Mid Term3 Review 15

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    Review for Mid-Term 3, 2015

    Based on material in lecture notes 6-8 inclusive.

    Mid-term exam will be in the recitation period,

    November 23rd, 7-8:20 pm.

    Split into two rooms as usual:

    Engineering 143 (A-M), Frey 119 (N-Z)

    Bring ID, calculator, pencil/pen, etc.

    2

    Phase Transition

    Aphase transition is the spontaneous transformation of one

    phase into another at a characteristic T & p.

    How do we know the phase of a substance at a given p and T?

    Using the Gibbs energy: spontaneous processes have negative

    changes in Gibbs free energy. For example, if you want to know the

    most stable phase of water at a given pressure and temperature, you

    can calculate the Gibbs energies for each of the phases, compare

    them and pick the lowest one

    Chemical potentials are equivalent to molarGibbs energies for pure

    substances, so the phase with the lowest chemical potential will be the

    most stable phase

    At the transition temperature and pressure the two

    phases have the same chemical potential.

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    Phase Transition, pure substance

    Heat exchange without temperature change

    Constant pressure cooling curve for

    a substance. Calorimetry can spot a

    phase change even if there are no

    outward visible changes.

    A discontinuity indicates a phase

    change. Heat is going towards

    changing internal structure, entropy.

    4

    Temperature Dependence of a Phase

    Recall

    p

    GS

    T

    =

    m mdG SdT Vdp d S dT V dp= =

    ( ) ( ) ( )m m mS s S l S g

    mm

    p

    GS

    T

    =

    m

    p

    ST

    =

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    5

    Phase Diagram

    Phase Diagram: Shows regions of pressure and temperature where

    phases are thermodynamically stable

    Phase Boundaries: Separate regions and showp and T where two

    phases exist in equilibrium

    Sublimation

    Fusion

    Liquefaction

    Triple point,

    3 phases in equilibrium

    Critical Point (liquid and

    vapor have same density)

    6

    How to find a Phase Boundary?

    The thermodynamic criterion of equilibrium is:

    At equilibrium, the chemical potential of a substance is the same

    throughout a sample, regardless of how many phases are present

    If amount dn is transferred from 1 to 2, Gibbs

    energy changes by1dn in location 1 and by

    +2dn in location 2

    2 1dG dn =

    If 1 > 2, G is negative, process is spontaneous

    If 1 = 2, no change in G, system at equilibrium

    1 (p,T)= 2(p,T) p = p(trans)

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    7

    How to find a Phase Boundary?

    Since

    How to predict the

    phase boundary?

    d= dd

    - d

    = 0

    So

    exact expression for the slope of a phase boundary -

    applies to phase equilibrium of any pure substance

    dp

    dT=

    trs

    Sm

    trsV

    m

    =

    trs

    Hm

    TtrsV

    m

    , ,m md S dT V dp =

    , ,m md S dT V dp =

    So

    We have

    , , , ,m m m md d S S dT V V dp =

    0trs m trs mS dT V dp= =

    8

    Liquid-Vapor Boundary

    Va p

    Va p

    Hdp

    dT T V

    =

    Clapeyron equation

    1 1ln ln

    Va p

    f i

    f i

    Hp p

    R T T

    =

    2

    ln Va pHd p

    dT RT

    =

    Vapor pressure

    Clausius-Clapeyron Equation.

    Integrate:

    )ln(since xdx

    dx

    =

    Large and positive

    Dominated by the gas

    V=RT/p

    Approximate, because

    weve assumed pV=nRT

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    Clapeyron Equation Example

    Calculate the slope of the solid-liquid phase boundary curve for

    benzene. At the melting point (278.7 K) at one atmosphere pressure:

    1-3

    1

    .mol10.3cm(benzene)

    kJ.mol95.9(benzene)

    =

    =

    mfus

    mfus

    V

    H

    From:

    mtrs

    mtrs

    VT

    H

    dT

    dp

    =

    1-

    1-1-

    -1-13

    33

    1

    atm.K2.34

    .K8.314J.mol

    .K.atm.moldm08206.0

    dm103.10K7.278

    J.mol9950

    =

    =

    dT

    dp

    p

    T

    solid

    liquid

    very steep!

