Thermodynamics and Kinetics

Lecture 14

Properties of Mixtures

Raoult’s Law

Henry’s Law

Activity

NC State University

Measures of concentration There are three measures of concentration:

molar concentration per unit volume (molarity)

c = n/(liter of solution)

molar concentration per unit mass (molality)

m = n/(kg of solution)

mole fraction

x j =n j

n ii

Ideal solutions Raoult's law states

Pj = xjPj*

where Pj* is the vapor pressure of pure component j.

The vapor pressure of component j in an ideal solution

is given by the product of its mole fraction and P j*.

The chemical potential can be expressed as:

mj = mj* + RTln(Pj/Pj

*)

where Pj* is vapor the pressure of the pure component j in

the standard state.

Ideal solutions The significance of this expression is that we can now

consider the equilibrium between vapor and solution to write:

mjsoln = mj

vap = mj0 + RTln(Pj/Pj

0)

but for the vapor Pj0 = 1 bar. In the limit that the vapor

becomes the pure vapor we have:

mj* = mj

vap* = mj0 + RTln(P*j/Pj

0)

Thus

mjsoln = mj

* + RTln(Pj/Pj*)

keeping in mind the notation * means the pure component.

Ideal solutions The equation below is central equation of binary solution

mixtures.

mjsoln = mj

* + RTln(Pj/Pj*)

Using Raoult's law xj = Pj/Pj* we see that the chemical

potential can be expressed as:

mjsoln = mj

* + RTln(xj)

This equation defines an ideal solution.

The free energy of mixing The free energy for formation of a solution from individual

components is given by

Since

Gsoln = n1m1 + n2m2 , G1* = n1m1* and G2* = n2m2*.

we have that

for an ideal solution there is no enthalpy of mixing. The

volume of the mixture also does not change for an ideal

solution. The entropy change can be obtained from

in agreement with a previous derivation.

Gmix = Gsoln

– G1

*– G2

*

Gmix = n1m1 + n2m2 – n1m1* – n2m2

*

= nRT x1ln x1 + x2ln x2

Two component phase diagrams The total vapor pressure over an ideal solution is given by

Ptotal = P1 + P2 = x1P1* + x2P2

* = (1 - x2)P1* + x2P2

*

= P1* + x2(P2

* - P1*)

A plot of the total pressure has the form of a straight line.

Liquid composition: specific example

Consider the example in the book of 1-propanol and

2-propanol, which have P1* = 20.9 torr and P2

* = 45.2 torr,

respectively. So in this example, the phase diagram has the

appearance:

The vapor mole fraction The value of the mole fraction in the vapor is not necessarily

equal to that of the liquid. In the vapor phase the relative

numbers of moles is given by Dalton's law. Applying Dalton's

law we find:

y1 = P1/Ptotal = x1P1*/Ptotal

or

y2 = P2/Ptotal = x2P2*/Ptotal

The vapor curve is not the same as the liquid line.

Deriving the vapor curve Using the equation of the liquid line we find:

When substituted into y2 = x2P2*/Ptotal we have

Solving for Ptotal we find:

x2 =Ptotal – P1

*

P2

*– P1

*

y2 =P2

*

Ptotal

Ptotal – P1

*

P2

*– P1

*

y2Ptotal = P2

* Ptotal – P1

*

P2

*– P1

*=

P2

*Ptotal

P2

*– P1

*–

P2

*P1

*

P2

*– P1

*

P2

*– y2 P2

*– P1

*Ptotal = P2

*P1

*

Ptotal =P2

*P1

*

P2

*– y2 P2

*– P1

*

The vapor curve The is shown in the Figure below. The purple line was

calculated using the Dalton's law expression. What lies

between the blue and purple lines? This is the two phase

region.

Two

Phase

Region

The two phase region If we pick a composition and pressure that is inside this

region then we can use a tie line to indicate the composition

of each phase.

Boiling in a two component system If we reduce the pressure above a two component mixture

of 1-propanol and 2-propanol with a mole fraction of 0.6

1-propanol what is the composition of the vapor?

What is the composition of the vapor? a. y2 = 0.60

b. y2 = 0.48

c. y2 = 0.78

d. y2 = 0.98

What is the composition of the vapor? a. y2 = 0.60

b. y2 = 0.48

c. y2 = 0.78

d. y2 = 0.98

Boiling in a two component system If we continue to reduce the pressure the system moves into

the two-phase region. In the two-phase region the

composition of the liquid and the vapor is not the same.

