1 / 53

Gibbs Free Energy Temperature and Spontaneity Free Energy and Equilibrium

Learn about Gibbs free energy, entropy, and temperature in thermodynamics, and how to predict the spontaneity of reactions using DG0 values.

lillier
Télécharger la présentation

Gibbs Free Energy Temperature and Spontaneity Free Energy and Equilibrium

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. Gibbs Free Energy Temperature and Spontaneity Free Energy and Equilibrium Thermodynamics of Living Systems

  2. Thermodynamics Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv > 0 surrounding Equilibrium : DSuniv = 0 system Nonspontaneous process: DSuniv < 0 If something happens the total entropy of the universe increases! All we need to know is DSuniv

  3. Predicting Spontaneity - S mS0(reactants) S nS0(products) = DS0 rxn If we mix reactants and products together will the reaction occur? Equilibrium : DSuniv = 0 surrounding Nonspontaneous process: DSuniv < 0 system Spontaneous process: DSuniv > 0 If we know DSsurr we can calculateDSuniv DSuniv = DSsys + DSsurr Can calculate from:

  4. Predicting Spontaneity - S mS0(reactants) S nS0(products) = DS0 rxn If we know DSsurr we can calculateDSuniv DSuniv = DSsys + DSsurr Can calculate from: surrounding Surrounding = everything in the universe except the system. DSsurr DS0rxn • “Impossible” to measure! system

  5. Heat and Entropy Endothermic Process Exothermic Reaction • +DHsys • 0 <DHsys • -DHsys • 0 >DHsys Surroundings - heat = ↓ S Surroundings + heat = ↑ S • DSsurr < 0 • DSsurr > 0 • DSsurr∝ -DHsys

  6. Heat and Entropy Heat released by the system increases the disorder of the surroundings. • DSsurr∝ -DHsys The effect of -DHsys on the surroundings depends on temperature: • At high temperature, where there is already considerable disorder, the effect is muted • At low temperature the effect is much more significant • The difference between tossing a rock into a calm pool (low T) and a storm-tossed ocean (high T) • -DHsys • DSsurr = T

  7. Predicting Spontaneity DSuniv = DSsys + DSsurr • -DHsys • DSsurr = • -DHsys DSuniv = DSsys + Substitution: T T • -TDSuniv = -TDSsys + DHsys Multiply by -T: -TDSuniv = DHsys- TDSsys Rearrange: This equation relates DSuniv to DHsysandDSsys. Both in terms of the system.

  8. Gibbs Free Energy -TDSuniv = DHsys- TDSsys DG = DHsys- TDSsys • Gibbs free energy (DG)- • AKA Free energy. • Relates S, H and T of a system. • Can be used to predict spontaneity. • G is a state function. Josiah Willard Gibbs (1839-1903) First American Ph.D. in Engineering (Yale, 1863) Praised by Albert Einstein as "the greatest mind in American history"

  9. Gibbs Free Energy For a constant temperature and constant pressure process: -TDSuniv = DHsys- TDSsys DG = DHsys- TDSsys Gibbs free energy (DG)- Can be used to predict spontaneity. DG < 0 The reaction is spontaneous in the forward direction. • -DG = -T(+DSuniv) • DSuniv > 0 DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. • +DG = -T(-DSuniv) • DSuniv < 0 DG = 0 The reaction is at equilibrium. • DG = -T(-DSuniv) = 0 • DSuniv = 0

  10. Gibbs Free Energy DG = DH- TDS • If you know DG for reactants and products then you can calculate if a reaction is spontaneous. • If you know DG for two reaction then you can calculate if the sum is spontaneous. • If you know DS, DH and T then you can calculate spontaneity. • Can predict the temperature when a reaction becomes spontaneous. • If you have DHvap or DHfus and DS you can predict boiling and freezing points. • If you have DHvap or DHfus and T you can predict the entropy change during a phase change. • Can predict equilibrium shifts.

  11. The standardfree-energy of reaction (DG0 )is the free-energy change for a reaction when it occurs under standard-state conditions. rxn Standard Free Energy Changes aA + bBcC + dD - [ + ] [ + ] = - mDG0 (reactants) S S = f DG0 DG0 rxn rxn Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f DG0 of any element in its stable form is zero. f dDG0 (D) nDG0 (products) aDG0 (A) cDG0 (C) bDG0(B) f f f f f

  12. Standard Free Energy Changes aA + bBcC + dD S = - S mDG0 (reactants) f DG0 rxn Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f nDG0 (products) f DG is a state function so free energy can be calculated from the table of standard values just as enthalpy and entropy changes.

