1. Spontaneity and Equilibrium isolated system : Isothermal process  Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system.

2. Calculate the maximum work that can be obtained from the combustion of 1 mole of methane at 298 K. Given  H o and  S o of the the combustion of methane.

3. Transformation at constant temperature and pressure  Maximum work, over and above pV-work, obtained in a process at constant temperature and pressure is equal to the decrease in the Gibbs energy of the system.

4. Special case: No work over and above pV-work w non p-V =0 Calculate the maximum non-pV work that can be obtained from the combustion of 1 mole of methane at 298 K.

6. Fundamental equations of Thermodynamics Maxwell Relations

8. Transformation at constant temperature

9. Chemical potential 

10. The value of  G f o of Fe(g) is 370 kJ/mol at 298 K. If  H f o of Fe(g) is 416 kJ/mol (assumed to be constant in the range 250-400 K), calculate  G f o of Fe(g) at 400 K. Transformation at constant pressure

11. G dependence on n H2OH2O H2OH2O Given a system consisting of two substances:

12. If there is no change in composition:

13. System at constant T and p  dn 1 Each subsystem is a mixture of substances.  Chemical Equilibrium

14.  Equilibrium is established if chemical potential of all substances in the system is equal in all parts of the system.  Matter flows from the part of system of higher chemical potential to that of lower chemical potential.

15.  Pure H 2  N 2 + H 2 Pd membrane Equilibrium never reached constant T & p

16.  G and  S of mixing of gases

19. Chemical reactions CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g)

20. Heat of Formation Formation reaction: reaction of forming 1 mole of product from the elements in their stable form at 25ºC and 1 atm. Heat of formation =  H of formation reaction =  F H Standard heat of formation =  Hº of formation reaction =  F Hº  F Hº(NO (g) ): ½ N 2(g) +½ O 2(g) → NO (g)  Hº  F Hº(CO (g) ): C graphite(s) +½ O 2(g) → CO (g)  Hº  F Hº(O (g) ): ½ O 2(g) → O (g)  Hº  F Hº(C diamond(s) ): C graphite(s) → C diamond(s)  Hº  F Hº(O 2(g) ): O 2(g) → O 2(g)  Hº=0  F Hº(C graphite(s) ): C graphite(s) → C graphite(s)  Hº=0

21.  G of Formation Gibbs energy of formation =  G of formation reaction =  F G Standard Gibbs energy of formation =  Gº of formation reaction =  F Gº  F Gº(NO (g) ): ½ N 2(g) +½ O 2(g) → NO (g)  Gº  F Gº(CO (g) ): C graphite(s) +½ O 2(g) → CO (g)  Gº  F Gº(O (g) ): ½ O 2(g) → O (g)  Gº  F Gº(C diamond(s) ): C graphite(s) → C diamond(s)  Gº  F Gº(O 2(g) ): O 2(g) → O 2(g)  Gº=0  F Gº(C graphite(s) ): C graphite(s) → C graphite(s)  Gº=0

22. Applying Hess’s Law

23. Chemical reactions CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g) As the reaction proceeds:  The number of moles of involved substances changes.  G of system will change:   Extent of reaction  Reaction advancement  Degree of reaction

24. As the forward reaction proceeds:  grows, d  positive, d  > 0

25.  even though G of products larger than G of reactants, the reaction proceeds!!!!!!!!!! Reason:  G mix

26. For a mixture:  G more negative if   G pure is small   G mix largely negative

27. Pure R Pure P G pure  G mix G total

28. Equilibrium constant R, P: Ideal gases

31. K p relation to K x p total in atm

32. K p relation to K c c in mol/L R=0.0821 atmL/mol.K

33. Consider the reaction N 2 O 4(g) → 2 NO 2(g)  F Gº(NO 2(g) )=51.31 kJ/mol  F Gº(N 2 O 4(g) )=102.00 kJ/mol Assume ideal behavior, calculate nono n eq nini x eq pipi N2O4N2O4 11-x 1+x (1-x)/(1+x)(1-x)/(1+x)*P tot NO 2 02x2x/(1+x)2x/(1+x)*P tot

34. Temperature dependence of K p

36. For the reaction N 2 O 4(g) → 2 NO 2(g)  F Hº(NO 2(g) )=51.31 kJ/mol  F Hº(N 2 O 4(g) )=102.00 kJ/mol K p (25 o C)=0.78 atm, calculate K p at 100 o C? For a given reaction, the equilibrium constant is 1.80x10 3 L/mol at 25 o C and 3.45x10 3 L/mol at 40 o C. Assuming  H o to be independent of temperature, calculate  H o and  S o.

37. Heterogeneous Equilibria

38.  F Gº kJ/mol  F Hº kJ/mol CaCO 3(s) -1128.8-1206.9 CaO (s) -604.0-635.1 CO 2(g) -394.4-393.51 Calculate The pressure of CO 2 at 25oC and at 827 o C?  Gº =130.4 kJ  º =178.3 kJ ln(K p )=ln(p CO2 )=-52.6 p CO2 =1.43x10 -23 atm At 1100 K: ln(p CO2 )=0.17 p CO2 =0.84 atm

39. Vaporization Equilibria Clausius-Clapeyron Equation Derive the above relations for the sublimation phase transition!

40. Mass Action Expression (MAE) For reaction: aA + bB cC + dD Reaction quotient –Numerical value of mass action expression –Equals “Q” at any time, and –Equals “K” only when reaction is known to be at equilibrium

41. 41 Calculate [X] equilibrium from [X] initial and K C Ex. 4 H 2 (g) + I 2 (g) 2HI (g) at 425 °C K C = 55.64 If one mole each of H 2 and I 2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H 2, I 2 and HI? Step 1. Write Equilibrium Law

42. 42 Ex. 4 Step 2. Concentration Table Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial2.00 0.000 Change Equil’m Initial [H 2 ] = [I 2 ] = 1.00 mol/0.500L =2.00M Amt of H 2 consumed = Amt of I 2 consumed = x Amt of HI formed = 2x – x +2x – x– x 2.00 – x

43. 43 Ex. 4 Step 3. Solve for x Both sides are squared so we can take square root of both sides to simplify

44. 44 Ex. 4 Step 4. Equilibrium Concentrations Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial2.00 0.00 Change Equil’m [H 2 ] equil = [I 2 ] equil = 2.00 – 1.58 = 0.42 M [HI] equil = 2x = 2(1.58) = 3.16 – 1.58 +3.16 – 1.58 +3.160.42

45. 45 Calculate [X] equilibrium from [X] initial and K C Ex. 5 H 2 (g) + I 2 (g) 2HI (g) at 425 °C K C = 55.64 If one mole each of H 2, I 2 and HI are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H 2, I 2 and HI? Now have product as well as reactants initially Step 1. Write Equilibrium Law

46. 46 Calculate [X] equilibrium from [X] initial and K C Ex. 6 CH 3 CO 2 H (aq) + C 2 H 5 OH (aq) CH 3 CO 2 C 2 H 5 (aq) + acetic acidethanol ethyl acetate H 2 O ( l ) K C = 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?

47. 47 Calculating K C Given Initial Concentrations and One Final Concentration Ex. 2a H 2 (g) + I 2 (g)  2HI (g) @ 450 °C Initially H 2 and I 2 concentrations are 0.200 mol each in 2.00L (= 0.100M); no HI is present At equilibrium, HI concentration is 0.160 M Calculate K C To do this we need to know 3 sets of concentrations: initial, change and equilibrium