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Thermodynamics

Thermodynamics. First Law – energy can never be created nor destroyed. (from CH 9) We want to be able to predict spontaneity: under given conditions, occurs without outside intervention Different than fast or slow!. Enthalpy to Predict Spontaneity?.

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Thermodynamics

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  1. Thermodynamics • First Law – energy can never be created nor destroyed. (from CH 9) • We want to be able to predict spontaneity: under given conditions, occurs without outside intervention • Different than fast or slow!

  2. Enthalpy to Predict Spontaneity? • C (s) + O2 (g)  CO2 (g) ΔHo = -393.5 kJ • This process is exothermic and spontaneous. • Most exothermic processes are spontaneous. • H2O (s)  H2O (l) ΔHo = 6.01 kJ • This process is endothermic and spontaneous. • Dissolving most ionic compounds in water is endothermic and spontaneous. • However, many (most?) endothermic processes are not spontaneous. • Enthalpy alone will not predict spontaneity.

  3. DH = 0 Spontaneous Non-spontaneous

  4. Entropy • Entropy – S – How energy is distributed among energy levels of particles in universe • disorder of a system • Consider placing cards in order • If you have 2 cards, there is a 1 in 2 chance they will be in order. (2!) • If you have 13 cards (1 suit), there is a 1 in 6 x 109 chance they will be in order. (13!) • If you have 52 cards, there is a 1 in 8 x 1067 chance they will be in order. (52!)

  5. Probability and Microstates Room of unnamed S117 student More than disorder: Which would require more energy starting with a empty room: to put everything in order orto make it look exactly like this?

  6. Probability and Microstates • Positional Probability • Gas expansion • Phase change • solubility

  7. Entropy and Positional Probability • ΔS = Sf – Si • Sgas >> Sliquid > Ssolid • S increases when a solute (solid or liquid) is dissolved. • Sdecreases when a solute (gas) is dissolved. • Sincreases as temperature increases.

  8. Processes Which Increase Entropy

  9. Change in Entropy • For these systems, predict the sign of the change in entropy (based positional prob.) • N2 (g) + 3H2 (g)  2NH3 (g) • AgCl (s)  Ag+ (aq) + Cl- (aq) • Unfolded protein  folded protein

  10. Isothermal Expansion of Ideal Gas • DE = 0 • q = -w • Work and heat are path dependent • One step: free expansion • One step: against mass of 1/4m • Two step, etc

  11. Work obtained from Expansion What is work for 1 step expansion against m = 1/4mi? Answer: -3/4 PDV 2 steps 6 steps W = -PDV W = -1.4PDV

  12. Work Required for Compression Draw PV diagram for 1 step against M, then 2 step against M/2 followed by M P P V V W = 3PDV W = 1.4PDV W = 2PDV

  13. Reversible Process • Reversible process: infinite number of steps (100 % efficient) • One step: 3PV - .75PV to get back to same place • Surroundings gain 2.25PV of work in form of heat • Ordered energy into disordered energy

  14. Real Process • More work into system than you can get out • But no change in universe, so that means more heat into surroundings • Cycle: increase entropy of universe Expansion: heat Compression heat

  15. Entropy Entropy is statistical DS = k ln Entropy is macroscopic DS = nRln DS = k = R/Na From

  16. Change of State:Equilibrium Process DS = DS = DS =

  17. Second Law • Second Law – In any spontaneous process, the entropy of the universe increases. • ΔSuniv= ΔSsystem + ΔSsurroundings > 0 • ΔSuniv > 0 , forward reaction is spontaneous • ΔSuniv < 0 , reverse reaction is spontaneous • ΔSuniv = 0 , equilibrium

  18. Temperature Dependence • ΔSuniv= ΔSsystem + ΔSsurroundings • Entropy change in the surroundings primarily due to heat flow • ΔSuniv= ΔSsystem + ΔHsurroundings/T Exothermicity is more important at low temperature. Explain this conceptually. Exothermicity is a major driving force

