2. Instructor: Dr. Upali Siriwardane e-mail: upali@chem.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m. Test Dates : March 25, April 26, and May 18; Comprehensive Final Exam: May 20,2009 9:30-10:45 am, CTH 328. March 30, 2009 (Test 1): Chapter 13 April 27, 2009 (Test 2): Chapters 14 & 15 May 18, 2009 (Test 3): Chapters 16, 17 & 18 Comprehensive Final Exam: May 20,2009 : Chapters 13, 14, 15, 16, 17 and 18 Chemistry 102(01) spring 2009

3. Review of Chapter 6. Energy and Chemical Reactions 6.1 The Nature of Energy 6.2 Conservation of Energy 6.3 Heat Capacity 6.4 Energy and Enthalpy 6.5 Thermochemical Equations 6.6 Enthalpy change for chemical Rections 6.7 Where does the Energy come from? 6.8 Measuring Enthalpy Changes: Calorimetry 6.9 Hess's Law 6.10 Standard Enthalpy of Formation 6.11 Chemical Fuels for Home and Industry 6.12 Food Fuels for Our Bodies

4. Thermochemistry Heat changes during chemical reactions Thermochemical equation. eg. H 2 (g) + O 2 (g) ---> 2H 2 O(l)  H =- 256 kJ;  is called the enthalpy of reaction. if  H is + reaction is called endothermic if  H is - reaction is called exothermic

5. Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding system surroundings universe

6. Types of Systems Isolated system no mass or energy exchange Closed system only energy exchange Open system both mass and energy exchange

7. Universe = System + Surrounding Why is it necessary to divide Universe into System and Surrounding

8. What is the internal energy change (  U) of a system?  U is associated with changes in atoms, molecules and subatomic particles E total = E ke + E pe +  U  U = heat (q) + w (work)  U = q + w  U = q -P  V; w =- P  V

9. What forms of energy are found in the Universe? mechanicalthermalelectricalnuclear mass: E = mc 2 others yet to discover

10. What is 1 st Law of Thermodynamics Eenergy is conserved in the Universe All forms of energy are inter-convertible and conserved Energy is neither created nor destroyed.

11. What exactly is  H? Heat measured at constant pressure q p Chemical reactions exposed to atmosphere and are held at a constant pressure. Volume of materials or gases produced can change. Volume expansion work = -P  V  U = q p + w;  U = q p -P  V q p =  U + P  V;w = -P  V  H =  U + P  V;q p =  H(enthalpy )

12. Heat measured at constant volume q v Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -P  V= 0  U = q v + w q v =  U + o;w = 0  U = q v =  U(internal energy ) How do you measure  U?

13. What is Hess's Law of Summation of Heat? To heat of reaction for new reactions. Two methods Two methods? 1st method: new  H is calculated by adding  Hs of other reactions. 2nd method 2nd method: Where  H f (  H of formation) of reactants and products are used to calculate  H of a reaction.

14. Method 1: Calculate  H  H for the reaction: SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) ----> H 2 SO 4 (l)  H = ? Other reactions: SO 2 (g) ------> S(s) + O 2 (g) ;  H = 297kJ H 2 SO 4 (l)------> H 2 (g) + S(s) + 2O 2 (g);  H = 814 kJ H 2 (g) +1/2O 2 (g) -----> H 2 O(g);  H = -242 kJ

15. SO 2 (g) ------> S(s) + O 2 (g);  H 1 = 297 kJ - 1 H 2 (g) + S(s) + 2O 2 (g) ------> H 2 SO 4 (l);  H 2 = -814 kJ - 2 H 2 O(g) ----->H 2 (g) + 1/2 O 2 (g) ;  H 3 = +242 kJ - 3 ______________________________________ SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l);  H = ?  H =  H 1 +  H 2 +  H 3  H = +297 - 814 + 242  H = -275 kJ Calculate  H for the reaction

16. Calculate Heat (Enthalpy) of Combustion: 2nd method C 7 H 16 (l) + 11 O 2 (g) -----> 7 CO 2 (g) + 8 H 2 O(l) ;  H o = ?  H f (C 7 H 16 ) = -198.8 kJ/mol  H f (CO 2 ) = -393.5 kJ/mol  H f (H 2 O) = -285.9 kJ/mol  H f O 2 (g) = 0 (zero) What method?  H o =  n   H f o products –  n   H f o reactants n = stoichiometric coefficients 2 nd method

