Gibbs Free Energy

1. Gibbs Free Energy • For a spontaneous reaction, the entropy of the universe must increase, Suniv (+) • Reactions with large negative enthalpy values, H << 0, are spontaneous. • How are S and H related? How can a spontaneous reaction be predicted? • The Gibbs free energy, G, of a state is defined as: • G = H - TS • For a process occurring at constant temperature • G = H - TS.

2. Gibbs Free Energy • There are three important conditions: • If G < 0 then the forward reaction is spontaneous. • If G = 0 then reaction is at equilibrium and no net reaction will occur. • If G > 0 then the forward reaction is not spontaneous. (However, the reverse reaction is spontaneous.) If G > 0, energy must be supplied from the surroundings to drive the reaction. • The progress of a reaction has the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.

3. Gibbs Free Energy

4. Gibbs Free Energy • Standard Free-Energy Changes • Standard free-energies of formation, Gf are tabulated just as standard enthalpies and entropies of formation. • Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and G = 0 for elements. • G for a process is given by • The quantity G for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (G > 0) or products (G < 0).

5. Calculating  Go from Enthalpy and Entropy Values Problem: Potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and matchheads, undergoes a solid-state redox reaction when heated, in which the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation): Use Hof and So values to calculate Gosys (Gorxn) at 25oC for this reaction. Plan: To solve for Go, use Hof values to calculate Horxn(Hosys), use So values to calculate Sorxn(Sosys), and apply Equation 20.6. 4 KClO3 (s) 3 KClO4 (s) + KCl(s)

6. Calculating  Go from Enthalpy and Entropy Values Solution: Calculating  Hosys from Hof values (with Equation 6.8): 4 KClO3 (s) 3 KClO4 (s) + KCl(s) Hosys = Horxn = • mHof(products) • nHof(reactants) = [3 mol KClO4 ( Hof of KClO4) + 1 mol KCl ( Hof of KCl)] - [4 mol KClO3 ( Hof of KClO3)]

7. Calculating  Go from Enthalpy and Entropy Values Hosys = [3 mol (-432.8 kJ/mol) + 1 mol (-436.7 kJ/mol)] - [4 mol (-397.7 kJ/mol)] = -144 kJ Calculating Sosys from So values (with Equation 20.3): Sosys = Sorxn = [3 mol KClO4 ( So of KClO4) + 1 mol KCl ( So of KCl)] - [4 mol KClO3 (So of KClO3)] = [3 mol (151.0 J/mol K) + 1 mol (82.6 J/mol K)] - [4 mol (143.1 J/mol K)] = - 36.8 J/K . . . Calculating Gosys at 298 K: Gosys = Hosys - TSosys = -144 kJ - [(298 K)(-36.8 J/K)(1kJ/1000 J)] The reaction is spontaneous which is consistent with Go < 0 Gosys = -133 kJ

8. Free Energy and Temperature • Focus on G = H - TS: • If H < 0 and S > 0, then G is always negative. • If H > 0 and S < 0, then G is always positive. (That is, the reverse of 1.) • If H < 0 and S < 0, then G is negative at low temperatures. • If H > 0 and S > 0, then G is negative at high temperatures. • Even though a reaction has a negative G it may occur too slowly to be observed. • Thermodynamics gives us the direction of a spontaneous process, it does not give us the rate of the process.

9. Reaction Spontaneity and Values of Go = Ho - TSo Ho So -T So Go Description - + - - Spontaneous at all T + - + + Nonspontaneous at all T + + - + or - Spontaneous at higher T; Nonspontaneous at lower T - - + + or - Spontaneous at lower T; Nonspontaneous at higher T Table 20.1 (p. 879)

10. Free Energy and Temperature

11. Effect of Temperature on Reaction Spontaneity Calculate the sign of Go and predict whether the reaction is spontaneous or nonspontaneous at all temperatures or only high temperature or only low temperature: 2 H2O2 (l) 2 H2O(l) + O2 (g)Ho = -196 kJ and So = 125 J/K 3 O2 (g) 2 O3 (g) Ho = 286 kJ and So = - 137 J/K 2 N2O(g) + O2 (g) 4 NO(g)Ho = 197.1 kJ and So = 198.2 J/K 2 Na(s) + Cl2 (g) 2 NaCl(s)Ho = - 822.2 kJ and So = - 181.7 J/K

12. Effect of Temperature on Reaction Spontaneity Temperature-independent cases (opposite signs) 1. Reaction is spontaneous at all temperatures:Ho < 0, So > 0 2 H2O2 (l) 2 H2O(l) + O2 (g)Ho = -196 kJ and So = 125 J/K 2. Reaction is nonspontaneous at all temperatures:Ho > 0, So < 0 3 O2 (g) 2 O3 (g) Ho = 286 kJ and So = - 137 J/K Temperature-dependent cases (same signs) 3. Reaction is spontaneous at higher temperature:Ho > 0 and So > 0 4. Reaction is spontaneous at lower temperature:Ho < 0 and So < 0 2 N2O(g) + O2 (g) 4 NO(g)Ho = 197.1 kJ and So = 198.2 J/K 2 Na(s) + Cl2 (g) 2 NaCl(s)Ho = - 822.2 kJ and So = - 181.7 J/K