    Vapor Pressure Example

    Find the vapor pressure of a liquid at some temperature

    1 1ln ln

    Va p

    f i

    f i

    Hp p

    R T T

    =

    At normal boiling point, the vapor pressure is one atmosphere.

    For benzene, T* (b.p.) is 80 C, 353 K. vapH = 30.8 kJ mol-1. What is

    the vapor pressure at 20 C, 293 K?

    = 353

    1

    293

    1

    314.8

    30800

    )0.1ln(ln fp

    atm117.0=fp or, about 90 torr

    UNITS!

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    Clausius-Clapeyron Example

    For water vapH = 40.65 kJmol-1 and the normal boilingpoint (at one atm) is 373.15 K. At what temperature does

    the vapor pressure of liquid water equal 380.0 Torr?

    1 1ln ln

    vap

    f i

    f i

    Hp p

    R T T

    =

    1atm = 760 Torr, 380 T = 0.5 atm

    Find Tf= 354.3 K

    Vapor pressure at different T?

    Whats the vapor pressure of water at 25 C and 1 atm?

    =

    if

    vap

    ifTTR

    Hpp

    11)/ln(

    Clausius-Clapeyron Eqn. Lets find

    the vapor pressure at 298.15K. We

    have vapH (100 C) = 40.7 kJmol-1 for

    H2O(l).

    3001.3

    15.373

    1

    15.298

    1

    314.8

    40700)1/ln( 298

    =

    =p So p298.15 = 0.03688 atm or 28.03 torr

    How does this change when the pressure changes from 1 to 10 atm?

    Need to think about how the chemical potential changes with pressure.

    For a gas (ideal): For a liquid:

    p

    RTdpgd =)( dPVld m=)(

    Where dP is the applied P

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    Vapor pressure change with T (contd)

    Liquid and vapor in equilibrium, we can equate these

    eqns. Integrate:

    Here, we have pi= 28.03 torr, Vm= 18.069 cm3 and we increase pressure

    from 1 to 10 atm.

    dPlVldp

    RTdpgd m )()()( === =

    f

    i

    f

    i

    p

    p

    m

    p

    p

    dPlVp

    RTdp)(

    PlVpRT mp

    pf

    i= )(ln

    Quite a small change!

    (Increase)

    torr22.28006669.103.28

    006647.015.298314.8

    910325.10110069.18

    03.28ln

    36

    ==

    =

    =

    f

    f

    p

    p

    Example

    trs

    trs m

    H

    T V

    =

    trs

    trs m

    dp S

    dT V

    =

    Clapeyron Equation:VT

    H

    dT

    dp

    sub

    sub

    sub =

    )(

    etc.

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    Example, Contd

    V

    V

    H

    H

    dT

    dpdTdp

    sub

    vap

    vap

    sub

    vap

    sub

    =

    )(

    )( subV =vapV=

    volume of gas

    78.7kJ/mol6.0

    kJ/mol)0.67.40(

    )(

    )( =

    =

    =

    H

    H

    dT

    dpdT

    dp

    vap

    sub

    vap

    sub

    Mixtures

    16

    Energy of simple mixture: non-reactive, binary

    * ?A A

    = Change of due to mixing

    Liquids, solutions, Raoults and Henrys laws.

    Component A conventionally the solvent, B the solute. Ideal,

    Ideal Dilute solutions.

    Approach this material from several angles...

    Changes in chemical potential in the mixture

    drive all processes.

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    17

    Generalized fundamental equation

    ...A A B BdG dn dn = dG SdT Vdp

    = At constant n

    At constant p,T

    However at constant T, p, dG also equal to maximum non-expansion work

    ,max ...add A A B Bdw dn dn =

    This work can be achieved by chemical rxn, e.g. via a battery.

    Electrical work results from changing chemical composition.

