1 / 36

Entropy and Free Energy (Kotz Ch 20) - Lecture #2

Entropy and Free Energy (Kotz Ch 20) - Lecture #2. Spontaneous vs. non-spontaneous thermodynamics vs. kinetics entropy = randomness (S o ). Gibbs free energy ( G o ) G o for reactions - predicting spontaneous direction thermodynamics of coupled reactions G rxn versus G o rxn

Download Presentation

Entropy and Free Energy (Kotz Ch 20) - Lecture #2

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. Entropy and Free Energy(Kotz Ch 20) - Lecture #2 • Spontaneous vs. non-spontaneous • thermodynamics vs. kinetics • entropy = randomness (So) • Gibbs free energy (Go) • Go for reactions - predicting spontaneous direction • thermodynamics of coupled reactions • Grxn versus Gorxn • predicting equilibrium constants from Gorxn Entropy & Free Energy (Ch 20) - lect. 2

  2. Entropy and Free Energy ( Kotz Ch 20 ) • some processes are spontaneous; others never occur. WHY ? • How can we predict if a reaction can occur, given enough time? THERMODYNAMICS 9-paper.mov 20m02vd1.mov • Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur. • To predict if a reaction can occur at a reasonable rate, one needs to consider: KINETICS Entropy & Free Energy (Ch 20) - lect. 2

  3. Product-Favored Reactions In general, product-favored reactions are exothermic. E.g. thermite reaction Fe2O3(s) + 2 Al(s)  2 Fe(s) + Al2O3(s) DH = - 848 kJ Entropy & Free Energy (Ch 20) - lect. 2

  4. Non-exothermic spontaneous reactions But many spontaneous reactions or processes are endothermic . . . NH4NO3(s) + heat  NH4+ (aq) + NO3- (aq) Hsol = +25.7 kJ/mol or have H = 0 . . . Entropy & Free Energy (Ch 20) - lect. 2

  5. PROBABILITY - predictor of most stable state WHY DO PROCESSES with  H = 0 occur ? Consider expansion of gases to equal pressure: This is spontaneous because the final state, with equal # molecules in each flask, is much more probable than the initial state, with all molecules in flask 1, none in flask 2 SYSTEM CHANGES to state of HIGHER PROBABILITY For entropy-driven reactions - the more RANDOM state. Entropy & Free Energy (Ch 20) - lect. 2

  6. Standard Entropies, So • Every substance at a given temperature and in a specific phase has a well-defined Entropy • At 298o the entropy of a substance is called • So - with UNITS of J.K-1.mol-1 • The larger the value of So, the greater the degree of disorder or randomness • e.g. So (in J.K-1mol-1) : Br2 (liq) = 152.2 • Br2 (gas) = 245.5 • For any process: So =  So(final) -  So(initial) So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1 Entropy & Free Energy (Ch 20) - lect. 2

  7. Vapour Ice Water Entropy and Phase S (gases) > S (liquids) > S (solids) So (J/K•mol) H2O(gas) 188.8 H2O(liq) 69.9 H2O (s) 47.9 Entropy & Free Energy (Ch 20) - lect. 2

  8. Entropy and Temperature The entropy of a substance increases with temperature. Molecular motions of heptane at different temps. • Higher T means : • more randomness • larger S 9_heptane.mov 20m04an2.mov Entropy & Free Energy (Ch 20) - lect. 2

  9. So (J/K•mol) CH4 248.2 C2H6 336.1 C3H8 419.4 Entropy and complexity Increase in molecular complexity generally leads to increase in S. 9_alkmot.mov 20m04an3.mov Entropy & Free Energy (Ch 20) - lect. 2

  10. ion pairs So (J/K•mol) MgO Mg2+ / O2- 26.9 NaF Na+ / F- 51.5 Entropy of Ionic Substances • Ionic Solids : Entropy depends on extent of motion of ions. This depends on the strength of coulombic attraction. • Entropy increases when a pure liquid or solid dissolves in a solvent. NH4NO3(s)  NH4+ (aq) + NO3- (aq) Ssol = So(aq. ions) - So(s) = 259.8 - 151.1 = 108.7 J.K-1mol-1 Entropy & Free Energy (Ch 20) - lect. 2

  11. Entropy Change in a Phase Changes For a phase change, DS = q/T where q = heat transferred in phase change H2O (liq)  H2O(g) For vaporization of water: H = q = +40,700 J/mol Entropy & Free Energy (Ch 20) - lect. 2

  12. Calculating S for a Reaction DSo = S So (products) - S So (reactants) Consider 2 H2(g) + O2(g)  2 H2O(liq) DSo = 2 So (H2O) - [2 So (H2) + So (O2)] DSo = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)] DSo = -326.9 J/K Note that there is a decrease in Sbecause 3 mol of gas give 2 mol of liquid. If S DECREASES, why is this a SPONTANEOUS REACTION?? Entropy & Free Energy (Ch 20) - lect. 2

