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Physical Chemsitry

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  • Chemical Potential

    inTPj

    jn

    G

    ,,

    = Potential Chemical Diffusion from high to low potential.

    Chemical potential is a Partial Molar Quantity

    )jnT,f(P,G :components-multiFor =

    Sum of moles of components

  • Chemical Potential of a Binary (A & B) Mixture

    )n , n , T , f(P G BA=

    BnTPB

    AnTPAnnPnnT

    dnn

    Gdnn

    GdTTGdP

    PGdG

    ABBABA

    +

    +

    +

    =

    ,,,,,,,,

    dPV dTS

    ij

    nTPjdn

    n

    G

    i,,

    iiii nSVj

    nSPjnVTjnTPjj

    n

    Un

    Hn

    An

    G

    ,,,,,,,,

    =

    =

    =

    =

    Chem. Potential applied to other variables:

  • Measures of Composition s = solute ; A = solvent; V = Tot. Vol. of solution. Weight %:

    Mole Fraction:

    100% xww

    ww

    As

    s

    s +=

    s

    snn

    n

    +=

    Molarity:

    Molality:

    Ass

    nn +=

    VnM s

    s=

    Akgn

    m ss

    =

    Different Composition Equations for different Laws

  • Partial molar Quantities Let us suppose that a solution is formed by

    mixing n1,n2, moles of substances 1,2,.. havingmolar volumes V*m,1, V*m,2,. at temp T andpressure P. Let V* be the total volume of theunmixed (pure) components at T and P.unmixed (pure) components at T and P.

    V* = n1V*m,1 + n2V*m,2+.. nrV*m,r= niV*m,i After mixing, we find that volume V of solution is

    not in general equal to the unmixed volume : V V*. e.g., 50 cm3 of ethanol added to 50 cm3 of water

    gives 96.5 cm3 at 20C and 1 atm RANJANRANJANRANJANRANJAN DEYDEYDEYDEY

  • (1) Difference betweenintermolecular forces in

    the solution and those in

    The difference between V of solution and V* results from:

    the solution and those in

    the pure component

    (2) Diff between the packingof molecules in solution andpacking in pure components

    RANJANRANJANRANJANRANJAN DEYDEYDEYDEY

  • Other Partial Molar Quantities

    jnTPii

    n

    VV,,

    =

    Partial Molar Volume:

    jnTPii

    n

    HH,,

    =

    jnTPii

    n

    SS,,

    =

    Partial Molar Enthalpy:

    Partial Molar Entropy:

  • Solution Volume and Partial Molar Volume V = n f(T,P, x1,x2,) where n = ni and f is some

    function of T, P and mole fractions. Differentiation of above equation gives : d V = f(T,P, x1,x2,) dn = Vidni xi = ni/n or ni = xin dV = x V dn for constant T, P, x dV = xiVidn for constant T, P, xi Comparison of expressions for dV gives: V = niVi for one phase system Where the V on the LHS is sometimes referred

    to as the mean molar volume, Vm of the solution.

    Change in volume of mixing, is given by:mixV =V-V*

  • Calculation for Partial Molar Volumes

    BnTPAA

    n

    VV,,

    =

    AnTPBB

    n

    VV,,

    =

    V = f(nA , nB) @ constant P & T

    dnVdnVdV

    +

    = BnTPB

    AnTPA

    dnn

    Vdnn

    VdVAB

    +

    =

    ,,,,

    BBAA dnVdnVdV +=

    BBAA VnVnV +=

    Integrate @ constant composition

  • Slope of V vs nB curve at any composition gives VB for that comp.mixV = V V* = i ni ( Vi V*m,i )

    RANJANRANJANRANJANRANJAN DEYDEYDEYDEY

  • Ideal Solution Ideal solution will be one where the molecules of

    the various species are so similar that replacing molecule of one species with another will not change the spatial structure or intermolecular interaction energy in the solution.

    For this the molecules must be essentially of same For this the molecules must be essentially of same size and shape.

    mixG = G G* = nii ni*i For an ideal solution : mixG = RT ni ln xi Equating : nii = ni(*i + RT ln xi ) i = *i (T,P) + RT ln xi (#1)

  • Last eqn on previous slide is adopted as definition of an ideal solution

    As xi0, i - . As xi increases, i increases at fixed T and P, reaching chemical pot of the pure i, *i in the limit xi =1

    RANJANRANJANRANJANRANJAN DEYDEYDEYDEY

  • Thermodynamic properties of Ideal Solutions

    Standard state of each component i of the ideal liquid solution is defined to be pure liquid i at the temp T and pressure P of the solution.

    G = RT n ln x mixG = RT ni ln xi mixV = 0 mixS = -R ni ln xi mixS = -( mixG / T)P,ni mixH = 0

    RANJANRANJANRANJANRANJAN DEYDEYDEYDEY