    ...A A B BdU TdS pdV dn dn = Similarly:

    A

    TnVSA Bn

    U=

    ...,,,

    Then:

    ...A A B BdG SdT Vdp dn dn =

    Combining the two

    A

    VSpTAn

    G=

    ,,,

    Gibbs-Duhem Equation

    18

    A A B BG n n = Since and theJdepend on composition, for abinary mixture G may change by

    However, we know that

    A B A A B B

    A B

    G GdG dn dn dn dn

    n n

    = =

    A A B BdG d n d n = A A B B A A B Bn dn ddn nd =

    So 0A A B Bn d n d =

    Gibbs-Duhem Equation(for a binary mixture at constant p, T)

    BA B

    A

    nd d

    n = .....BA B

    A

    ndV dV etc

    n=

    It is true for any molar quantities, for binary mixture, we have

    More general form of G-D Eq. is

    0j j

    j

    SdT Vdp n d =http://en.wikipedia.org/wiki/Gibbs-Duhem_relation

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    Gibbs-Duhem Equation

    19

    0A A B Bn d n d =

    Gibbs-Duhem Equation

    (for binary mixture at constant p, T)

    Significance is that the chemical potential of one component of

    a mixture cannot change without affecting the chemical

    potentials of the other components.

    http://en.wikipedia.org/wiki/Gibbs-Duhem_relation

    Review: Partial Molar Quantities

    Concept of partial molar quantity and the relation can be applied to

    any extensive state function: G, A, H, U, V, S, Cp etc at constant T,p

    http://www.chem.neu.edu/Courses/1382Budil/PartialMolarQuantities.htm

    Integral

    V=nAVA+nBVB+..

    G=nAA+nBB+..

    S=nASA+nBSB+..

    M=nAMA+nBMB+..Mass

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    Change of due to mixing General expression of Gibbs energy for mixture

    21

    f A A B BG n n = 0A A B Bn d n d =Gibbs-Duhem

    At const T, p

    What is the difference between in its pure state (J* )and in

    mix (J) ?

    * *

    i A A B BG n n =

    * lnJ J JRT x = Ideal solution,

    * ln lnoA A A A ART a RT a = =

    *

    *ln lnoBB B B B B

    B

    KRT a RT a

    p = =

    Real solution

    What is the standard Gibbs energy change for mixing?

    mix ln lnideal A A B BG nRT x x x x =

    mix mix

    E idealG G G= Use excess function

    Chemical Potential and Mixing

    *

    *ln JJ J

    J

    pRT

    p

    =

    Chemical potential drops for mixture

    22

    Real solutions

    (chi) is the mole fraction here.

    Summary, more details later

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    G-D equation, Rauolt, Henrys Law

    *

    pa x

    p= =

    pa x

    K= =

    Treat as solvent

    Treat as solute

    *

    *ln AA A

    A

    pRT

    p =

    *

    *ln BB B

    B

    pRT

    p =

    p p x=

    How to find?

    0A A B Bn d n d =

    0A A B Bx d x d =

    B BA B

    B B

    d dx x

    dx dx

    =

    Using Gibbs-

    Duhem relation

    Change of due to mixing General expression of Gibbs energy for mixture

    24

    f A A B BG n n = 0A A B Bn d n d =Gibbs-Duhem

    At const T, p

    What is the difference between in pure state ( )and in mixture () ?

    * ?A A =

    * *

    i A A B BG n n =

    *

    *ln AA A

    A

    pRT

    p

    =

    How to express this in terms of mole fractions: xA, xB,as we did for ideal gas mixtures?

    Goal-I

    component

    A in solution

    component

    A pure liq.

    component A above

    soln. and abovepure liq. (ideal gas

    assumed)At equilibrium, the chemical potential of each

    component has the same value in each phase

    where it is present

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    25

    Ideal solution: Raoults Law

    Ideal solutions: both solute and

    solvent obey Raoults LawMixture of benzene and toluene

    *

    *

    AA A A A

    A

    pp x p or x

    p= =

    Ideal solutions obey Raoults Law:

    i.e. the partial vapor pressure relative to

    p* equals the molar fraction in the liquid.

    pA*

    *

    *ln AA A

    A

    pRT

    p

    =

    * lnA ART x=

    mole fraction in liq.

    Raoults Law Example

    The vapor pressure of benzene is 53.3 kPa at 60.6 C but it fell

    to 49.2 kPa when 51.2 g of an involatile organic compound was

    dissolved in 500 g benzene. Calculate the molar mass of the

    compound, assume the solution is ideal.

