Lecture 8' - Entropy, Free Energy, And Equilibrium

8/14/2019 Lecture 8' - Entropy, Free Energy, And Equilibrium

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Lecture 8- Entropy, Free Energy, and

Equilibrium


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Introduction

Last lecture, we approached thermodynamics from an

energetic basis:

Conservation of Energy (The 1st Law)

Work (w), Heat (q), and Internal Energy (E).

We identified several State Variables:

Internal Energy, E

Enthalpy, H = E + PV

we discussed prediction of energy changes (E, H, q, w),

resulting from a change in system state (a process).

and the relative probabilities of observing two states, at equilibrium.

Given knowledge of their energetic difference, E.

However, we made no attempt to predict the direction of

system changes.


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Example

Consider the spontaneous dissociation of ATP:

The 1st Law can be applied to this reaction:

reaction = change of chemical state.

analysis yields the resulting heat transfer and work

however, the 1st

law tells us nothing about the reactions direction. is it spontaneous, or not?

Intuitively, we expect an equilibrium to exist,

b/w ATP, ADP, and Pi.

but the 1st Law does not tell us where it lies.


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Example: Dialysis Experiment

A more intuitive example: A Dialysis Experiment

bag filled with sucrose, placed in water.

permeable to sucrose and H20.

clearly, system not at equilibrium.

Intuitive Expectation:

after some time, we expect net movement:

of sucrose out of the bag;

of H20 into the bag

until concentrations (in and out) are equal:

the point of equilibrium.

Why is the equilibrium state favored?


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Dialysis Experiment: Statistical View

Why is the equilibrium state favored?

not mechanicaltwo states nearly iso-energetic

minimal interaction b/w sucrose molecules.

Lets take a statistical view

Imagine a vast ensemble of such systems:

distributed over system microstates:

each a unique way to distribute sucrose over the solution volume.

assumption: each system microstate equally likely.

Then the most probable macroscopic system state:

the state of uniform concentration:

it maximizes the number of ways to distribute N particles:

each way = a different, but indistinguishable microstate.


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The Entropy

One determinant of the likelihood of a system state:

Is the number of ways, W to distribute particles, while maintaining

that state

either over energy levels, or over volume elements;

This expresses the inherent randomness of the system state.

For this reason, we define another system state variable:

The Entropy, S = kBlnW,

where kB = R/NA is the Boltzmann constant;

R = molar gas constant.

NA = Avogadros number


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Entropy is an Extensive Property

Here, lnW is chosen so that S is an extensive property.

Consider a system composed of two independent subsystems:

1. with individual entropies, S1 and S2, respectively.

2. And with no mixing of particles between the two.

Overall entropy = sum over subsystem entropies...

S = kBlnW = kBln(W1W2) = kBlnW1 + kBlnW2

= S1 + S2.

Thus, when combining quantities of a pure substance:

Entropies are additive.

This statistical thermodynamic definition of entropy:

Is quite clear, and also useful for analysis

but not clearly related to measurable physical quantities:


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Reversible and Irreversible Process

Last time, we classified processes as of two types:

reversible processes:

everywhere close to equilibrium: An equilibrium transition = series of equilibrium states.

Occupancies amongst the particle states thus distributed:

according to a Boltzmann distribution at each point in the

transition.

irreversible processes:

deviate from equilibrium.

transitional states indeterminate (not well defined)

First, lets consider reversible processes.


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S for a Reversible Process

Consider an infinitesimal, reversible process

between two equilibrium states, A and B.

which results in an exchange of heat, qrev

with the surroundings.

If we know the work done by the system, we can estimate E.

Question: what is the accompanying entropy change, S?

For convenience, lets choose a path such that work, w = 0:

in this case, E = q (from the 1st Law)

furthermoreno deformation of internal energies:

initial and final distributions over thesame set of particle states.

Distinguishes a transition which does no work

Note: results will apply equally to reversible processes with w != 0:

since work transfers zero entropy.


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S for a Reversible Process (cont.)


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Example: Isothermal, Reversible

Expansion of an Ideal Gas

Reversible, Isothermal expansion of an Ideal Gas:

1 mole of gas (n = 1), from V1 to V2.

for an ideal gas, energy independent of volume:

E = constant.

1st Law: dE = 0 = dq - dw dq = dw = PdV.

for an ideal gas,

PV = nRT dq = RT(dV/V), and

Thus, the entropy of an ideal gas increases upon expansion...

and also: heat is absorbed

note the lack of dependence on the (constant) Temperature. note also, for n moles:S = n R ln(V2/V1)


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Dialysis Experiment (Revisited)

Our dialysis bag can now be modeled as:

the reversible expansion of an ideal gas:

if we neglect the diffusion of water

Then (for n = 1 mole):

Where C1 and C2 express sucrose concentration.

