Thermodynamics. Entropy and its Applications. Lecture 11. NC State University

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1 Thermodynamics Entropy and its Applications Lecture 11 NC State University

2 System and surroundings Up to this point we have considered the system, but we have not concerned ourselves with the relationship between the system and the surroundings. When we consider heat flow In the state function, entropy, we need to carefully account for how the heat flow takes place in order to determine the effect. We will show that for reversible processes the entropy change is zero. However, for irreversible processes the entropy change can be positive (spontaneous) or negative (reverse process is spontaneous). For isolated systems the entropy change is also zero. However, when there is heat flow between the system and surroundings the entropy can be non-zero. In the case that the heat flow is irreversible the entropy change will be non-zero.

3 System and surroundings both play in role in the entropy DS sys > 0 Isolated system: Entropy increases for any spontaneous process

4 System and surroundings both play in role in the entropy DS sys + DS surr > 0 DS surr > 0 DS sys < 0 or DS sys > 0 If the system is not isolated, then the entropy can have either sign for the System as long as the surroundings can compensate. Closed system in contact with surroundings: Total entropy increases for any spontaneous process

5 Problem solving We can identify the following main types of problems that involve entropy change: Isothermal expansion/compression (reversible/irreversible) Temperature change Equilibration Mixing Phase Transition Statistical or Conformational Entropy Adiabatic (trick question) if q = 0 then DS = 0. Whenever you are solving an entropy problem remember to consider both system and surroundings. The system is always calculated along a reversible path.

6 Isothermal compression Calculate the entropy for an reversible compression of oxygen gas. The initial pressure of the gas is 1 bar in a volume of 100 L. The final pressure of the of the gas is 10 bar and the temperature is 400 K. Solution: Note that you need to obtain the number of moles. The problem does not ask you for a molar entropy. Write down the expression for the entropy: DS = nr ln V 2 V 1 = nr ln P 1 P 2 We need either the ratio of volumes or the ratio of pressures. We are given the pressures so we can use those. P 2 /P 1 = 10 bar/ 1 bar = 10.

7 Isothermal compression We obtain the number of moles using the ideal gas law: n = P 1V 1 RT = 10 5 Pa 0.1 m J/mol K 400 K = 1 bar 100 L L b ar/mol K 400 K = 3.00 moles Now we can substitute into the entropy expression:

8 Equilibration We have seen a simple example where there are two metal blocks, both made of the same material. However, this need not be the case. For any arbitrary materials 1 and 2 in contact we need to know the initial temperatures and heat capacities to calculate the final temperature. Now we solve for the equilibrium temperature T eq.

9 Equilibration: Calculating the entropy Once you have obtained the equilibrium temperature, the entropy is easily calculated from: In such problems, you are assuming that the two objects in thermal contact are a closed system. The overall entropy for heat flow should be positive since heat flow from a hotter to a colder body is a spontaneous process. Example: The coffee cup problem. A hiker uses an aluminum coffee cup. If the mass of the cup is 4 grams and the ambient temperature (i.e. the cup s temperature) is -10 o C and the hiker pours 50 ml. of coffee with a temperature of 90 o C into the cup: A. Calculate the equilibrium temperature. B. Calculate the entropy change.

10 Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature of the transition. D fus S = q P,fus T fus = D fush T fus Example, using the data in the Table calculate the entropy of vaporization for the following compounds. Compound D vap H kj/mol T b K Solution: Use the following relation C 6 H C 2 H CCl CH Br H 2 S D vap S = D vaph T vap

11 Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature of the transition. D fus S = q P,fus T fus = D fush T fus Example, using the data in the Table calculate the entropy of vaporization for the following compounds. Compound D vap H kj/mol T b K D vap S J/mol-K C 6 H C 2 H CCl CH Br H 2 S Note the similarity in the values for the entropy of vaporization. This is known as Trouton s rule.