17.  H = [  n (  H o f ) Products] - [  n (  H o f ) reactants]  H = [ 7(- 393.5 + 8 (- 285.9)] - [-198.8 + 11 (0)] = [-2754.5 - 2287.2] - [-198.8] = -5041.7 + 198.8 = -4842.9 kJ = -4843 kJ Calculate  H for the reaction

18. Why is  H o f  H o f of elements is zero?  H o f, Heat formations are for compounds Note:  H o f of elements is zero

19. Chapter 18. Thermodynamics: Directionality of Chemical Reactions 18.1Reactant-Favored and Product-Favored Processes 18.2Probability and Chemical Reactions 18.3Measuring Dispersal or Disorder: Entropy 18.4Calculating Entropy Changes 18.5Entropy and the Second Law of Thermodynamics 18.6Gibbs Free Energy 18.7Gibbs Free Energy Changes and Equilibrium Constants 18.8Gibbs Free Energy, Maximum Work, and Energy Resources 18.9Gibbs Free Energy and Biological Systems 18.10Conservation of Gibbs Free Energy 18.11Thermodynamic and Kinetic Stability

20. Chemical Thermodynamics spontaneous reaction – reaction which proceed without external assistance once started chemical thermodynamics helps predict which reactions are spontaneous

21. ThermodynamicsThermodynamics Will the rearrangement of a system decrease its energy? If yes, system is favored to react — a product- favored system. Most product-favored reactions are exothermic. Often referred to as spontaneous reactions. “Spontaneous” does not imply anything about time for reaction to occur. Kinetic factors are more important for certain reactions.

22. Thermodynamics and Kinetics Diamond  graphite Thermodynamically product favored Slow Kinetics Paper burns Thermodynamically product favored Fast Kinetics In this chapter we only look into thermodynamic factors

23. Bases on Energy: Product- Favored Reactions In general, product-favored reactions are exothermic. H) (Negative  H) In general, reactant-favored reactions are endothermic. H) (Positive  H)

24. Product-Favored Reactions But many spontaneous reactions or processes are endothermic or even have H = 0. But many spontaneous reactions or processes are endothermic or even have  H = 0. NH 4 NO 3 (s)  NH 4 NO 3 (aq);  H = +

25. Direction of Reaction Product favored reactions are always a transformation of a reactants favored reaction. Product Favored Reaction 2Na (s) + 2Cl 2(g) => 2NaCl (s) Reactant Favored Reaction 2NaCl (s) => 2Na (s) + 2Cl 2(g) However, non spontaneous reactions could be made to take place by coupling with energy source: another reaction or electric current Electrolysis 2NaCl (l) => 2Na (s) + 2Cl 2(g)

26. Expansion of a Gas The positional probability is higher when particles are dispersed over a larger volume Matter tends to expand unless it is restricted

27. Gas Expansion and Probability

28. Entropy, S The thermodynamic property related to randomness is ENTROPY, S. Product-favored processes: final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. Reaction of K with water

29. S[H 2 O(l)] > S[H 2 O(s)] at 0  C.

30. Entropies of Solid, Liquid and Gas Phases S (gases) > S (liquids) > S (solids) S (gases) > S (liquids) > S (solids)

31. Entropy, S Entropies of ionic solids depend on coulombic attractions. S o (J/Kmol) MgO26.9 NaF51.5 S o (J/Kmol) MgO26.9 NaF51.5

32. Entropy and Molecular Structure

33. Entropy and Dissolving

34. Qualitative Guidelines for Entropy Changes Entropies of gases higher than liquids higher than solids Entropies are higher for more complex structures than simpler structures Entropies of ionic solids are inversely related to the strength of ionic forces Entropy increases when making solutions of pure solids or pure liquids in a liquid solvent Entropy decrease when making solutions of gases in a liquid

35. Entropy of a Solution of a Gas

36. Phase Transitions H 2 O (s) => H 2 O (l)  H > 0;  S > 0 H 2 O (l) => H 2 O (g)  H > 0;  S > 0 spontaneous at high temperatures H 2 O (l) => H 2 O (s)  H < 0;  S < 0 H 2 O (g) => H 2 O (l)  H < 0;  S < 0 spontaneous at low temperatures

