) 1 microstate 6 microstates (most probable distribution) S = k ln W  S = k ln W f - k ln W i if W f > W i,  S > 0 & entropy increases.

13. 13 Changes in Entropy,  S entropy change is favorable when the result is a more random system (State C: higher entropy ).  S is positive (  S > 0) Some changes that increase the entropy are: rxns where products are in a more disordered state. (solid > liquid > gas) less order (solid< liquid < gas)  larger S (disorder) reactions which have larger numbers of product molecules than reactant molecules. increase in temperature (more movement) solids dissociating into ions upon dissolving

14. 14 Changes in Entropy in a System (melting) Particles fixed in space Particles can occupy many positions

15. 15 Changes in Entropy in a System (vaporization) Particles occupy more space (larger volume)

16. 16 Changes in Entropy in a System (solution process) Structure of solute and solvent disrupted (also more solute particles)

17. 17 Predict entropy change for a process/reaction For which process/reaction is  S negative? Freezing ethanol Mixing CCl 4 with C 6 H 6 Condensing bromine vapor 2O 3 (g)  3O 2 (g) 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) 2Li(s) + 2H 2 O(l)  2LiOH(aq) + H 2 (g) 2NH 3 (g)  N 2 (g) + 3H 2 (g)  entropy dec

18. 18 The 2nd Law of Thermodynamics The entropy of the universe increases in a spontaneous process.  S universe =  S system +  S surroundings > 0  S universe =  S system +  S surroundings = 0 (equilibrium) If  S system >> 0,  S surroundings 0! If  S system > 0 for  S universe > 0! the increase in  S surroundings often comes from the heat released in an exothermic reaction,  system < 0.

19. 19 The 3 rd Law of Thermodynamics S = k ln W S = k ln W = k ln 1 = 0  S = S f – S i ; where S i = 0 @ 0 K the absolute entropy of a substance is always (+) positive at the new T - allows determination of entropy of substances. (W = 1, there is only one way to arrange the particles to form a perfect crystal) the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K

20. 20 Standard Entropies S° entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution Values can be used to calculate the standard entropy change for a reaction,  S o rxn (=  S o sys )

22. 22 Trends: Standard Entropies Molar Mass For monatomic species, the larger the molar mass, the larger the entropy available energy states more closely spaced, allowing more dispersal of energy through the states

23. 23 Trends: Standard Entropies States the standard entropy of a substance in the gas phase is greater than the standard entropy of the same substance in the solid or liquid phase at a particular temperature Substance S°, (J/mol∙K) H 2 O (l)70.0 H 2 O (g)188.8

24. 24 Trends: Standard Entropies Allotropes the more highly ordered form has the smaller entropy -different forms of an element

25. 25 Trends: Standard Entropies Molecular Complexity (inc # of atoms) larger, more complex molecules generally have larger entropy more available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) Ar (g)39.948154.8 NO (g)30.006210.8

26. 26 Trends: Standard Entropies Dissolution dissolved solids generally have larger entropy distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO 3 (s)143.1 KClO 3 (aq)265.7

27. 27 Q. Arrange the following in order of increasing entropy @ 25 o C! (lowest to highest) Ne(g), SO 2 (g), Na(s), NaCl(s) and H 2 (g) Na(s) < NaCl(s) < H 2 (g) < Ne(g) < SO 2 (g) Q. Which has the larger entropy in each pair? a)Li(s) or Li(l) b) C 2 H 5 OH(l) or CH 3 OCH 3 (l) c) Ar(g) or Xe(g) d) O 2 (g) or O 3 (g)

28. Calculate  S  for the reaction 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (g)  S is +, as you would expect for a reaction with more gas product molecules than reactant molecules standard entropies from Appendix IIB  S, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SS S  NH3, S  O2, S  NO, S  H2O, Substance S , J/mol  K NH 3 (g)192.8 O2(g)O2(g)205.2 NO(g)210.8 H 2 O(g)188.8

29. Consider 2 H 2 (g) + O 2 (g)  2 H 2 O(liq) @ 25 o C ∆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 (130.6 J/Kmol) + 1 mol (205.0 J/Kmol)] ∆S o = -326.4 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating ∆S o for a Reaction ∆S o =  S o (products) -  S o (reactants)

30. The more negative  H syst and the lower the temperature the higher (more positive)  S surr 30 Temperature Dependence of  S surroundings  system 0 )  system > 0 (endothermic), it takes heat from the surroundings, decreasing the entropy of the surroundings (  S surroundings < 0 )

31. 2 H 2 (g) + O 2 (g)  2 H 2 O(liq) @ 25 o C ∆S o system = -326.4 J/K Calculating ∆S o for the surroundings ∆S o surroundings = +1917 J/K Can calculate ∆H o system = ∆H o rxn = -571.7 kJ (also from tabulated data) (also from tabulated data)

32. 2 H 2 (g) + O 2 (g)  2 H 2 O(liq) @ 25 o C ∆S o system = -326.4 J/K ∆S o surroundings = +1917 J/K ∆S o universe = +1591 J/K The entropy of the universe is increasing, so the reaction is spontaneous ( product- favored). (see slide # 18)  S universe =  S system +  S surroundings Given S o surr, S o sys and T, determine S o univ and predict if the reaction will be spontaneous. Given  S o surr,  S o sys and T, determine  S o univ and predict if the reaction will be spontaneous.

