Chemistry 433Chemistry 433

Lecture 14Lecture 14Gibbs Free Energy

and Chemical Potential

NC State University

The internal energy expressed in terms of its natural variables

We can use the combination of the first and second laws to derive an expression for the internal energy in terms of its natural variables. If we consider a reversible process:

dU = δq + δwδw = -PdV (definition of work)

S ( )δq = TdS (second law rearranged)Therefore, dU = TdS - PdV

This expression expresses the fact that the internal energyU has a T when the entropy changes and a slope -P whenthe volume changes We will use this expression to derivethe volume changes. We will use this expression to derivethe P and T dependence of the free energy functions.

The Gibbs energy expressed in terms of its natural variables

To find the natural variables for the Gibbs energy we beginTo find the natural variables for the Gibbs energy we beginwith the internal energy:dU = TdS - PdV

d b i iand substitute into:dH = dU + PdV + VdPto find:dH = TdS + VdP (S and P are natural variables of enthalpy)and using:dG = dH - TdS - SdTdG = dH - TdS - SdTwe find:dG = -SdT + VdP (T and P are natural variables of G)O i h G i f l It t l i blOnce again we see why G is so useful. Its natural variablesare ones that we commonly experience: T and P.

The variation of the Gibbs energy with pressure

We shown that dG = VdP - SdT. This differential can be used to determine both the pressure and temperature dependence of the free energy. At constant temperature:p gy pSdT = 0 and dG = VdP. The integrated form of this equation is:

For one mole of an ideal gas we have:ΔG = VdP

P1

P2

ΔGm = RT dPPP1

P2

= RT ln P2

P1

Note that we have expressed G as a molar quantity Gm = G/n.

The variation of the Gibbs energy with pressure

We can use the above expression to indicate the pfree energy at some pressure P relative to the pressure of the standard state P = 1 bar.

P

G0(T) is the standard molar Gibb’s free energy for a gas. G ’ f

Gm T = G0 T + RT ln P1 bar

As discussed above the standard molar Gibb’s free energy is the free energy of one mole of the gas at 1 bar of pressure. The Gibb’s free energy increases logarithmically with pressure. gy g y pThis is entirely an entropic effect.Note that the 1 bar can be omitted since we can write:

RT ln P1 bar = RT ln P – RT ln 1 = RT ln P

The pressure dependence of G for liquids and solids

If we are dealing with a liquid or a solid the molar volume g qis more or less a constant as a function of pressure. Actually, it depends on the isothermal compressibility, κ = 1/V(∂V/∂P) but κ is very small It is a number of theκ = -1/V(∂V/∂P)T, but κ is very small. It is a number of the order 10-4 atm-1 for liquids and 10-6 atm-1 for solids. Wehave discussed the fact that the density of liquids is

t t l ff t d b Th ll l fnot strongly affected by pressure. The small value of κis another way of saying the same thing.For our purposes we can treat the volume as a constant p pand we obtain

Gm T = G0 T + Vm P – 1m m

QuestionWhat is the isothermal compressibility κ = -1/V(∂V/∂P)T for an ideal gas?

A. κ = -1/VB. κ = 1/PC. κ = 1/TD. κ = nRT/P

QuestionWhat is the isothermal compressibility κ = -1/V(∂V/∂P)T for an ideal gas?

A. κ = -1/VB. κ = 1/PC. κ = 1/TD. κ = nRT/P

QuestionCompare the following two equations for the pressure dependence of the Gibbs free energy. Which one applies t th f ti f di d f hit ?to the formation of diamond from graphite?

PA. Gm T = G0 T + RT ln P1 bar

B. Gm T = G0 T + Vm P – 1

QuestionCompare the following two equations for the pressure dependence of the Gibbs free energy. Which one applies t th f ti f di d f hit ?to the formation of diamond from graphite?

