thermodynamics and kineticsthermodynamics and...

Thermodynamics and KineticsThermodynamics and Kinetics

Free Energy Applications

NC State University

The variation of the Gibbs energy with pressure

We have 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:G = VdP

P1

P2

Gm = 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, = 1/V(V/P) but is very small It is a number of the = -1/V(V/P)T, but 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 is 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

The formation of diamond

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) rGo = + 2.90 kJ/moland the densities are:(graphite) = 2.26 and (diamond) = 3.51calculate the pressure required to transform carbon intodiamonddiamond.

Solution: Graphite will be in equilibrium with diamond when:

0 = G0 + Vm P – 1

P = 1 – G0

= 1 – G0

P = 1 –Vm

= 1 – Md

– Mgr

Example: the formation of diamond p

Plugging the values we find:

0 = G0 + Vm P – 1G0 G0

P = 1 – GVm

= 1 – GMd

– Mgr

2900 J/ l= 1 – 2900 J/mol

0.012 kg/mol 13510 – 1

22603510 2260= 1.5 x 109 Pa = 15000 bar

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:

= G Rate of change of G as number ofmoles of i changes with all other 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 = dn + dn

N2O4

dG = NO2dnNO2

+ N2O4dnN2O4

dG = 2NO2dn – N2O4

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 rxnG:

rxnG = dGdn T,P

This is rxnG but it is not rxnGo! Note that we will use rxnGand G 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: i T = i

0 T + RT ln Pi

1 bar

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

Application of definitions to the chemical reactionchemical reaction

We can write the Gibbs energy as:

and use the chemical potentials:G = 2NO2

– N2O4

p

NO2= NO2

0 + RT ln PNO2

= 0 + RT ln P

to obtain the following:

N2O4= N2O4

+ RT ln PN2O4

G = 2NO20 – N2O4

0 + 2RT ln PNO2– RT ln PN2O4

G = G0 + RT lnPNO2

2

G0 = 2 0 0G = G0 + RT ln 2

PN2O4

, G0 = 2NO20 – N2O4

0

Calculating G and Go

The free energy is

G = Go + RT ln QG = Go + RT ln Q

where Q is defined from the sample reaction

As

Under standard conditionsUnder standard conditions

dandG = Go

Note the significance of G and Gog

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

On the other hand Go is the standard molar Gibbs energyOn the other hand G 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 G = 0 so:

PNO2

0 = G0 + RT ln K , G0 = –RT ln K , K =PNO2

PN2O4

The pressure dependence of KWe can see from the gas phase form of the equilibriumconstant that the free energy depends on pressure as well.For the general gas phase reactionFor the general gas phase reaction,

we can write the equilibrium constant as

And the free energy is

From Dalton’s law

The pressure dependence of KIf we substitute these mole fractions and total pressure intothe equilibrium constant we have

Which depends on the total pressure unless z – c – d = 0.This expression shows that, in general, the free energydoes depend on the total pressure.p p

Pressure dependence problemN2O4(g) 2 NO2(g)

K = 0.25

What pressure of NO2 do you need to start with to formneed to start with to form

1.0 atm of N2O4?


K = 0.25

Our condition is

So we have


K = 0.25

Solve for the mole fraction:


K = 0.25Solve for the mole fraction:

Since the mole fraction of NO2 is 1 initially, the pressure of2 y, pNO2 must 1.5 atm initially.

Temperature dependence of Go

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 ftemperature dependence of the free energy.

G0 = –RT ln(K)0 0 0G0 = H0 – TS0

ln(K) = – H0

RT + S0

RRT Rln(K2) – ln(K1) = H0

R1T1

– 1T2

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

Van’t Hoff plotsp

Slope = -Ho/RNote: Ho > 0Note: Ho > 0

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

Van’t Hoff plot for drug bindingVan t Hoff plot for drug bindingA practical example of the application of the

’t H ff ti b f d i d bi divan’t Hoff equation can be found in drug binding.The equilibrium constant for drug binding to an

ti it b d b flactive site can be measured by fluorescence, NMR, etc. at various temperatures. Then one may plot ln K vs 1/T and fit the result to a linemay plot ln K vs. 1/T and fit the result to a line.In most cases the binding will be exothermic to that Ho < 0 and then slope of the line will bethat H 0 and then slope of the line will be positive rather than negative as shown in the previous slide.

Van’t Hoff plot for drug bindingVan t Hoff plot for drug binding

Slope = Ho/RSlope = -H /RNote: Ho < 0

In this example, the slope is positive because the enthalpy of binding is negative (i.e. binding is exothermic).

Thermodynamics of glycolysisReaction kJ/molD-glucose + ATP D-glucose-6-phosphate + ADP Go = -16.7D-glucose-6-phosphate D-fructose-6-phosphate Go = +1.7g p p p pD-fructose- 6-diphosphate + ATP D-fructose-1,6-diphosphate + ADP

Go = -14.2D-fructose-1,6-diphosphate glyceraldehyde-3-phosphate +

dihydroxyacetone phosphate Go = +23.8dihydroxyacetone phosphate glyceraldehyde-3-phosphate Go = + 7.5 glyceraldehyde-3-phosphate + phosphate + NAD+

