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Spontaneous Processes

Spontaneous Processes. The Second Law: D S  0 The entropy of a closed system can only increase. If a process will decrease entropy in a closed system, then it does not occur spontaneously. Its opposite will occur spontaneously. But we very rarely work with closed systems.

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Spontaneous Processes

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  1. Spontaneous Processes The Second Law: DS  0 The entropy of a closed system can only increase. If a process will decrease entropy in a closed system, then it does not occur spontaneously. Its opposite will occur spontaneously. But we very rarely work with closed systems...

  2. Spontaneous Processes …except for the universe as a whole! DSsys + DSsurr 0 The entropy of a system can spontaneously decrease, as long as the entropy of the surroundings increases by at least as much. Do we have to keep calculatingDSsurr ? Not necessarily!

  3. Spontaneous Processes Let’s stay at constant T and P: Now everything is in terms of the system. The criterion for spontaneity given by the Second Law becomes:

  4. Gibbs Free Energy The Gibbs Free Energy is a new state function, defined as: At constant temperature and pressure, DG is From the previous slide, we end up with Josiah Willard Gibbs (1839-1903)

  5. Gibbs Free Energy G is extremely useful for chemistry and biochemistry, since so much takes place at constant temperature and pressure. The condition of constant T and P is very importantwhen using G. Otherwise, the entropy change of the surroundings might be different leading to a different result. • G is still defined and can be calculated for any change of state, including changing P and T. • We can also define another state variable, Helmholtz Free Energy (A = E - TS), which has similar characteristics as G, but relates more directly to constant T and V processes. At constant T and P, consideration of DG will answer the question “Will a given reaction be spontaneous?”

  6. Gibbs Free Energy Summary The Gibbs Free Energy is a direct measure of spontaneity: G = H - TS It sums up, in a way, the competition between energy considerations and “configurational” barriers. We have also learned that a process is spontaneous if G < 0 Thus, if H < 0 the process is exothermic (downhill) S > 0 the process is increases disorder So H dominates spontaneity at low temperatures S dominates spontaneity at high temperatures

  7. Calculation of DG In many cases, we can build on calculations we have already done in order to get DG. DH = 6.75 kJ mol–1 DS = 45.5 J K–1 mol–1 so DG = 6750 – (373)(45.5) = –10.2 kJ mol–1. Example: ice melting at 100° C In other cases, like chemical reactions, standard values have been found. We need only to add them up properly.

  8. Calculation of DG H2O(g)  H2(g) + 1/2 O2(g) Is So298 greater than, less than, or equal to zero? So298= -(188.82) + 130.684 + 1/2 (205.14) J/(K mol) = 44.4 J / (K mol) Spontaneous? Ho298=-(-241.82) + 0 + 1/2 (0) kJ/mol = 241.82 Go298=Ho298-T So298 = 241.82 kJ/mol - (298 K)*0.0444 kJ/(K mol) = 228.56 kJ/mol The process is non-spontaneous!

  9. Calculation of DG Example: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DG°f –50.7 0 –394.36 –228.6 DG°r = –800 kJ mol–1

  10. Calculation of DG We can perform calculus on G directly: dG = dH – TdS – SdT = dE + pdV + Vdp – TdS – SdT but dE = –pdV +TdS (dw = -pdV and dqrev = TdS) In other words, constant T and constant P

  11. Calculation of DG A puzzle: at constant T and p, Assume everything is reversible. so but Hence, ??? According to this, we can’t ever haveDG < 0 if everything is reversible at constant T and p. But what about all those chemical reactions? Surely they can be run reversibly! But Where is the mistake?

  12. Calculation of DG A puzzle: at constant T and p, Assume everything is reversible. so but Hence, ??? Not all work is PV work! For example, electrochemical, mechanical, etc. “Free” means free to do non-PV work!

  13. Temperature Dependence Simplest approximation: DG = DH - T DS assume both DH and DS do not change for moderate changes of T DG(T) = DG(298 K) - (T - 298 K)DS If we are assuming DH and DS independent of T, why not just ignore DG dependence? Explicit dependence on T in DG, whereas only implicit for DH and DS

  14. Temperature Dependence General expression (reversible path, only PV work): dG = dH - TdS - SdT = dE +PdV + VdP - TdS - SdT but dE = –PdV +TdS since dw = -PdV and dqrev = TdS and dE = dw + dq (we are on a specified path so this is ok) so … dG = (-PdV + PdV) + (TdS - TdS) + VdP - SdT = VdP - SdT At constant pressure: dG = -SdT

  15. Temperature Dependence Gibbs-Helmholtz equation If DH changes little with temperature See text p. 93 for mathematical derivation.

  16. Pressure Dependence For the pressure dependence we hold T constant: dG = VdP-SdT=VdP Thus, For solids and liquids, V does not vary with temperature so, and for an ideal gas:

  17. Pressure Dependence Example Can we force graphite to diamond by increasing the pressure? We will use: and the fact that molar volumes of graphite and diamond are known: Where we have used a conversion factor to convert from cm3 atm to kJ. Now, we want the pressure that makes G = 0: Why? 0 = 2.84 - 1.935 10-4 (P-1) kJ/mol P = 15,000 atm Experimentally, the required pressure is more! Why?

  18. Example: Reversible Process 100 °C H2O (l) H2O (g) What is the free energy change? Well, this is at constant T and P. Its reversible. So, G°vap= 0 = H°vap-T S°vap Note that this implies:

  19. DG of Mixing Consider the isobaric, isothermal mixing of two gases: Gas A at 1 Atm Gas B at 1 Atm Gas A+B at 1 Atm (total) Is this reaction spontaneous? Well, both P2,A and P2,Aare less than 1 atm…so yes, this process is spontaneous.

  20. Protein Unfolding Proteins have a native state. (Really, they tend to have a tight cluster of native states.) Denaturation occurs when heat or denaturants such as guanidine, urea or detergent are added to solution. Also, the pH can affect folding. When performing a denaturation process non-covalent interactions are broken. Ionic, van der-Waals, dipolar, hydrogen bonding, etc. Solvent is reorganized.

  21. Protein Unfolding heat Let’s consider denaturation with heat. We can determine a great deal about the nature of the protein from such a consideration. The experimental technique we use for measuring thermodynamic changes here is the differential scanning calorimeter. Basic experiment: Add heat to sample, measure its temperature change.

  22. Protein Unfolding T1 Heat T2 Protein + Solvent Solvent T1-T2 In differential scanning calorimetry you have two samples: Your material of interest Control You put in an amount of heat to raise the temperature of the control at a constant rate, then measure the rate of change in temperature of the other sample as a function of the input heat. This is a measure of the heat capacity!

  23. Protein Unfolding pH8.0 Bacillus stearothermophilus Rabbit pH6.0 E. coli Data for glyceraldehyde-3-phosphate dehydrogenase. Is the protein more stable at pH 8 or 6? Why is B. stear. more stable?

  24. Protein Unfolding We are given the following data for the denaturation of lysozyme: 10 25 60 100 °C G° kJ/mol 67.4 60.7 27.8 -41.4 H° kJ/mol 137 236 469 732 S° J/ K mol 297 586 1318 2067 TS° kJ/mol 69.9 175 439 771 Where is the denaturation temperature? What then is special about the temperature at which the denaturation is spontaneous?

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