Solutions

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M ole fraction of com ponent A = xAM ass Fraction of com ponent A = m A

Volum e Fraction of com ponent A = fA

Typically we m ake a binary blend, A + B, w ith m ass fraction, m A, and want volum e fraction, fA, or m ole fraction , xA.

fA = (m A/rA)/((m A/rA) + (m B/rB))

xA = (m A/M WA)/((m A/M WA) + (m B/M W B))

Solutions

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Three ways to get entropy and free energy of m ixing

A) Isotherm al free energy expression, pressure expression

B) Isotherm al volum e expansion approach, volum e expression

C) From statistical therm odynam ics

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M ix two ideal gasses, A and B

p = pA + pB pA is the partial pressure pA = xAp

For single com ponent m olar G = µµ0 is at p0,A = 1 barAt pressure pA for a pure com ponentµA = µ0,A + RT ln(p/p0,A) = µ0,A + RT ln(p)

For a m ixture of A and B with a total pressure ptot = p0,A = 1 bar and pA = xA ptot

For com ponent A in a binary m ixture

µA(xA) = µ0.A + RT xAln (xA ptot/p0,A) = µ0.A + xART ln (xA)

Notice that xA m ust be less than or equal to 1, so ln xA m ust be negative or 0So the chem ical potential has to drop in the solution for a solution to exist.

Ideal gasses only have entropy so entropy drives m ixing in this case.

This can be written, xA = exp((µA(xA) - µ0.A)/RT)W hich indicates that xA is the Boltzm ann probability of finding A

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M ix two real gasses, A and B

µA* = µ0.A if p = 1

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For isotherm al DU = CV dT = 0Q = W = -pdV

For ideal gas Q = W = nRTln(Vf/Vi)

Q = DS/TDS = nRln(Vf/Vi)Consider a process of expansion of a gas from VA to Vtot

The change in entropy is DSA = nARln(Vtot/VA) = - nARln(VA/Vtot)

Consider an isochoric m ixing process of ideal gasses A and B.

A is originally in VA and B in VB

Vtot is VA + VB

The change in entropy for m ixing of A and B is

Dsm ixing A and B = - nARln(VA/Vtot) - nBRln(VB/Vtot) = - nR(xAlnxA + xBlnxB)For an isotherm al, isochoric m ixture of ideal gasses (also isobaric since P ~ T/V)

For ideal gasses DH m ixing = 0 since there is no interationDG m ixing = DH m ixing - TDSm ixing = - TDSm ixing = nRT(lnxA + lnxB)So the m olar Gibbs Free energy for m ixing is DG m ixing = RT(xAlnxA + xAlnxB)

Ideal Gas M ixing

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Statistical Therm odynam ics

Boltzm ann’s Law: DS = kBlnWW Is the num ber of states

For m ixing of nA and nB with n total m olecules

W = n!/(nA! nB!)Sterling ’s approxim ation for large n, n! ~ n lnn – n

We assum e that n is large then

lnW = -(nA ln(nA/n) + nB ln(nB/n))DS = -kB (nA ln(nA/n) + nB ln(nB/n)) = -nkB (xA ln(xA) + xB ln(xB))

DG m ixing = DH m ixing - TDSm ixing = - TDSm ixing = nRT(xA lnxA + xB lnxB)

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Som e Types of Entropy

Therm odynam ic entropy m easured experim entally, Q /T

Configurational also called Com binatorial Conform ationalTranslational and Rotational Entropy (Brownian m otion)

Vibrational entropyConform ational entropy com puted in internal or Cartesian coordinates (which can even be different from each other)Conform ational entropy com puted on a latticeEntropy associated with organization on m ixing (Hyrophobic effect and

m any other unexpected features on m ixing)

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An “Ideal Solution” m eans:

The change on m ixing:DS = -nkB (xA ln(xA) + xB ln(xB))Since (ln x) is always negative or 0, DS is always positive for ideal solutionsDG = -T DS

Since (ln x) is always negative or 0, DG is always negative (or 0) and ideal solutions always m ixDH is 0, there is no interaction in ideal m ixtures, there is no excluded volum e, particles are ghosts to each otherDV = (dDG/dp)T = 0, there is no loss or gain of volum e com pared to the sum m ed volum e

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Real Solutions

M olar excess functions or departure functions:

Difference between real value and ideal value

Excess DG m ixing = DG m ixing - RT(xAlnxA + xAlnxB)Excess DSm ixing = DSm ixing + R(xAlnxA + xAlnxB)Excess DH m ixing = DH m ixingExcess DVm ixing = DVm ixing

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Real Solutions

xA becom es aA the activity so

DG m ixing = RT(xAlnaA + xAlnaB)

Excess DG m ixing = DG m ixing - RT(xAlnxA + xBlnxB)= RT(xAln(aA/xA) + xBln(aB/xB) )

= RT(xAln(gA) + xBln(gB))g Is the activity coefficient

Excess DSm ixing = -R(xAln(gA) + xBln(gB))

M ethod to use departure functions for calculations (PREOS.xls)1) Calculation of properties in the ideal state is sim ple

2) W ith an equation of state the departure function can be calculated3) For any transition first calculate the departure function to the ideal state4) Then carry out the desired change as an ideal m ixture or gas5) Then use the departure function to return to the real state

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DG m ixing = nRT(xA lnaA + xB lnaB)

= RT(xAln(gA) + xAln(gB))

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Get the activity coefficient at infinite dilution by

extrapolating the slope for pure com ponent to x = 0This is used for Henry ’s Law and a few other places.