    * and BB B B B

    B A

    np x p x

    n n= =

    where B = benzene and A

    = solute

    * *( ) orB B B B BB A

    A B B

    n p n p pp n

    n n p

    = =

    Hence:A

    A

    A

    mn

    M=

    * *( ) ( )

    A B B A BA

    B B B B B B

    m p M m pM

    n p p m p p= =

    Subs. In:

    MA=96.0 g mol-1

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    Vapor Pressure Example

    Consider a mixed solution of 1-propanol (1) and 2-propanol (2). Itsideal at all concentrations. Given p1

    *=20.9 torr, p2*=45.2 torr at 25 C,

    calculate the total vapor pressure and composition of the vapor for

    x2=0.75 at this temperature.

    Ideal, so p(total) = p1 + p2 = x1p1* + x2p2

    *

    = (0.25)(20.9) + (0.75)(45.2) = 39.1 torr

    Daltons law of partial pressures:

    y1(mole fraction of 1 in vapor) = p1/p(total) = x1p1*/p(total)

    = (0.25)(20.9)/(39.1) = 0.13Similarly y2 = (0.75)(45.2)/(39.1) = 0.87

    Note: y1+y2=1 and the vapor has more of the more volatile component

    Another example

    The Henrys law constant for a solute B in a solvent A is

    8.21x103 kPa. What is the vapor pressure of B in a

    solution of molality = 0.25 mol/kg? The molar mass of

    the solvent A is 74.1 g/mol.

    BBB Kxp = Weve been given K in mole fraction (pressure)units and we need to convert it to solve the

    problem.BBB Kbp =

    0.25 mol/kg is 0.25 moles of B in 1000g of A. 1000g of A

    contains 1000/74.1 or 13.5 moles. Mole fraction of B is hence

    .0182.0)5.1325.0(

    25.0 ==Bx BBB Kxp =

    Therefore kPa1491021.80182.0 3 ==Bp

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    29

    Departure from Raoults Law & Ideal Dilute Solutions

    When mixture components are structurally different, see strong

    deviation from Raoults law. But when1) The major component is nearly pure (its x1), it still obeys Raoults law

    2) The minor component is nearly absent (its x0), its pressure obeys Henrys law

    then this solution is called an ideal dilute solution.

    Note: even when solute x 0, some solutions do not follow Henrys law

    KB is some empirical

    constant

    30

    Henrys Law

    For an ideal-dilute solution: the solvent obeys

    Raoults law and the solute obeys Henrys law

    For real solutions at low concentrations,

    although the vapor pressure of the solute is

    proportional to the mole fraction, the

    constant of proportionality is not the vapor

    pressure of the pure substance*when 0,B B B B B Bp x K x K p=

    KB is a measured slope as xB0 with

    dimensions of pressure.

    The solute behaves entirely differently from when its pure state,

    since its molecules are surrounded by solvent molecules.

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    Units

    Henrys law constants are often expressed in terms of

    molality:

    Molality is the molar concentration per mass of

    solvent, has units of moles/kg.

    Compare Molarity, the molar concentration per

    volume and has units of moles/dm3, for example.

    Mole fraction is dimensionless, KB has

    units of pressure.B B B

    p x K=B (solute)

    B (solute) BBB Kbp = Molality bB has dimensions of mol kg-1

    ,KB has units of Pa.kg.mol-1

    Henrys Law Example

    The mole fraction of CO2 in the atmosphere has risen from

    0.000314 in 1960 to 0.000397 in 2013. Use Henrys law to

    calculate the change in concentration in the oceans (in molality)

    at 25 C, assuming K(CO2) = 3.01x103 kPa.kg.mol-1 in water at

    25 C.

    Henrys Law: pB = bBKB (concentration in molality)

    The partial pressure of solute in vapor is its concentration in solution times

    its Henrys law constant. So:

    p(CO2)= 0.000314 x 101.3 kPa = 0.0318 kPa.

    b(CO2) = 0.0318 kPa/ (3.01x103

    kPa.kg.mol-1

    ) = 0.01057x10-3

    mol.kg-1

    For the higher concentration:

    p(CO2)= 0.000397 x 101.3 kPa = 0.0402 kPa.

    b(CO2) = 0.0402 kPa/ (3.01x103 kPa.kg.mol-1) = 0.01336x10-3 mol.kg-1

    Increase of about 26%.