12 Phase Transition Compare the entropy of sublimation of water to the entropy of vaporization at 0 o C. Which is larger? Why? Solution: The entropy of a phase transition is given by: DS phase = DH phase T phase From the information in Atkins (page 61) we find that D vap H = kj/mol and D sub H = kj/mol. Using these values and the temperature of 273 K we find. DS vap = DS sub = J/mol 273 K J/mol 273 K = J/mol K = J/mol K

13 Conformational entropy The entropy of a polymer or a protein depends on the number of possible conformations. This concept was realized first more than 100 years ago by Boltzmann. The entropy is proportional to the natural logarithm of the number of possible conformations, W. S = k B ln W For a polymer W = M N where M is the number of possible conformations per monomer and N is the number of monomers. For a typical polypeptide chain in the unfolded state M could be a number like 6 where the conformations include different f,y angles and side chain angles. On the other hand, when the protein is folded the conformational entropy is reduced to W = 1 in the theoretical limit of a uniquely folded structure. Thus, we can use statistical considerations to estimate the entropy barrier to protein folding.

14 Conformational entropy Statistical (conformational) entropy S = k B ln W Molar entropy S = R ln W Two important cases for calculating W: For a polymer or macromole W = M N where M is the number of possible conformations per monomer and N is the number of monomers. This is the called combinatoric. Number of ways to distribute particles in a given set of energy levels. This is called the factorial.

15 Entropy is proportional to the number of ways that energy can be distributed Energy of molecule (E m ) is quantized (arbitrary units U)

16 Distribute 6 units of E among 3 molecules config a config b W = 3 W = 3 6 total ways to distribute 6 units of E

Counting ways to distribute energy In the example we have 3 particles in each configuration. Such that For each configuration we have a different set of n i.

17 Counting ways to distribute energy In the example we have 3 particles in each configuration. Such that For each configuration we have a different set of n i. If we have N total particles (here N=3), then the total number of ways to distribute the particles among the allowed energy levels is determine by the factorial.

18 Statistical distribution among energies In the example we have 3 particles in each configuration. Such that where the exclamation sign is the factorial. For example, 4! = 4 x 3 x 2 x 1 = 24. In the above examples we can calculate the number of ways to distribute the energy.

19 Statistical distribution among energies In this case once we have determined the occupation numbers we can just plug those Into the factorial. For configuration a We have n 0 = 2, n 1 = 0 and n 2 = 1. Although the actual occupied levels are not the number of occupied levels is statistically the same, n 0 = 1, n 1 = 2 and n 2 = 0.

20 Total E: 30 U W = 12! (5!) (0!) (0!) (4!) (0!) (0!) (3!) = 27720

21 Conformational entropy of a protein Estimate the conformational entropy of myoglobin. Myoglobin has 150 residues with 6 possible conformations per residue. Assume there is a unique conformation for the folded structure. Solution: The definition S = R lnw is known as the statistical entropy. R is the gas constant and W is the number of possible conformations for a structure and W = M N. For the unfolded protein: W = and S = 150R ln(6). For the folded protein: W = 1 and S = 0. There conformational entropy is D fold S - D unfold S and is therefore 2233 J/mol or 2.2 kj/mol.

Energy The Levinthal Paradox The Levinthal paradox assumes that all of the possible conformations will be sampled with equal probability until the proper one (N = native) is found.

22 Energy The Levinthal Paradox The Levinthal paradox assumes that all of the possible conformations will be sampled with equal probability until the proper one (N = native) is found. Thus, the funnel surface looks like a hole in a golf course. The paradox states that if a protein samples Conformational Entropy all 6M conformations it will take a time longer than the age of the universe to find the native fold (N) for a polypeptide where M = 100 and if it takes seconds to sample each possible conformation!

23 Energy The Pathway Model Imagine that the a unique pathway winds through the surface to the hole. The path starts at A and the folding goes through a unique and well-defined set of conformational changes. Here the entropy must decrease rapidly since the number of degrees of Conformational freedom in the folding pathway is quite Entropy small compared to 6M. On this diagram the configurational entropy is given by the width of the funnel and the relative by the height relative to the bottom (folded state). The vertical axis is energy NOT free energy.

