1. Spontaneous Processes The Second Law:  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...

2. Spontaneous Processes …except for the universe as a whole! The entropy of a system can spontaneously decrease, as long as the entropy of the surroundings increases by at least as much.  S sys +  S surr  0 Do we have to keep calculating  S surr ? 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: Josiah Willard Gibbs (1839-1903) At constant temperature and pressure,  G is From the previous slide, we end up with

5. Gibbs Free Energy The condition of constant T and P is very important when using G. Otherwise, the entropy change of the surroundings might be different leading to a different result. G is extremely useful for chemistry and biochemistry, since so much takes place at constant temperature and pressure. At constant T and P, consideration of  G will answer the question “ Will a given reaction be spontaneous?” 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.

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

8. Calculation of  G H 2 O(g)  H 2 (g)+1/2 O 2 (g) Is  S o 298 greater than, less than, or equal to zero?  S o 298 = -(188.82) + 130.684+1/2 (205.14) J/(K mol) = 44.4 J / (K mol) Spontaneous?  H o 298 =-(-241.82) +0+1/2 (0) kJ/mol = 241.82  G o 298 =  H o 298 -T  S o 298 = 241.82 kJ/mol - (298 K)*0.0444 kJ/(K mol) = 228.56 kJ/mol The process is non-spontaneous!

9. Calculation of  G Example: CH 4 (g) + 2O 2 (g)CO 2 (g) + 2H 2 O(g)  G° f –50.7 0 –394.36 –228.6  G° r = –800 kJ mol –1

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

11. Calculation of  G A puzzle: Hence, so but Assume everything is reversible. at constant T and p, ??? According to this, we can’t ever have  G < 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  G A puzzle: Hence, so but Assume everything is reversible. at constant T and p, ??? Not all work is PV work! For example, electrochemical, mechanical, etc. “Free” means free to do non-PV work!

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

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 dq rev = 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  H 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 cm 3 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 H 2 O (l)H 2 O (g) 100 °C 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.  G 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 P 2,A and P 2,A are 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 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. heat

22. Protein Unfolding 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! Protein + Solvent Heat T1 T2 T1-T2

23. Protein Unfolding Data for glyceraldehyde-3-phosphate dehydrogenase. pH8.0 pH6.0 Bacillus stearothermophilus E. coli Rabbit 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: 102560100 °C  G ° kJ/mol67.460.727.8-41.4  H ° kJ/mol137236469732  S ° J/ K mol29758613182067 T  S ° kJ/mol69.9175439771 Where is the denaturation temperature? What then is special about the temperature at which the denaturation is spontaneous?