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Chem 340 - Lecture Notes 8 – Fall 2013 – Gas phase Mixtures, Free Energy

In chemistry we are most interested in mixtures of components because we want to study

reactions, to minimally go from reactant to product. This introduces another variable, xi, the

partial molar fraction, xi = ni/nTot, in this case nTot = i ni is a simple sum, it makes no

difference what else is in mix, just sum all the components, and compare each to the total.

This concept of partial molar contribution is not limited to number of moles, but with other

properties it can be affected by intermolecular interactions.

Ideal gases, PTot = i Pi and Pi = xi PTot brings partial pressures, works because ideal

gases do not interact, each contributes to total pressure in same way. Same works for

volume: VTot = i Vi but if we talk about mixtures of interacting substances this changes.

Pure liquid, e.g. water, if add 1 mole to a beaker of water the volume increases by Vm = 18ml

However if we mix 1 mole water to a beaker of ethanol, V = 14 ml, so incremental volume,

or partial molar volume, depends on the other constituents in mix, due to interactions

Write this as: Vi = (V/ni)P,T,n’ where we indicate that everything else is constant, P,T, & n’s

Consider two component mix, A+B, as example, ask how V changes with addition: dnA or dnB

dV = (V/nA)P,T,nB dnA + (V/nB)P,T,nAdnB = VA dnA + VB dnB

Here VA and VB are partial molar volumes, (Under these conditions!)

Integrate and see V = ∫VA dnA + ∫VB dnA = VA nA + VB nA

Looks like the partial pressure and volume scheme above, but recall: VA = (V/nA)P,T,nB

so coefficients of nA and nA are not just mole fraction times V, or NOT xAVTot

With this definition, for a pure compound, Vm = (V/nA)P,T , i.e. molar volume

but for a mixture it is specifically, incremental change in volume on addition of nA

can be smaller than Vm, as for water-ethanol, and even negative (e.g. MgSO4 in H2O)

Free energy of a mixture:

Gibbs free energy is also extrinsic, and varies with mixture, express as Chemical Potential

In chemistry we do not have fixed compositions, the number of moles of reactant and product

change continuously during a reaction. Need to express G as G(T,P,n1,n2,n3, . . . )

This brings us to the concept of chemical potential, i, defined as:

Since this is slope of G(T,P,n1,n2,n3, . . . ) vs ni

it is Gibbs free energy per mole of component i

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Fundamental equation (dG = - SdT + VdP ) then is modified to :

At constant T, P the first two terms are zero, so:

Since is a molar quantity and G is extensive, for a pure substance A: A = GmA

But for mixture with interaction, this will not be true, A will depend on B, C etc.

Since const T,P eliminates PV work from dG, non-PV work (non-expansion) is left, mix A&B:

dG = dwadd,max = A dnA + B dnB

e.g. electrochemical cell does non-PV work, due to change of composition at electrode

Potential concept comes from driving force: let iI >i

II for species i

in region I and II of a vessel at constant T,P. Then if dni defined as

material moved from I to II, it is (-) in I and (+) in II, but same value:

dG = -iIdni +i

IIdni = (iII – i

I)dni < 0

transfer is spontaneous until dG = 0

or material is transferred from region of higher to lower potential

Note: i for each component i is same throughout the mixture

General! - other energy state functions relate to chemical potential as well –

G = H –TS = U + PV – TS U = G + TS – PV or

dU = dG - PdV - VdP + TdS + SdT but dG = VdP – SdT + A dnA + B dnB . . .

dU = - PdV + TdS + A dnA + B dnB . . .

but since dU = - PdV + TdS (const. compos., dn=0) means: A = (U/nA)V,S,nB

same thing can be derived for H and A: A = (H/nA)P,S,nB and A = (A/nA)V,T,nB

but for different conditions (constraints) – chem potential shows variation with composition

