Spontaneity Entropy and Free Energy

1. Spontaneity Entropy and Free Energy WHAT DRIVES A REACTION TO BE SPONTANEOUS?

2. ENTHALPY (H) heat content (exothermic reactions are generally favored)

3. 1st Law of Thermodynamics The total energy in the universe is constant. DE = q (heat) + w (work) -heat out -work on surroundings + heat in + work on system Calculating DH: • stoichiometry—using info given with equation • H rxn = ∑ n DHf° products -- ∑ n DHf° reactants • Hess’s Law—adding up equations and DH’s • calorimetry q = mCDT or q = CDT • bond energies ∑ bonds broken -- ∑ bonds formed • Note: pure elements have DHf° values of 0

4. ENTROPY ‘S’The Second Law of Thermodynamics The universe is constantly increasing disorder. DSuniv = DSsystem + DSsurroundings ΔSsurroundingsbased on heat flow exothermic + DSsurr endothermic -DSsurr

5. ENTROPY (S) Disorder of a system (more disorder is favored) Nature tends toward chaos! Think about your room at the end of the week! Factors that can indicate entropy changes: • Phase changes • Temperature changes • Volume changes • Mixing substances • Change in number of particles • Change in moles of gas

6. Spontaneous reactions are those that occur without outside intervention. They may occur fast OR slow (that is kinetics). Some reactions are very fast (like combustion of hydrogen) other reactions are very slow (like graphite turning to diamond)

7. Predicting the entropy of a system based on physical evidence: • The greater the disorder or randomness in a system, the larger the entropy. • The entropy of a substance always increases as it changes from solid to liquid to gas. • When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases (carbonates are an exception! --they interact with water and actually bring MORE order to the system)

8. Predicting the entropy of a system based on physical evidence: • When a gas molecule escapes from a solvent, the entropy increases. • Entropy generally increases with increasing molecular complexity (crystal structure: KCl vs CaCl2) since there are more MOVING electrons! • Reactions increasing the number of moles of particles often increase entropy.

9. Exercise 2 Predicting Entropy Changes Predict the sign of the entropy change for each of the following processes. A: Solid sugar is added to water to form a solution. B: Iodine vapor condenses on a cold surface to form crystals.

10. Solution: A: +∆S Sugar molecules have been randomly dispersed in the water. Greater entropy + DS B: -∆S Gaseous iodine is forming a solid. Less entropy - DS

11. Sample Problem A Which of the following has the largest increase in entropy? a) CO2(s)  CO2(g) b) H2(g) + Cl2(g)  2 HCl(g) c) KNO3(s)  KNO3(l) d) C(diamond) C(graphite)

12. Answer: • the substance changes from a highly organized state to a more disorganized state.

13. Choose the substance expected to have the greater absolute entropy. • Pb(s) or Cgraphite • He(g) at 1 atmosphere or He(g) at 0.5 atmosphere • H2O(l) or CH3CH2OH(l) at the same temperature • Mg(s) at 0°C or Mg(s) at 150 °C both at the same pressure

14. Pb has greater molar entropy. Pb, with metallic bonding, forms soft crystals with high amplitudes of vibration; graphite has stronger (covalent) bonds is more rigid and thus more ordered. • He(g) at 0.05 has greater molar entropy. At lower pressure (greater volume) He atoms have more space in which to move so are more random. • CH3CH2OH has greater molar entropy. Ethanol molecules have more atoms and thus more vibrations; water exhibits stronger hydrogen bonding • Mg(s) at 150 °C has greater molar entropy. At the higher temperature the atoms have more kinetic energy and vibrate faster and, thus, show greater randomness.

15. ENTROPYThe Third Law of Thermodynamics • The entropy of a perfect crystal at 0 K is zero. • not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even elements

16. This means the absolute entropy of a substance can then be determined at any temperature higher than 0 K. (Handy to know if you ever need to defend why G & H for elements = 0. . . BUT S does not!) So what?

17. BIG MAMMA, verse 2 S°rxn = S°(products) - S°(reactants) S is + when disorder increases (favored) S is – when disorder decreases Units are usually J/K mol (not kJ ---tricky!)

18. Sample Problem B Calculate the entropy change at 25C, in J/K for: 2 SO2(g) + O2(g)  2 SO3(g) Given the following data: SO2(g) 248.1 J/K mol O2(g) 205.3 J/Kmol SO3(g) 256.6 J/K mol Entropy change = -188.3 J/K

19. ENTROPY CHANGES FOR REVERSIBLE PHASE CHANGES (that’s a phase change at constant temperature) ΔS = heat transferred = q temperature at which change occurs T **where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvins

20. EX: water (l @ 100 → water (g @ 100) the entropy will increase. Taking favored conditions into consideration, the equation above rearranges into: S = - H T Give signs to ΔH following exo/endo guidelines! (If reaction is exo.; entropy of system increases—makes sense!)

