1 / 46

Ch. 16: Spontaneity, Entropy, and Free Energy

Ch. 16: Spontaneity, Entropy, and Free Energy. 16.1 Spontaneous Processes and Entropy. Spontaneous. Occurs without outside intervention Thermodynamics tell us the _____________ not the speed Thermodynamics only consider initial and final states, not pathway

Download Presentation

Ch. 16: Spontaneity, Entropy, and Free Energy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Ch. 16: Spontaneity, Entropy, and Free Energy 16.1 Spontaneous Processes and Entropy

  2. Spontaneous • Occurs without outside intervention • Thermodynamics tell us the _____________ not the speed • Thermodynamics only consider initial and final states, not pathway • Must use kinetics and thermodynamics to understand reaction completely • What makes a reaction spontaneous?

  3. Entropy • S - measure of disorder or ____________ • the driving force for a spontaneous reaction is an increase in entropy • Natural tendency: low to high S • Entropy also describes the number of possible positions of a molecule

  4. Example • Which has higher positional S? • Solid CO2 or gaseous CO2? • N2 gas at 1 atm or N2 gas at 0.01 atm? • Predict the sign of DS • Solid sugar is added to water • Iodine vapor condenses on cold surface to form crystals

  5. Ch. 16: Spontaneity, Entropy, and Free Energy 16.2 Entropy and 2nd law of thermodynamics

  6. 2nd Law of Thermodynamics • In any spontaneous reactions, there is always an increase in entropy of the universe ∆Suniv = ∆Ssys + ∆Ssurr • If ∆Ssys is negative, it can still be spont. as long as the ∆Ssurr is larger and positive • ∆Suniv > 0 : • ∆Suniv = 0 : • ∆Suniv < 0 :

  7. Ch. 16: Spontaneity, Entropy, and Free Energy 16.3 The Effect of Temperature on Spontaneity

  8. H2O(l)  H2O(g) • ∆Ssys has ____ sign b/c of the increase in # of positions • ∆Ssurr is determined mostly by heat flow • Vaporization is _______________ so it removes heat from the surroundings • So ____________ random motion of surroundings • ____________ ∆Ssurr

  9. Temperature’s Effects • If the ∆Ssys and ∆Ssurr have different signs, the ______________ determines the ∆Suniv • For vaporization of water • Above 100°C, ∆Suniv is ____________ • Below 100°C, ∆Suniv is ____________ • Impact of the transfer of heat will be greater at lower temperatures

  10. Determining ∆Ssurr • Sign of ∆Ssurr • depends on direction of heat flow • ∆Ssurr____ for exothermic reactions • ∆Ssurr____ for endothermic reactions • Magnitude of ∆Ssurr • Depends on temperature • Heat flow = ∆H at constant P • Very small at high T, increases as T decreases

  11. Summary

  12. Example • A process has a DH of +22 kJ and a DS of -13 J/K. At which temperatures is the process spontaneous? • if there is no subscript, DS = DSsys • DSuniv___ 0 to be spontaneous DSsys + DSsurr > 0

  13. Example • For methanol, the enthalpy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 213 J/K. What is the normal boiling point of methanol? • DSsys= 213 J/K and DH =71.8 kJ/mol K • at the boiling point, the vaporization begins to be spontaneous • DSuniv = 0 to be at bpt DSsys + DSsurr = 0

  14. Ch. 16: Spontaneity, Entropy, and Free Energy 16.4: Free Energy

  15. Free Energy • G: helps you determine the temperature dependence of spontaneity • G ≡ H – TS definition of G for constant T

  16. Free Energy and Entropy • So what sign would the ∆G of reaction with a + ∆Suniv have? So for a reaction at constant T and P:

  17. Example • H2O(s)  H2O(l) • If ∆H°=6.03x103 J/mol and ∆S°=22.1 J/mol.K and the temperature is at -10°C, is it spontaneous?

  18. Example 1 • What could be another way to check for spontaneity of a reaction? • Check to see if the ∆Suniv is positive • How can we solve for ∆Ssurr? • use ∆Ssurr= - ∆H°/T • At -10°C, is it spontaneous?