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    33

    More detailed summary for real solution

    B B Bp x K=*AA Ap x p=

    Ideal dilute solution: xA1 and xB0

    The activity is an effective mole fraction

    *

    AA A A

    A

    pa x

    p= =

    * *ln lnA A A A A ART a RT x = =

    Activity:

    Activity coefficient:Aa

    A

    Solvent

    As 1, , 1A A A Ax a x *o

    A A =Standard state Limiting behavor

    * *

    *ln lnAA A A A

    A

    pRT RT x

    p = =

    * *

    * *ln lnB B BB B B

    B B

    p K xRT RT

    p p = =

    lno

    B BRT x= *

    *lno BB B

    B

    KRT

    p = wherelno

    A ART x= *o

    A A =where

    B

    B B BB

    p

    a xK = =

    Activity:

    Activity coefficient:Ba

    B

    ln lno o

    B B B B B BRT a RT x = =

    *

    *lno BB B

    B

    KRT

    p =

    Standard state Limiting behavor

    As 0, , 1B B B Bx a x

    Solute

    34

    The Validity of Raoults & Henrys Laws

    Can you read the figure?

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    35

    Colligative Properties

    There is no reference in these properties to the identities of the

    solutes and/or solvents - strictly dependent upon composition

    The simplest example is 1+2 system:

    (1) a pure solvent phase, which may be vapor, liquid, or solid

    (2) a solution phase.(3) an interface between the two phases that is not crossed by the solute

    Increase of boiling point

    Osmosis(water purification)

    Decrease of freezing point

    Solubility(dissolve salt in water)

    Eq. of solvent in: pure s & mix l

    Eq. of solvent: in pure g & mix lEq. of solute in: pure s & mix l

    Eq. of solvent in: pure l & mix l

    , ,( ) ( ) ln AA s A lT T RT x =

    , ,( ) ( ) ln AA g A lT T RT x = , ,( ) ( ) ln BB s B lT T RT x =

    * *

    , ,( ) ( ) ln

    AA Ap p RT x

    =

    Chemical Potential of Solvent

    Presence of solute decreases

    (solvent), resulting in reduced

    freezing point and increased boiling

    point.

    It is an entropy effect, because it

    occurs even for ideal solutions. When

    solute is present, there is an

    additional contribution to the entropy

    of the liquid. Hence a smaller changein entropy on vaporization.

    For b.p. change, look for the temperature at which chemical potentials

    of vapor and solvent molecules are the same at 1 atm pressure. Same

    for melting

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    37

    Elevation of Boiling point

    Boiling at T* Boiling at T

    *

    * *

    ( ) ( ) 1 1vap vapvap m

    T TH

    T T T T

    =

    *1 1

    ln vap m

    A

    Hx

    R T T

    =

    * * *

    , ,( ) ( )

    A g A lT T =

    *( ) 0vap T =Pure liquid:

    *

    ( ) ( ) ln AA l A lT T RT x = ( )

    ln vap

    A

    Tx

    RT

    =In solution:

    2

    p

    H

    T T T

    =

    2

    vap vap m

    p

    H

    T T T

    =

    Gibbs-Helmholtz equation:

    2

    ln vapA Hd x

    dT RT

    =

    Integral:

    *

    ( )A g

    T=

    *2vap m

    BH Tx

    RT

    Further simplification gives:

    *2

    ;B

    vap m

    RTT Kx K

    H = =

    See

    justification

    5.1

    39

    Depression of freezing point

    T depends on the solvent, change is biggest for solvents with high boiling

    points. For practical applications, mole fraction of B is proportional to molality, b

    (mol kg-1), in dilute solutions, and Kb is the empirical boiling-point constant

    T =Kbb;K

    b =

    RT *2

    fusH

    mA

    A =mol/kg for

    solvent

    *

    , ,( ) ( ) ln

    AA g A lT T RT x =

    boiling Same argument applies also for freezing

    *

    , ,( ) ( ) ln AA s A lT T RT x =

    (same ideas)

    *2

    ;B

    fus m

    RTT Kx K

    H = =

    constant, K

    *2

    ;f ffus m

    RTT K b K

    H A = =

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    40

    Osmosis

    A+B AWhat is in balance?

    pp+

    gh =

    *

    ,( ) ,( )( ) ( )A l A lp p = *

    ,( )( ) lnA l Ap RT x=

    * *

    ,( ) ,( )( ) ( )

    p

    A l A l m m

    p

    p p V dp V

    =

    lnm A BV RT x RTx = /BB RT B n V = =

    Vant Hoff Eq.