What is the composition? a. x2 = 0.40 , y2 = 0.60

b. x2 = 0.50 , y2 = 0.70

c. x2 = 0.50 , y2 = 0.50

d. x2 = 0.40 , y2 = 0.70

What is the composition? a. x2 = 0.40 , y2 = 0.60

b. x2 = 0.50 , y2 = 0.70

c. x2 = 0.50 , y2 = 0.50

d. x2 = 0.40 , y2 = 0.70

What is the composition? a. x2 = 0.40 , y2 = 0.60

b. x2 = 0.50 , y2 = 0.70

c. x2 = 0.50 , y2 = 0.50

d. x2 = 0.40 , y2 = 0.70

Boiling in a two component system If we continue to reduce the pressure the system reaches the

boundary with pure vapor.

What is the composition?

a. x2 = 0.40 , y2 = 0.60

b. x2 = 0.50 , y2 = 0.70

c. x2 = 0.50 , y2 = 0.50

d. x2 = 0.40 , y2 = 0.70

What is the composition?

a. x2 = 0.40 , y2 = 0.60

b. x2 = 0.50 , y2 = 0.70

c. x2 = 0.50 , y2 = 0.50

d. x2 = 0.40 , y2 = 0.70

The tie line The tie line shown in Figure (red line) is at a total pressure

of 30 torr. You can read the x2 and y2 values from the plot

(or calculate them using the equations above used to generate

the blue and purple curves in the composition-pressure plot.

The lever rule The tie line can be used to define the quantity of each phase

present in the two phase region. The total composition xa

can be used together with x2 and y2.

za

y2 - za

za

The lever rule

za

y2 - za

za

The lever rule states that the amount of each phase present

is inversely proportional to the length of distance along the

tie line from the phase boundary to the total composition, za.

The lever rule

za

y2 - za

za

The lever rule states:

n liquid

nvapor=

y2 – za

za – x2

The total composition is za.

Non-ideal solutions Many solutions are not ideal. For ideal solutions the role of

intermolecular interactions can be ignored. This may be

because they are small or because two components have

the same interaction with each other that they have with

themselves. In other words similar solvents will form ideal

solutions. However, in many cases, intermolecular interactions

cause deviations from Raoult's law. We can consider the "like"

interactions between molecules of same species and "unlike"

interactions between molecules of different species. If the

unlike-molecule interactions are more attractive than the like

molecule interactions, the vapor pressure above a solution will

be smaller than we would calculate using Raoult's law. If the

unlike-molecule interactions are more repulsive, then the

vapor pressure is greater than for the ideal solution.

Henry’s law The statement of Henry’s law is:

P1 = x1kH,1

It looks like Raoult’s law except that you have this

funny constant kH,1 instead of P1* (the vapor pressure of

component 1). This law is only valid for dilute solutions,

i.e. when component 1 is the solute. Under these conditions

the vapor pressure of component 1 really does not matter,

because component 1 is mostly surrounded by component 2

and so its properties really quite different from the properties

of pure 1.

Henry’s law can be applied to mixtures of solvents

And also to solutions of gases in liquids. For example, see

the problems on the concentration of O2 and N2 in water at

the end of the lecture.

Henry’s law Attractive interactions between unlike molecules leads to

negative deviations from Raoult's law (lower vapor pressure

than ideal) and repulsive interactions lead to positive

deviations (higher vapor pressure than ideal).

As any solution approaches a mole fraction of one (i.e.

approaches a pure solution of one component) it becomes

an ideal solution. In other words, P1 x1P1* as x1 1.

However, as x1 0 the component is surrounded by unlike

molecules and the solution has the maximum deviation from

ideal behavior. For this case we define Henry's law,

P1 x1kH,1 as x1 0. In this expression kH,1 is the Henry's

law constant. Although we have focused on component 1 the

same holds true for component 2.

Non-ideal solutions For the example shown in the plot above we have assumed

that P1* = 100 torr. Note that as x1 1 the slope approaches

the ideal slope obtained from Raoult's law. However, as

x1 0 (and therefore x2 1) the slope is quite different

from ideal behavior. Note that the slope has the value of the

Henry's law constant kH,1. This is depicted in the Figure below

(purple line). The Henry's law value can be quite different from

the ideal value.

Activity The activity in non-ideal solutions corresponds to mole

fraction in ideal solutions.

The activity replaces mole fraction in the expression for the

chemical potential.