  13. DG° Calculations S = mDG0 (reactants) - S f DG0 rxn nDG0 (products) f Appendix 3 To predict spontaneity of any rxn: Pick any reactants and products. Write a balanced equation. Calculate DG0rxn. Is the reaction spontaneous?

  14. Standard Free Energy Changes - mDG0 (reactants) S S = f DG0 rxn nDG0 (products) f • Calculate the standard free-energy change for the following reaction: From appendix 3: 2KClO3(s) 2KCl(s)+ 3O2(g) • KClO3(s)DGf = -289.9 kJ/mol • KCl(s)DGf = -408.3 kJ/mol • O2(g)DGf = 0 kJ/mol DG0rxn= [2(408.3 kJ/mol) + 3(0)]  [2(289.9 kJ/mol)] DG0rxn = 816.6  (579.8) DG0rxn = 236.8 kJ/mol Is the reaction spontaneous? DG0rxn < 0 Yes!

  15. Another Example From appendix 3: + O2 • C, diamond(s)DGf = 2.9 kJ/mol • O2(g)DGf = 0 kJ/mol • CO2(g)DGf = -394.4kJ/mol ΔG°rxn = [ΔG°f (CO2)] - [ΔG°f (C, diamond) + ΔG°f (O2)] C(s, diamond) + O2(g) CO2(g) DG°rxn = -397.3 kJ Is the reaction spontaneous? very slowly Therefore, diamonds are contributing to climate change!

  16. More DG° Calculations Similar to DH°, one can use the DG° for various reactions to determine DG° for the reaction of interest (a “Hess’ Law” for DG°) Hess’ Law-states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.  What is the DG° for this reaction: Given: C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ C(s, graphite) + O2(g) CO2(g) DG° = -394 kJ

  17. More DG° Calculations What is the DG° for this reaction: Given: C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ C(s, graphite) + O2(g) CO2(g) DG° = -394 kJ CO2(g) C(s, graphite) + O2(g) DG° = +394 kJ C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ CO2(g) C(s, graphite) + O2(g) DG° = +394 kJ C(s, diamond) C(s, graphite) DG° = -3 kJ DG°rxn < 0…..rxn is spontaneous

  18. Alternative DG Calculation Is the following reaction spontaneous at 298 K? (assume standard conditions) Given: Given DG°rxn = DH°rxn- TDS°rxn Then Find First Calculate

  19. Alternative DG Calculation

  20. Alternative DG Calculation DS°rxn = -36.8 J/K DH°rxn = -144 kJ Entropically unfavorable Enthalpically favorable What about at 5000 K? (Assuming the same DH and S) DG°rxn = 50 kJ DG°rxn > 0…rxn is nonspontaneous at 5000 K DG°rxn < 0…..rxn is spontaneous at 298 K

  21. Gibbs Free Energy Temperature and Spontaneity Free Energy and Equilibrium Thermodynamics of Living Systems

  22. DG Temperature Dependence Spontaneous DG°rxn = -133 kJ at 298 K DG°rxn = 50 kJ at 5000 K Nonspontaneous Reaction spontaneity is a temperature dependent phenomenon! H2O(l) H2O(s) DGrxn = DHrxn- TDSrxn

  23. DG Temperature Dependence Enthalpy: DHrxn < 0 The reaction is enthalpically favorable. Entropy: DSrxn > 0 The reaction is entropically favorable. Need both to predict spontaneity. And sometimes temperature! DG < 0 Spontaneous DG = DH- TDS DG > 0 Nonspontaneous DG = 0 Equilibrium 1) If DH < 0 and DS > 0, then DG is negative at all T 2) If DH > 0 and DS > 0, then DG depends on T 3) If DH < 0 and DS < 0, then DG depends on T • 4) If DH > 0 and DS < 0, then DG is positive at all T

  24. DG Temperature Dependence DG < 0 Spontaneous DG = DH- TDS DG > 0 Nonspontaneous DG = 0 Equilibrium 3) DH < 0 and DS < 0 (Enthalpically favorable, entropically unfavorable) If DH < TDS, then DG is positive. If DH > TDS, then DG is negative. Nonspontaneous at high T Spontaneous at low T (Enthalpically unfavorable, entropically favorable) 2) DH > 0 and DS > 0 If DH < TDS, then DG is negative. If DH > TDS, then DG is positive. Spontaneous at high T Nonspontaneous at low T

  25. DG Temperature Dependence DG = DH - TDS

  26. Predicting T from Gibbs Equation Spontaneous DG°rxn = -133 kJ at 298 K DG°rxn = 50 kJ at 5000 K Nonspontaneous Reaction spontaneity is a temperature dependent phenomenon! At what temperature will the reaction become spontaneous? Find T where DG changes from positive to negative. I.e. when DG =0. DG = DH – TDS = 0 T = DH/DS

