Physical ChemistryPhysical Chemistry

The Second LawEntropy

Applications of Entropy pp py

NC State University

Spontaneity of Chemical Reactionsp yOne might be tempted based on the results of thermo-chemistry to predict that all exothermic reactions would y pbe spontaneous. The corollary this would be the statement that no endothermic reactions are spontaneous. However, this is not the case There are numerous examples ofthis is not the case. There are numerous examples of endothermic reactions that are spontaneous. Of course, heat must be taken up from the surroundings in order for

h t N th l th th l f thsuch processes to occur. Nonetheless, the enthalpy of the reaction does not determine whether or not the reaction will occur, only how much heat will be required or , y qgenerated by the reaction. The observation that gases expand to fill a vacuum and that different substances spontaneously mix when introduced into the same vesselspontaneously mix when introduced into the same vessel are further examples that require quantitative explanation.

Spontaneity of Chemical Reactionsp yAs you might guess by now, we are going to define a new state function that will explain all of these observations pand define the direction of spontaneous processes. This state function is the entropy.Entropy is related to heat and heat flow and yet heat is notEntropy is related to heat and heat flow and yet heat is nota state function. Recall that q is a path function. It turnsout that the state function needed to describe spontaneoush i th h t di id d b th t tchange is the heat divided by the temperature.

Here we simply state this result.

S = qrevS = T

EnginesHistorically, people were interested in understanding the efficiency with which heat is converted into work. This was a very important question at the dawn of the industriala very important question at the dawn of the industrial revolution since it was easy to conceive of an engine powered by steam, but it turned out to be quite difficult to build one that was efficient enough to get anything done! In an engine, there is a cycle in which fuel is burned to heat gas inside the piston. The expansion of the piston leads to g p p pcooling and work. Compression readies the piston for the next cycle. A state function should have zero net change for the cycle It is only the state that matters to such afor the cycle. It is only the state that matters to such a function, not the path required to get there. Heat is a path function. As we all know in an internal combustion engine ( t i ) th i t l f h t(or a steam engine), there is a net release of heat. Therefore, we all understand that q 0 for the cycle.

A cyclic heat engine

I1 2 Phase Transition Path Condition

I. 12 Isothermal w = -qII 23 Adiabatic U = wI II. 23 Adiabatic U = wIII. 34 Isothermal w = -qIV. 41 Adiabatic U = w

IIIVThe work isw = wI + wII + wIII + wIVIIIV

34q = qI + qIII

= - wI - wIII

F th di b ti tFor the adiabatic stepsqII = qIV = 0For the isothermal steps

IIIp

U = 0

Work and Heat for the CycleNeither the work nor the heat is a state function. Neither oneis zero for the cycle as should be the case for a state function.The ork isThe work is:w = wI + wII + wIII + wIV=-nRThotln(V2/V1)–Cv(Tcold–Thot)–nRTcoldln(V4/V3)–Cv(Thot – Tcold) w = -nRThotln(V2/V1) – nRTcoldln(V4/V3) [since wII = - wIV]w = -nRThotln(V2/V1) – nRTcoldln(V1/V2) [since V4/V3 = V1/V2]w = -nRThotln(V2/V1) + nRTcoldln(V2/V1) [property of logarithms]w nRThotln(V2/V1) nRTcoldln(V2/V1) [property of logarithms]The heat is:q = qI + qIII [since wII = wIV = 0]

= w w [since dU = 0 for isothermal steps]= - wI - wIII [since dU = 0 for isothermal steps]q = nRThotln(V2/V1) + nRTcoldln(V4/V3) q = nRThotln(V2/V1) + nRTcoldln(V1/V2) [since V4/V3 = V1/V2]q = nRThotln(V2/V1) - nRTcoldln(V2/V1) [property of logarithms]

A new state function: Entropy

The heat is not a state function. The sum qI + qIII is not zero.From this point on we will make the following definitions:From this point on we will make the following definitions:

qI = qhot qIII = qcold

RT l V2 RT l V2 0q = qhot + qcold = nRThotlnV2

V1– nRTcoldln

V2

V1 0

qrev = qhot + qcold = nRln V2 – nRln V2 = 0

However, the heat divided by temperature is a state function.

