• driving force for formation of a new phase

- more complicated case involving changes in comp - in forming a small amount of a new phase, Θ from a soln phase α (xB

α) - as long as the amount of the α phase is large, we can take A and B without changing values of GA

α & GBα, namely, without changing

the org comp of the α phase - if Gm

θ falls on the tangent line drawn from Gmα at xB

α → D for formation of θ is zero - therefore, Gm

θ should fall below the tangent line drawn from Gmα

)()()( BmBBBBAAm xGxGxxGxG

Fig. 7.17 (a) Molar Gibbs energy diagram. (b) Method for evaluation of the driving force for the formation of a new phase from a supersaturated solution.

- as the process of forming the Θ phase continues, the supersaturation decreases gradually and so will the

m

NPTNPT

GNGG

Dii

,,,,

- to compare with Fig. 7.15(b) (letting x as Nθ)

)0()(

)(

//

2

2

BB

B

mBBBm

xPx

dxGd

xxxVPD

: gradual decrease in supersaturation & driving force

)()()( BmBBBBAAm xGxGxxGxG

- finally, obtaining x’Bα in equil with Θ

in the transf of β → α when is it possible, xB β = xB

α ?as long as xB lies to the left of To,id, the region where Gm

α < Gmβ

xxx

eq

BBif

/

happens to be

, then

α β

xxB eq /

xB x B

0

local equil

0/ BBeq xxx

α

β

T0(xB)mGD

• partitionless transf under local equil

( 分配係數 )

without partitioning 분배계수 = 1=

BB xxK //

BB xx /

same comp btw &

• molar Gibbs energy diagram for ternary sys

in the binary the shape of Gmα (Gibbs energy surface)

in the ternary a surface shape

equil btw α & β → tangential line in the binary → tangential in the ternary and many planes

CCC

BBB

AAA

GG

GG

GG

Fig. 7.25(a) Molar Gibbs energy diagram for a two-phase equilibrium in a ternary system. The two-phase field is created by the common tangent-plane rolling under the two surfaces.

general equil cond btw α & β is

if the phase β is a line compound phase of θ, then the equil btw α & θ is

GGGGGG

CBCBBaCc

CACAAaCc

caca

caca

here, f = v = c + 2 - P - (2) = 3 + 2 – 2 – 2 = 1 (at const T & P)

in other words, variables are one of yA & yB (xA or xB) of the

phase θ

+ two among xA, xB, xC of the phase α – two eqs to be satisfied

∴ f = v = 3 – 2 = 1

Fig. 7.25(b) Molar Gibbs energy diagram for a ternary system with an ordinary solution phase, , and a line compound, . In this diagram a+c=1 because the whole diagram is for one mole of atoms.

• solubility product ( 溶解度積 ) a ternary stoichiometric compound φ (AlBmCn, l+m+n=1)

CBAm

iim

nml

x

GG

- equil btw α & Φ being described as follows:

CBAanamal

RTGnGmGlG

then

aRTGGnGmGlGG

CBAm

iiiCBAm

lnlnln ,

)ln(

standard Gibbs energy of formation for mG

nC

mB

lA

m aaaRTG

exp

if α is rich in A and dilute in B & C 1 AA xa

nC

mB

nC

mB

m xxffRTG

exp solubility product for α(A)log expression is m lnxB + n lnxC

= …a linear form

for the pure compound (comp is constant)