  27. Predicting T from Gibbs Equation At what T is the following reaction spontaneous? Given:DH°= 30.91 kJ/mol DS°= 93.2 J/mol.K Br2(l) Br2(g) DG = DH – TDS = 0 T = DH/DS • H > 0, S > 0 • The reaction will be spontaneous when T > 331.7 K T = (30.91 kJ/mol) /(93.2 J/mol.K) T = 331.7 K

  28. Gibbs Free Energy Temperature and Spontaneity Free Energy and Equilibrium Thermodynamics of Living Systems

  29. The standardfree-energy of reaction (DG0 )is the free-energy change for a reaction when it occurs under standard-state conditions. rxn Free Energy and Equilibrium aA + bBcC + dD aA + bBcC + dD - mDG0 (reactants) S S = f DG0 rxn nDG0 (products) f DG° < 0 favors products spontaneously DG° > 0 favors reactants spontaneously Does not tell you it will go to completion! DG° fwd DG° fwd = -DG° rev DG° rev The value of G° calculated under the standard conditions characterizes the “driving force” of the reaction towards equilibrium.

  30. Free Energy and Equilibrium DG° = -R T lnK equilibrium constant (Kp, Kc, Ka, Ksp, etc.) • Arguably most important equation in chemical thermodynamics! • It allows us to calculate the extent of a chemical reaction if its enthalpy and entropy changes are known. • The changes in enthalpy and entropy can be evaluated by measuring the variation of the equilibrium constant with temperature. • This relationship is only valid for the standard conditions, i.e. when the activities of all reactants and products are equal to 1. standardfree-energy (kJ/mol) temperature (K) gas constant (8.314 J/Kmol)

  31. Free Energy and Equilibrium Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent FORWARD REACTION REVERSE REACTION Forward reaction goes to completion; essentially no reverse reaction DG° = -R T lnK R is constant so at a given temperature: DG0(kJ) K Significance 200 9x10-36 100 3x10-18 50 2x10-9 10 2x10-2 1 7x10-1 0 1 -1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035

  32. Using DG° and K - mDG0 (reactants) S S = f DG0 rxn nDG0 (products) f Using the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C. From appendix 3: • H2(g)DGf = 0 kJ/mol • Cl2(s)DGf = 0 kJ/mol • HCl(g)DGf = -95.3 kJ/mol • 2HCl(g)H2(g)+ Cl2(g) DG0rxn= [1(0) + 1(0)] - [2(95.3 kJ/mol)] Non-spontaneous! Favors reactants! DG0rxn = 190.6 kJ/mol DG° = -R T ln K

  33. Using DG° and K - mDG0 (reactants) S S = f DG0 rxn nDG0 (products) f Using the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C. • 2HCl(g)H2(g)+ Cl2(g) DG0rxn= [1(0) + 1(0)] - [2(95.3 kJ/mol)] Non-spontaneous! Favors reactants! DG0rxn = 190.6 kJ/mol DG° = -R T ln K • 190.6 kJ/mol =  (8.314 J/K·mol)(25C) lnKP correct units • 190.6 kJ/mol =  (8.314 x 103 kJ/K·mol)(298 K) lnKP KP = 3.98 x 1034 Favors reactants!

  34. Free Energy and Equilibrium DG° = -R T lnK DG° DH - TDS - - • ln K = • ln K = Rearrange: R T R T Substitution: DG = DH- TDS DH TDS - • ln K = Rearrange: + R T R T 1 DH • DS () - • ln K = + R T R

  35. Free Energy and Equilibrium DG° = -R T lnK Measure equilibrium with respect to temperature: DG° DH - TDS - - • ln K = • ln K = Rearrange: R T R T Substitution: DG = DH- TDS DH TDS - • ln K = Rearrange: + R T R T 1 DH • DS () - • ln K = + R T R y = m • x + b

  36. Free Energy and Equilibrium Find the DS and DH of the following: kobs kf rateAB = kobs [A] rateBA = kf [B] At equilibrium: rateAB = rateBA kobs [A] = kf [B] A B kobs [B] = Ku = kf [A]

  37. Free Energy and Equilibrium Find the DS and DH of the following: kobs kf DH° = 60 kJ/mol DS° = 200 J/Kmol • DS R Slope = -7261.1 K DH - R

  38. DG° vsDG You have already delta ΔG°,  ΔS° and ΔH°in which the ° indicates that all components are in their standard states.  Definition of “standard”: • Even if we start a reaction at standard conditions (1 M) the reaction will quickly deviate from standard. • DG° indicates whether reactants or products are favored at equilibrium. • DG at any give time is used to predict the direction shift to reach equilibrium. • If a mixture is not at equilibrium, the liberation of the excess Gibbs free energy (DG) is the “driving force” for the composition of the mixture to change until equilibrium is reached. *There is no "standard temperature", but we usually use 298.15 K (25° C).