T Thot+ Tcold

nRln V1nRln V1

0

, y pThis reasoning leads to the idea of a state function called theentropy. We can write:

qS = qrev

T

Thermodynamics of an EngineThe cycle just described could be the cycle for a piston in a steam engine or in an internal combustion engine. g gThe hot gas that expands following combustion of a small quantity of fossil fuel drives the cycle. If you think about the fact that the piston is connected to the crankshaft youthe fact that the piston is connected to the crankshaft you will realize that the external pressure on the piston is changing as a function of time and is helping to realize

i th t l t id l ibl ian expansion that as close to an ideal reversible expansionas the designers can get. If we ignore friction and assume that the expansion is perfectly reversible we can apply the p p y pp yabove reasoning to your car. The formalism above for the entropy can be used to tell us the thermodynamic efficiency of the engineof the engine.

Thermodynamic EfficiencyWe define the efficiency as the work extracted divided by the total heat input. p

efficiency = work doneheat used

= |wtotal|qhot

= |nR(Tcold – Thot)ln (V2 / V1)|nRThotln (V2 / V1)

= Thot – Tcold

Thot

The efficiency defined here is the ideal best case. It assumesa reversible process with no losses due to friction. The

qhot nRThotln (V2 / V1) Thot

ptemperature Thot is the temperature of the expansion in theengine. The temperature Tcold is the temperature of the exhaust T cannot be less than the temperature of theexhaust. Tcold cannot be less than the temperature of the surroundings.

QuestionYour car has an operating temperature of 400 K. If the ambient temperature is 300 K, what is the thermodynamic efficiency of the the engine?

A. 75%A. 75%B. 50%C. 25%D 5%D. 5%

QuestionYour car has an operating temperature of 400 K. If the ambient temperature is 300 K, what is the thermodynamic efficiency of the the engine?

A. 75%A. 75%B. 50%C. 25%D 5%D. 5%

= |wtotal| = Thot – Tcold = 1 Tcold = 1 300K = 0 25 = | total|qhot

= hot cold

Thot= 1 – cold

Thot= 1 – 300K

400K = 0.25

QuestionThe thermodynamic cycle was derived for reversible expansions. What are the consequences if the cycle is not perfectly reversible?

A. The work of expansion will decreaseA. The work of expansion will decreaseB. The work of compression will decreaseC. There will be no adiabatic expansionD There can be no cycleD. There can be no cycle

QuestionThe thermodynamic cycle was derived for reversible expansions. What are the consequences if the cycle is not perfectly reversible?

A. The work of expansion will decreaseA. The work of expansion will decreaseB. The work of compression will decreaseC. There will be no adiabatic expansionD There can be no cycleD. There can be no cycle

The Thermodynamic Temperature ScaleTemperature Scale

The definition of entropy is qhot/Thot + qcold/Tcold = 0. W it thiWe can write this as

qhot/Thot = - qcold/TcoldSince qcold is negative we can combine the minus sign qcold g gwith qcold and write the expression as

|qhot|/Thot = |qcold|/Tcoldand finallyand finally

|qhot|/|qcold| = Thot/TcoldThe ratio of the heats is equal to the ratio of temperatures f t t i th d i l Thi d fifor two steps in a thermodynamic cycle. This defines a temperature scale and allows one to measure temperature as well (i.e. this scheme represents a thermometer). Both ( )this expression and the thermodynamic efficiency further imply that there is an absolute zero of temperature.