  19. Spontaneity of Ice Melting: An Endothermic Process

  20. J W Gibbs • Gibbs thought it would be easier to examine only the properties of the system to predict spontaneity. (Rather than worrying about the entire universe!) • He rewrote the 2nd Law

  21. Gibbs called this term “free energy” ΔSsurr = qsurr / T At constant T and P, qsurr = -ΔHsys ΔSsurr = -ΔHsys / T ΔSuniv = ΔSsys + ΔSsurr ΔSuniv = ΔSsys +[ -ΔHsys / T] -TΔSuniv = ΔHsys - TΔSsys ΔG= ΔHsys - TΔSsys

  22. Moves Right Moves Left 0 ΔG= ΔHsys - TΔSsys ΔG = -TΔSuniv ΔG < 0 ΔG > 0

  23. Third Law • Unlike H or E, the absolute entropy of a compound can be measured. • Third Law – the entropy of a pure crystalline substance @ 0 K is 0. (all motion stops) • S = q / T (J/K) • Appendix 3 lists So • These can be used to calculate the ΔSo for a reaction. • ΔSorxn = Σ ΔSo (products) - Σ ΔSo (reactants)

  24. Conceptual Understanding of absolute entropy • What is the cause of the trends you see? • Why isn’t entropy of aluminum = zero, like heat of formation?

  25. Standard Free Energy of Formation • Standard Free Energy of Formation – ΔGof– The free energy change for the formation of 1 mole of a substance in the standard state. • Can calculate ΔGof ΔGo= ΣΔGof (products) - ΣΔGof (reactants) • Can calculate from ΔHoand ΔSo ΔGorxn= ΔHorxn– TΔSorxn Can calculate from Hess’s Law

  26. Example Problem • HCl (g) + NH3 (g)  NH4Cl (s) • Find ΔGo @ 298 K

  27. 2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s) Find ΔGo at 298K Example Problem

  28. Example Problem • 2 SO2 (g) + O2 (g)  2 SO3 (g) • Find ΔGo at 298 K and 1500 K What assumption is made in the second problem?

  29. Free Energy and Keq • ΔGo is at standard state (P = 1 atm, conc = 1 M, 298 K) • At non-standard conditions, • ΔG = ΔGo + RT lnQ • R = 8.314 J/(molK) = 8.314 x 10-3 kJ/(molK)

  30. Example Problem • 2 SO2 (g) + O2 (g)  2 SO3 (g) • ΔGo at 298 K is –141.77 kJ • What is ΔG at 298 K if PSO2 = 0.50 atm, PO2 = 1.0 atm and PSO3 = 0.10 atm ?

  31. Relationship of Free Energy and Equilibrium • ΔG = ΔGo + RT lnQ • At equilibrium • Q = Keq • ΔG = 0 • 0 = ΔGo + RT lnKeq • ΔGo = - RT lnKeq • Keq= e -ΔGo /RT

  32. Determination of Equilibrium Constant • 2 SO2 (g) + O2 (g)  2 SO3 (g) • ΔGo at 298 K is –141.77 kJ • What is Keq at 298K for this reaction? Answer: Keq= 7.1 x 1024 Rule of Thumb: What change in free energy will lead to a 10-fold change in equilibrium?

  33. Free Energy Diagrams • “Uphill” and “downhill” • Condition dependent • Hydrolysis of ATP has DGo = -30.5kJ/mol • Does the reaction have more potential for energy release at phyisological conditions of [ATP]=3.5mM, [ADP] = 1.8 mM, [Pi] = 5.0mM

  34. Application • Glutamine is an important carrier of nitrogen in metabolism. It can be made from glutamate and ammonia, a reaction with a standard free energy of 14 kJ at 25 oC. What is the equilibrium constant for this reaction? What is the equilibrium constant for this reaction if it is coupled to ATP hydrolysis?

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