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    5.60 Spring 2008 Lecture #22 page 1

    Two-Component Phase Equilibria IIIIdeal and Non-Ideal Solutions

    Free energy change Gmix in ideal solutions

    A (liq) B (liq) A + B (liq)G

    * ( ) * G ( ) mix( ) mix G ( ) n x n x ( ) n x n x ( ) 1 A A A B B B 2 A A A B B B

    G G ( )G ( )mix 2 1A

    A( ) A B B

    ( ) B nxA A nxB B

    nx RTlnx nx RTlnx ( ) ( )G nRT x lnx x lnxmix A A B B

    Purely entropic, as in gas mixture

    G = Vdp SdT G

    mix T mix

    p

    A A B BS nR x lnx x lnx

    Hmix Gmix mix T S 0

    No enthalpy change, all of G is due to entropy of mixing

    Change in volume Vmix

    Vmix

    Gpmix

    0 T

    No volume change, just like ideal gas.

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    5.60 Spring 2008 Lecture #22 page 2

    Non-Ideal Solutions:

    In reality, molecules interact:

    A A A

    A

    B

    B B B

    uAA < 0 uBB< 0 uAB uABPURE MIXED

    u = 2uAB- (uAA + uBB)

    This difference determines how solutions depart from ideality.

    I. Positive Deviations : u > 0 (most common)

    Mixing is energetically not favorable in liquid phase.

    H = U + (PV) U

    G u nRT x lnx x lnx G idealmixn4

    A A B B mix

    e.g. acetone & carbon disulphide

    CH3C = O + S = C = S

    CH3

    BA

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    5.60 Spring 2008 Lecture #22 page 3

    * *acetone Ap p

    2

    *BCSp p

    p = pA + pB

    pB

    pA

    1xCS2 = xB0

    p Real

    Raoult

    *

    pCS2 xCS2pCS2 *pacetone xacetone pacetone

    ptot real ptot Raoult

    vapor pressure is higher than expected by Raoults Law

    II. Negative Deviations : u < 0

    e.g. acetone & chloroform

    CH3

    CH3C = O +

    BACl

    Cl

    H - C - Cl

    H-bondingattraction

    Mixing is energetically favorable in liquid phase.

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    *CHCl

    B

    A B

    5.60 Spring 2008 Lecture #22 page 4

    Realp

    0 xCHCl3 = x

    p = p + p

    pB

    pA

    *pB* * pacetone pA

    Raoult

    1

    *pCS2 xCS2pCS2

    *pCHCl3 xCHCl3pCHCl3

    p real p Raoulttot tot

    Ideal Dilute Solutions and Henrys Law:

    Non-ideal solutions are difficult to describe completely

    Describe limits x B 1 and xB 0 Ideal Dilute Solution

    I. x CS2 = xB 1 (B is the solvent)

    Then Raoults Law applies for CS 2

    CS2 molecules see mostly other CS 2 molecules

    *pCS2 xCS2pCS2

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    5.60 Spring 2008 Lecture #22 page 5

    II. x B 0 (B is the solute)

    Then Henrys Law applies: pCS2 xCS2KCS2 pB = xBKB

    Henrys Law constant

    KB Henrys Law constant, depends on the solvent/solute mixtureand the temperature.

    Labeled just K B, even though it depends on A also.

    2KCS K

    Positive deviation BK*Bp

    Henry*

    (Negative deviation would2

    pCS p

    have BK*Bp )

    1xCS2 = xB

    p

    Raoult

    Real

    Ideal dilute solution:

    Solvent, e.g. A: x A ~ 1 Raoults Law Ap*

    AAx p

    Solute, e.g. B: x B ~ 0 Henrys Law pB = xBKB

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    5.60 Spring 2008 Lecture #22 page 6

    Total phase diagram:

    Ideal Real p

    Liquidp

    Liquid **phase pB phase pB

    * *pA pAgas gasphase phase

    0 xB,yB 1 0 xB,yB 1

    Stronger non-ideality: Azeotropes

    Liquidp phase

    At this pointpB xB = yB

    *pA gasphase

    0 xB,yB 1

    Similar for constant p phase diagramCan have positive or negative azeotropes

    T T

    phase

    gas

    T A* T A*

    T B*LiquidLiquidphasephase

    0 xB,yB 1 0 xB,yB 1

    Cannot distillpast azeotropexB = yB

    gasphase

    T B*