Mixtures, Equilibrium& Chemical Potential µ

Single compartment• Total G -> minumum

equlibrium

Two compartments • Chemical potential µA of each component is equal in both compartments in equilibrium

A (gas)

A (liquid)

• Solutions• Suspensions• Colloids (milk,

nanoparticles, colloidal silver, sulphur, gold)

From minimizing Gibbs Free Energy to the Equilibrium Constant (K)

Chemical Potential µ• G = H-TS is an extensive variable• For a pure substance J the chemical potential is

defined as molar Gibbs energy: µJ ≡ Gm=G/n

• Chemical Potential is molar Free Energyof a particular chemical ingredient in amixture

• In a mixture of n1,n2,..,nJ

µJ =∂G∂nJ

⎛

⎝⎜

⎞

⎠⎟p,T ,ni≠J

Dependence of S,µ on P,T,V• Boltzmann: S = kB ln N, or per

mole R ln (n), where n is the number of microstates of 1 molecule

• n ~ volume V

• GasV~P-1

vacuum

• Solution• 1

V~C-1€

Si = Si0 + R lnVi

Vi0

µ =Gm = Hm −TSm

€

µgasi = µi

g0 − RT lnVi

Vi0

µgasi = µi

g0 + RT ln PiPi0

µsolutioni = µi

s0 + RT lnCi

Ci0

The fundamental equation

dG = V×dP – S×dT + µA×dnA + µB×dnB + …Fundamental Equation of Chemical Thermodynamics

JinTpJJ n

G

¹

÷÷ø

öççè

涶

º,,

µ

• A general formula for the Gibbs energy change of a mixture can be derived

• µA is the slope of G vs nA

AB

The chemical potential in solution

µi = µic0 + RT ln ci

c0

Chemical potential (free energy per particle i) in solution increases with log(concentration) for entropic reasons (less states):

For non-ideal solutions: ci becomes effective concentration ai

The chemical activity ai , an effective particle concentration, replaces ci for non-ideal solutions: ai = gi·ci

• ai º activity of species i;

• gi º activity coefficient, gi = 1 , ai = ci for ideal solution;

• ci = concentration of species i, c0 defines ‘standard’ state

µ gasi = µi

P0 + RT ln PiP0

dG = V×dP – S×dT + µA×dnA .. (fundamental eq.)

GP,T = µA×nA + µB×nB + …

µi = µic0=1 + RT lnci

We may choose units and standard states, eg 1 M

**Chemical Equilibrium• Consider the general chemical reaction:

aA + bB + … « cC + dD + …• When is this system at equilibrium?– When none of its thermodynamic properties change

with time (T,P,V, [A],[B],[C]..).– When the rate of the forward reaction exactly

equals the rate of the reverse equilibrium– When the total Gibbs Free energy is at its minimum

Only reactantsOnly products

Mixture of both

**Thermodynamic Criterion Of Equilibrium

If drG = 0, reaction is in chemical equilibrium.

If drG < 0, reaction will proceed in direction written.

If drG > 0, reaction will proceed in reverse direction.

dG = µA×dnA + µB×dnB +µC×dnC + µD×dnD …

In forward reactionnA , nB go down, dni is negativenC , nD go up, dni is positive

aA + bB + … « cC + dD + …

**ExampleStandard conditions include:• 1 atm pressure • 1M concentration (solutions) • Pure solid (if a solid) or pure

liquid (if a liquid) • For elements, the standard

Gibbs energy of formation, DGf

0 of an element in its normal state is 0

• There is no standard state for temperature. G will vary with temperature. – 298K (i.e. 25°C) is a common

temperature chosen for standard reference values of G

**Changes of the number of molecules in a reaction (dni)

dndn ii n=

Let ni be a stochiometry coefficient of molecule i in a reaction (e.g. -1, -3, and 2 for N2+3H2↔2NH3, or –N2-3H2+2NH3 =0).

Every time the reaction shifts the molecules of each type appear and disappear according to this relation

e.g. if dn = 1, it will tell us that one molecule of N2 and 3 molecules of H2 disappeared and 2 molecules of NH3 appeared:

n is the number of ‘generalized’ reaction product, ‘reaction coordinate’

**Equilibrium Constant of a Reaction

iii cRT ln0 += µµdni =ν idnegA↔ 2B :vA = −1,vB = 2; dnA = −dn, dnB = 2dn

Assuming standard concentration is c0 = 1 Mwhere ci is the current molar concentration:

Concentrations for minimal free energy G:

µidni =i∑ ν idn(µi

0 + RT lncii∑ ) = 0

RT ln ciνi

i∏ = − ν iµi

0

i∑ ≡−ΔGreaction

0

RTGK reaction0

ln D-=

Equilibrium Concentrations & DG0

• Reaction

• Equilibrium constant

• Reaction Free Energy

• Equilibrium concentrations are related to DG0

11

RTGK reaction0

ln D-=

K ≡Cproduct1

ν p1 Cproduct2ν p2 ...

Creac t1νr1 Creact2

νr 2 ...ΔG0

≡ ν piµi0

products∑ − ν rjµ j

0

reactants∑

ν r1R1 +ν r2R2 +..↔ν p1P1 +ν p2P2 +..

The Equilibrium Constant, K and concentration ratios. Examples

For reaction: aA + bB « cC + dDthe equilibrium constant is written as: b

BaA

dD

cC

ccccK =

• Units of K depend on the reaction stoichiometry,• Examples: • A « B : K = CB / CA K is dimensionless• D + R « DR : K = CDR / (CD CR ) K [M-1]• DR « D + R : K = (CD CR ) / CDR K [M]

For reaction: A « B : K = CB / CA

Relationship between Gibbs free energy of reaction and the equilibrium concentrations is defined by constant K

lnK =−ΔreactionG

o

RT

Review: concentrations and energies

Where G0 comes from chem. potentials (molar free energies) at standard 1M concentrations, while K is calculated at equilibrium concentrations.