37. Entropy Changes for Phase Changes For a phase change, S SYS = q SYS /T For a phase change,  S SYS = q SYS /T (q = heat transferred) Boiling Water H 2 O (liq)  H 2 O(g) H = q = +40,700 J/mol  H = q = +40,700 J/mol

38. Phase Transitions Heat of Fusion energy associated with phase transition solid-to- liquid or liquid-to-solid  G fusion = 0 =  H fusion - T  S fusion 0 =  H fusion - T  S fusion  H fusion = T  S fusion Heat of Vaporization energy associated with phase transition gas-to- liquid or liquid-to-gas  H vaporization = T  S vaporization

39. Qualitative prediction of  S of Chemical Reactions Look for (l) or (s) --> (g) If all are gases: calculate  n  n =  n (gaseous prod.) -  n(gaseous reac.) N 2 (g) + 3 H 2 (g) --------> 2 NH 3 (g)  n = 2 - 4 = -2 If  n is -  S is negative (decrease in S) If  n is +  S is positive (increase in S)

40. Entropy Change Entropy (  S) normally increase (+) for the following changes: + i) Solid ---> liquid (melting) + + ii) Liquid ---> gas + most + iii) Solid ----> gas most + + iv) Increase in temperature + + v) Increasing in pressure(constant volume, and temperature) + + vi) Increase in volume +

41. Predict  S! 2 C 2 H 6 (g) + 7 O 2 (g)--> 4 CO 2 (g) + 6H 2 O(g) 2 CO(g) + O 2 (g)-->2 CO 2 (g) HCl(g) + NH 3 (g)-->NH 4 Cl(s) H 2 (g) + Br 2 (l) --> 2 HBr(g)

42. 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 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 = 2 mol (69.9 J/Kmol) – [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)] S o = -326.9 J/K  S o = -326.9 J/K There is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating  S for a Reaction Based on Hess’s Law second method: S o =  S o (products) -  S o (reactants)  S o =  S o (products) -  S o (reactants) Based on Hess’s Law second method: S o =  S o (products) -  S o (reactants)  S o =  S o (products) -  S o (reactants)

43. Third Law of Thermodynamics Provides reference point for absolute entropy Entropy of a perfectly crystalline substance at absolute zero (T= 0 K) is zero. Unlike  H entropy values are positive above temperatures above absolute zero.

44. Standard Molar Entropy Values

45. Substance S o (J/K. mol) C (diamond) 2.37HBr (g) 198.59 C (graphite) 5.69HCl (g) 186.80 CaO (s) 39.75HF (g) 193.67 CaCO 3 (s ) 92.9HI (g) 206.33 C 2 H 2 (g) ` 200.82H 2 O (l) 69.91 C 2 H 4 (g) 219.4H 2 O (g) 188.72 C 2 H 6 (g) 229.5NaCl (s) 72.12 CH 3 OH (l) 127O 2 (g) 205.03 CH 3 OH (g) 238SO 2 (g) 248.12 CO (g) 197.91SO 3 (g) 256.72 Standard Entropies at 25 o C

46. Entropy & Spontaneity How can water boil and freeze spontaneously? Enthalpy change can not predict spontaneity! Some endothermic processes are spontaneous Need another thermodynamic property.

47. Laws of Thermodynamics First : The total energy of the universe is constant Second : The total entropy (S) of the universe is always increasing Third : The entropy(S) of a pure, perfectly formed crystalline substance at absolute zero is zero First : The total energy of the universe is constant Second : The total entropy (S) of the universe is always increasing Third : The entropy(S) of a pure, perfectly formed crystalline substance at absolute zero is zero

48. Dissolving NH 4 NO 3 in water—an entropy driven process. NH 4 NO 3 (s)  NH 4 NO 3 (aq);  H = + 2nd Law of Thermodynamics

49. Second Law of Thermodynamics In the universe the ENTROPY cannot decrease for any spontaneous process The entropy of the universe strives for a maximum in any spontaneous process, the entropy of the universe increases for product-favored process  S universe = ( S sys + S surr ) > 0  S univ = entropy of the Universe  S sys = entropy of the System  S surr = entropy of the Surrounding  S univ =  S sys +  S surr

50. Entropy of the Universe  S univ =  S sys +  S surr  S univ  S sys  S surr + + + + +(  S sys >  S surr) - + - + (  S surr >  S sys)

