Chapter 17

1. Chapter 17 Free energy and Thermodynamics suroviec Spring 2014

2. I. Spontaneous Change and Equilibrium • Chemical changes – reactions • Physical changes – expansions, heat, precipitation • Spontaneous change – occurs without outside intervention • Does not tell us about rate of change • Does indicate that we are headed to equilibrium

3. II. Heat and Spontaneity • HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq) DH = ? • CH4 (g) + O2 (g)  CO2 (g) + H2O (g) DH = ? • NH4NO3 (s)  NH4+ (aq) + NO3- (aq) DH = ? • gas expanding : ? • Melting of ice : DH = ?

4. III. Dispersal of Energy • Entropy – S • Second law of thermodynamics • In a spontaneous process the entropy of the universe increases

5. A. Dispersal of Energy • The dispersal of energy over as many different energy states is the KEY to entropy • Thermal energy has been transferred

8. B. Dispersal of Matter • Since it is difficult to calculate to exact energy levels and dispersal of total energy among them lets initially look at the dispersal of matter among a system.

9. B. Dispersal of Matter • Looking at the flask again, the entropy of a state increases with the number of energetically equivalent ways to arrange the components of the system. • 2nd law states that for any spontaneous process, the entropy of the universe increases.

10. C. Energy change and change in state • Entropy of a sample of matter changes as it changes state.

11. IV. Heat transfer • There can also be spontaneous processes that occur even when entropy seems to decrease.

12. B. Temperature dependence of DSsurr • If freezing increases entropy of surroundings, why would water not freeze at all temps?

13. C. Quantifying entropy changes in the surrounding • When a system exchanges heat with the surroundings, it changes the entropy of the surroundings • At constant pressure we can use qsys to quantify the change in entropy for the surroundings.

14. C. Quantifying entropy changes in the surrounding • We also know that the higher the temperature the lower the magnitude of DSsurr

15. Example Consider the combustion reaction below: C3H8 (g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔHrxn = -2044 kJ • Calculate DSsurr at 25oC • Determine the sign of DSsys • Determine the sign of DSuniv

16. Example • Given DH°rxn = -125kJ and ΔS°rxn = 253J/K at 25°C, what is the ΔSuniv?

17. V. Gibbs Free Energy • Knowing that: • We can use that to rearrange and solve for Gibbs free energy.

18. V. Gibbs Free Energy • The change in Gibbs Free Energy for a process occurring at a constant temperature and pressure is proportional to the negative of ΔSuniv

19. A. Effect of ΔH, ΔS and T • These terms are all related, but how?

20. VI. So how do we know if a reaction is going to be spontaneous?

21. Example Given: C2H4(g) + H2(g)  C2H6(g) Where ΔH = -137.5kJ and ΔS = -120.5 J/K • Calculate ΔG at 25°C • How does T affect ΔG

22. Example Given: 2Ca(s) + o2(g)  2CaO(s) Where ΔH°rxn = -1269.8 kJ and ΔS°rxn = -364.6 J/K • Calculate ΔG°rxn at 25°C • Is it spontaneous?

23. VI. Entropy changes in chemical reactions • As a reminder: ΔH°rxn is change in enthalpy for a process in which reactants and products are in their std. states.

24. A. S° and the 3rd Law of Thermodynamics • Remember that we said that ΔH°f for an element in its std. state is zero. • However since S is temperature dependent, we can better define the zero of entropy.

25. B. Relative standard entropies • As we have seen, the entropy of a substance depends on its state.

26. C. Calculating ΔS°rxn • To calculate ΔS°rxn we use the following equation:

27. Example Compute ΔS°rxn for the following reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

28. Example Compute ΔS°rxn for the following reaction: 2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)

29. VI. Calculating ΔG°rxn • Free energy changes • We can use ΔH°rxn and ΔS°rxn

30. Example • Calculate the ΔG°rxn for the following reaction at 25°C 2SO2(g) + O2(g)  2SO3(g)

31. Example • What is the previous reaction was not at 25°C but instead was at 125°C?

32. B. ΔG°rxn using free energies • Since ΔG°rxn is the change in free energy and fee energy is a state function, we can solve ΔG°rxn as we did for ΔH°rxn and ΔS°rxn

33. Example • Use standard free energies of formation to determine ΔG°rxn at 25°C CH4(g) + 8O2(g)  CO2(g) + 2H2O(g) + 4O3(g)

34. C. Determining ΔG°rxn for a series of reactions • To determine ΔG°rxn for a series of reactions, we use the same rules that we did with ΔG°rxn

35. Example • Determine the ΔG°rxn for the following reaction: 2NO(g) + O2(g)  2NO2(g) ΔG°rxn = -71.2 kJ N2(g) + O2(g)  2NO(g) ΔG°rxn = +175.2 kJ 2N2O(g)  2N2(g) + O2(g) ΔG°rxn = -207.4 kJ

36. VII. DGo, K and product favorability • K = equilibrium constant • How does K relate to ΔG°rxn ? • ΔG°rxn is the increase or decrease in free energy as the reactants are completely converted to products • Reactants are not always converted completely to products.

37. VII. DGo, K and product favorability • When reactants are mixed, they proceed to position of lowest ree energy and then reach equilibrium.

38. Fig. 19-13, p.929

39. Combining Everything that we have learned • The free energy at equilibrium is lower than the free energy of reactants or products • DGorxn gives the position of the equilibrium • DGorxn describes the direction of the reaction

40. 17_13.JPG