  13. 2nd Law of Thermodynamics A reaction is spontaneous (product-favored) if S for the universe is positive. DSuniverse = DSsystem + DSsurroundings DSuniverse > 0 for product-favoredprocess First calc. entropy created by matter dispersal (DSsystem) Next, calc. entropy created by energy dispersal (DSsurround) Entropy & Free Energy (Ch 20) - lect. 2

  14. Calculating S(universe) 2 H2(g) + O2(g)  2 H2O(liq) DSosystem = -326.9 J/K Can calculate that DHorxn = DHosystem = -571.7 kJ DSosurroundings = +1917 J/K Entropy & Free Energy (Ch 20) - lect. 2

  15. Calculating S(universe) (2) 2 H2(g) + O2(g)  2 H2O(liq) DSosystem = -326.9 J/K(less matter dispersal) DSosurroundings = +1917 J/K (more energy dispersal) DSouniverse = +1590 J/K The entropy of the universe increases so the reaction is spontaneous. Entropy & Free Energy (Ch 20) - lect. 2

  16. The Laws of Thermodynamics 0. Two bodies in thermal equilibrium are at same T 1. Energy can never be created or destroyed.  E = q + w 2. The total entropy of the UNIVERSE ( = system plus surroundings) MUST INCREASE in every spontaneous process.  STOTAL =  Ssystem +  Ssurroundings > 0 3. The entropy (S) of a pure, perfectly crystalline compound at T = 0 K is ZERO. (no disorder) ST=0 = 0 (perfect xll) Entropy & Free Energy (Ch 20) - lect. 2

  17. Gibbs Free Energy, G DSuniv = DSsurr + DSsys Multiply through by -T -TDSuniv = DHsys - TDSsys -TDSuniv = change in Gibbs free energy for the system = DGsystem Under standard conditions — The Gibbs Equation DGo = DHo - TDSo Entropy & Free Energy (Ch 20) - lect. 2

  18. Standard Gibbs Free Energies, Gof • Every substance in a specific state has a • Gibbs Free Energy, G = H - TS • recall: only H can be measured. Therefore: • there is no absolute scale for G • only G values can be determined • DGof the Gibbs Free Energy of formation (from elements) is used as the “standard value” • We set the scale of G to be consistent with that for H - DGof for elements in standard states = 0 Entropy & Free Energy (Ch 20) - lect. 2

  19. Ho-TS < 0 and Go = Ho-TS Sign of DG for Spontaneous processes 2nd LAW requirement for SPONTANEITY is : STOTAL = Ssystem + Ssurroundings > 0 Multiply by T TSsystem + TSsurroundings > 0 andSsurroundings = Hosystem/T Thus TSsystem - Hosystem > 0 Multiply by -1 (->reverse > to <), drop subscript “system” DGo< 0for all SPONTANEOUS processes Entropy & Free Energy (Ch 20) - lect. 2

  20. Sign of Gibbs Free Energy, G DGo = DHo - TDSo • change in Gibbs free energy = (total free energy change for system - free energy lost in disordering the system) • If reaction isexothermic (DHo is -ve)and entropy increases (DSo is +ve), then DGomust be -ve and reaction CAN proceed. • If reaction is endothermic (DHo is +ve), and entropy decreases (DSo is -ve), then DGomust be +ve; reaction CANNOT proceed. Entropy & Free Energy (Ch 20) - lect. 2

  21. Gibbs Free Energy changes for reactions DGo = DHo - TDSo DHoDSoDGo Reaction exo (-) increase(+) - Product-favored endo(+) decrease(-) + Reactant-favored exo (-) decrease(-) ? T dependent endo(+) increase(+) ? T dependent Spontaneous in last 2 cases only if Temperature is such that Go < 0 Entropy & Free Energy (Ch 20) - lect. 2

  22. Methods of calculating G DGo = DHo - TDSo Two methods of calculating DGo a) Determine DHorxn and DSorxn and use Gibbs equation. b) Use tabulated values of free energies of formation, DGfo. Gorxn = SGfo (products) - S Gfo (reactants) Entropy & Free Energy (Ch 20) - lect. 2

  23. Calculating DGorxn EXAMPLE: Combustion of acetylene C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g) From standard enthalpies of formation:DHorxn = -1238 kJ From standard molar entropies: DSorxn = - 0.0974 kJ/K Calculate DGorxnfrom DGo = Ho - TSo DGorxn = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1209 kJ Reaction is product-favored in spite of negative DSorxn. Reaction is “enthalpy driven” Entropy & Free Energy (Ch 20) - lect. 2

  24. Calculating DGorxn for NH4NO3(s) EXAMPLE 2: NH4NO3(s)  NH4NO3(aq) Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven? 9_amnit.mov 20 m07vd1.mov Entropy & Free Energy (Ch 20) - lect. 2

  25. DGorxn for NH4NO3(s)  NH4NO3(aq) From tables of thermodynamic data we find DHorxn = +25.7 kJ DSorxn = +108.7 J/K or +0.1087 kJ/K DGorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ Reaction is product-favored . . . in spite of positive DHorxn. Reaction is “entropy driven” Entropy & Free Energy (Ch 20) - lect. 2