    = molar concn.xB=nB/nA, approx.

    solventparticle

    Liquid(solid)/vapor

    Solution/vapor:boiling point

    Solution/solid:freezing point

    Solution/solid: solubility

    Solution/solvent: osmosis

    , ,( , ) ( , ) lno o

    A A o

    pT p T p RT

    p

    =

    2

    ln vapHd p

    dT RT =lnvap

    o

    pR

    T p

    =

    Clausius-Clapeyron equation for

    sublimation or vaporization

    Vant Hoff equation

    Applicability: dilute solution

    Applicability: dilute solution

    Applicability: dilute solution

    Applicability: dilute solution

    *

    , ,( , ) ( , ) ln AA AT p T p RT x =

    ( )ln

    vap

    A

    TR x

    T

    = 2

    ln vapA Hd x

    dT RT

    =

    *

    , ,( , ) ( , ) ln AA AT p T p RT x =

    ( )( )ln

    fusconA

    TTR x

    T T

    = = 2

    ln fusA Hd x

    dT RT

    =

    *

    , ,( , ) ( , ) ln BB BT p T p RT x =

    ( )( )ln

    fusconB

    TTR x

    T T

    = =

    2

    ln fusB Hd x

    dT RT

    =

    * *

    , ,( ) ( ) ln AA Ap p RT x =

    *

    , ( ) lnm AA p V RT x=

    lnm A BV RT x R Tx =

    Similarities

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    Quiz (9th November, 2012)

    What is the freezing point of a 0.2 molal solution of

    sucrose in water? [fusH (H2O)=6.02 kJmol-1,

    Tfus(H2O)=273.15K]

    What is the value of Kffor an organic solvent, molar mass

    = 84g, whose Tfus= 279.6 K and fusH = 2.68 kJ mol-1? If a

    0.2 molal solution of hexane in this solvent was prepared,

    calculate the freezing point depression.

    Solutions

    Approximate for small T-T* and small xB, then:

    BfB

    mfus

    xKxH

    RTT =

    =

    2*

    This is our working equation. But we see the amount of solute is given in

    mole fraction here, and we need to convert to molal concentration, i.e. mBmoles/kg of solvent. Remember:

    1-kgg10001000AB

    BA

    BB

    Mm

    mM

    mx

    = For dilute solutions, where MA is the molar

    mass of the solvent in g and mB

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    Solutions, Contd

    Now we can just substitute in the information we have:.

    We find the freezing point depression in a 0.2 molal solution of sucrose is -0.37 K,

    so the freezing point is 273.15 - 0.37 = 272.78 K.

    For the organic solvent, we find

    Kf= 8.314 x 279.62/2680 = 242.5 K

    For 0.2 molal solution, T = 0.2x242.5x84/1000 = 0.2x20.3 K = 4.1 K, so

    Tfus = 275.5K

    Example

    The addition of 100g of a compound (A) to 750g of CCl4 (B)lowered the freezing point of the solvent by 10.5K.Calculate the molar mass of the compound.

    Note:Kf (CCl4) = 30 K kg mol-1

    B

    BB

    n

    mM = Depression of freezing point :

    AH

    RTK

    bKT

    mfus

    f

    f

    =

    =

    2*

    b is molality (mol/kg) of solute,

    A=mol/kg of solvent

    BAB bmn =

    Hence:

    Tm

    KmM

    A

    fB

    B

    = and g/mol108.3K5.10kg75.0

    )molkgK30(g100 21

    =

    =

    BM

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    Solubility example

    The enthalpy of fusion of napthalene is 18.80 kJ mol

    1

    and its meltingpoint is 81C. Calculate its ideal solubility in benzene at 25C.

