1. Chapter 17 Lesson 2 Free Energy and Thermodynamics

2. 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

3. ∆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.

4. 4 ∆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

5. 5  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

6. 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”

7. 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

8. 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

9. 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

10. 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)

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12. 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)

13. 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)

14. Copyright McGraw-Hill 2009 Free Energy and Equilibrium  G =  G° + RT ln Q  G = non-standard free energy  G° = standard free energy (from tables) R = 8.314 J/K·mole T = temp in K Q = reaction quotient

15. ∆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

16.  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

17. 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!!

18. 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)

19. Copyright McGraw-Hill 2009 Relationship Between  G° and K

20. 20 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°

21. 21 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)

22.  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

23. 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

24. 24 High Entropy Low Entropy Chemistry In Action: The Thermodynamics of a Rubber Band

25. Copyright McGraw-Hill 2009 18.6 Thermodynamics in Living Systems Coupled reactions Thermodynamically favorable reactions drives an unfavorable one Enzymes facilitate many nonspontaneous reactions

26. 26 The Structure of ATP and ADP in Ionized Forms ATP  ADP  G° = -31 kJ/mol

27. Copyright McGraw-Hill 2009 ATP-ADP Interconversions C 6 H 12 O 6 oxidation:  G° =  2880 kJ/mol ATP  ADP  G° = -31 kJ/mol Synthesis of proteins: (first step) alanine + glycine  alanylglycine  G° = 29 kJ/mol Protein synthesis is now favored.