  26. Calculating DGorxn DGorxn = SDGfo (products) - SDGfo (reactants) EXAMPLE 3: Combustion of carbon C(graphite) + O2(g)  CO2(g) DGorxn = DGfo(CO2) - [DGfo(graph) + DGfo(O2)] DGorxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. DGorxn = -394.4 kJ Reaction is product-favored as expected. Entropy & Free Energy (Ch 20) - lect. 2

  27. Free Energy and Temperature 2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g) DHorxn = +467.9 kJDSorxn = +560.3 J/K DGorxn = 467.9 kJ - (298K)(0.560kJ/K) = +300.8 kJ Reaction is reactant-favored at 298 K At what T does DGorxnjust change from (+) to (-)? i.e. what is T for DGorxn = 0 = DHorxn - TDSorxn If DGorxn = 0 then DHorxn = TDSorxn so T = DHo/DSo ~ 468kJ/0.56kJ/K = 836 K or 563oC Entropy & Free Energy (Ch 20) - lect. 2

  28. Go for COUPLED CHEMICAL REACTIONS Reduction of iron oxide by CO is an example of using TWO reactions coupled to each other in order to drive a thermodynamically forbidden reaction: Fe2O3(s)  4 Fe(s) + 3/2 O2(g) DGorxn = +742 kJ with a thermodynamically allowed reaction: 3/2 C(s) + 3/2 O2 (g)  3/2 CO2(g) DGorxn = -592 kJ Overall : Fe2O3(s) + 3/2 C(s)  2 Fe(s) + 3/2 CO2(g) DGorxn= +301 kJ @ 25oC BUTDGorxn < 0 kJ for T > 563oC See Kotz, pp933-935 for analysis of the thermite reaction Entropy & Free Energy (Ch 20) - lect. 2

  29. Other examples of coupled reactions: Copper smelting Cu2S (s)  2 Cu (s) + S (s) DGorxn= +86.2 kJ (FORBIDDEN) Couple this with: S (s) + O2 (g)  SO2 (s) DGorxn= -300.1 kJ Overall: Cu2S (s) + O2 (g)  2 Cu (s) + SO2 (s) DGorxn= +86.2 kJ + -300.1 kJ = -213.9 kJ (ALLOWED) Coupled reactions VERY COMMON in Biochemistry : e.g. all bio-synthesis driven by ATP  ADP for whichDHorxn = -20 kJ DSorxn = +34 J/K DGorxn = -30 kJ @ 37oC Entropy & Free Energy (Ch 20) - lect. 2

  30. Thermodynamics and Keq • Keq is related to reaction favorability. • If DGorxn < 0, reaction is product-favored. • DGorxn is the change in free energy as reactants convert completely to products. • But systems often reach a state of equilibrium in which reactants have not converted completely to products. • How to describe thermodynamically ? Entropy & Free Energy (Ch 20) - lect. 2

  31. Grxn versus Gorxn Under any condition of a reacting system, we can define Grxn in terms of the REACTION QUOTIENT, Q Grxn =Gorxn + RT ln Q If Grxn < 0 then reaction proceeds to right If Grxn > 0 then reaction proceeds to left At equilibrium, Grxn = 0. Also, Q = K. Thus DGorxn = - RT lnK Entropy & Free Energy (Ch 20) - lect. 2

  32. Thermodynamics and Keq (2) 2 NO2 N2O4 DGorxn = -4.8 kJ • pure NO2 has DGrxn < 0. • Reaction proceeds until DGrxn = 0 - the minimum in G(reaction) - see graph. • At this point, both N2O4 and NO2 are present, with more N2O4. • This is a product-favored reaction. 9_G_NO2.mov 20m09an1.mov Entropy & Free Energy (Ch 20) - lect. 2

  33. Thermodynamics and Keq (3) N2O42 NO2DGorxn = +4.8 kJ • pure N2O4 has DGrxn < 0. • Reaction proceeds until DGrxn = 0 - the minimum in G(reaction) - see graph. • At this point, both N2O4 and NO2 are present, with more NO2. • This is a reactant-favored reaction. 9_G_N2O4.mov 20m09an2.mov Entropy & Free Energy (Ch 20) - lect. 2

  34. Thermodynamics and Keq (4) Keq is related to reaction favorability and so to DGorxn. The larger the value of DGorxn the larger the value of K. DGorxn = - RT lnK where R = 8.31 J/K•mol Entropy & Free Energy (Ch 20) - lect. 2

  35. Thermodynamics and Keq (5) DGorxn = - RT lnK Calculate K for the reaction N2O4  2 NO2DGorxn = +4.8 kJ DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K K = 0.14 When Gorxn > 0, then K < 1 - reactant favoured When Gorxn < 0, then K >1 - product favoured Entropy & Free Energy (Ch 20) - lect. 2

  36. Entropy and Free Energy(Kotz Ch 20) • Spontaneous vs. non-spontaneous • thermodynamics vs. kinetics • entropy = randomness (So) • Gibbs free energy (Go) • Go for reactions - predicting spontaneous direction • thermodynamics of coupled reactions • Grxn versus Gorxn • predicting equilibrium constants from Gorxn Entropy & Free Energy (Ch 20) - lect. 2

More Related