The entropy thus increases on expansion.


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Example: Mixing of 2 Pure Substances


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S due to Heating


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Condition for Reversibility

Reversible Processes:

performed everywhere close to equilibrium.

can we define a condition for reversibility?

Consider our isothermal expansion of an (n =1) ideal gas:

For the system (i.e., the gas), we have: ;

For the surroundings: ;

The total S is the sum: Stotal = Ssys + Senv = 0.

This is the condition for reversibility:

Suniv = 0.


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Irreversible Processes

Irreversible Processes: are those which depart from equilibrium.

we can precisely define these, as well.

Ex.: Isothermal expansion of an ideal gas, in a vacuum. Entropy change of the gas (n = 1 mole):

no heat exchange occurs: q = 0.

however, S is path independent, and: .

Entropy change of the surroundings:

for the vacuum: Ssurr= 0;

Then the total entropy change is given by the sum:

This is the condition for irreversibility: Suniv > 0 .


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The 2nd Law of Thermodynamics

Combining these two conditions yields:

the 2nd Law of Thermodynamics:

Stotal >= 0,

= applies to a reversible process.

> applies to an irreversible process.

the total entropy change of a process is never negative.

although eitherSsys orSsurr can be negative.


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The Condition for Equilibrium

For a reversible process, the 1st Law can be written: dE = TdS PdV (since dS = dqrev/T).

Now we can answer our original question:

what is the criterion for equilibrium?

In particular, for an isolated system (constant E and V): from the 1st Law, we have:

reversibility dS = dqrev/T TdS = 0 (From the 1st Law)

dS = 0 (away from T = 0)

therefore, at equilibrium, S is an extremum of the system:

closer analysis reveals that S must be a maximum.

Condition for equilibrium (E, V constant):

An isolated system will reach equilibrium, when S is maximized. away from equilibrium, system entropy will spontaneously increase.


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The Gibbs Free Energy

Butbiological systems generally at constant T and P

what is the condition for equilibrium, in this case?

Lets define the Gibbs Free Energy: G = H TS. Then, dG = dH TdS SdT

= (dE + PdV + VdP) TdS SdT

for a reversible process, we have: dE = TdS PdV, and

dG = VdP SdT.

Finally, for a reversible process at constant P and T, we then have:

dG = 0

So, G is an extremum: in this case, a minimum.

Condition for equilibrium (P, T constant):

A system at constant P, T will reach equilibrium when G is a minimum.


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Example: Protein Denaturation

Consider the denaturation of a single folded polypeptide:

Gden = Hden - TSden

How will the two terms compensate? We expect Sden > 0,

since Sden = RT ln(Wden/Wnative) and Wden/Wnative >> 1.

We also expect Hden > 0 ,

since denaturation removes favorable interactions. Since both H and S are > 0, they will oppose

with the dominant factor determined by T.


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Phase Transition

The favorability of denaturation (biopolymer melting):

will increase with increasing T

The point of neutral favorability defined by

G = 0: this will occur at Tm = H/S.

the Proteins melting temperature.

For T > Tm , Gden < 0:

the denaturation process is favored. For T < Tm, Gden > 0:

the reverse reaction is favored

formation of the native state (folding).

Together, this describes a melting transition: centered at T = Tm, the protein melting temperature.


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Protein Denaturation: The

Equilibrium Distribution

We can also take a statistical perspective:

in order to determine the equilibrium distribution vs. T.

i.e.,nden/nnative vs. T.

Assume a very simplified model of protein energetics:

a single native conformation,

which has energyEn.

a large number of equivalent, denatured conformations:

each of which has energy,Ed.


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Protein Denaturation: The

Equilibrium Distribution


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Conclusion

In this Lecture, we identified a new state variable:

The Entropy, S.

The Entropy change (S) accompanying a change of state was

investigated, and used to:

distinguish reversible (Stotal = 0) from irreversible (Stotal > 0)

processes,

As expressed by the 2nd Law of Thermodynamics.

discuss the condition for equilibrium for a process at constant E, V.

Another state variable was then defined:

The Gibbs Free Energy, G.

G was used to investigate the condition for equilibrium at constant T,

P;

and to re-express the Boltzmann distribution in a more familiar form.

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Lecture 8' - Entropy, Free Energy, And Equilibrium