Integral form of fund. equation, const P, T: dG = A dnA + B dnB

G = A nA + B nB suggests write total derivative as:

dG = A dnA + B dnB + nA dA + nB dB

These two forms of dG must be equal, so

dG = A dnA + B dnB = A dnA + B dnB + nA dA + nB dB

or 0 = nA dA + nB dB

Gibbs-Duhem equation: i ni dI = 0 sum of changes in

chemical potential must balance. Binary mix: dA = -(nB/nA)dA or

increase in chem. pot. A match by decrease in B,B

More general: works for partial molar volumes and other quantities

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Thermodynamics of Mixing

Consider the two section volume above with pure H2 on one side and H2 in mix on the other

If there is a permeable membrane so H2 can pass through, then an equilibrium forms

H2pure = H2

mix and recall G(T,P) = Go(T) + nRT ln(P/Po)

H2pure = o

H2pure(T) + RT ln(PH2/P

o) =H2mix from pure = Gm and n=1 (molar value)

So see chemical potential depends on partial pressure, do not need membrane (general),

can discuss any component, A, not just H2, substitute PA = xAP

Amix = o

A(T) + RT ln(P/Po) + RT ln xA

Amix = A

pure(T,P) + RT ln xA where pureA(T,P) = o

A(T) + RT ln(P/Po)

Since xA < 1, Amix < A

pure(T,P), or G (-) for mixing, always spontaneous for gases, const P

Consider separated volumes, same T, P, what is Gmix? Need evaluate G initial and G final

Gi = nAGoA + nBGo

B + nCGoC + nDGo

D + . . .

Gf = nA(GoA + RT ln xA) +nB(Go

B + RT ln xB) +nC(GoC + RT ln xC) +nD(Go

D + RT ln xD) . .

so take the difference to get Gmix

Gmix = nART ln xA +nBRT ln xB +nCRT ln xC +nDRT ln xD . .

or Gmix = i niRT ln xi = nRT i xi ln xi use ni = n xi

again see that since xi < 1, Gmix < 0

mixing ideal gases is spontaneous

since S = -(G/T)P = - nR i xi ln xi

can compare behavior, consider binary, xA=1-xB

Gmix = nRT x ln x + (1-x)ln (1-x)]

Minimum Gmix is at x = 0.5 , Gmix = 0, x=1 or 0

Conversely entropy maximum at x = 0.5

each component gets max partial volume

Example. Consider four compartments, all T=298 K, p+1 bar. Four

gases, He nHe =1, Ne nNe =3, Ar nAr = 2, Xe nXe =2.5. Calculate Gmix

Gmix = nART ln xA +nBRT ln xB +nCRT ln xC +nDRT ln xD

Gmix = nRT i xi ln xi

Gmix = 8.5 mol x 8.314 J/molK x 298 K x

[(1/8.5)ln(1/8.5)+(3/8.5)ln(3/8.5)+ (2/8.5)ln(2/8.5)+ (2.5/8.5)ln(2.5/8.5)]

Gmix = - 28 kJ spontaneous

Entropy Smix = - nR i xi ln xi = - Gmix /T = 28kJ/298K = 93 J/K

Driving force is all entropy here, the enthalpy contribution is zero,

Hmix = 0, since no interaction in ideal gases liquids not true!

Hmix energy for one species surround another, if H<TS then G<0

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Chemical reaction

In case of a reaction, all moles of reactant that get used up show up as product, weighted by

the stoichiometric coefficients, i, if we express the reaction of reactants and products (Xi) as

A + B + . . C + D +. . then short form 0 = Xi (where for react -, prod +)

if reaction proceeds infinitesimally, then dni balance out, dG only depend on that, const T,P

dG = dni if let ξ = measure of extent of reaction: ni = niinit

+ ξ dn =dξ

dG = ( ) dξ = Greact dξ or rearrange: (G/ξ)T,P = ( ) =Greact

at const T,P Greact < 0 spontaneous reaction, see depend on other components

(G/ξ)T,P < 0 reaction - spontaneous as written

(G/ξ)T,P > 0 reaction - spontaneous in opposite direction than written

(G/ξ)T,P = 0 reaction – at equilibrium, no change

obviously important to determine ξ at equilibrium, mathematically straightforward

(G/ξ)T,P = 0 = ( ) so substitute in chemical potentials and solve

Where is ξ? Consider Amix = A

pure(T,P) + RT ln xA each depends on ξ

Plug these into the sum and get terms at standard conditions Goreact

And get terms in the log forms depend on xi ‘s related to the ξ

Example: Consider : 2 NO2 (g) N2O4 (g)

Assume (2-2 ξ) moles NO2 & ξ moles N2O4 at P = 1 bar, T = 298 K.

