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Thermodynamics

Thermodynamics. Spontaneity, Entropy & Free Energy Chapter 16 On the A.P. Exam: 5 Multiple Choice Questions Free Response—Every Year. Standard State Conditions. Indicated by a degree symbol Gases at 1 atm Liquids are pure. Solids are pure. Solutions are at 1 M.

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Thermodynamics

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  1. Thermodynamics Spontaneity, Entropy & Free Energy Chapter 16 On the A.P. Exam: 5 Multiple Choice Questions Free Response—Every Year

  2. Standard State Conditions • Indicated by a degree symbol • Gases at 1 atm • Liquids are pure. • Solids are pure. • Solutions are at 1 M. • Temperature is 25o C (298 K).

  3. First Law of Thermodynamics • Conservation of Energy—Energy can be neither created nor destroyed. • The amount of energy in the universe is constant. • Energy can change forms, but the amount does not change.

  4. Example • Chemical (potential) energy is stored in the bonds of methane gas. • Combustion releases the energy stored in the bonds as heat. • Energy exchange between system & environment = enthalpy

  5. Spontaneity • Spontaneous processes need no outside cause; they happen naturally. • Spontaneous = Fast • (Thermodynamics is not concerned with how a reaction happens—just whether it will happen.)

  6. What Causes Spontaneity?? • Exothermicity?? (Tendency to reach lowest potential energy) • Partial explanation, but does not always hold true • Probability—how likely are arrangements of atoms (randomness or entropy)

  7. Entropy • The amount of disorder or randomness in a system • Higher entropy is favored because random arrangements are more likely. • Ex.: Gas molecules are more likely to be evenly dispersed than clumped up in one place.

  8. Which has higher entropy? • 1 mole of gas or 1 mole of liquid? • Pure water or a 0.5 M salt solution? • Gas at 25o C or gas at 100o C? • 1 mole of gas or 5 moles of gas?

  9. Entropy Increases: • In state changes--solid to liquid, liquid to gas or solid to gas • When solids dissolve • When temperature increases • When volume of a gas increases (or pressure decreases) • When a reaction occurs that produces more moles of gases

  10. Second Law of Thermodynamics • In any spontaneous process, there is an increase in entropy of the universe. • Think of the universe as being inclusive of a system and its surroundings—combined entropy must increase (Cell)

  11. DSsurr • Sign depends on direction of heat flow—heat added to surroundings increases entropy & heat absorbed from surroundings decreases entropy

  12. DSsurr • Magnitude—depends on temperature—greater impact of exothermic system if temperature is low • Analogy—giving $50 to a millionaire or to you—to whom does it mean more?

  13. DSsurr • Equal to –DH/T • negative—because the surroundings are opposite from the system • DH –change in heat in the system (from system’s perspective) • T—temperature in Kelvin

  14. What’s the Point??? • Dssurris larger at lower temperatures because the denominator of the fraction is smaller . • The effect of heat flow for a process is more significant when temperature is lower.

  15. Free Energy • G • Uses enthalpy, entropy and temperature to predict the spontaneity of a reaction • If DG is negative, then a process is spontaneous.

  16. Calculating DG • DG = DH – TDS • DGis free energy • DH is change in enthalpy • DSis change in entropy • All terms are from the point of view of the system.

  17. DG = DH – TDS • Mathematical relationships: • If DH is negative and DS is postive—always spontaneous • If DH is positive and DS is negative—never spontaneous • If both DH and DS have the same sign—temperature dependent

  18. Example • At what temperature is the following process spontaneous? Br2(l) Br2(g) DH = 31.0 kJ/mol & DS = 93.0 J/K mol • Calculate temperature when DG is zero. (DG = DH – TDS)

  19. DG and Boiling Point • At the exact boiling point, DG is zero. • Set known values of DH, DS and/or T equal to zero and solve for unknowns.

  20. Example • The enthalpy of vaporization of mercury is 58.51 kJ /mol and the entropy of vaporization is 92.92 J/K mol. What is mercury’s normal boiling point?