33. The reaction C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) has  H rxn = -2044 kJ at 25°C. Calculate the entropy change of the surroundings. combustion is largely exothermic, so the entropy of the surroundings should increase (inc in # gas mol)  H system = -2044 kJ, T = 298 K  S surroundings, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SS T,  H

34. Spontaneous or Not? originally: originally:  S universe =  S system +  S surroundingsbut but  S universe =  S system –  system /T  system  S system Spontaneous? Exothermic  H sys < 0 Less order  S sys > 0 Spontaneous under all conditions;  S univ > 0 Exothermic  H sys < 0 More order  S sys < 0 Favorable at low T Endothermic  H sys > 0 Less order  S sys > 0 Favorable at high T Endothermic  H sys > 0 More order  S sys < 0 Not spontaneous under any conditions  S univ < 0

35. 35 Without doing any calculations, determine the sign of S sys and S surr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous. 2CO(g) + O 2 (g)  2CO 2 (g)  rxn = -566.0 kJ 2NO 2 (g)  O 2 (g) + 2NO(g)  rxn = +113.1 kJ -/T)  S universe =  S system + ( -  sys /T)  S system = ( -); 3 mol gas form 2 mol gas  S surr = (+); spontaneous @ low T  S system = (+); 2 mol gas form 3 mol gas  S surr = ( -) ; spontaneous @ high T -/T)  S surr = ( -  sys /T)

36. 36 Without doing any calculations, determine the sign of  S sys and  S surr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous. 2H 2 (g) + O 2 (g)  2H 2 O(g)  rxn = -483.6 kJ CO 2 (g)  C(s) + O 2 (g)  rxn = +393.5 kJ -/T)  S universe =  S system + ( -  sys /T)  S system = ( -) ; 3 mol gas form 2 mol gas  S surr = (+) ; spontaneous @ low T  S system = ( -) ; complicated gas forms a solid & gas  S surr = ( -) ; nonspontaneous @ all T -/T)  S surr = ( -  sys /T)

37. 37 At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings, if  H o rxn = -127 kJ and  S o rxn = 314 J/K. Plan: set  S o rxn =  S o surr and solve for T; convert kJ to J rxn implies system!!! Ans: T = +404 K

38. Gibbs Free Energy, G = H -TS 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 hence, ∆ G = ∆ H -T ∆ S hence, ∆ G = ∆ H -T ∆ S Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys ∆S univ = ∆S surr + ∆S sys J. Willard Gibbs 1839-1903

39. ∆G = ∆H - T∆S Gibbs free energy change = total energy change for system - energy lost in disordering the system If the reaction is exothermic (negative ∆H) and entropy increases (positive ∆S o ) then ∆ G must be NEGATIVEthen ∆ G must be NEGATIVE the reaction is spontaneous (and product- favored) at ALL temperatures.

40. 40 ∆G > 0)  G will be positive ( ∆G > 0) when  H is positive (endothermic) and  S is negative (more ordered). So the change in free energy will be positive at all temperatures. The reaction will therefore be nonspontaneous at ALL temperatures W hen  G = 0 the reaction is at equilibrium ∆G = ∆H - T∆S

41. 41  G =  H – T  S Spontaneous or Not? A decrease in Gibbs free energy (  G < 0) corresponds to a spontaneous process An increase in Gibbs free energy (  G > 0) corresponds to a nonspontaneous process

42. Calculating ∆G o ∆G o ∆∆S o Calculating ∆G o : ∆G o = ∆H -T∆S o Combustion of acetylene @ 25 o C 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 Use standard molar entropies to calculate ∆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 (spontaneous) Reaction is product-favored in spite of negative ∆S o rxn. Reaction is “enthalpy driven”

43. From tables of thermodynamic data we find ∆H o rxn = +25.7 kJ (endothermic) ∆H o rxn = +25.7 kJ (endothermic) ∆S o rxn = +108.7 J/K or +0.1087 kJ/K (disorder) ∆G o rxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ (spontaneous) = -6.7 kJ (spontaneous) Reaction is product-favored in spite of positive ∆H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq) Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? Calculating ∆G o ∆G o ∆∆S o Calculating ∆G o : ∆G o = ∆H -T∆S o

44. The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = +95.7 kJ and  S = +142.2 J/K at 25°C. Calculate  G and determine if it is spontaneous. Since  G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.  H = +95.7 kJ,  S = 142.2 J/K, T = 298 K  G, kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GGT,  H,  S

