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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

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

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  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

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