    301.0

    20.115.298

    1

    15.354

    1

    314.8

    108.1811ln

    20.1

    3

    *

    ==

    =

    =

    =

    ex

    TTR

    Hx

    B

    fus

    B

    B here is the solute,

    naphthalene

    Naphthalene dissolves until its chemical potential equalizes in the solution

    and the vapor. Hence we can use the Gibbs-Helmholtz/modified Clausius-

    Clapeyron equation:

    Assume x(Nap) n(Nap)/n(Bz), so in 1000 g of solvent:

    n(Nap) (0.301)(1000/78) = 3.85 mol. Hence molality is 3.85.

    M(Nap) = 128 g, so 3.85 mol = 493 g in 1000g benzene.

    Actually, xB=0.3 corresponds to closer to 5.5 mol/1000g

    47

    Constituents and Components

    Constituent:A chemical species that is present in a system. For example,

    a mixture of water and ethanol has 2 constituents

    Component:A chemically independent component of the system. The

    number of components in a system, C, is the minimum

    number of independent species needed to define the

    composition of all of the phases present in the system

    When no reaction takes place, Constituents = Components When a reaction can occur, the number of components is the

    minimum number of species which specifies the composition

    of all of the phases (or: smallest number of independently variable

    chemical species to describe the composition of each phase)

    Number of Phases,P = 2

    Number of Constituents = 3

    Number of Components, C = 2*

    *CaCO3

    can be expressed in terms of 2 components in two different

    phases from the stoichiometry of the reaction

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    48

    Examples

    N.B., if additional HCl or NH3 were added to the system, decomposition

    of the NH4Cl would not give the correct gas phase compositions, and

    either HCl or NH3 would be invoked as a 2nd component

    *NH3 and HCl are fixed in stoichoimetric proportions by the reaction, and

    compositions of both phases can be specified by NH4Cl

    Number of Phases, P = 2

    Number of Constituents = 3

    Number of Components, C = 1*

    Consider the thermal decomposition of ammonium chloride:

    Number of Phases, P = 2 Number of Constituents = 3

    Number of Components, C = 3*

    *At room temperature, O2 (g) and H2 (g) do not react to form water, so

    they are not in equilibrium: regarded as independent constituents

    49

    Variance and Degrees of Freedom

    The variance, F, is the number of intensive variables in a system that

    can be changed independently without disturbing the number of phases

    in equilibrium.

    In a single-component, single-phase system (C=1, P=1) the pressure

    and temperature may be changed independently without disturbing the

    number of phases in equilibrium:

    F = 2, system is bivariant, or has two degrees of freedom

    If two phases are in equilibrium in a single-component system (C=1,

    P=2) (e.g., a liquid and its vapor), the temperature (or pressure) can be

    changed, but there must be an accompanying dependent change inpressure (or temperature) to preserve the phases in equilibrium

    F = 1, system has one degree of freedom

    If three phases are in equilibrium (C=1, P=3) (e.g.: s, l and g), neither the

    T nor p can be changed, otherwise the equilibrium is not preserved.

    F = 0, system has no degrees of freedom

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    50

    Phase Rule

    2F C P=

    By J. W. Gibbs

    1. Count the number of intensive variables (p and T count as 2)

    2. Specify composition of the phase by mole fractions of C-1 components

    x1, + x2, ++ xC,=

    Since there are P phases, total number of composition variables is P(C - 1),

    and total number of intensive variables is P(C - 1) + 2

    3. At equilibrium, chemical potential of J must be same in every phase

    J, = J, = .... for P phases

    There are P - 1 equations of this kind for each component J. If there are C

    components, total number of equations is C(P - 1)

    4. Each equation reduces our freedom to vary any of the P(C - 1) + 2 intensive

    variables, so the number of degree of freedom is

    F = P(C - 1) + 2 - C(P - 1) = C - P + 2

    Degrees of freedom for several phases in equilibrium

    51

    Understand the changes given phase diagram

    a In gas phase, cooled at constant p, F = 2

    (for fixed p, F=1,T can be varied at will)

    b Liquid appears at the phase transition

    (at Tb), and F = 1. (for fixed p, F=0, T = Tb)

    c Lowering T takes liquid to single phase

    liquid region, F = 2, (for fixed p, F=1,T

    can be varied at will)

    d Liquid-solid phase transition at Tf, with

    F =1, and at constant p, T is not under ourcontrol

    e Lowering T further results in single solid

    phase with F = 2, (for fixed p, F=1,

    T can be varied at will)

    Consider pure waterWhat happens going from a to e?

