Topic+5 Solution

7/30/2019 Topic+5 Solution

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Topics

Ideal solutions

Simple mixtures

Partial molar volumes

Surface Tension

Capillary rise

Adhesive force

Equation of Young and Laplace

Colligative properties


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What is ideal solution?

When two substances whose molecules are

very similar form a liquid solution, the vapor

pressure of the mixture is very simply

related to the vapor pressures of the puresubstances.

An ideal solution orideal mixture is a solution with thermodynamic properties

analogous to those of a mixture ofideal gases.

i) The Gibbs energy of mixing 2 liquids to form an ideal solution is calculated in

the same way as for two perfect gases. The enthalpy of mixing is zero and

Gibbs energy is due entirely to the entropy of mixing.

ii) A regular solution is one in which the entropy of mixing is the same as for an

ideal solution but the enthalpy of mixing is non-zero.
http://en.wikipedia.org/wiki/Solutionhttp://en.wikipedia.org/wiki/Ideal_gashttp://en.wikipedia.org/wiki/Ideal_gashttp://en.wikipedia.org/wiki/Solution


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Figure 1. Vapor-liquid equilibria for (a) pure toluene; (b) a mixture of equal

amounts of toluene and benzene: and (c) pure benzene. In the solution (b) only

half the molecules are benzene molecules, and so the concentration of benzene

molecules in the vapor phase is only half as great as above pure benzene. Note

also that although the initial amounts of benzene and toluene in the solution were

equal, more benzene than toluene escapes to the gas phase because of

benzenes higher vapor pressure
http://chemed.chem.wisc.edu/chempaths/index.php?option=com_awiki&view=chemprime&Itemid=850&article=File:Vapor-Liquid_Equilibrium_in_Mixtures_and_Pure_Substances.jpg


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The partial vapor pressure of A above the liquid mixture, pA, will then be

the vapor pressure of pure A, PA*, multiplied by the fraction of the

molecules in the liquid which are of type A, that is, the mole fraction of A,

xA. In equation form

pA = xAPA* (Raoults law) (1a)

Similarly for component B

pB = xBPB* (1b)

Adding these two partial pressures, we obtain the total vapor pressure

P=pA +pB = xAPA* + xBPB

* (2)

Liquid solutions which conform to Eqs. (1) and (2) are said to obey

Raoults law and to be ideal mixtures or ideal solutions.


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Simple mixtures

Chemistry mixtures

Mingled together

They may react or not

The simplest is mixture of two unreacted substances, e.g. A and B

1 BA xx


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Partial molar volume

Adding 1 mol of water to a huge volume of pure water at STP. How muchvolume increased?

18 cm3/mol

Adding 1 mol of water to huge amount ethanol. Volume increase

14 cm3/mol

The volume change depend on the surrounding molecules.

Partial molar volume of ethanol in water is 14 cm3

/mol

Partial molar volume of a substance A in a mixture is the change in volume per

mole of A added in large volume of the mixture.


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For a mixture of A and B, partial molar

volume changes as the compositionchange from pure A to pure B due to

change in molecular environment and

interaction forces between molecules.

This results in change of thermodynamic

properties

(Atkins and Paula, 2008)

VJ

Partial molar volume of a substance , VJ

The subscript n signify the amount of other

Substances present are constant


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Thus VJ is the slope of plot ofVvs n

(Atkins and Paula, 2008)


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The volume of the mixture changes with the addition of A and B. Take

infinitesimal amount, thus

dV=

Thus for a constant composition of A and B with nA and nB being the

number of moles respectively

V =

=

(eq. 1)

(eq. 2)


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Measurement of partial molar volume

Measure the volume at different composition, plot the graph and fit to

certain function Example: Mixture of ethanol/water containing 1.000 kg of water

32 028256.036394.06664.5493.1002 xxx Where= volume, V (cm3) andx = nE/mol

DIY: Calculate VE at 50% by mass of ethanol water

mixture with density 0.914 g cm-3

.VW = 17.4 cm3 mol-1.

