simple mixtures thermodynamic description of mixtures aryo abyoga a (080358395) gerald mayo l...

SIMPLE MIXTURES

THERMODYNAMIC DESCRIPTION OF MIXTURES

ARYO ABYOGA A (080358395)

GERALD MAYO L (0806472212)

LEONARD AGUSTINUS J (0806472225)

Simple Mixtures Often in chemistry, we encounter mixtures

of substances that can react together. Chapter 7 deals with reactions, but let’s

first deal with properties of mixtures that don’t react.

We shall mainly consider binary mixtures – mixtures of two components.

xA xB 1

Dalton’s Law

The total pressure is the sum of all the partial pressure.

We already used mole fraction to descrice the partial pressure of mixtures of gases which refers to a total pressure

p j x j p

pA pB (xA xB )pp

The partial molar volume is the contribution that one component in amixture makes to the total volume of a sample

H2O EtOHAdd 1.0 mol H2O Add 1.0 mol H2O

Volume increasesby 18 cm3 mol-1Volume increasesby 14 cm3 mol-1

Molar volume of H2O:18 cm3 mol-1

Partial molar volume ofH2O in EtOH: 14 cm3 mol-1

The different increase in total volume in the H2O/EtOH example dependson the identity of the molecules that surround the H2O. The EtOHmolecules pack around the water molecules, increasing the volume byonly 14 cm3 mol-1Partial molar volume of substance A in a mixture is the change in volumeper mole of A added to the large volume of the mixture

partial molar volume

Partial Molar Volumes Imagine a huge volume of pure water at

25 °C. If we add 1 mol H2O, the volume increases 18 cm3 (or 18 mL).

So, 18 cm3 mol-1 is the molar volume of pure water.

Partial Molar Volumes Now imagine a huge volume of pure

ethanol and add 1 mol of pure H2O it. How much does the total volume increase by?

Partial Molar Volumes When 1 mol H2O is added to a large

volume of pure ethanol, the total volume only increases by ~ 14 cm3.

The packing of water in pure water ethanol (i.e. the result of H-bonding interactions), results in only an increase of 14 cm3.

Partial Molar Volumes The quantity 14 cm3 mol-1 is the partial

molar volume of water in pure ethanol. The partial molar volumes of the

components of a mixture varies with composition as the molecular interactions varies as the composition changes from pure A to pure B.

VJ VnJ

p,T ,n'

The partial molar volume of components of a mixture vary as themixture goes from pure A to pure B - that is because the molecularenvironments of each molecule change (i.e., packing, solvation, etc.)Partial molar volumes of a water-ethanolbinary mixture are shownat 25 oC across all possibleCompositions. The Partial molar volume, Vj, of a substance j define as :

The partial molar volume is the slope of a plot of total volume as theamount of J in the sample is changed (volume vs. composition)Partial molar volumes vary

with composition (different slopes atcompositions a and b) - partial molar volume at b is negative (i.e., theoverall sample volume decreases as A is added)

dV VnA

p,T ,nB

dnA VnB

p,T ,nA

dnB

When a mixture is changed by When a mixture is changed by dndnAA of of A and A and dndnBB of B, then the total volume of B, then the total volume changes by: changes by:

Partial Molar Volumes

If partial molar volumes are known for the twocomponents, then at some temperature T, thetotal volume V (state function, always positive)of the mixture is

dV VnA

p,T ,nB

dnA VnB

p,T ,nA

dnB

dV VAdnA VBdnB

V VAdnA0

nA VBdnB0

nB


V VAdnA0

nA VBdnB0

nBV VA dnA0

nA VB dnB0

nBV VAnA VBnB


V (NaCl, aq) = volume of sodium chloride solution

V_

(NaCl, aq) = partial molar volume of sodium chloride

in water


Partial Molar Volumes How to measure partial molar volumes? Measure dependence of the volume on

composition. Fit a function to data and determine the

slope by differentiation.

Partial Molar Volumes Ethanol is added to 1.000 kg of water. The total volume, as measured by

experiment, fits the following equation:

V 1002.93 54.6664x 0.36394x 2 0.028256x 3

x nE


VE VnE

p,T ,nw

Vx

p,T ,nw

dV

dx54.6664 (2)0.36394x (3)0.028256x 2

Partial Molar Volumes Molar volumes are always positive, but

partial molar quantities need not be. The limiting partial molar volume of MgSO4 in water is -1.4 cm3mol-1, which means that the addition of 1 mol of MgSO4 to a large volume of water results in a decrease in volume of 1.4 cm3.

Partial Molar Gibbs energies

The concept of partial molar quantities can be extended to any extensive state function.

For a substance in a mixture, the chemical potential is defined as the partial molar Gibbs energy.