PA. Gm T = G0 T + RT ln P1 bar

B. Gm T = G0 T + Vm P – 1

Systems with more than one component

Up to this point we have derived state functions for pure p p psystems. (The one exception is the entropy of mixing).However, in order for a chemical change to occur we musthave more than one component present We need generalizehave more than one component present. We need generalizethe methods to account for the presence of more than onetype of molecule. In the introduction we stated that we wouldd thi i tit ll d th h i l t ti l Thdo this using a quantity called the chemical potential. Thechemical potential is nothing more than the molar Gibbsfree energy of a particular component. Formally we write itgy p p ythis way:

μ = ∂G Rate of change of G as number ofmoles of i changes with all otherμ i = ∂ni T,P, j ≠ i

moles of i changes with all othervariables held constant.

Example: a gas phase reactionp g p

Let’s consider a gas phase reaction as an example. We willt tb k luse a textbook example:

N2O4 (g) = 2 NO2 (g)We know how to write the equilibrium constant for this reaction.q

K =PNO2

2

PN2O4

At constant T and P we will write the total Gibbs energy as:

dG = μ dn + μ dn

N2O4

dG = μNO2dnNO2

+ μN2O4dnN2O4

dG = 2μNO2dn – μN2O4

dnWe use the reaction stoichiometry to obtain the factor 2 for NO2.

Definition of the Gibbs free energy change for chemical reactionchange for chemical reaction

We now define ΔrxnG:

Δ rxnG = dGdn T,P

This is ΔrxnG but it is not ΔrxnGo! Note that we will use ΔrxnGand ΔG interchangably.If we now apply the pressure dependence for one componentIf we now apply the pressure dependence for one component,

t lti t t

Gm T = G0 T + RT ln P1 bar

to a multicomponent system:μ i T = μ i

0 T + RT ln Pi

1 bar

These two expressions are essentially identical. The chemical potential, μi, is nothing more than a molar free energy.

QuestionHow should one think of the chemical potential, μof component j?of component j?

A. It is the potential energy of that componentB It i l f f th t tB. It is a molar free energy of that componentC. It is potential entropy of that componentD. It is a molar entropy of that component

QuestionHow should one think of the chemical potential, μof component j?of component j?

A. It is the potential energy of that componentB It i l f f th t tB. It is a molar free energy of that componentC. It is potential entropy of that componentD. It is a molar entropy of that component

Single component:G T = G0 T + RT ln P

Multiple components each have a μj:

Gm T = G T + RT ln 1 bar

Pμ i T = μ i0 T + RT ln Pi

1 bar

Application of definitions to the chemical reactionchemical reaction

We can write the Gibbs energy as:

and use the chemical potentials:ΔG = 2μNO2

– μN2O4

p

μNO2= μNO2

0 + RT ln PNO2

μ = μ0 + RT ln P

to obtain the following:

μN2O4= μN2O4

+ RT ln PN2O4

ΔG = 2μNO20 – μN2O4

0 + 2RT ln PNO2– RT ln PN2O4

ΔG = ΔG0 + RT lnPNO2

2

ΔG0 = 2 0 0ΔG = ΔG0 + RT ln 2

PN2O4

, ΔG0 = 2μNO20 – μN2O4

0

Note the significance of ΔG and ΔGog

The change ΔG is the change in the Gibbs energy function.It h th ibl f lIt has three possible ranges of value:ΔG < 0 (process is spontaneous)ΔG = 0 (system is at equilibrium)( y q )ΔG > 0 (reverse process is spontaneous)

On the other hand ΔGo is the standard molar Gibbs energyOn the other hand ΔG is the standard molar Gibbs energychange for the reaction. It is a constant for a given chemicalreaction. We will develop these ideas for a general reactionl t i th F l t’ id th t tlater in the course. For now, let’s consider the system atequilibrium. Equilibrium means ΔG = 0 so:

PNO2

0 = ΔG0 + RT ln K , ΔG0 = –RT ln K , K =PNO2

PN2O4

QuestionWhich statement is true at equilibrium if K = 2?