45. The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = +95.7 kJ and  S = +142.2 J/K. Calculate the minimum T @ which it will be spontaneous. The temperature must be higher than 673K for the reaction to be spontaneous (i.e. 674 K)  H = +95.7 kJ,  S = 142.2 J/K,  G < 0  Answer: Solution: Concept Plan: Relationships: Given: Find: T  G,  H,  S

46. Gibbs Free Energy, G Calculating ∆G o Calculating ∆G o (two ways) a) Determine ∆H o rxn and ∆S o rxn and use Gibbs equation (at various temps). b) b) Use tabulated values of free energies of formation, ∆G f o @ 25 o C ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

47. 47

48. Calculate  G  at 25  C for the reaction CH 4(g) + 8 O 2(g)  CO 2(g) + 2 H 2 O (g) + 4O 3(g) standard free energies of formation from Appendix IIB  G , kJ Solution: Concept Plan: Relationships: Given: Find: GG  G  f of prod & react Substance  G  f, kJ/mol CH 4 (g)-50.5 O2(g)O2(g)0.0 CO 2 (g)-394.4 H 2 O(g)-228.6 O3(g)O3(g)163.2 (spontaneous)

49. The reaction SO 2(g) + ½ O 2(g)  SO 3(g) has  H  = -98.9 kJ and  S  = -94.0 J/K at 25°C. Calculate  G  at 125  C and determine if it is spontaneous. Since  G is -ve, the rxn is spontaneous at this temperature, but less spontaneous than at 25  C (-127 kJ)  H  = -98.9 kJ,  S  = -94.0 J/K, T = 398 K  G , kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GG T,  H ,  S  (PRACTICE PROBLEM)

50. ∆G, ∆G˚, and K eq ∆G is the 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, the reaction may be spontaneous or nonspontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

51.  FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products, both at standard conditions. ∆G˚ rxn 1 (∆G˚ rxn < 0).  Therefore, both ∆G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq Summary: Summary: ∆G˚ = - RT ln K

52. Calculate K for the reaction @ 25 o C 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 = e –1.94 = 0.14 (reactant favored) When ∆G o rxn > 0 (nonspontaneous), then K 0 (nonspontaneous), then K < 1!!

53. Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g)  G° = -RT lnK +46400 J = -(8.314 J/K)(700 K) lnK ln K = -7.97 K = e − 7.97 = 3.45  10 − 4 small!!! since K is << 1, the position of equilibrium favors reactants        H° = [ 2(-46.19)] − [ 0 +3( 0)] = -92.38 kJ = -92380 J  S° = [2 (192.5)] − [(191.50) + 3(130.58)] = -198.2 J/K  G° = -92380 J - (700 K)(-198.2 J/K)  G° = +46400 J (nonspontaneous)

54. 54 Calculate  G at 427°C for the reaction below if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g) Q = P NH3 2 P N2 1  P H2 3 (2.0 atm) 2 (33.0 atm) 1 (99.0) 3 = = 1.2  10 -7  G =  G° + RTlnQ spontaneous  G = +46400 J + (8.314 J/K)(700 K)(ln 1.2  10 -7 )  H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J  S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K  G° = -92380 J - (700 K)(-198.2 J/K)  G° = +46400 J (nonspontaneous)  G = 46400 J − 92700 J = -46300 J = − 46 kJ  −  G°

55. 55 Q. Rank the following in order of increasing molar entropy (S o ) @ 25 o C! a) Cl 2 (g), I 2 (g), Br 2 (g), and F 2 (g) b) H 2 O(g), H 2 O 2 (g), H 2 S(g) F 2 (g) < Cl 2 (g) < Br 2 (g) < I 2 (g) H 2 O(g) < H 2 S(g) < H 2 O 2 (g)

56. 56 C)  S surr = +114 kJ/K, reaction is spontaneous A)  S surr = +114 kJ/K, reaction is not spontaneous B)  S surr = +321 J/K, reaction is spontaneous D)  S surr = -355 J/K, reaction is not spontaneous E)  S surr = +321 J/K, it is not possible to predict the spontaneity of this reaction without more information. Use only the information provided here to determine the value of  S surr @ 355 K. Predict whether this reaction shown will be spontaneous @ this temperature, if  H rxn = -114 kJ.

57.  G under Nonstandard Conditions   G =  G  only when the reactants and products are in their standard states  there normal state at that temperature  partial pressure of gas = 1 atm  concentration = 1 M  under nonstandard conditions,  G =  G  + RTlnQ  Q is the reaction quotient  at equilibrium  G = 0   G  = − RTlnK and  G° =  H  − T  S°  H  − T  S° = − RTlnK, by rearranging RTlnK = −  H  − T  S°, and dividing by R T

58. Temperature Dependence of K for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant RTlnK −  H  T  S° ———— = ———— + ———— RT RT R T