  19. Example 1 • Is it spontaneous at 0°C?

  20. Example 1 • Is it spontaneous at +10°C?

  21. Example 1 When ∆Suniv° is 0, will ∆G° always be 0 too? How do the signs of ∆Suniv° and ∆G° relate?

  22. Enthalpy and Entropy • When H and S are in opposition, the spontaneity depends on T • In opposition: +∆S and +∆H OR –∆S and -∆H • Exothermic direction is spontaneous at low T

  23. Example 2 • At what T is this reaction spontaneous, at 1 atm of pressure? • Br2(l)  Br2(g) (aka Boiling) • ∆H=31.0 kJ/mol, ∆S=93.0 J/molK • Looking for boiling point: ∆G=0= equilibrium

  24. Ch. 16: Spontaneity, Entropy, and Free Energy 16.5: Entropy Changes in Chemical Reactions

  25. What about Chemical Changes? • We have been working with only physical changes so far… • Compare using # of independent states • N2(g) + 3H2(g)  2NH3(g) • Entropy increases/decreases • Compare using # molecules in higher entropy states • 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) • Entropy increases/decreases

  26. Example • Predict the sign of ∆S for the following reactions: • CaCO3(s)  CaO(s) + CO2(g) • 2SO2(g) + O2(g)  2SO3(g)

  27. Assigning S values • Normally deal with changes in S but can assign actual S values to substances • 3rd law of thermodynamics: _______ for a perfect crystal at 0 K • A standard value to help us set S values • Appendix 4 contains S° for common substances at 298 K and 1 atm • Since S is state function, can find ∆S by subtracting final - initial

  28. Example 3 • Find the ∆S° at 25°C for: • 2NiS(s) + 3O2(g)  2SO2(g) + 2NiO(s) 53 205 248 38

  29. Ch. 16: Spontaneity, Entropy, and Free Energy 16.6: Free Energy and Chemical Reactions

  30. Standard Free Energy Change • ∆G°: change when reactants and products are both in standard states • Cannot measure directly but can calculate it from other measured values • Standard states: p. 260 chart • The more negative ∆G° is, the further a reaction’s ____________ position lies to the _________

  31. 3 Ways to Calculate ∆G° • _________________________________ • Can calculate ∆ H ° and ∆ S ° using appendix and products – reactants • Then put in the equation to find ∆G° • ____________________ • Rearrange equations to get the goal equation • Add the ∆G° values for the equations • ___________ : standard free energy of formation

  32. Ch. 16: Spontaneity, Entropy, and Free Energy 16.7: Free Energy and Pressure

  33. Dependence on Pressure • ∆H is not dependent on the pressure of the system but ∆S is- Why? • ∆ G is dependent on both ∆ H and ∆ S so is also dependent on pressure • can calculate the ∆G of a substance at a pressure other than 1 atm using: • where • ∆G° is the change in free energy at 1 atm and ∆G is the change in free energy at the P • where R (_______________) and T is Kelvin

  34. Dependence on Pressure • Can also determine the ∆G for a reaction using the reaction quotient (Q) • where R is gas law constant (8.314 J/molK) and Q is the reaction quotient in pressures (gases only) or concentrations • aka [Products/Reactants]

  35. Example • Find the ∆G for the following reaction at 25°C and with CO at 5.0 atm and H2 at 3.0 atm. CO(g) + 2H2(g)  CH3OH(l) • must find ∆G° first • using product and reactants values from appendix • calculate Q from pressure

  36. Example 1 CO(g) + 2H2(g)  CH3OH(l) • ∆G°= • ∆G= ∆G° + RT ln Q

  37. What does ∆G mean? • a negative ∆G means • _ • _

  38. Ch. 16: Spontaneity, Entropy, and Free Energy 16.8: Free Energy and Equilibrium

  39. Equilibrium • no matter how much of the reactants or products are initially mixed, at a given set of conditions, the equilibrium position (K) will be the same • at equilibrium, ∆G= 0 and Q= K

  40. Equilibrium • ∆G°=0: • K=1 • G°react= G°prod • ∆G°<0: • K>1 • G°react> G°prod • ∆G°>0: • K<1 • G°react< G°prod

  41. Example 2 N2(g) + 3H2(g)  2NH3(g) • For the reaction above, ∆G° =-33.3 kJ/mol. For each of the mixtures below at 25°C, predict the direction in which the system will shift to reach equilibrium • pNH3 = 1.00 atm, PN2 = 1.47, PH2 = 0.0100 • ∆G = ∆G° + RT ln K • ∆G =

  42. Example 2 • pNH3 = 1.00 atm, PN2 = 1.00, PH2 = 1.00 • ∆G = ∆G° + RT ln K • ∆G=

  43. Temperature Dependence of K • A plot of ln K vs. 1/T is linear where • y = ln K and x = 1/T (in Kelvin) • m = -∆H°/R • b = ∆ S°/R

  44. Ch. 16: Spontaneity, Entropy, and Free Energy 16.9: Free Energy and Work

  45. Work • Can calculate the maximum amount of work that can be done by a process • wmax = ∆G • if the ∆G is +, then w is the amount of work expended to make a process occur • impossible to achieve maximum work because of wasted energy

More Related