A ΔG RT l (2) ΔGo 0A. ΔG = -RT ln(2), ΔGo = 0B. ΔG = 0, ΔGo = -RT ln(2)C. ΔG = ΔGo = -RT ln(2)B. ΔG = 0, ΔGo = 0

QuestionWhich statement is true at equilibrium if K = 2?

A ΔG RT l (2) ΔGo 0A. ΔG = -RT ln(2), ΔGo = 0B. ΔG = 0, ΔGo = -RT ln(2)C. ΔG = ΔGo = -RT ln(2)B. ΔG = 0, ΔGo = 0

Temperature dependence of ΔGo

The van’t Hoff equationThe van t Hoff equation We use two facts that we have derived to determine thet t d d f th f Htemperature dependence of the free energy. Here are weare skipping some derivations that are not often used andderiving a very useful expression in biology.

ΔG0 = –RT ln(K)ΔG0 = ΔH0 – TΔS0

ln(K) = – ΔH0

RT + ΔS0

RΔH0 1 1ln(K2) – ln(K1) = ΔH0

R1T1

– 1T2

If we plot ln(K) vs 1/T the slope is -ΔHo/R. This is a useful expression for determining the standard enthalpy change.

Example: the formation of diamond p

Graphite and diamond are two forms of carbon. Given that th f f f ti f di d ithe free energy of formation of diamond is:C(s, graphite) C(s, diamond) ΔrGo = + 2.90 kJ/moland the densities are:ρ(graphite) = 2.26 and ρ(diamond) = 3.51calculate the pressure required to transform carbon intodiamonddiamond.

Solution: Graphite will be in equilibrium with diamond when:

0 = ΔG0 + ΔVm P – 1

P = 1 – ΔG0

= 1 – ΔG0

P = 1 –ΔVm

= 1 – Mρd

– Mρgr

Example: the formation of diamond p

Plugging the values we find:

0 = ΔG0 + ΔVm P – 1ΔG0 ΔG0

P = 1 – ΔGΔVm

= 1 – ΔGMρd

– Mρgr

2900 J/ l= 1 – 2900 J/mol

0.012 kg/mol 13510 – 1

22603510 2260= 1.5 x 109 Pa = 15000 bar

Protein folding example:T t t d lTwo state model

kf

U Fku

Unfolded Foldedku

K = [F]/[U]

K = ff/(1-ff)( )Fraction folded ff Fraction unfolded 1-ff

Thermodynamic modelyff = 1/(1+K)K ΔGo/RTK = e-ΔGo/RT

ff = 1/(1 + e-ΔGo/RT)ff = 1/(1 + e-ΔHo/RTeΔSo/R)

The temperature at which the protein is 50% folded can be defined as Tis 50% folded can be defined as Tmthe melt temperature. At Tm , ΔGo = 0 or Tm = ΔHo/ΔSo.At Tm , ΔG 0 or Tm ΔH /ΔS .

Equilibrium melt curvesq

Mostly folded o o

Mostly unfoldedTm

Mostly unfolded

In this case: Tm = 300 K = ΔHo/ΔSo

Van’t Hoff plotsp

Slope = -ΔHo/R

The standard method for obtaining theThe standard method for obtaining the reaction enthalpy is a plot of ln K vs. 1/T

QuestionWhat determines the steepness of the equilibrium melt curve?A. The melt temperatureB. The equilibrium constantC. The ratio of the enthalpy to the entropyD. The magnitude of the enthalpy and entropy

QuestionWhat determines the steepness of the equilibrium melt curve?A. The melt temperatureB. The equilibrium constantC. The ratio of the enthalpy to the entropyD. The magnitude of the enthalpy and entropy

QuestionWhat can you say about the enthalpy of reaction for the process shown in the figure below?p gA. The reaction is exothermic.B. The reaction is endothermic.C. The reaction is spontaneous.D. None of the above.

QuestionWhat can you say about the enthalpy of reaction for the process shown in the figure below?p gA. The reaction is exothermic.B. The reaction is endothermic.C. The reaction is spontaneous.D. None of the above.