51. Can calc. that H o rxn = H o system = -571.7 kJ Can calc. that  H o rxn =  H o system = -571.7 kJ 2 H 2 (g) + O 2 (g)  2 H 2 O(liq) S o sys = -326.9 J/K  S o sys = -326.9 J/K Entropy Changes in the Surroundings = +1917 J/K 2nd Law of Thermodynamics

52. 2 H 2 (g) + O 2 (g)  2 H 2 O(liq) S o sys = -326.9 J/K  S o sys = -326.9 J/K S o surr = +1917 J/K  S o surr = +1917 J/K S o uni = +1590. J/K  S o uni = +1590. J/K The entropy of the universe is increasing, so the reaction is product-favored. 2nd Law of Thermodynamics

53. Gibbs Free Energy, G S univ = S surr + S sys  S univ =  S surr +  S sys Multiply through by (-T) -TS univ = H sys - TS sys -T  S univ =  H sys - T  S sys -TS univ = G system -T  S univ =  G system Under standard conditions — G o = H o - TS o  G o =  H o - T  S o  S univ =  H sys T +  S

54. Gibbs Free Energy, G G o = H o - T S o   G o =  H o - T  S o Gibbs free energy change = difference between the enthalpy of a system and the product of its absolute temperature and entropy predictor of spontaneity Total energy change for system - energy lost in disordering the system Total energy change for system - energy lost in disordering the system

55. Predict the spontaneity of the following processes from  H and  S at various temperatures. Predict the spontaneity of the following processes from  H and  S at various temperatures. a)  H = 30 kJ,  S = 6 kJ, T = 300 K b)  H = 15 kJ,  S = -45 kJ,T = 200 K

56. a)  H = 30 kJ  S = 6 kJT = 300 K  G =  H sys -T  S sys or  G =  H - T  S.  H = 30 kJ  S= 6 kJ T = 300 K  G = 30 kJ - (300 x 6 kJ) = 30 -1800 kJ  G = -1770 kJ b)  H = 15 kJ  S = -45 kJT = 200 K  G =  H sys -T  S sys or  G =  H - T  S.  H = 15 kJ  S = -45 kJ T = 200 K  G = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJ  G = 15 + 9000 kJ = 9015 kJ

57.  G The sign of  G indicates whether a reaction will occur spontaneously. +Not spontaneous 0 At equilibrium -Spontaneous  S  G The fact that the effect of  S will vary as a function of temperature is important. This can result in changing the sign of  G. Free energy

58. Predicting Whether a Reaction is Product Favored using  G Sign of  H system Sign of  S system Product-favored? Negative (exothermic)PositiveYes Negative (exothermic)NegativeYes at low T; no at high T Positive (endothermic)PositiveNo at low T; yes at high T Positive (endothermic)Negative No

59. Predict  G at different  H,  S, T  G =  H - T  S. T - - all T + T + + all T – T - /+ + high/low T + T -/+ - low/high T -

60. Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o H o S o G o Reaction  H o  S o  G o Reaction exo(-)increase(+)-Prod-favored endo(+)decrease(-)+React-favored exo(-)decrease(-)?T dependent endo(+)increase(+)?T dependent

61. G o = H o - TS o  G o =  H o - T  S o

62.  G f o Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states.  G values can then be calculated from:  G o =  n p  G f o products –  n r  G f o reactants Standard free energy of formation,  G f o

63. Substance  G f o Substance  G f o C (diamond) 2.832HBr (g) -53.43 CaO (s) -604.04HF (g) -273.22 CaCO 3 (s) -1128.84 HI (g) 1.30 C 2 H 2 (g) 209H 2 O (l) -237.18 C 2 H 4 (g) 86.12H 2 O (g) -228.59 C 2 H 6 (g) -32.89NaCl (s) -384.04 CH 3 OH (l) -166.3O (g) 231.75 CH 3 OH (g) -161.9SO 2 (g) -300.19 CO (g) -137.27SO 3 (g) -371.08 All have units of kJ/mol and are for 25 o C Standard free energy of formation

64. How do you calculate  G There are two ways to calculate  G for chemical reactions. i)  G =  H - T  S. ii)  G o =   G o f (products) -   G o f (reactants)

65. Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o Two methods of calculating G o Two methods of calculating  G o (a) Determine H o rxn and S o rxn and use Gibbs equation. (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. (b) Use tabulated values of free energies of formation,  G f o.  G o rxn =   G f o (products) -   G f o (reactants)