  21. Third Law of Thermodynamics • The entropy of a perfect crystal at 0 K is zero. • Standard entropy values have been determined—Appendix 4

  22. DS • To determine entropy change for a reaction, subtract the entropy values • DS = SnpSoproducts – SnrSoreactants • n = number of moles (State functions depend on amount of substance present.)

  23. Example • Calculate DSo at 25o C for the reaction: 2NiS(s) + 3O2 (g)  2SO2 (g) + 2NiO (s) • So in J/K mol: • SO2 248 J/K mol • NiO 38 J/K mol • O2 205 J/K mol • NiS 53 J/K mol

  24. Example • Calculate DSo for the reduction of aluminum oxide by hydrogen gas: Al2O3(s) + 3H2(g) 2 Al(s) + 3 H2O(g) Note: equal number of gas mol Ans: 179 /K (large & positive because of water’s asymmetry)

  25. Free Energy • Cannot be directly measured • Can be calculated using enthalpy and entropy changes (DG = DH – TDS) • Can be calculated from already determined standard free energy values: DG = SnpGoproducts – SnrGoreactants

  26. Example: • Calculate DH, DS, and DG using the following information for the reaction 2SO2 (g) + O2 (g)  2SO3(g). • SO2: DH = -297 kJ/mol; So = 248 J/K mol • SO3: DH = -396 kJ/mol; So = 257 J/K mol • O2: DH = 0 kJ/mol; So = 205 J/K mol

  27. Example: • Using the following data, calculate DG for the reaction Cdiamond (s)  C graphite (s) • Cdiamond (s) + O2 (g) CO2 (g) DG = -397 kJ • Cgraphite (s) + O2 (g)  CO2 (g) DG = -394 kJ

  28. Reverse one equation (change sign of DG) and add them together.

  29. Example • Given the following free energies of formation, calculate the DG for the reaction 2CH3OH (g) + 3O2(g) 2 CO2(g) + 4 H2O (g) CH3OH (g): DG = -163 kJ/mol O2 (g): DG = 0 kJ/mol CO2 (g): DG = -394 kJ/mol H2O (g): DG = -229 kJ/mol

  30. Free Energy & Pressure • Entropy depends on pressure because changes in pressure affect volume. • G = Go + RTln(P) • G = free energy at new pressure • Go = free energy at 1 atm • R = gas constant (8.3145 J/K*mol) • T = Kelvin temperature • P = new pressure

  31. DG at Nonstandard Conditions • From the previous equation, a relationship between DG and Q or K can be derived. • (See p. 807) • DG = DGo + RT ln (Q)

  32. Example • CO (g) + 2H2 (g)  CH3OH (l) • Calculate DG for this process where CO is at 5.0 atm and H2 is at 3.0 atm. • Calculate DGo using standard values for reactants & products. • Then use DG = DGo + RT ln (Q)

  33. DG and Equilibrium • At equilibrium, a system has used all of the free energy available. • DG equals zero because forward & reverse reactions have the same free energy. • DG = DGo + RT ln (Q) • 0 = DGo + RT ln (K) • DGo = - RT ln (K)

  34. Example • For the synthesis of ammonia, DGo = -33.33 kJ/mol of N2 consumed at 25o C. • Predict how each system below will shift in order to reach equilibrium: • NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2 atm • NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm

  35. NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2atm • DG = DGo + RT ln (Q)

  36. NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm • DG = DGo + RT ln (Q)

  37. Example • 4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s) • Use the data to calculate K for this reaction at 25o C. • Fe2O3: DH = -826 kJ/mol; So = 90 J/K mol • Fe: DH = 0 kJ/mol; So = 27 J/K mol • O2: DH = 0 kJ/mol; So = 205 J/K mol

  38. Calculate DH

  39. Calculate DSo

  40. Calculate DGo

  41. Calculate K : DG = -RTlnK

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