Heat TransferTo examine the function that we have just defined, let us imagine that we place to identical metal bricks in contact g pwith one another. If one of the bricks is at equilibrium at 300 K and the other at 500 K, what will the new equilibrium temperature be? Intuitively you would say 400 K and youtemperature be? Intuitively, you would say 400 K and you would imagine that heat flows spontaneously from the warmer brick to the colder brick. The entropy function

k th id tit timakes these ideas quantitative. S = q2

T2+ q1

T1

q1 q2

T = 500 K T = 300 K1 2 Using this definitionof entropy change asy gthe heat flow dividedby the temperature.

Heat TransferLet’s assume that heat flows from the hot body to the coldbody. Then q1 is negative (the flow from the hot body) andy q1 g ( y)q2 is positive (the flow into the cold body). Moreover,

q2 = -q1 = q This means that we can substitute in q to obtain:This means that we can substitute in q to obtain:

S = qT2

– qT1

= q 1T2

– 1T1

= q 1300 – 1

500 = 0.0013 qT2 T1 T2 T1 300 500q

U i thi d fi itiT = 500 K T = 300 K1 2Using this definitionS > 0, which saysthat the process is spontaneous.

Calculating reversible and i ibl thirreversible paths

It is important to reiterate that the calculation of the entropy of the system always follows a reversible path. You might ask, well what happens if the process is not reversible?reversible? To consider this let us the example of expansion of gas in a cylinder. The process can occur along different paths. a constant pressure expansion w = P V = P (V V)a. constant pressure expansion, w = -PextV = - Pext(Vf – Vi)b. reversible isothermal expansion the work w = -nRTln(Vf/Vi).

For both a. (irreversible) and b. (reversible) we will calculate the entropy of the system along a reversible path.Ssys = q/T = -w/T = nRln(Vf/Vi)Ssys q/T w/T nRln(Vf/Vi)For the reversible path, we can use the fact that Ssurr = - Ssysto obtain Ssurr = - nRln(Vf/Vi).

System and surroundingsy gThe heat transfer example shows us that we always mustconsider the surroundings Any time the system releasesconsider the surroundings. Any time the system releasesheat it goes into the surroundings and this contributes to theoverall entropy change. Thus, the total entropy is:

S S + SStotal = Ssurr + SsysFor a reversible process the total entropy is zero. If a process is irreversible and spontaneous the entropy changeis positive. This implies that we must treat the system andthe surroundings differently when we calculate entropy.The rule is: always calculate the entropy of the system alongThe rule is: always calculate the entropy of the system alonga reversible path.If the process is truly reversible then:

S = 0 and S = SStotal = 0 and Ssurr = - Ssys

Understanding the irreversible pathg p

For the irreversible path we use the actual work of the constant pressure expansion, w = -PextV = - Pext(Vf – Vi) to calculateSsurr = -q/T = w/T = -Pext(Vf – Vi)/T where T is the same temperature we used for the isothermal expansion. Note thattemperature we used for the isothermal expansion. Note thatthe sign is opposite since the heat is flowing into the surround-ings (and out of the system).We see that in this case the entropy change S is smaller inWe see that in this case the entropy change Ssurr is smaller in magnitude than Ssys. We know this from the first law wherewe saw that the irreversible work of expansion is always less than the reversible work of expansion. Thus,

Stotal > 0for the irreversible process.for the irreversible process.

The dependence of the entropy on volumeentropy on volume

For a constant temperature (isothermal)expansion we have:S / /dS = qrev/T = - wrev/T.

The logic behind this statement is that the internal energy change is zero for a constant temperature process and g p pso qrev = - wrev. To calculate the reversible work we simply plug in wrev = -PdV. According to the ideal gas law P = nRT/V so dS = nRdV/VP = nRT/V so dS = nRdV/V.

dSS1

S2

= nR dVVV1

V2

The result of this equation is that S = nRln(V2/V1) at constant temperature. p

The dependence of the entropy on temperatureentropy on temperature

The entropy change as a function of the temperature isfderived at constant volume using the fact that

dU = qV = nCVdT.The reversible heat in this case, qrev, is a constant volume, qrev,heat and so it can be replaced by dS = qrev/T = nCvdT/T at constant volumeTo obtain S we need to integrate both sidesTo obtain S we need to integrate both sides

W bt i

dSS1

S2

= nCvdTTT1

T2

We obtain:S = S2 – S1 = nCvln(T2/T1).