When considering a non-ideal solution the above expressions

hold and thus the mole fraction xj is no longer equal to Pj/Pj*.

However, as the mole fraction approaches unity (a pure

substance) the solution becomes ideal.

Thus, as xj 1, aj xj.

a j =Pj

Pj

*

m jsoln = m j

* + RT ln a j

The activity coefficient

We can define an activity coefficient g1 such that

g1 = a1/x1.

The property of the activity coefficient is that it

approaches a value of 1 (ideal behavior) as the

composition approaches the pure solvent:

a1 x1 and g 1 1 as x1 1. Thus, the solution

This definition is based on a Raoult's law standard

state, which is also known as a solvent standard state.

The activities or chemical potentials are meaningless

unless we know the standard state

Henry’s law constant and the

solubility of gases

Henry’s law constants in H2O

(atm x 103)

He 131

N2 86

CO 57

O2 43

Ar 40

CO2 1.6

Problem: Divers get the bends if

bubbles of N2 form in their blood

because they rise too rapidly.

Calculate the molarity of N2 in

water (i.e. blood) at sea level and

100 m below sea level.

At sea level: aN2 = PN2

/kH,N2 = 0.8 atm/86 x 103 atm = 9.3 x 10-6

cN2 = 55.6 aN2

= 5 x 10-4 mol/L

At 100 m: aN2 = PN2

/kH,N2 = 9.8 atm/86 x 103 atm = 1.1 x 10-4

cN2 = 55.6 aN2

= 6 x 10-3 mol/L

A note on conversion from

mole fraction to molarity The conversion from mole fraction to molarity can be

solved analytically.

Note that for water as solvent (component 1) x1~ 1 and the

Concentration of water is c1 ~ 55.6 so that the conversion

For a dilute solute such as a gas is c2 ~ 55.6 x2

x1 =n1

n1 + n2

=c1

c1 + c2

since n1 = c1V and V cancels

x1(c1 + c2) = c1

x1c2 = c1(1 – x1)

x1c2 = c1x2

c2 = c1

x2

x1

Question Henry’s law constants in H2O

(atm x 103)

He 131

N2 86

CO 57

O2 43

Ar 40

CO2 1.6

A species of fish requires a

concentration of O2 > 100 mM.

A marine biologist is trying to

determine the depth profile for O2

in sea water. The first step is to

calculate the concentration of O2

in sea water at sea level.

As an assistant you perform the calculation and find that the

O2 concentration is:

A. 250 mM

B. 430 mM

C. 760 mM

D. 4300 mM

As an assistant you perform the calculation and find that the

O2 concentration is:

A. 250 mM

B. 430 mM

C. 760 mM

D. 4300 mM

Question Henry’s law constants in H2O

(atm x 103)

He 131

N2 86

CO 57

O2 43

Ar 40

CO2 1.6

Most fish require a concentration

of O2 that is greater than 100 mM.

A marine biologist is trying to

determine the depth profile for O2

in sea water. The first step is to

calculate the concentration of O2

in sea water at sea level.

aO2 = PO2

/kH,O2 = 0.2 atm/43 x 103 atm = 4.7 x 10-6

cO2 = 55.6 aO2

= 2.5 x 10-4 mol/L

The Henry’s law constant is

an equilibrium constant Gases dissolve in liquids to form solutions. This dissolution

is an equilibrium process for which an equilibrium constant

can be written. For example, the equilibrium between oxygen

gas and dissolved oxygen in water is O2(aq) <--> O2(g).

The equilibrium constant for this equilibrium is:

K’H = p(O2)/c(O2).

The form of the equilibrium constant shows that the

concentration of a solute gas in a solution is

directly proportional to the partial pressure of that

gas above the solution. The form of the equation can be

Rearranged to give:

p(O2) = c(O2)K’H = x(O2) c(H2O)KH

The Henry’s law constant is

tabulated for c and x The Henry’s law constant can be tabulated for mole fraction

(as seen in the previous problem) or for molarity. If the units

of the Henry’s law constant are atm then it is valid for mole

fraction. If the units are atm/(mol/L) then K’H is tabulated

for molarity. The relationship between the two is:

p(O2) = c(O2)K’H = x(O2) c(H2O)KH

So that K’H = 55.6 (mol/L)KH since the concentration of water

is c(H2O) = 55.55 molar. For example, K’H is 757 atm/(mol/L).

KH =KH

55.6= 757

55.6= 13.6