J GnJ

p,T ,n'


For a pure substance:

GnJGJ ,m

J GnJ

p,T ,n'

nJGJ ,m

nJ

p,T ,n '

GJ ,m

Partial Molar Gibbs energies Using the same arguments for the derivation of partial molar volumes,

Assumption: Constant pressure and temperature

GnAA nBB


Fundamental equation of chemical thermodynamics

dGVdp SdT AdnA BdnB

dGAdnA BdnB (at constant p and T)

dwadd,max AdnA BdnB (at constant p and T)

Chemical Potential

GH TS U pV TSU pV TS GdU pdV Vdp SdT TdS dGdU pdV Vdp SdT TdS (Vdp SdT AdnA BdnB )

dU pdV TdS (AdnA BdnB )

dU AdnA BdnB (at constant S and V)

J UnJ

S,V ,n '

Chemical Potential

J HnJ

S,p,n'

J AnJ

V ,T ,n'

Gibbs-Duhem equation

GnAA nBBdGnAdA nBdB AdnA BdnB

dGAdnA BdnB (at constant p and T)

nAdA nBdB 0

nJdJ 0J

Gibbs-Duhem equation

nAdA nBdB 0

dB nAnBdA

Molarity and Molality

Molarity, c, is the amount of solute divided by the volume of solution. Units of mol dm-3 or mol L-1.

Molality, b, is the amount of solute divided by the mass of solvent. Units of mol kg-1.

Using Gibbs-Duhem

The experimental values of partial molar volume of K2SO4(aq) at 298 K are found to fit the expression:

B 32.280 18.216x1 2

B VK 2SO4

x molality of K2SO4

Using Gibbs-Duhem

nAdVA nBdVB 0

dA nBnAdB

dA A*

A nBnAdB

A A* nB

nAdB

A A*

nBnAdB

Using Gibbs-Duhem

A A*

nBnAdB

B 32.280 18.216x1 2

dB

dx9.108x 1 2

dB 9.108x 1 2dx

Using Gibbs-Duhem

A A*

nBnAdB

dB 9.108x 1 2dx

A A* nB

nA9.108x 1 2dx A

* 9.108nBnAx 1 2dx

Using Gibbs-Duhem

A A*

nBnA

9.108x 1 2dx A* 9.108

nBnAx 1 2dx

nBnA

nB1 kg MA

nBMA

1 kgxMA

A A* 9.108 xMA x

1 2dx A* 9.108MA x1 2dx

A A* 9.108MA x1 2dx

0

bBA A

* 2 3 9.108MAbB3 2

Using Gibbs-Duhem

A A* 2 3 9.108MAbB

3 2

A* 18.079 cm3 mol 1

MA 0.01802 kg mol-1

A 18.079 0.1094bB3 2

Thermodynamics of mixing

So we’ve seen how Gibbs energy of a mixture depends on composition.

We know at constant temperature and pressure systems tend towards lower Gibbs energy.

When we combine two ideal gases they mix spontaneously, so it must correspond to a decrease in G.


Gm Gm RT ln

p

p

RT lnp

p

RT ln p


RT ln p

Gi nA A RT ln p nB B

RT ln p G f nA A

RT ln pA nB B RT ln pB

mixGnART lnpAp

nBRT lnpBp


mixGnART lnpAp

nBRT lnpBp

pAp

xApAp

xB

mixGnART ln xA nBRT ln xBxAn nA xBn nBmixGnRT(xA ln xA xB ln xB )

mixG 0

Gibbs energy of mixing

A container is divided into two equal compartments. One contains 3.0 mol H2(g) at 25 °C; the other contains 1.0 mol N2(g) at 25 °C. Calculate the Gibbs energy of mixing when the partition is removed.


Two processes: 1) Mixing2) Changing pressures of the gases.

Gi 3.0 H2

RT ln 3p 1.0 N2

RT ln p G f 3.0 H2

RT ln 3 2 p 1.0 N2

RT ln1 2 p mixG3.0(RT ln

32 p

3p) 1.0(RT ln

12 p

p)

mixG3.0(RT ln 12) 1.0(RT ln 1

2)

mixG 6.9 kJ


mixGnRT(xA ln xA xB ln xB )

mixG3.0(RT ln34

) 1.0(RT ln14

)

mixG 2.14 kJ 3.43 kJ

mixG 5.6 kJ

p

p

Other mixing functions

GT

p,n

S

mixS mixG

T

T

nRT(xA ln xA xB ln xB ) nR(xA ln xA xB ln xB )

mixS nR(xA ln xA xB ln xB )

Other mixing functions

mixGnRT(xA ln xA xB ln xB )

mixS nR(xA ln xA xB ln xB )

GH TSnRT(xA ln xA xB ln xB ) H T( nR(xA ln xA xB ln xB )

nRT(xA ln xA xB ln xB ) H nRT(xA ln xA xB ln xB )

H 0 (constant p and T)

simple mixtures thermodynamic description of mixtures aryo abyoga a (080358395) gerald mayo l...

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