K =CD

νDCEνE ..

CAνACB

νB ..

ΔreactionGo = ΔHo −TΔSo

Concentrations and Heats

• DG0 contains Enthalpic (DH0) and Entropic (-TDS0) contributions. How do we measure them?

• Example: Levosalbutamol vs Salbutamol (racemic)

• Chiral transition: ln(CR/CL) = -(GR-GL)/(RT)

RTGKo

reactionD-=ln

“As a bronchodilator, it is used to treat asthma and Chronic obstructive pulmonary disease (COPD). In general, levosalbutamol has similar pharmacokinetic and pharmacodynamic properties to salbutamol; however, its manufacturer, Sepracor, has implied (although not directly claimed) that the presence of only the R-enantiomer produces fewer side effects.” Wikipedia

Enthalpy and Entropy from Equilibrium at several temperatures: van’t Hoff Equation

RS

RTHK

or

or D

+D

-=ln

RS

RTHK

RS

RTHK

or

or

or

or

D+

D-=

D+

D-=

22

11

ln

ln

÷÷ø

öççè

æ-

D=÷÷

ø

öççè

æ

122

1 11lnTTR

HKK o

r

Equilibrium constant at two different temperatures:

Subtracting:

Substituting DG for DH-TDS:

Jacobus Henricus van’t Hoff (1852-1911),a Dutch chemist

Alternative to Calorimetry. Idea:• Equilibrium concentrations depend on T• Higher T will favor the reaction direction with gain of entropy.• Recipe: Measure K ( T )

Plotting ln K vs 1/T

)10ln(3.2,3.23.2

log =D

+D

-=RS

RTHK

or

or

RHor /D-

RS

RTHK

or

or D

+D

-=ln

Van’t Hoff Equation with log10

The slope isThe intercept is RSor /D

Example of van’t Hoff Plot

1/T(K)0.0020 0.0022 0.0024 0.0026 0.0028 0.0030 0.0032

log

K

8

10

12

14

16

18Nd(OH)3(cr) + 3H+ = Nd3+ + 3H2O(l)

Review• The chemical potential of

component J:– Gas– Liquid mixture– ΔG and entropy of mixing.

• The chemical equilibrium– K via concentrations and

reaction stoichiometry– From K, to ΔGo

– From K at T1 and T2, to ΔHo and ΔSo, Van’t Hoff lnK = −

ΔrGo

RT

÷÷ø

öççè

æ-

D-=÷÷

ø

öççè

æ

212

1 11lnTTR

HKK o

0

ln0

ccRT ic

ii += µµ

]][[][..,

1 BABAKgeaK

n

iii

•==Õ

=

n

µig = µi

P0 + RT ln PiP0

ai below may also be molar fraction xi or concentration ci depending on the standard state and ideality

RS

RTHK

or

or D

+D

-=ln

Gas-Solution Equilibrium for Each Solution Ingredient

• i may be water or drug

• xi - Solute (or solvent) molar fraction

• * - pure (saturated) ingredient

µi,liq = µi,vap ; µi,liq* = µi,vap

*

µi,liq* + RT ln ci

ci*

⎛

⎝⎜

⎞

⎠⎟= µi,vap

* + RT ln pipi*

⎛

⎝⎜

⎞

⎠⎟

RT ln cici*

⎛

⎝⎜

⎞

⎠⎟= RT ln

pipi*

⎛

⎝⎜

⎞

⎠⎟

cici*

⎛

⎝⎜

⎞

⎠⎟i

=pipi*

⎛

⎝⎜

⎞

⎠⎟

Raoult’s Law

• Solvent (eg water) pressure vs molar fraction of non-volatile solute

• Vapor pressure of a solution is decreased as the solute concentration is increased

• P*water = 0.23 bar at 20oC (100C?)

P = xw P*w= (1- xsolute) P*

w

P*w-P=DP = xsolute P*

w

French physicist François-Marie Raoult

• Water pressure will be lower as you add salt• Salty water will boil at higher temperature

xwi ≡ nwi / ntotal

w = pgi / pi*g pure i is water

Henry’s Law (gas in solvent)

• Gas dissolves in liquid proportionally to its pressure. Example: Oxygen in blood

• Here K is an empirical constant, slope of the tangent to the experimental curve.

• Pi = xi P0 - Dalton’s law

Pi,gas = xi,sol×KxH

Henry’s Law

Raoult’s Law for xsolvent®1Henry’s Law for xsolute®0Mixtures that obey

andare called ideal-dilute solutions.

xliqi = pgi / pi

*g pure i is gas component (eg oxygen)

Air Pressure vs O2 in Blood• oxygen (O2) : KH=769.2 L·atm/mol• carbon dioxide (CO2) : KH=29.4 L·atm/mol• hydrogen (H2) : KH=1282.1 L·atm/mol

8,848 m (M.Everest)4,421 m (M.Whitney)

Temp (C) P(kPa) P(mmHg)0 0.6 4.53 0.8 6.05 0.9 6.88 1.1 8.310 1.2 9.012 1.4 10.514 1.6 12.016 1.8 13.518 2.1 15.819 2.2 16.520 2.3 17.521 2.5 18.722 2.6 19.823 2.8 21.124 3.0 22.425 3.2 23.826 3.4 25.227 3.6 26.728 3.8 28.429 4.0 30.030 4.2 31.532 4.8 36.035 5.6 42.040 7.4 55.550 12.3 92.360 19.9 149.370 31.2 234.180 47.3 354.990 70.1 525.9100 101.3 760.0

Water pressure vs T

Ph = P0 • e -M g h/RT