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Transcript of 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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Colligative Properties of
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.
Colligative Properties of
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Colligative Properties of
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.