  39. Free Energy and Equilibrium DG° = -R T ln K At equilibrium: DG =DG° + R T ln Q reaction quotient At any time: temperature (K) standardfree-energy (kJ/mol) non-standard free-energy (kJ/mol) gas constant (8.314 J/Kmol) • The sign of DG tells us that the reaction would have to shift to the left to reach equilibrium. DG < 0, reaction will shift right • DG > 0, reaction will shift left • DG = 0, the reaction is at equilibrium • The magnitude of DG tells us how far it has to go to reach equilibrium.

  40. Free Energy and Equilibrium DG =DG° + R T ln Q reaction quotient temperature (K) standardfree-energy (kJ/mol) non-standard free-energy (kJ/mol) gas constant (8.314 J/Kmol) If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0)

  41. Another Example For the following reaction at 298 K: From appendix 3: H2(g)+ Cl2(g)2 HCl(g) • H2(g)DGf = 0 kJ/mol • Cl2(s)DGf = 0 kJ/mol • HCl(g)DGf = -95.3 kJ/mol Given: • H2 = 0.25 atm • Cl2 = 0.45 atm HCl = 0.30 atm Which way will the reaction shift to reach equilibrium? DG =DG° + R T ln Q calculate constant calculate given

  42. Another Example For the following reaction at 298 K: From appendix 3: H2(g)+ Cl2(g)2 HCl(g) • H2(g)DGf = 0 kJ/mol • Cl2(s)DGf = 0 kJ/mol • HCl(g)DGf = -95.3 kJ/mol Given: • H2 = 0.25 atm • Cl2 = 0.45 atm HCl = 0.30 atm Which way will the reaction shift to reach equilibrium? DG =DG° + R T ln Q • G° = [2(95.27 kJ/mol)]  [0 + 0] • = 190.54 kJ/mol

  43. Another Example For the following reaction at 298 K: From appendix 3: H2(g)+ Cl2(g)2 HCl(g) • H2(g)DGf = 0 kJ/mol • Cl2(s)DGf = 0 kJ/mol • HCl(g)DGf = -95.3 kJ/mol Given: • H2 = 0.25 atm • Cl2 = 0.45 atm HCl = 0.30 atm Which way will the reaction shift to reach equilibrium? DG =DG° + R T ln Q • G° = 190.54 kJ/mol Q = 0.80 constant given G = 190,540 J/mol + (8.314J/K·mol)(298 K) ln (0.80) G = 191.09 kJ/mol Because ΔG < 0, the net reaction proceeds from left to right to reach equilibrium.

  44. Gibbs Free Energy Temperature and Spontaneity Free Energy and Equilibrium Thermodynamics of Living Systems

  45. “Uphill” Reactions Synthesis of proteins: (first step) alanine + glycine alanylglycine G° = 29 kJ/mol Because ΔG > 0, the reaction is non-spontaneous. No reaction! alanine glycine Need to couple two reactions!

  46. Coupled Reactions Coupled Reactions- using a thermodynamically favorable reaction (G° < 0) to drive an unfavorable one (G° > 0) . Example: Industrial ore separation- Zinc Metal Major applications in the US Galvanizing (55%) Alloys (21%) Brass and bronze (16%) Miscellaneous (8%) Sphalerite ore White pigment (ZnO) Fire retardant (ZnCl2) Vitamin supplement (Zn2+) Reducing agent (Zn(s)) We need 2000 tones of the zinc metal per year!

  47. Coupled Reactions Coupled Reactions- using a thermodynamically favorable (G° < 0) reaction (G° < 0) to drive an unfavorable one (G° > 0) . Example: Industrial ore separation- Unfavorable reaction (G° > 0) 95 % of Zinc is produced by this method

  48. Coupled Reactions in Biology G° < 0 G° > 0 glucose + Pi → glucose-6-phosphate ATP + H2O → ADP + Pi glucose + ATP → glucose-6-phosphate + ADP

  49. Coupled Reactions in Biology ? Food Structural motion and maintenance Coupled reactions Fats and Carbohydrates ATP and NADPH Chemical Batteries for the Body Stored bond energy

  50. Coupled Reactions in Biology Digestion/respiration: Generation of ATP: Burning Glucose Low Energy Higher Energy

More Related