Exactly the same reasoning applies at constant pressure, y g pp p ,so that S = nCpln(T2/T1).

Summary of entropy calculationsIn the last lecture we derived formula for the calculation ofthe entropy change as a function of temperature and volume changes These are summarized in the table belowvolume changes. These are summarized in the table below.

Constant temperature S = nRln(V2/V1)Constant volume S nC ln(T /T )Constant volume S = nCvln(T2/T1)Constant pressure S = nCpln(T2/T1)

Sometimes we only have pressure information and the entropy change can be rewritten as follows.

P VP1V1 = P2V2 P1

P2= V2

V1

S R l V2 R l P1S = nR ln 2

V1= nR ln 1

P2

Entropy of mixingWhen two substances can mix there is a spontaneous tendency for this occur. We quantify this using the entropy state function. If we consider two containers separated bystate function. If we consider two containers separated bya stopcock. It N2 gas is in one and Br2 gas is in the otherwe know from experience that the gases will mix once thestopcock is openedstopcock is opened.

N2 Br2N2 Br2

N2 + Br2 N2 + Br22 2 2 2

Entropy of mixingFor each gas we can describe the mixing as a volumechange. The N2 gas is originally on the right side containedin volume V at pressure P After opening the stopcock thein volume V at pressure P. After opening the stopcock, the available volume is 2V and the partial pressure of P1 = x1P.The same is true for Br2. Its initial pressure is P2 and finalpressure is P2 = x2P. We treat the entropy as the sum oftwo expansions (i.e. an expansion for each gas).

mixS = expS1 + expS2

mixS = – n1R ln P1

P – n2R ln P2

PP PmixS = – n1R ln x1 – n2R ln x2

i S = – nR x1lnx1 + x2lnx2 Note that the mole fractionmixS nR x1lnx1 + x2lnx2 Note that the mole fraction applies to the final composition

Entropy of phase transitionThe entropy of phase transition can be calculated usingthe enthalpy and temperature of the transition.

fusS =qP,fus

Tfus= fusHTfus

Example, using the data in the Table calculate the entropyof vaporization for the following compounds.p g p

Compound vapHkJ/mol

TbK

C6H6 30.8 353.2

Solution: Use the following relation

C2H6 14.7 184.6CCl4 30.0 350.0CH4 8.18 111.7Br2 29.7 332.4 S

vapH

g

Br2 29.7 332.4H2S 18.7 212.8 vapS = vap

Tvap

Entropy of phase transitionThe entropy of phase transition can be calculated usingthe enthalpy and temperature of the transition.

fusS =qP,fus

Tfus= fusHTfus

Example, using the data in the Table calculate the entropyof vaporization for the following compounds.p g p

Note the similarity inCompound vapH

kJ/molTbK

vapSJ/mol-K

C H 30 8 353 2 87 2the values for theentropy of vaporization.This is known as

C6H6 30.8 353.2 87.2C2H6 14.7 184.6 79.6CCl4 30.0 350.0 85.7CH4 8.18 111.7 73.2B 29 5 332 4 88 5 This is known as

Trouton’s rule.Br2 29.5 332.4 88.5H2S 18.7 212.8 87.7

Conformational entropyThe entropy of a polymer or a protein depends on the numberof possible conformations. This concept was realized first morethan 100 years ago by Boltzmann The entropy is proportionalthan 100 years ago by Boltzmann. The entropy is proportionalto the natural logarithm of the number of possible conformations, W. S = R ln W

For a polymer W = MN where M is the number of possible conformations per monomer and N is the number of monomers.