For pure, unmixed components, just linear variation see graph

Gpure = (2-2 ξ)Gom(NO2 ) + ξ Go

m (N2O4)

If this was all, then at equilibrium ξ= 0 or 1, depending on lowest

But for mixed gases must add correction:

Gmix = Gpure + Gmix

If Gmix alone were responsible for Gmix then ξ ~ 0.55 is minimum

But must combine Gpure + Gmix, and in this case ξ ~ 0.72 is mini.

So if Greact ~ Gprod then Gmix determine ξ, if very different get

mostly reactant or product (go to completion)

Direction determined by Grxn : use tables

values likeHrxn or Srxn

Example: Determine Gorxn for: 6CO2 (g) + 6H2O (l) C6H12O6 (s) + 6O2(g) at T= 298 K

Gorxn = Go

f (C6H12O6) + 6Gof (O2) - 6Go

f (H2O) - 6Gof (CO2)

= [-910 + 6x0 – 6(-237) - 6(-394)]kJ/mol = 2879 kJ/mol

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Now what is Grxn at T = 310 K? -- use Gibbs Helmholtz

Grxn(T2) = T2 [Gorxn/298K + Ho

rxn(1/ T2 – 1/298K)]

Horxn = Ho

f (C6H12O6) + 6Hof (O2) - 6Ho

f (H2O) - 6Hof (CO2)

= [-1273 +6x0 – 6(-286) – 6(-394)]kJ/mol = 2803 kJ/mol

Grxn(310K) = 310[2879/298 + 2803(1/310 – 1/298)]kJ/mol = (2993-111)kJ/mol

= 2882 kJ/mol

So small change but also small temperature change, assumed Hrxn const

Equilibrium constant

Canonical example reaction:

Separate out terms in standard conditions, combine ln terms, note Po ref to std state: 1 bar

Where:

Products of pressures is reaction quotient of pressures (note all ref to Po)

simplifies to

So if start out with lots of reactant, large A and B, small C and D, then Q will be small <<1

And ln term will be negative and large, so Grxn < 0, spontaneous forward

if C,D large partial P, then Q large >>1, and ln term pos, so Grxn > 0, not spont. (reverse)

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Now at equilibrium: Grxn = 0, then 0 = Gorxn + RT ln QP or Go

rxn = - RT ln QP

Let QP = KP for Grxn = 0 pressure equilibrium constant

or KP = exp (-Go

rxn / RT )

Note KP does not depend on pressure, since each partial pressure is referenced to std Po

Example: calculate KP at T = 298 K for: CO + H2O CO2 + H2 , use Tables

Gorxn = Go

f (H2) + Gof (CO2) -Go

f (H2O) -Gof (CO)

= [-394 + 0 – (-237) – (-137)]kJ/mol = -20 kJ/mol

KP = exp(-Gorxn /RT) = exp( 20x103/8.314x298) = 3200 = 3.2 x 103

Since KP >>1 expect reaction to come to equilibrium with mostly products, little reactant

Example Engel 6.10 – computing partial pressure at equilibrium: dissociate Cl2 2 Cl

Trick here is to realize n, # moles, is changing as reaction goes on, table helps sort out

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KP depends on T via KP = exp(-Gorxn /RT) where remember std stateG

But if increase P, see it affects or , so it looks like KP affected but in the extent of reaction

In this case, increase P and see decreased, favor reactant (LeChatlier, respond to stress)

can summarize as increase # moles gas, higher P favor reactant, decrease n, favor product

Temperature variation of KP, use the Gibbs Helmholtz variation of G to get van’t Hoff eqn:

Use tabulatedGof values to get Go

rxn and compute KP at 298 K, then integrate to get KP(T):

assumes H not change over interval, which works

since change Hrxn depends on CP , if Hrxn ~const:

here PCl = xClP numerator

PCl2 = xCl2P denominator

KP in terms ξ amt Cl2 gone

= ξ/n0 divide by n02

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Engel example continued: dissociate Cl2 2 Cl use the van’t Hoff equation:

Not spontaneous at

T=298K, or favor Cl2

But since H > G so

S (+). Due to -TS,

increase in T should

decrease G and

eventually favor 2Cl•

KP so small, no product

formed at 800 K, but by

2000 K get product , but

amount not obvious from

KP –need extent, = ξ/n0

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Gases and Liquids (solids) in equilibria

If have reaction like: CaCO3(s) CaO(s) + CO2(g)

The pressure dependence of the KP or G for the reaction is

depends on the gas component

eq(CaO,s,P) ~eq(CaO,s) andeq(CaCO3,s,P) ~

eq(CaCO3,s)

So Grxn = = 0 at equilibrium

= eq(CaO,s,P) + eq(CO2,g,P) - eq(CaCO3,s,P)

= eq(CaO,s) +

eq(CO2,g) eq(CaCO3,s) + RT ln (PCO2/P

o)

Gorxn =

eq(CaO,s) + eq(CO2,g)

eq(CaCO3,s) = - RT ln (PCO2/Po)

So ln KP = ln (PCO2/Po) = -Go

rxn/RT

So for this example reaction KP seems to depend only on the partial pressure of CO2, works

because the reactant and products in solid state are in standard states, i.e. pure and

effectively independent of pressure. If in standard states, concentration not changing even if

amount must change if it reacts

Example: calculate CO2 pressure in equilibrium with CaCO3 at T= 1000 K

Look up values for Gof and Ho

f sum to get:Gorxn = 131 kJ/mol, Ho

rxn = 178 kJ/mol

At 298 K: ln (PCO2/Po) = ln KP = -Go

rxn /RT = -53

At 1000K: ln (PCO2/Po) = ln KP(1000K)

= ln KP – (Horxn /R)(1/1000 -1/298)

= -53 – (178x103/8.314)(-0.00235) = -2.35 PCO2 = 0.095 bar

Now can redo at T = 1100 K, PCO2 = 0.673 bar; T = 1200 K, PCO2 = 1.23 bar;

If all condensed phases can usually ignore P dependence, but if reaction invoves a volume

change then pressure can have an impact. Previous example: C(diamond) C(graphite)

Density change: 2.25 g/ml to 2.52 g/ml favor diamond at high pressure, so at equilibrium:

Grxn = G(graphite) - G(diamond) + [V(graphite) - V(diamond)]P or

0 = 0 – 2900J + [1/(graphite) – 1/(diamond)](P-1)

P = 1+2900J/[12x10-3 kg/mol (1/2.25x103 kgm-3 -1/3.52x103 kgm-3)] =(105+1.5x109)Pa

P = 1.51 x 104 bar

Same story, diamond not stable, but due to slow kinetics will last a long time (needs to break

bonds and totally rearrange to get to graphite)

Since PA = xAP, can substitute and do unit conversion of KP

The same works for molarity: ci = ni/V = Pi/RT, choose stand. conc. co

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These show KP and Kx and Kc are related but are not equal. In particular depend on

which gives rise to correction terms (P/Po) for Kx and (coRT/Po)for Kc

These results above then indicate how system changes with pressure and temperature if

number of gas molecules increases or decreases in reaction, LeChatlier principle

Says that for endothermic reaction (H +) increasing temperature will cause increase in KP

and thus more product will form, but if exothermic, (H -) increase T (T +) will reverse rxn

LeChatlier - temperature favor endothermic (take in heat), oppose exothermic (give off heat)

For pressure use: Kx = KP(P/Po)

KP independent of P, so issue is , if positive, products have more gas molecules than

reactants, the exponent is negative and as pressure increases reaction reverses, but if

reactants have more gas phase molecules, increased pressure favors more product form

LeChatlier - says pressure opposes more gas molecules, favors less