(Eq. 1)

Figure 5.3 The partial molar volume of ethanol as

expressed by the polynomial in the brief illustration.

InterActivity Using the data from the brief illustration,

determine the value ofb at which VE has a minimumvalue.
http://ebooks.bfwpub.com/pchem9e/sections/figurehttp://ebooks.bfwpub.com/pchem9e/sections/figure


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Partial molar Gibbs energies

',, nTpJ

Jn

G

The same concept can be extended to extensive state function.

Chemical potential is defined as partial molarGibbs energy

Thus for binary mixture

BBAA nnG

For a pure substance we can writeG = nJGJ,m

from eq 1 obtain J = GJ,m

(eq 1)


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Chemical potential, depend on composition, p and T. Thus Gibbs energy

changes when these variable changes. Thus for a system containing A, B

etc, fundamental equation of chemical thermodynamics

... BBAA dndnSdTVdpdG

... BBAA dndndG

For constant P and T,

Since max,adddwdG thus

...max, BBAAadd dndndw

Additional work can arise due to compositional change in a system.

Example: Electrochemical cell. Change in composition of electrolyte may

results in electrical work done by the cell.


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The chemical potential not only shows how G varies composition but can

also be proved to show how U, H and A change with composition

',, nVSJ

Jn

U

',, npSJ

JnH

',, nTVJ

J

n

A

Internal Energy

at constant

vol, entropy and n

Enthalpyat constant

pressure, entropy and n

Helmhotz energy

at constantvol, temperature and n


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Gibbs-Duhem equation

BBAA nnG

BBAABBAA dndndndndG

BBAA dndndG

Since for binary system

Thus

It has been shown that at constant p and T

Which implies at constant T and P

0 BBAA dndn

J

JJdn 0

Gibbs-Duhem eq.

(eq 5.5)

(refer slide 13)


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Concentration

1) Molarity, c is the amount of solute divided by volume of the solution

(mol dm-3 or mol L-1). c = 1 mol dm-3

2) Molality, b, is the amount of solute divided by the mass of solvents (mol kg-1)

b = 1 mol kg-1


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Thermodynamic of mixing

Key points of thermodynamic mixing

i) The Gibbs energy of mixing calculated by forming the

difference of Gibbs energies before and after mixing: the

quantity is negative for perfect gases at the same pressure.

ii) The entropy of mixing of perfect gases initially at the same

pressure is positive and the enthalpy of mixing is zero


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Thermodynamic of mixing

)ln()ln( pRTnpRTn

nnG

BBAA

BBAAi

Mixing of ideal gas DG = ?, DS = ?, DH = ?

nA, T, p

nB, T, p

nA+nBpA+pB=p

T

Before mixing After mixing

)ln()ln( BBBAAA

BBAAf

pRTnpRTn

nnG

)lnln(

lnln

lnln

BBAA

BBAA

BB

AA

ifmix

xxxxnRT

xRTnxRTn

p

pRTn

p

pRTn

GGG

D

xA, xB < 1 thusDmixG


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Example

3 mol H2(g) and 1.0 mol N2(g) at 25oC are placed in 2 equal volume

containers. The gases were mixed by opening the valve between the

containers. Assume ideal gas behaviour, calculate DmixG

3 mol H2, T,

3p

1 mol N2, T,

p

nA+nBpA+pB=2p

TEqual volume

kJ9.6ln13

ln3

ln3

ln

21

23

D

p

pRT

p

pRT

ppRTn

ppRTnG BBA

Amix

p(H2) =3/2p

p(N2) = 1/2p

P=4/2=2


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Question

3 mol He(g) and 5 mol N2

(g) at 25 oC are placed in 2 equal volume

containers. The gases were mixed by opening the valve between the

containers. Assume ideal gas behaviour, calculate DmixG

What is the value ofDmix

G if the pressure in the container are maintained

constant?