S R ln W

pFor a typical polypeptide chain in the unfolded state M could bea number like 6 where the conformations include different angles and side chain angles On the other hand when theangles and side chain angles. On the other hand, when theprotein is folded the conformational entropy is reduced to W = 1in the theoretical limit of a uniquely folded structure. Thus, we can use statistical considerations to estimate the entropybarrier to protein folding.

Problem solvingWe can identify the following main types of problemsthat involve entropy change:

Isothermal expansion/compression (reversible/irreversible)Temperature changeEquilibrationMixingPhase TransitionStatistical or Conformational EntropyAdiabatic (trick question) if q = 0 then S = 0.

Whenever you are solving an entropy problem rememberto consider both system and surroundings. The systemis always calculated along a reversible path.

Isothermal compressionCalculate the entropy for an irreversible compression ofoxygen gas. The initial pressure of the gas is 1 bar in avolume of 100 L The final pressure of the of the gas isvolume of 100 L. The final pressure of the of the gas is10 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:p py

S = nR ln V2

V1= nR ln P1

P2

We need either the ratio of volumes or the ratio of pressures.We are given the pressures so we can use those.P2/P1 = 10 bar/ 1 bar = 10.

Isothermal compressionWe obtain the number of moles using the ideal gas law:

P V 105 Pa 0 1 m3

n = P1V1

RT =10 Pa 0.1 m

8.31 J/mol–K 400 K

1 b 100 L=

1 bar 100 L

0.0831 L–bar/mol–K 400 K

Now we can substitute into the entropy expression:

= 3.00 moles

Now we can substitute into the entropy expression:

S = nR ln P1

P2= 3.0 mol 8.31 J/mol–K ln 10 = 249 J/KP2

EquilibrationWe have seen a simple example where there are twometal blocks, both made of the same material.However this need not be the case For any arbitraryHowever, this need not be the case. For any arbitrarymaterials 1 and 2 in contact we need to know the initialtemperatures and heat capacities to calculate the final temperature.

q1 = – q2

Cp,1 T1 – Teq = – Cp,2 T2 – Teq

Now we solve for the equilibrium temperature Teq.

p, eq p, eq

C T C T = C T + C TCp,1T1 – Cp,1Teq = – Cp,2T2 + Cp,2Teq

Cp,1T1 + Cp,2T2 = Cp,1 + Cp,2 Teq

C T + C TTeq =

Cp,1T1 + Cp,2T2

Cp,1 + Cp,2

Equilibration: Calculating the entropyOnce you have obtained the equilibrium temperature, theentropy is easily calculated from:

S1 = nCp,1 lnTeq

T1, S2 = nCp,2 ln

Teq

T2

In such problems, you are assuming that the two objectsin thermal contact are a closed system. The overallentropy for heat flow should be positive since heat flowpy pfrom a hotter to a colder body is a spontaneous process.

Example: The coffee cup problem A hiker uses an aluminum coffee cupExample: 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 oC and the hiker pours 50 ml. of coffee with a temperature of 90 oC into the cup: p pA. Calculate the equilibrium temperature.B. Calculate the entropy change.

Gas exchange in a green leafThe cells of the spongy layerare irregular in shape and g ploosely packed. Their main function seems to be the temporary storage of sugars and amino acids synthesized in the palisade layer. They also aid in the exchange of gases between the leaf and the environment. During the day, these

ll i ff d t t th i th tcells give off oxygen and water vapor to the air spaces that surround them. They also pick up carbon dioxide from the air spaces The air spaces are interconnected and openair spaces. The air spaces are interconnected and open to the outside through pores called stomata (singular, stoma).