66. Calculating G o rxn 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)

67. Calculating G o rxn Calculating  G o rxn Method (a) : 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  S o rxn = +108.7 J/K or +0.1087 kJ/K G o rxn = +25.7 kJ - (298 K)(+0.1087 J/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 product-favored in spite of negative  H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq)

68. Calculating G o rxn Calculating  G o rxn Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) Method (b) : G o rxn = G f o (CO 2 ) - [G f o (graph) + G f o (O 2 )]  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]  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  G o rxn = -394.4 kJ Reaction is product-favored G o rxn =  G f o (products) -  G f o (reactants)  G o rxn =   G f o (products) -   G f o (reactants)

69. We can calculate  G o values from  H o and  S o values at a constant temperature and pressure.Example. Determine  G o for the following reaction at 25 o C EquationN 2 (g) + 3H 2 (g) 2NH 3 (g)  H f o, kJ/mol0.00 0.00 -46.11 S o, J/K. mol 191.50 130.68 192.3 Calculation of  G o

70. Example. Calculate the  S o rxn at 25 o C for the following reaction. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) SubstanceS o (J/K. mol) CH 4 (g) 186.2 O 2 (g) 205.03 CO 2 (g) 213.64 H 2 O (g) 188.72 Calculation of standard entropy changes

71. Calculate the  S for the following reactions using  S o =   S o (products) -   S o (reactants) a) 2SO 2 (g) + O 2 (g) ------> 2SO 3 (g)  S o [SO 2 (g)] = 248 J/K mole ;  S o [O 2 (g)] = 205 J/K mole;  S o [SO 3 (g)] = 257 J/K mole b) 2NH 3 (g) + 3N 2 O (g) --------> 4N 2 (g) + 3 H 2 O (l)  S o [ NH 3 (g)] = 193 J/K mole ;  S o [N 2 (g)] = 192 J/K mole;  S o [N 2 O(g)] = 220 J/K mole;  S[ H 2 O(l)] = 70 J/K mole

72. a)2SO 2 (g) + O 2 (g------> 2SO 3 (g) a)2SO 2 (g) + O 2 (g------> 2SO 3 (g)  S o [SO 2 (g)] = 248 J/K mole ;  S o [O 2 (g)] = 205 J/K mole;  S o [SO 3 (g)] = 257 J/K mole  S o 496 205 514  S o =   S o (products) -   S o (reactants)  S o = [514] - [496 + 205]  S o = 514 - 701  S o = -187 J/K mole

73. Calculate the  G value for the following reactions using: Calculate the  G value for the following reactions using:  G o =   G o f (products) -   G o f (reactants) N 2 O 5 (g) + H 2 O(l) ------> 2 HNO 3 (l) ;  G o = ?  G f o [ N 2 O 5 (g) ] = 134 kJ/mole ;  G f o [H 2 O(g)] = -237 kJ/mole;  G f o [ HNO 3 (l) ] = -81 kJ/mole N 2 O 5 (g) + H 2 O(l) ------> 2 HNO 3 (l) ;  G o = ?  G f o 1 x 134 1 x (-237) 2 (-81) 134 -237 -162  G o =  G o f (products) - 3  G o f (reactants)  G o = [-162] - [134 + (-237)]  G o = -162 + 103  G o = -59 kJ/mole The reaction have a negative  G and the reaction is spontaneous or will take place as written.

74. 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  H o rxn = +467.9 kJ  S o rxn = +560.3 J/K G o rxn = +300.8 kJ  G o rxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does G o rxn change from (+) to (-)? At what T does  G o rxn change from (+) to (-)? Set G o rxn = 0 = H o rxn - TS o rxn Set  G o rxn = 0 =  H o rxn - T  S o rxn

75. Effect of Temperature on Reaction Spontaneity

76. How do you calculate  G at different T and P  G =  G o + RT ln Q Q = reaction quotient at equilibrium  G =   =  G o + RT ln K  G o = - RT ln K If you know  G o you could calculate K

77. Concentrations, Free Energy, and the Equilibrium Constant Equilibrium Constant and Free Energy  G =  G o + RT ln Q Q = reaction quotient 0 =  G o + RT ln K eq  G o = - RT ln K eq