13. Determining the Effect of Temperature on Go Problem: An important reaction in the production of sulfuric acid is the oxidation of SO2 (g) to SO3 (g): At 298 K, Go = -141.6 kJ; Ho = -198.4 kJ; and So = -187.9 J/K. (a) Use the data to decide if this reaction is spontaneous at 25oC and how Go will change with increasing T. (b) Assuming Ho and So are constant with T, is the reaction spontaneous at 900.oC? 2 SO2 (g) + O2 (g) 2 SO3 (g)

14. Determining the Effect of Temperature on Go Solution: (a) Since Go < 0, the reaction is spontaneous at 298 K: a mixture of SO2 (g), O2 (g), and SO3 (g) in their standard states (1 atm) will spontaneously yield more SO3 (g). With So < 0, the term -TSo > 0 and becomes more positive at higher T. Therefore, Go will be less negative, and the reaction less spontaneous, with increasing T. (b) Calculating Go at 900.oC (T= 273 + 900. = 1173 K): Go = Ho - TSo = - 198.4 kJ - [(1173 K)(-187.9 J/K)(1 kJ/1000 J)] = 22.0 kJ Since Go > 0, the reaction is nonspontaneous at higher T.

15. Fig. 20.11

16. Free Energy and the Equilibrium Constant • G and K (equilibrium constant) apply to standard conditions. • G and Q (equilibrium quotient) apply to any conditions. • It is useful to determine whether substances under any conditions will react:

17. Free Energy and the Equilibrium Constant • At equilibrium, Q = K and G = 0, so • From the above we can conclude: • If G < 0, then K > 1. • If G = 0, then K = 1. • If G > 0, then K < 1.

18. The Relationship Between Go and K at 25oC Go (kJ) K Significance 200 9 x 10 -36 Essentially no forward reaction; 100 3 x 10 -18 reverse reaction goes to 50 2 x 10 -9 completion. 10 2 x 10 -2 1 7 x 10 -1 0 1 Forward and reverse reactions -1 1.5 proceed to same extent. -10 5 x 101 -50 6 x 108 -100 3 x 1017 Forward reaction goes to -200 1 x 1035 completion; essentially no reverse reaction. Table 20.2 (p. 883)

19. Calculating G at Nonstandard Conditions Problem: The oxidation of SO2, which we discussed earlier, is too slow at 298 K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298 K and at 973 K. Go298 = -141.6 kJ/mol of reaction as written; using Ho and So values at 973 K, Go973 = -12.12 kJ/mol of reaction as written. (b) In experiments to determine the effect of temperature on reaction spontaneity, sealed containers are filled with 0.500 atm SO2, 0.0100 atm O2, and 0.100 atm SO3 and kept at 25oC and at 700oC. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate G for the system in part (b) at each temperature. Plan: (a) We know Go, T, and R, so we can calculate the K’s from Equation 12.10. (b) Calculate Q, and compare it with each K from part (a). 2 SO2 (g) + O2 (g) 2 SO3 (g)

20. Calculating G at Nonstandard Conditions 1000 J 1 kJ -141.6 kJ/mol x 1000 J 1 kJ -12.12 kJ/mol x Plan : Continued (c) Since these are not standard-state pressures, we calculate G at each T from Equation 20.11 with the values of Go (given) and Q [found in part (b)]. Solution: (a) Calculating K at the two temperatures: At 298 K, the exponent is Go = -RT ln K so K = e-( G/RT) . - ( Go/RT) = - = 57.2 8.31 J/mol K x 298 K K = e-( G/RT) = e57.2 = 7 x 1024 So At 973 K, the exponent is - ( Go/RT) = - = 1.50 . 8.31 J/mol K x 973 K K = e-( G/RT) = e1.50 = 4.5

21. Calculating G at Nonstandard Conditions (b) Calculating the value of Q : Since Q < K at both temperatures, the denominator will decrease and the numerator increase–more SO3 will form–until Q equals K. However, at 298 K, the reaction will go far to the right before reaching equilibrium, whereas at 973 K, it will move only slightly to the right. (c) Calculating G, the nonstandard free energy change, at 298 K and 973 K p2SO3 0.1002 0.5002 x 0.0100 Q = = = 4.00 p2SO2 x pO2 G298 = Go + RT ln Q = -141.6 kJ/mol + (8.31 J/mol K x x 298 K x ln 4.00) = -138.2 kJ/mol . 1 kJ 1000 J G973 = Go + RT ln Q = - 12.12 kJ/mol + (8.31 J/mol K x x 973 K x ln 4.00 = - 0.90 kJ/mol . 1 kJ 1000 J

22. Fig. 20.12

23. Free-Energy Changes for NH3

24. Fig. 20.D (p. 881)

25. The Coupling of a Nonspontaneous Reaction to the Hydrolysis of ATP Fig. 20.C (p. 881)

26. Why is ATP a “High-Energy” Molecule? Fig. 20.E (p. 882)