21. Calculate ∆Ssurr for each of these reactions at 25C and 1 atm. Sb2S3(s)+3 Fe(s)  2 Sb(s)+3 FeS(s) ∆H = -125kJ Sb4O6(s)+6 C(s)4 Sb(s)+6 CO(g) ∆H = 778kJ

22. Solution: ∆Ssurr= 419 J/K ∆Ssurr = -2.61 × 103 J/K

23. SUMMARY ENTROPY: S = + MORE DISORDER (FAVORED CONDITION) S = - MORE ORDER

24. Whether a reaction will occur spontaneously may be determined by looking at the S of the universe. ΔS system + ΔS surroundings = ΔS universe • IF ΔS universe is +, then reaction is spontaneous • IF ΔS universe is -, then reaction is NONspontaneous

25. Consider 2 H2 (g) + O2 (g) → H2O (g) ignite & rxn is fast! ΔSsystem = -88.9J/K Entropy declines (due mainly to 3 → 2 moles of gas!)

26. First law of thermodynamics demands that this energy is transferred from the system to the surroundings so... -ΔHsystem = ΔHsurroundings OR - (- 483.6 kJ) = + 483.6 kJ

27. ΔS°surroundings = ΔH°surroundings = + 483.6 kJ T 298 K = 1620 J/K

28. Now we can find ΔS°universe ΔS system + ΔS surroundings = ΔS universe (-88.9 J/K) + (1620 J/K) = 1530 J/K Even though the entropy of the system declines, the entropy change for the surroundings is SOOO large that the overall change for the universe is positive.

29. Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering.

30. FREE ENERGY • Calculation of Gibb’s free energy is what ultimately decides whether a reaction is spontaneous or not. • NEGATIVE G’s are spontaneous.

31. G can be calculated one of several ways: Gºrxn = Go(products) - Go(reactants) ΔG°f = 0 (for elements in standard state) G = H - TS

32. G = Go + RT ln (Q) Define terms: G = free energy not at standard conditions Go = free energy at standard conditions R = universal gas constant 8.3145 J/molK T = temp. in Kelvin ln = natural log Q = reaction quotient: (for gases this is the partial pressures of the products divided by the partial pressures of the reactants—all raised to the power of their coefficients) Q = [products] [reactants]

33. “RatLink”: G = -RTlnK Terms: Basically the same as above --- however, here the system is at equilibrium, so G = 0 and K represents the equilibrium constant under standard conditions. K = [products] still raised to power of coefficients [reactants]

34. “nFe”: G = - nFE Remember this!! Terms: Go = just like above—standard free energy n = number of moles of electrons transferred (look at ½ reactions) F = Faraday’s constant 96,485 Coulombs/mole electrons Eo= standard voltage ** one volt = joule/coulomb**

35. Sample Problem C: Find the free energy of formation for the oxidation of water to produce hydrogen peroxide. 2 H2O(l) + O2(g)  2 H2O2(l) Given the following information: G°f H2O(l) -56.7 kcal/mol O2(g) 0 kcal/mol H2O2(l) -27.2 kcal/mol Free energy of formation = 59.0 kcal/mol

36. Calculate ∆H, ∆S, and ∆G 2 SO2 (g) + O2 (g)  2 SO3 (g) carried out at 25C and 1 atm.

37. Solution: ∆H = -198 kJ ∆S = -187 J/K ∆G = -142 kJ

38. Hess’s law of summation Using the following data @ 25C, calculate Go Cdiamond (s) Cgraphite (s) Cdiamond (s) +O2 (g)CO2 (g) ∆G= -397 kJ Cgraphite (s) +O2 (g) CO2 (g) ∆G= -394 kJ ∆G = -3 kJ

39. G = -RTlnK ) The overall reaction for the corrosion (rusting) of iron by oxygen is 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) Calculate the equilibrium constant for this reaction at 25C. K = .548

40. Gibb’s equation can also be used to calculate the phase change temperature of a substance. During the phase change, there is an equilibrium between phases so the value of G is zero. Really just like what we did earlier in this unit with enthalpy and entropy!

41. Sample Problem: Find the thermodynamic boiling point of H2O(l)  H2O(g) Given the following information: Hvap = +44 kJ Svap = 118.8 J/K Thermodynamic BP = 370K

42. SUMMARY OF FREE ENERGY: G = + NOT SPONTANEOUS G = - SPONTANEOUS

43. Conditions of G: H S Result neg pos spontaneous at all temps pos pos spontaneous at high temps neg neg spontaneous at low temps pos neg not spontaneous, ever

44. Relationship to K and E: G K E 0 at equilibrium; K = 1 0 negative >1, products favored positive positive <1, reactants favored negative

45. Summary Signs are important! DS + increased disorder - decreased disorder DH + endothermic -exothermic DG + nonspontaneous - spontaneous 0 @ equilibrium

46. Summary of Thermodynamic terms

47. 3 equations D?rxn = ∑ ?f products - ∑ ?f reactants DG = DH - TDS DG = -RTlnK

49. Enthalpy change ΔH o = ∑ Δ Hof,products - ∑ ΔHof, reactants Δ Hrxn = ∑ bonds broken - ∑ bonds formed

50. Enthalpy ΔS = ∑ Sproducts - ∑ Sreactants ΔG = ΔH - TΔS At equilibrium ΔG = 0, therefore ΔS = ΔH / T