F i t f f t i iti ll t th t f i i


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Since ln x < 0 thus DmixS > 0 for all composition

Entropy increase when gas is dispersed

DmixH = 0

Can be proved

Because of no interactions between molecules

for ideal case

)lnln(,,

BBAA

nnp

mixmix xxxxnR

T

GS

BA

DD

For a mixture of perfect gases initially at the same pressure, entropy of mixing


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Liquid mixtures (Ideal solution)

(a) Raoults law provides a relation between the

vapour pressure of a substance and its mole

fraction in a mixture; it is the basis of the

definition of an ideal solution.

(b) Henrys law provides a relation between the

vapour pressure of a solute and its mole fraction

in a mixture; it is the basis of the definition of an

ideal-dilute solution.


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Liquid mixtures (Ideal solution)

A(g)+B(g)

A(l)+B(l)

A(g)

A(l)

Pure A in equilibrium with its vapour Mixtures of A, B in equilibrium with vapours

** ln AAA pRT

AAApRTln

Combine

*

*

lnA

AAA

p

pRT Raoults law

*

AAA pxp Thus the chemical potential of componentof an ideal solution

AAA xRTln*

Can be used as definition of ideal solution.

Ideal solution

follow both these

equations


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Henrys Law

BBB Kxp

Henrys law vapour pressure of solute is

proportional to its mole fraction (but theproportionality constant is not vapour pressure

of pure substance)

Ideal Dilute Solution - Mixture for which

solvent obeys Raoults Law and solute obeys

Henrys Law

KB has pressure unit. The value ofKB can be

obtained by extrapolating the graph toxB =1

In practical Henrys is expressed

BBB Kbp

bB = molality

xB is the mole fraction of the solute and KB is an empirical constant


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DIY

xc 0 0.20 0.40 0.60 0.80 1.00

pc/kPa 0 4.7 11 18.9 26.7 36.4

pA/kPa 46.3 33.3 23.3 12.3 4.9 0

The table shows vapour pressures of each component of a mixture of acetone

(A) and chloroform (C) at 35 oC

Show that the mixture conforms Raoults law for component in large excess

and Henrys law for minor component.

Determine Henrys Law constant.


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Question

Estimate the molar concentration of O2 in water at 25oC if its partial pressure

is 21 kPa,KO2 is 7.9 x 104 kPa kg mol-1 and rH2O=0.99709 kg dm-3


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Surface tension Energy required to increase the surface area of a liquid by a unit of

area. How does the energy of the droplet depend on its surface area???

A molecule at the surface of a liquid experiences only net inward

cohesive forces.dA = d

Where:A is the Helmholtz energy

is constant of proportionality also called as

the surface tension

is the unit element of area

Unit: energy/ area or J m-2

The work, w needed to change the surface

area, , of a sample

Definition of surface tension


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Capillary riseThe rise of a liquid that wets a tube up the inside of a small diameter tube

(i.e., a capillary) immersed in the liquid.

The liquid creeps up the inside of the tube (as a result of adhesiveforces between the liquid and the inner walls of the tube) until the

adhesive and cohesive forces of the liquid are balanced by the weight of

the liquid.

The smaller the diameter of the tube, the higher the liquid rises.

The capillary rise is given by:

h= 2 / gr


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Adhesive ForceForces of attraction between a liquid and a solid surface.

The difference in strength between cohesive forces and adhesive

forces determine the behavior of a liquid in contact with a solid

surface.

Cohesive forces are the forces that hold liquids

together. Adhesive forces are the forces between a liquid and

another surface.

Capillary rise implies that the:

Adhesive forces > cohesive forces

Capillary fall implies that the:

Cohesive forces > adhesive forces


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Rubber Membrane

Rubber membrane at the end of cylindrical tube. An

inner pressure Pi can be applied, which is differentthan the outside pressure Pa


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The Young-Laplace equation relates the pressure

difference between the two phaseD

P and thecurvature of the surface

R1 and R2 are the two principal radii of curvature. DP

is also called Laplace pressure, and equation abovealso referred to as the Laplace equation

D

21

11

. RRP


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Illustration of the curvature of a

cylinder and a sphere

For a cylinder of radius r a convenient choice is R1

= r and R2 = so that the curvature is 1/r + 1/ = 1/r

For a sphere with radius R we have R1 = R2 and the

curvature is 1/R + 1/R = 2/R


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Example

How large is the pressure in a spherical

bubble with a diameter of 2 mm and a

bubble of 20 nm diameter in pure water,

compared with the pressure outside?