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    52

    Two Component SystemFor two component system, C = 2, and F = 4 P.

    3 free variables for single phase, use T, P, x (mole fraction).

    If either T or p is held constant, remaining variance is F = 3 P, 2 variables Phase diagrams of P & x for stable phases at constant T

    Phase diagrams of T & x for stable phases at constant p

    pB*

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    54

    Composition of the Vapor

    Compositions of liquid and vapor in equilibrium are not necessarily the

    same, vapor should be richer in the more volatile component

    A Ap y p= B Bp y p=

    For an ideal mixture, partial pressures may be expressed in terms of

    mole fractions in the liquid (see equations on previous page)

    *

    * * * , 1

    ( )

    A AA B A

    B A B A

    x py y y

    p p p x= =

    If pA*>pB*, yA>xA (vapor richer in

    more volatile component).

    The total vapor pressure as

    composition of the vapor:

    * *

    * * *( )

    A B

    A B A A

    p pp

    p p p y=

    A(l)+B(l)

    A(g)+B(g)

    55

    Combining the Vapor & Liquid diagrams

    Better understand the phase transition between vapor and liquid by combining

    the two previous diagrams into one. The horizontal axis shows the overall

    composition, zA, of the system:

    Liquid & vapor

    coexist

    p

    A(l)+B(l)

    A(g)+B(g)

    * * *AB B A

    p p p p x=

    * *

    * * *( )

    A B

    A B A A

    p pp

    p p p y=

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    56

    What happens as p is varied

    Line of constant overall composition: isopleth

    pA*

    pB*

    All points above the solid diagonal correspond to a system under high pressure where

    there is only a liquid phase (zA = xA, composition of liquid)All points on lower curve correspond to a system with low pressure where only the

    vapor phase exists (applied pressure less than the vapor pressure, so zA = yA)

    Points between lines are systems where two phases exist: 1 liquid & 1 vapor

    57

    Interpretations of the Diagrams

    Lowering the pressure on a liquid by drawing out a piston:

    start at point a: Here, F = 2 and P = 1; only the liquid phase exists; changes to system

    do not affect overall composition, so system moves down vertical line that passes

    through a, until point a1 is reached (pressure reduced top1)

    The horizontal axis shows the overall composition, zA, of the system:

    p3

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    58

    The Lever Rule

    Can get n/n

    T-x and p-x diagrams summary

    59

    2 component system: F=2+2-Phase

    Variables: p, T, x

    p,T diagram

    TB*

    TA*

    T,x diagram

    p,T,x

    A(l)+B(l)

    A(g)+B(g)

    xA,yA,zA

    p,x diagram

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    60

    Critical Solution Temperatures

    Tuc

    61Weak compound

    Critical Solution Temperatures, contd

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    Phases

    a. Ideal solution behavior.

    b. A low (T)-boiling azeotrope

    c. A eutectic point

    d. Congruent melting

    e. Incongruent melting

    f. Partial miscibility in 1 or more phases

    g. Compound formation

    h. Complete miscibility in all phasesA+AB L+AB

    AB+BL+ B

    Solid-Liquid Phase Diagrams

    A2BAB2

    Eutectic Point

    for A+A2B

    system

    Compound

    formation

    L+A

    L+A2B AB2+B

    AB2 melts

    L+B

    peritectic point

    Peaks represent

    compound formation.

    Increasing temperature

    (up) along dotted line

    includes incongruent

    melting as it crosses thehorizontal line. Here the

    AB2 melts but there is

    excess B, so the melt

    contains L and B (solid).

    A2B+AB2

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    Completely Miscible Solids

    TB*

    TA*

    T,x diagram

    Looks like a gas/vapor diagram, but now solid/liquid

    liquid

    solid

    Liquid and solid in equilibrium with

    Proportions given by lever rule