78. K eq is related to reaction favorability and so to G o rxn. K eq is related to reaction favorability and so to  G o rxn. The larger the (-) value of G o rxn the larger the value of K. The larger the (-) value of  G o rxn the larger the value of K. G o rxn = - RT lnK  G o rxn = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

79. For gases, the equilibrium constant for a reaction can be related to  G o by:  G o = -RT lnK For our earlier example, N 2 (g) + 3H 2 (g) 2NH 3 (g) At 25 o C,  G o was -32.91 kJ so K would be: ln K = = ln K = 13.27; K = 5.8 x 10 5  G o -RT -32.91 kJ -(0.008315 kJ. K -1 mol -1 )(298.2K) Free energy and equilibrium

80. Calculate the  G for the following equilibrium reaction and predict the direction of the change using the equation: Calculate the  G for the following equilibrium reaction and predict the direction of the change using the equation:  G =  G o + RT ln Q ; [  G f o [ NH 3 (g) ] = -17 kJ/mole N 2 (g) + 3 H 2 (g) 2 NH 3 (g);  G = ? at 300 K, P N2 = 300, P NH3 = 75 and P H2 = 300 N 2 (g) + 3 H 2 (g) 2 NH 3 (g);  G = ?

81. To calculate  G o To calculate  G o Using  G o =   G o f (products) -   G o f (reactants)  G f o [ N 2 (g) ] = 0 kJ/mole;  G f o [ H 2 (g) ] = 0 kJ/mole;  G f o [ NH 3 (g) ] = -17 kJ/mole Notice elements have  G f o = 0.00 similar to  H f o N 2 (g) + 3 H 2 (g) 2 NH 3 (g);  G = ?  G f o 002 x (-17) 00-34  G o =   G o f (products) -   G o f (reactants)  G o = [-34] - [0 +0]  G o = -34  G o = -34 kJ/mole

82. To calculate Q To calculate Q Equilibrium expression for the reaction in terms of partial pressure: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) p 2 NH3 K=_________ p N2 p 3 H2 p 2 NH3 Q=_________ ; p N2 p 3 H2 Q is when initial concentration is substituted into the equilibrium expression 75 2 Q=_________ ; p 2 NH3 = 75 2 ; p N2 =300; p 3 H2 =300 3 300 x 300 3 Q =6.94 x 10 -7

83. To calculate  G o To calculate  G o  G =  G o + RT ln Q  G o = -34 kJ/mole R = 8.314 J/K mole or 8.314 x 10 -3 kJ/Kmole T = 300 K Q= 6.94 x 10 -7  G = (-34 kJ/mole) + ( 8.314 x 10 -3 kJ/K mole) (300 K) ( ln 6.94 x 10 -7 )  G = -34 + 2.49 ln 6.94 x 10 -7  G = -34 + 2.49 x (-14.18)  G = -34 -35.37  G = -69.37 kJ/mole

84. Calculate K (from  G 0 ) N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ 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 = +4800 J = - (8.31 J/K)(298 K) ln K G o rxn = - RT lnK  G o rxn = - RT lnK K = 0.14 G o rxn > 0 : K 0 : K < 1 G o rxn 1  G o rxn 1 K = 0.14 G o rxn > 0 : K 0 : K < 1 G o rxn 1  G o rxn 1 Thermodynamics and K eq

85. Concentrations, Free Energy, and the Equilibrium Constant The Influence of Temperature on Vapor Pressure H 2 O (l) => H 2 O (g) K eq = p water vapor p water vapor = K eq = e - G'/RT

86.  G as a Function of the Extent of the Reaction

87.  G as a Function of the Extent of the Reaction when there is Mixing

88. Maximum Work  G = w system = - w max (work done on the surroundings)

89. Coupled Reactions How to do a reaction that is not thermodynamically favorable? Find a reaction that offset the (+)  G Thermite Reaction Fe 2 O 3(s) => 2Fe (s) + 3/2O 2(g) 2Al (s) + 3/2O 2(g)  Al 2 O 3(s)

90. ADP and ATP

91. Acetyl Coenzyme A

92. Gibbs Free Energy and Nutrients

93. Photosynthesis: Harnessing Light Energy

94. Using Electricity for reactions with (+)  G: Electrolysis Non spontaneous reactions could be made to take place by coupling with energy source: another reaction or electric current Electrolysis 2NaCl (l) => 2Na (s) + 2Cl 2(g)