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Fundamental Implications If we know the shape of a liquid surface we knows

its curvature and we can calculate the pressuredifference

In the absence of external fields (e.g. gravity) the

pressure is the same everywhere in the liquid;otherwise there would be a flow of liquid to regionsof low pressure, thus DP is constant and Young-Laplace equation tells us that in this case thesurface of the liquid has the same curvature

everywhere

With the help of Young-Laplace equation it ispossible to calculate the equilibrium shape of aliquid surface (geometry)

A l i th Y L l


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Applying the Young-Laplace

Equation Applying Young-Laplace equation to simple geometries is

usually obvious at which side the pressure is higher

E.g. both inside a bubble and inside a drop, the pressure ishigher than outside

But in other case, it is not so obvious, because thecurvature can have an opposite sign

E.g. a drop hanging between the planar ends of twocylinders, then the two principal curvatures defined by

C1 = 1/R1 and C2 = 1/R2

Can have a different sign. We count it positive if theinterface is curved towards the liquid

The pressure difference is defined as DP = Pliquid - Pgas


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Colligative properties are properties of

solutions that depend solely on thenumber of particles dissolved in the

solution.

Colligative properties do not depend on

the kinds of particles dissolved.

Colligative properties are a physical

property of solutions


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Among colligative properties are

Vapor pressure lowering

Boiling point elevation

Melting point depression

Osmotic pressure


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Vapor Pressure Lowering

As solute molecules are

added to a solution,

the solvent becomeless volatile

(=decreased vapor

pressure).

Solute-solventinteractions contribute

to this effect.


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Vapor Pressure

Therefore, the vapor

pressure of a solution

is lower than that of

the pure solvent.

B ili P i t El ti d


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Boiling Point Elevation and

Freezing Point Depression

Solute-solvent

interactions also

cause solutions to

have higher boiling

points and lower

freezing points than

the pure solvent.


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Freezing Point Depression

The change in freezingpoint can be found

similarly:

DTf= Kfm

Here Kf is the molal

freezing point

depression constant ofthe solvent.

DTf is subtracted from the normal freezing

point of the solvent.

Boiling Point Ele ation and


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Boiling Point Elevation and

Freezing Point DepressionIn both equations, DT

does not depend on

what the solute is, but

only on how manyparticles are

dissolved.

DTb = Kb m

DTf= Kfm

Colligative Properties of


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ElectrolytesBecause these properties depend on the number of

particles dissolved, solutions of electrolytes (whichdissociate in solution) show greater changes than thoseof nonelectrolytes.

e.g. NaCl dissociates to form 2 ion particles; its limitingvant Hoff factor is 2.



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ElectrolytesHowever, a 1 Msolution of NaCl does not show

twice the change in freezing point that a 1 M

solution of methanol does.

It doesnt act like there are really 2 particles.


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vant Hoff Factor

One mole of NaCl in

water does not

really give rise to

two moles of ions.


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The vant Hoff Factor

Reassociation is more

likely at higher

concentration.

Therefore, thenumber of particles

present is

concentration

dependent.


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Osmosis

Semipermeable membranes allow some

particles to pass through while blocking

others.

In biological systems, most

semipermeable membranes (such as

cell walls) allow water to pass through,

but block solutes.


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OsmosisIn osmosis, there is

net movement ofsolvent from the areaofhigher solventconcentration (lowersoluteconcentration)

to the are oflowersolventconcentration (highersoluteconcentration).

Water tries to equalize the concentration on both sides untilpressure is too high.


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Osmotic Pressure

The pressure required to stop osmosis,

known as osmotic pressure, , is

nV

=( )RT = MRT

where Mis the molarity of the solution

If the osmotic pressure is the same on both sides of a membrane

(i.e., the concentrations are the same), the solutions are isotonic.

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