(Activity coefficient is a/x)

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We want the infinite dilution activity coefficients to write a

function to predict the activity and the activity coefficient.

This is one of the param eters for the functions.

W ith the a function for activity we can predict the free energy and the m iscibility.

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Henry’s law for dilute solutions

Vapor pressure of solute = ki xiki is the Henry’s Law Constantxi is the solute m olar fraction (low ~0)

Raoult ’s Law for solvent or ideal m ixturesVapor pressure = p* xjxj is the solvent m olar fraction (high ~1)p* is the vapor pressure of the pure solvent

Solute Solvent or ideal m ixture

If a solution is ideal then xA = aA and gA = 1At infinite dilution a solvent is ideal (follows Raoult ’s law) so

(daA/dxA)xA =>1 = 1 = (dgAxA/dxA)xA =>1 = gA + xA(dgA/dxA)xA =>1But gA = 1 at xA = 1, then Raoult ’s law is followed if (dgA/dxA)xA =>1 => 0 (See next slide)

A solute follows Henry ’s Law if

(daB/dxB)xB =>0 = gB infinite dilution= (dgBxB/dxB)xB =>0 = (gB + xB(dgB/dxB))xB =>0

So(xB(dgB/dxB))xB =>0 = 0

This isn’t that useful

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Activity of a com ponent in a solution m ust be defined relative to a standard state, either

Henrian (extrapolated) or Raoultian (m easured, x = 1 :: a = 1)

You would use Henrian standard state if the solute had lim ited solubility like PCBs or oil in water and you wanted the infinite dilution of the water com ponent

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Analytic Solution M odels

Want to calculate the m iscibility of com ponentsNeed to generate a phase diagramCalculation of the vapor pressure

Sim plest approach could be a polynom ial. However this generally doesn’t inherently have any real m eaning. The polynom ial just reproduces existing data within experim ental lim its. Extrapolation is dangerous.

Sym m etric Feature: Phase behavior

Asym m etric feature: vapor pressure

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Excess DSm ixing = -R(xAln(gA) + xAln(gB))

Excess DG m ixing = RT(xAln(gA) + xAln(gB))

Interaction coefficientsZeroth order infinite dilution (trivial)

First order e1B self-interaction coefficient

This is sim ilar to a virial expansion

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Thallium /M ercury

Lower freezing point of M ercury for therm om eter and switches 8.5% -60°C versus -40°C

Also Rat poison

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Hildebrand Regular Solution M odel

The change on m ixing:

DS = -nkB (xA ln(xA) + xB ln(xB)) Ideal SolutionSince (ln x) is always negative or 0, DS is always positive for ideal solutionsDG = DH -T DSSince (ln x) is always negative or 0, DG is positive or negative depending on DH :: can m ix or dem ixDepending on the sign of DH

DV = (dDG/dp)T = 0, there is no loss or gain of volum e com pared to the sum m ed volum e

DH = n W xAxBW is the interaction coefficient or regular solution constant

M olar Gibbs free energy of m ixing

DG m = RT(xA ln(xA) + xB ln(xB)) + W xAxB

W = zN A[uAB – (uAA+uBB)/2]

The equation is sym m etric

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Hildebrand solubility param eter, d

DH = n W xAxBW is the interaction coefficient or regular solution constant

M olar Gibbs free energy of m ixing

G m = RT(xA ln(xA) + xB ln(xB)) + W xAxB

W = zN A[uAB – (uAA+uBB)/2]

Flory-Huggins chi param eter, ~W /kT

Hansen solubility param eters

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Asym m etric equations for asym m etric phase diagram

Sub-regular solution m odel

Redlich-Kister Expression

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Non-ideal entropy of m ixingQuasi-regular solution m odel

This is a non-com binatorial entropy

This would occur if there were ordering on m ixing, say when you add oil to water. The enthalpy of m ixing

favors m ixing!!! Water organizes at the surface of oil to a great extent, this reduces entropy and m akes oil and water dem ix. Water really likes oil.This is called the “hydrophobic effect” and is an im portant term for protein folding.

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Hildebrand Solubility Param eter, d

Two m aterials w ith sim ilar d are m iscibleLuo Y, Chen X, Wu S, Ca S Luo Z, Shi Y, M olecular Dynam ics Sim ulation Study on Two-Com ponent

Solubility Param eters of Carbon Nanotubes and Precisely Tailoring the Therm odynam ic Com patibility between Carbon Nanotubes and Polym ers, Langm uir 36 9291-9305 (2020).

Flory-Huggins Equation

c ~ 1/T

Hildebrand and Scratchard

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Graphical M ethod to Estim ate Chem ical Potential

Does dG/dxi = dG/dni = µ i?NodG/dni = dG/dxi dxi/dnidxi/dni = d(ni/(ni+nj))/dni = 1/(ni+nj) – ni/(ni+nj)2

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Solving the Regular Solution M odel

We know that the excess Gibbs free energy for m ixing is given by:

G E/RT = xA ln gA + xB ln gB Generic expression using activity coefficientAnd we have defined for the regular solution m odel that G E/RT = W xA xB Hildebrand Regular Solution expression

If we propose the answer ln gA = W xB2

We find by substitution generic expression that it equals the regular solution expression since xB + xA = 1(Solving this directly seem s to be extrem ely difficult to m e.)

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Use of the Gibbs-Duhem Equation to determ ine the activity of a com ponent

Constant p, T

If you know gA you can obtain gB by integration

Restatem ent of Gibbs-Duhem for Solutions

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