Entropy O2 mixing in a the stomata of a leafthe stomata of a leaf

During photosynthesis O2 is produced in the thylakoidmembrane of green leaves The gas meets the atmospheremembrane of green leaves. The gas meets the atmospherein the stomata of the leaf. Assuming that the followingmole fractions exist calculate the molar entropy of mixing.I id 0 5 d 0 5Inside xO2_in = 0.5 and xN2_in = 0.5Outside xO2_out = 0.2 and xN2_out = 0.8Assume that the stomata closes with an equal volume of outside air and inside gas present in an enclosed space.Solution: Assume that the initial state is premixed gas bothinside and outside Call oxygen compound 1 and nitrogeninside and outside. Call oxygen compound 1 and nitrogencompound 2. The final composition is:xf1 = (xi1 + xo1)/2 = (0.5 + 0.2)/2 = 0.35x = (x + x )/2 = (0 5 + 0 8)/2 = 0 65xf2 = (xi2 + xo2)/2 = (0.5 + 0.8)/2 = 0.65

Entropy O2 mixing in a the stomata of a leafthe stomata of a leaf

Write down the total entropy change as products (final mixedcomposition) minus reactants (premixed air from inside andcomposition) minus reactants (premixed air from inside andoutside). There is an entropy of mixing of each of the two“reactants” that must be subtracted from the entropy of thefi l i N t th t th i h lf h f h f thfinal mix. Note that there is half as much of each of thereactant gases as the product.

S S S SStotal = Sfinal_mix – Smix_in – Smix_out

= –nR xf1ln xf1 + xf2ln xf2

R R

Since the problem asks for the molar entropy we really need

+nR2 xo1ln xo1 + xo2ln xo2 + nR

2 xi1ln xi1 + xi2ln xi2

Since the problem asks for the molar entropy we really needSfinal_mix/n so we can write the formula as:

Entropy O2 mixing in a the stomata of a leafthe stomata of a leaf

Write down the total entropy change as products (final mixedcomposition) minus reactants (premixed air from inside andcomposition) minus reactants (premixed air from inside andoutside). There is an entropy of mixing of each of the two“reactants” that must be subtracted from the entropy of thefi l i N t th t th i h lf h f h f thfinal mix. Note that there is half as much of each of thereactant gases as the product.

x x x xStotal,m = – R xf1ln xf1 + xf2ln xf2 – xo12 ln xo1 + xo2

2 ln xo2 – xi12 ln xi1 + xi2

2 ln xi2

= (8.31 J / mol–K)(0.05) = 0.421 J / mol–K

This is a small entropy. Part of the reason is that the gaseswere already mixed If they had been completely unmixed

(8.31 J / mol K)(0.05) 0.421 J / mol K

were already mixed. If they had been completely unmixedthe entropy would have been ~ 5.4 J/mol-K.

Phase TransitionCompare the entropy of sublimation of water to the entropyof vaporization at 0 oC Which is larger? Why?of vaporization at 0 C. Which is larger? Why?

Solution: The entropy of a phase transition is given by:

Sphase =Hphase

Tphase

From the information in Atkins (page 61) we find thatvapH = 45.07 kJ/mol and subH = 51.08 kJ/mol.Using these values and the temperature of 273 K we findUsing these values and the temperature of 273 K we find.

Svap = 45070 J/mol273 K = 165.1 J/mol–K

Ssub = 51080 J/mol273 K = 187.1 J/mol–K

Conformational entropy of a proteinpy pEstimate the conformational entropy of myoglobin. Myoglobin has 150 residues with 6 possibleMyoglobin has 150 residues with 6 possible conformations per residue. Assume there is a unique conformation for the folded structure.qSolution: 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 Wnumber of possible conformations for a structure and W = MN. For the unfolded protein: W = 6150 and S = 150R ln(6).p ( )For the folded protein: W = 1 and S = 0.There conformational entropy is foldS - unfoldS and is therefore 2233 J/mol or 2.2 kJ/mol.

