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Thermodynamics

Production of quicklime. Thermodynamics. Liquid benzene. ⇅. Solid benzene. Chapter 19. CaCO 3 (s) ⇌ CaO + CO 2. Gibbs Energy. For a constant-pressure & constant temperature process:. Gibbs energy (G). D G = D H sys - T D S sys.

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Thermodynamics

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  1. Production of quicklime Thermodynamics Liquid benzene ⇅ Solid benzene Chapter 19 CaCO3 (s) ⇌ CaO + CO2

  2. Gibbs Energy For a constant-pressure & constant temperature process: Gibbs energy (G) DG = DHsys - TDSsys DG < 0 The reaction is spontaneous in the forward direction DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction DG = 0 The reaction is at equilibrium

  3. Fig 19.17 Analogy between Potential Energy and Free Energy

  4. Fig 19.18 Free Energy and Equilibrium

  5. Standard free-energy of reaction (DGorxn)≡ free-energy change for a reaction when it occurs under standard-state conditions. aA + bB cC + dD - mDG° (reactants) S S = f DG° rxn nDG° (products) f • Standard free energy of formation (DG°) • Free-energy change that occurs • when 1 mole of the compound • is formed from its elements • in their standard states. f

  6. What’s “Free” About Gibbs Energy? • ΔG ≡ the theoretical maximum amount of work that can be • done by the system on the surroundings at constant P and T • ΔG = − wmax Fig 19.19 Energy Conversion

  7. What’s “free” about the Gibbs energy? • “Free” does not imply that the energy has no cost • For a constant-temperature process, “free energy” • is the amount available to do work • e.g., Human metabolism converts glucose to • CO2 and H2O with a ΔG° = -2880 kJ/mol • This energy represents approx. 688 Cal • or about two Snickers bars worth...

  8. The Haber process for the production of ammonia involves the equilibrium • Assume that ΔH° and ΔS° for this reaction do not change with temperature. • Predict the direction in which ΔG° for this reaction changes with increasing temperature. • (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C. Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity

  9. DG = DHsys - TDSsys • The temperature dependence of ΔG° comes from the entropy term. • We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. • Because ΔS° is negative, the term –T ΔS° is positive and grows larger with increasing temperature. • As a result, ΔG° becomes less negative (or more positive) with increasing temperature. • Thus, the driving force for the production of NH3 becomes smaller with increasing temperature.

  10. The Haber process for the production of ammonia involves the equilibrium • Assume that ΔH° and ΔS° for this reaction do not change with temperature. • Predict the direction in which ΔG° for this reaction changes with increasing temperature. • (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C. Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity

  11. DGo = DHsys - TDSsys • The reaction is spontaneous at 25 oC • The reaction is nonspontaneous at 500 oC

  12. Gibbs Free Energy and Chemical Equilibrium • We need to distinguish between ΔG and ΔG° • During the course of a chemical reaction, not all • products and reactants will be in their standard states • In this case, we use ΔG • When the system reaches equilibrium, the sign of ΔG° • tells us whether products or reactants are favored • What is the relationship between ΔG and ΔG°?

  13. Gibbs Free Energy and Chemical Equilibrium When not all products and reactants are in their standard states: ΔG = ΔG° + RT lnQ R ≡ gas constant (8.314 J/K•mol) T ≡ absolute temperature (K) Q ≡ reaction quotient = [products]o / [reactants]o At Equilibrium: Q = K ΔG = 0 0 = ΔG° + RT lnK ΔG° = −RT lnK

  14. DG° = -RT lnK or Table 19.5

  15. Example Calculate ΔG° for the following process at 25 °C: BaF2(s)⇌ Ba2+(aq) + 2 F−(aq); Ksp = 1.7 x 10-6 ΔG = 0 for any equilibrium, so: ΔG° = − RT ln Ksp ΔG° = − (8.314 J/mol∙K) (298 K) ln (1.7 x 10-6) ΔG° = + 32.9 kJ/mol ΔG° ≈ + 33 kJ/mol Equilibrium lies to the left

  16. Thermodynamics in living systems • Many biochemical reactions have a positive ΔGo • In living systems, these reactions are coupled to a • process with a negative ΔGo (coupled reactions) • The favorable rxn drives the unfavorable rxn

  17. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O (l) Metabolism of glucose in humans ΔG° = -2880 kJ/mol • Does not occur in a single step as it would in simple • combustion • Enzymes break glucose down step-wise • Free energy released used to synthesize ATP from ADP:

  18. Fig 19.20 Free Energy and Cell Metabolism ΔG° = +31 kJ/mol ADP + H3PO4→ ATP + H2O (Free energy stored)

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