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

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Presentation on theme: "1 Chapter 19 Principles of Reactivity: Entropy & Free Energy."— Presentation transcript:

1 1 Chapter 19 Principles of Reactivity: Entropy & Free Energy

2 2 I.Spontaneity A.Spontaneous processes B.Factors affecting spontaneity II.Entropy A.Entropy changes B.The third law of thermodynamics C.Calculating entropy changes in chemical reactions D.The second law of thermodynamics III.Gibbs Free Energy A.Definition B.Calculating free energy changes in chemical reactions C.Standard free energy of formation D.Free energy and temperature E.Free energy and equilibrium

3 3 I. Spontaneity A. Spontaneous process = one that, once started, continues on its own without input of energy nonspontaneous = needs continual input of energy Chemical Thermodynamics Chapter 6: Thermochemistry heat of reaction,  H first law: conservation of energy Thermodynamics: energy changes entropy (  S), free energy (  G) second and third laws  spontaneity of reaction, position of equilibirum

4 4 I. Spontaneity B. Factors affecting spontaneity 1. decrease in potential energy (  H < 0) apple falling water flowing downhill 2H 2 + O 2  2H 2 O HCl + NaOH  NaCl + H 2 O thermite 2. increase in disorder (tendency toward higher probability) deck of cards (6 x 10 9 combinations) gas molecules in a container salt dissolving in water ice melting, water evaporating

5 5 II. Entropy, S = degree of disorder (probability) A. Entropy changes Entropy change,  S = S f - S i 1. examples expansion of a gas  S dissolution of a solid  S H 2 O(s)  H 2 O(l)  H 2 O(g)  S H 2 O(g)  H 2 O(l)  H 2 O(s)  S H 2 CO 3  CO 2 + H 2 O  S 2H 2 + O 2  2H 2 O  S CaO + SO 2  CaSO 3  S remove partition NaCl(s) NaCl(aq) H2OH2O

6 6 II. Entropy A. Entropy changes 2. calculating  S for fusion and vaporization S  qS  qincrease KE  increase in disorder S S  1T1T 10 K  20 K  S 1 q 300 K  310 K  S 2 q  S 1 >  S 2 e.g., For H 2 O(s)  H 2 O(l),  H fus = kJ/mol at 0ºC and for H 2 O(l)  H 2 O(g),  H vap = kJ/mol at 100ºC. What are  S fus and  S vap ? S =S = qTqT

7 7 II. Entropy B. The third law of thermodynamics “For a pure, crystalline substance at 0 K, S = 0” Increase T  increase in:translational motion vibrational motion rotational motion increase in S Standard state entropy, Sº = entropy of a substance at 25ºC, 1 atm (Table 19.1, Appendix L) (Sº  0 for elements!)

8 8 II. Entropy C. Calculating entropy changes in chemical reactions Standard entropy change,  Sº  Sº rxn =  Sº products -  Sº reactants e.g., 2H 2 + O 2  2H 2 O  Sº rxn = [2Sº(H 2 O)] - [2Sº(H 2 ) + Sº(O 2 )] = [2(188.8)] - [2(130.6) + (205.0)] = J/K (note units)

9 9 II. Entropy C. Calculating entropy changes in chemical reactions e.g., Calculate the standard entropy change for the reaction, 2Al(s) + 3Cl 2 (g)  2AlCl 3 (s)Sº(Al) = 28.3 J/mol·K Sº(Cl 2 ) = J/mol·K Sº(AlCl 3 ) = J/mol·K

10 10 II. Entropy D. The second law of thermodynamics “In any spontaneous process, the entropy of the universe must increase (  S universe > 0). system surroundings universe  S universe = (  S system +  S surroundings ) > 0 e.g.,erecting a building 6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 CFCs

11 11 III. Gibbs Free Energy A. Definition  S universe =  S system +  S surroundings  S surroundings = – q system T = –  H system T  S universe =  S system –  H system T  S universe =  H system – T  S system  G system =  H system – T  S system Gibbs free energy state function relates spontaneity to heat of reaction (  H) and entropy (  S) for a process to be spontaneous,  S universe > 0 or  G system < 0

12 12 III. Gibbs Free Energy A. Definition  G < 0: exergonic, spontaneous  G > 0: endergonic, nonspontaneous  G = 0: system is at equilibrium (no free energy remains) G G  Reactants Products Equilibrium  G rxn < 0 “spontaneous” R  E  G < 0spontaneous P  E  G < 0spontaneous E  R  G > 0nonspontaneous E  P  G > 0nonspontaneous

13 13 III. Gibbs Free Energy B. Calculating free energy changes in chemical reactions e.g., C(s) + ½O 2 (g) + 2H 2 (g)  CH 3 OH(l) at 25º  Hº rxn =  H f º(products) –  H f º(reactants) = –238.7 kJ/mol  Sº rxn =  Sº products –  Sº reactants = [Sº(CH 3 OH)] – [Sº(C) + ½Sº(O 2 ) + 2Sº(H 2 )] = [126.8] – [(5.7) + ½(205.1) + 2(130.7)] = –242.9 J/mol·K  Gº rxn =  Hº rxn – T  Sº rxn = (–238.7 kJ/mol) – (298 K)(–242.9 J/mol·K) = –166.3 kJ/mol =  G f º(CH 3 OH)

14 14 III. Gibbs Free Energy C. Using standard free energies of formation  Gº rxn =  G f º(products) -  G f º(reactants) standard free energies of formation (Table 19.3, Appendix L; note:  G f º(element) = 0) e.g., CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  Gº rxn = [  G f º(CO 2 ) + 2  G f º(H 2 O)] - [  G f º(CH 4 )] = [(-394.4) + 2(-237.1)] - [(-50.7)] = kJ

15 15 III. Gibbs Free Energy C. Using standard free energies of formation e.g., If  Gº rxn for the following reaction is kJ, what is  Gº f for TiCl 2 ? TiCl 2 (s) + Cl 2 (g)  TiCl 4 (s)(  Gº f (TiCl 4 ) = kJ/mol)

16 16 III. Gibbs Free Energy D. Free energy and temperature  G =  H – T  S  H  S  G – + –at all temperatures + – +at all temperatures + + –at higher temperatures, + at lower temperatures – – –at lower temperatures, + at higher temperatures e.g., C 12 H 22 O O 2  12CO H 2 O (produces heat)  H 0   G < 0 at any temperature (for reverse reaction,  G > 0 at any temperature; requires input of energy) NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 – (aq) (gets cold)  H > 0,  S > 0   G < 0 at higher temperatures H2OH2O

17 17 III. Gibbs Free Energy D. Free energy and temperature e.g., CH 3 –CH 3  CH 2 =CH 2 + H 2  Hº = +137 kJ  Sº = J/mol·K at 25ºC,  Gº = kJ nonspontaneous Above what temperature does the reaction become spontaneous?

18 18 III. Gibbs Free Energy E. Free energy and equilibrium  G =  Gº + RTlnQ At equilibrium,  G = 0, Q = K  =  Gº + RTlnK  Gº 1equilibrium lies to the right  Gº > 0  K < 1equilibrium lies to the left  Gº = 0  K = 1equilibrium lies in the middle  Gº = –RTlnK G  RPRPRP  Gº > 0  Gº < 0  Gº = 0

19 19 III. Gibbs Free Energy E. Free energy and equilibrium  Gº = –RTlnK or K = e –  Gº/RT e.g., 2SO 2 + O 2 2SO 3 What is K at 25ºC?  Gº f (SO 2 ) = kJ/mol  Gº f (SO 3 ) = kJ/mol


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