The Levinthal Paradox

The Levinthal paradoxpassumes that all of thepossible conformationswill be sampled with ne

rgy

will be sampled withequal probability untilthe proper one (N = native)

f fE

nis found. Thus, the funnelsurface looks like a holein a golf course. The paradox

ConformationalE tg p

states that if a protein samplesall 6M conformations it will take a time longer than the age of the universe to find the native fold (N) for a polypeptide where

Entropy

the universe to find the native fold (N) for a polypeptide whereM = 100 and if it takes 10-11 seconds to sample each possibleconformation!

The Pathway ModelImagine that the a uniquepathway winds throughthe surface to the hole.The path starts at A andthe folding goes through ne

rgy

the folding goes througha unique and well-definedset of conformational changes Here the entropy

En

changes. Here the entropymust decrease rapidly sincethe number of degrees of Conformational

Entropfreedom in the folding pathway is quitesmall compared to 6M. On this diagram the configurational entropy is given by the width of the funnel and the relative

Entropy

entropy is given by the width of the funnel and the relativeby the height relative to the bottom (folded state). The verticalaxis is energy NOT free energy.

Evidence for folding pathwaysOne piece of evidence for folding pathways comes from trappingOne 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. 33 231, 1978

Beyond pathwaysKim et al. showed that some of the previous data and interpretations were wrong. The major 2-DS

i t i th t ti DS'species contains the two native DS's, 30-51 and 5-55. The third disulfide is formed quite slowly because itis quite buried It is possible tois quite buried. It is possible to isolate a stable species with only the first two disulfides formed and the third remaining in the reduced formthird 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

Kim (1993) Nature 365 185observations and others like it can be used to the idea of a pathway into question.

The Folding Funnel

The folding funnel shown here represents the change in energy for a largenumber of folding paths ne

rgy

number of folding paths that lead to the native configuration. There are no energy barriers This

En

no energy barriers. This implies that all paths havean equal probability leading Conformational

E tto the folded state. The funnelshown here has no energy barriers and all paths lead directly tothe native state. Thus, this funnel is consistent with two state

Entropy

e a e s a e us, s u e s co s s e o s a efolding behavior.

Dill and Chan, Nature Struct. Biol. (1997), 4:10-19

Barriers and misfoldingThe energy surface doesnot have to be a smootht j t Th btrajectory. There can bebarriers that will trap intermediate states. These ne

rgy

states may then be observedto determine aspects of the folding trajectory Whenever

En

folding trajectory. Wheneveran intermediate is observed there will be a question as to

h th thi i t f thConformational

whether this is part of a pathway or whether a funnel descriptionis more applicable. Note that the funnel provides the possibility

Entropy

for misfolding. This will typically result in multiple minima inthe energy landscape. Dill and Chan, Nature Struct. Biol. (1997), 4:10-19

The hydrophobic effectThe hydrogen bonding network of water is disrupted bya hydrophobic solute. The O-H bond vector will tend topoint away from the hydrophobic solute and toward anoxygen lone pair in order to reach the lowest enthalpyconfiguration The water molecules surrounding theconfiguration. The water molecules surrounding the hydrophobic solute are forced into a more ordered structure.Another way to realize this is to understand that the watermolecules have fewer possible hydrogen bond partners sincemolecules have fewer possible hydrogen bond partners sincethey cannot hydrogen bond to the hydrophobic solute. Theresult is a negative entropy of solvation for hydrophobicmolecules. If hydrophobic solutes cluster they can decreasetheir exposed surface area and thus S > 0 for aggregationof hydrophobic solutes in water.of hydrophobic solutes in water.

Hydrogen bonding in water

Hydrophobic interactions

Hydrophobic interactionsFor methane solvated by water the magnitude of the hydrophobic effect has been estimated to be 80 entropyunits (i.e. -80 J/mol-K). We can estimate the entropiccontribution due to exposed hydrophobic side chains inproteins When this is done the entropic effect is aproteins. When this is done the entropic effect is a dominant contribution to the driving force in protein folding.