24 Evidence for folding pathways One piece of evidence for folding pathways comes from trapping disulfide intermediates. This method was pioneered by Creighton using BPTI (and has only been used on other proteins). Creighton et al. Prog. Biophys. Mol. Biol , 1978

25 Beyond pathways Kim et al. showed that some of the previous data and interpretations were wrong. The major 2-DS species contains the two native DS's, and The third disulfide is formed quite slowly because it is quite buried. It is possible to isolate a stable species with only the first two disulfides formed and the third remaining in the reduced form. Studies with a mutant in which the third DS was replaced by 2 Ala, and which folded at a similar rate to the wild type, support the idea that the trapped disulfide species have partial native-like structure. These observations and others like it can be used to the idea of a pathway into question. Kim (1993) Nature

Energy The Folding Funnel The folding funnel shown here represents the change in energy for a large number of folding paths that lead to the native configuration. There are no energy barriers.

26 Energy The Folding Funnel The folding funnel shown here represents the change in energy for a large number of folding paths that lead to the native configuration. There are no energy barriers. This implies that all paths have an equal probability leading Conformational to the folded state. The funnel Entropy shown here has no energy barriers and all paths lead directly to the native state. Thus, this funnel is consistent with two state folding behavior. Dill and Chan, Nature Struct. Biol. (1997), 4:10-19

Energy Barriers and misfolding The energy surface does not have to be a smooth trajectory. There can be barriers that will trap intermediate states.

27 Energy Barriers and misfolding The energy surface does not have to be a smooth trajectory. There can be barriers that will trap intermediate states. These states may then be observed to determine aspects of the folding trajectory. Whenever an intermediate is observed there will be a question as to whether this is part of a pathway or whether a funnel description Conformational Entropy is more applicable. Note that the funnel provides the possibility for misfolding. This will typically result in multiple minima in the energy landscape. Dill and Chan, Nature Struct. Biol. (1997), 4:10-19

28 The hydrophobic effect The hydrogen bonding network of water is disrupted by a hydrophobic solute. The O-H bond vector will tend to point away from the hydrophobic solute and toward an oxygen lone pair in order to reach the lowest enthalpy configuration. The water molecules surrounding the hydrophobic solute are forced into a more ordered structure. Another way to realize this is to understand that the water molecules have fewer possible hydrogen bond partners since they cannot hydrogen bond to the hydrophobic solute. The result is a negative entropy of solvation for hydrophobic molecules. If hydrophobic solutes cluster they can decrease their exposed surface area and thus DS > 0 for aggregation of hydrophobic solutes in water.

29 Hydrogen bonding in water

30 Hydrophobic interactions

Hydrophobic interactions For methane solvated by water the magnitude of the hydrophobic effect has been estimated to be 80 entropy units (i.e. -80 J/mol-K).

31 Hydrophobic interactions For methane solvated by water the magnitude of the hydrophobic effect has been estimated to be 80 entropy units (i.e. -80 J/mol-K). We can estimate the entropic contribution due to exposed hydrophobic side chains in proteins. When this is done the entropic effect is a dominant contribution to the driving force in protein folding. Figure shows the tendency of hydrophobic molecules to Aggregate in solution in order to minimize their surface area.

32 Entropy-driven drug binding If a drug binds to a receptor and loses tightly bound water molecules, the overall change in entropy can be positive because of the increase in degrees freedom of the waters of solvation. An example is observed in the binding of etorphine binding to m-opioid receptors (ΔH = kj/mol and ΔS = J/mol-K). In general, m-opioid drugs, most of which are in the opiate class, are rigid and relative hydrophobic alkaloids. European Journal of Pharmacology(1985) 108, Pages

33 The natural agonist met-enkephalin has degrees of freedom in solution Met-enkephalin Floppy, high entropy in solution Morphine Rigid, low entropy in solution

34 Estimating the conformational entropy There are 4 peptide bonds and we can assume each set of f,y angles has 4 possible conformations. Thus, W = 4 4. The molar entropy is S = R ln(w) = 4R ln(4) = 46 J/mol-K.

Second Law Applications. NC State University

Second Law Applications. NC State University Chemistry 433 Lecture 11 Second Law Applications NC State University Summary of entropy calculations In the last lecture we derived formula for the calculation of the entropy change as a function of temperature