1 3 di h h l t NADH H Go 6 31,3-diphosphoglycerate + NADH + H+ Go = + 6.3 1,3-diphosphoglycerate + ADP 3-phosphoglycerate + ATP Go = -18.8 3-phosphoglycerate 2-phosphoglycerate Go = +4.6 2 phosphoglycerate 2 phosphoenolpyruvate + H O Go = +1 72-phosphoglycerate 2-phosphoenolpyruvate + H2O Go = +1.7 2- phosphoenolpyruvate + ADP pyruvate + ATP Go = -31.4 pyruvate + NADH + H+ lactate + NAD+ Go = -25.1 pyruvate acetaldehyde + CO2 Go = -19.8pyruvate acetaldehyde CO2 G 19.8 acetaldehyde + NADH + H+ ethanol + NAD+ Go = -23.7

Phosphorylation of glucoseD-glucose + ATP D-glucose-6-phosphate + ADP Go = -16.7

The reaction can be decomposed into two reactions.p

D-glucose + phosphate D-glucose-6-phosphate + H2O Go = +14.3

ATP + H2O ADP + phosphate Go = -31.0

The sum of the two reactions results in an overall negative free energyh d t d d diti I thi th t lchange under standard conditions. In this manner the strongly

spontaneous hydrolysis of ATP is coupled to the otherwise unspontaneousglucose phosphorylation. This reaction is typical of the role played by ATPin the cellin the cell.

Note that the values for Go assume a concentration of 1 M. Clearly, theconcentrations in the cell are often quite different from the standard stateconcentrations in the cell are often quite different from the standard stateand this will have profound consequences for the direction of spontaneouschange.

The role of enzymesAll of the reactions in the glycolytic pathway are catalyzed by enzymes.For example, the reaction considered on the previous slide is catalyzedby hexokinase. The role of the enzyme is to speed up the reaction, buty y p p ,the enzyme does not change thermodynamics of the process. The roleof enzymes is the same as that of any catalyst. Catalysts affect the kinetics of the reaction, but not the thermodynamics. We will considerthe role of catalysts in the second half of the course.

Notice that Go for certain steps is positive. For example,D l 6 h h t D f t 6 h h t Go 1 7D-glucose-6-phosphate D-fructose-6-phosphate Go = +1.7is catalyzed by phosphoglucose isomerase.

The equilibrium constant for this process isThe equilibrium constant for this process is K = exp{-Go/RT} = exp{-1700/8.31/310} ~ 0.5The concentration of D-fructose-6-phosphate at equilibrium will be less than that of D-glucose-6-phosphate.less than that of D glucose 6 phosphate.

ExampleWhat is the equilibrium concentration of D-glucose-6-phosphate givenAn initial concentration of 6 mM in the growth medium.

D-glucose-6-phosphate D-fructose-6-phosphate Go = +1.7

Solution:K = exp{-Go/RT} = exp{-1700/8.31/310} ~ 0.5

D-gluc-6P D-fruc-6PInitial 0.006 0Change –x xFinal 0.006 – x x

ExampleWhat is the equilibrium concentration of D-glucose-6-phosphate givenAn initial concentration of 6 mM in the growth medium.

D-glucose-6-phosphate D-fructose-6-phosphate Go = +1.7

Solution:K = exp{-Go/RT} = exp{-1700/8.31/310} ~ 0.5

D-gluc-6P D-fruc-6PInitial 0.006 0Change –x xFinal 0.004 0.002

Intracellular conditions are not equilibrium conditions

If th b t t i i f ti i hi hl tIf the subsequent step in a series of reactions is highly spontaneousthis will tend to deplete the product for the previous reaction. Thus,more of the product will tend to be formed by L’Chateliers principle.

We can observe this quantitatively by considering the value of Q, thereaction quotient. Since,

G = Go + RT ln QG G RT ln Q the value of G may not be zero. In other words, the coupled seriesof reactions in the cell are not at equilibrium but rather they proceedunder steady state conditions where the concentrations are not at 1 M,ybut are poised so that the overall of effect on a series of reactions isto produce a net spontaneous change.

Sample Problem in MetabolismThe enzyme aldolase catalyzes the conversion of fructose 1,6-diphosphate (FDP) to dihydroxyactone phosphate (DHAP) andglyceraldehyde-3-phosphoate (GAP). Under physiological conditionsg y y p p ( ) p y gthe concentrations of these species in red blood cells (erythrocytes) are [FDP] = 35 M, [DHAP] = 130 M and [GAP] = 15 M. Will theconversion occur spontaneously under these conditions?

Solution: The standard free energy change for the reaction isFDP DHAP + GAP Go = +23.8 kJ

dand Q = [DHAP][GAP]/[FDP] = (130 x 10-6)(15 x 10-6)/ (35 x 10-6)

= 5.57 x 10-5

G = Go + RT ln Q = 23800 J/mol + (8.31 J/mol-K)(310 K)ln(5.57 x 10-5)= -1434 J/mol or -1.43 kJ/mol

The reaction will occur spontaneously under the conditions of the cell.

Sample problemThe thermal unfolding of a serine protease (SP) from a thermophilic bacterium was studied and gave the followingTh d i d tThermodynamic data: Go = -15.4 kJ/mol and Ho = -75.0 kJ/mol for the process

SP (random coil) SP (native)Determine the melt temperature.

A. 82 oCB. 92 oCC 102 oC

So = (Ho - Go)/T = (-75.0+15.4)/298 x1000 = -200 J/mol-K

G 0C. 102 oCD. 112 oC

The melt temperature occurs when Go = 0.T = Ho/So = -75,000/(-200) = 375 K = 102 oC!

thermodynamics and kineticsthermodynamics and...

Documents