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Chemistry 433Chemistry 433
L t 22Lecture 22
Non‐ideal behavior
NC State University
Free energy and entropy of mixingWe can calculate the free energy of mixing for an ideal
solution based on the chemical potential. The free
energy of mixing is
ΔmixG = Gsln(T,P,n1,n2) ‐ G1*(T,P,n1) ‐ G2*(T,P,n2)mix ( , , 1, 2) 1 ( , , 1) 2 ( , , 2)
ΔmixG = n1μ1sln + n2μ2
sln ‐ n1μ1* ‐ n2μ2
*
ΔmixG = nRT(x1lnx1 + x2lnx2)ΔmixG nRT(x1lnx1 x2lnx2)The free energy of mixing is always negative because x1and x2 are always less than one An ideal solution willand x2 are always less than one. An ideal solution will
always form spontaneously. This is due entirely to the
entropy of mixing The entropy is given byentropy of mixing. The entropy is given by
ΔmixS = ‐(∂ΔmixG/∂T) = ‐ nR(x1lnx1 + x2lnx2)
Definition of ideal behaviorDefinition of ideal behavior
This result is exactly the same as the mixing entropy forThis result is exactly the same as the mixing entropy for
an ideal gas. For an ideal solution we find further that
Δ V = (∂Δ G/∂P) = 0ΔmixV = (∂ΔmixG/∂P) = 0and
Δ H = Δ G + T Δ S = 0ΔmixH = ΔmixG + T ΔmixS = 0.The volume change and enthalpy of mixing are zero for
an ideal solution because the size and shape andan ideal solution because the size and shape and
intermolecular potential of the two species being
i d l th th imixed are nearly the same as the pure species.
A model of molecular interactionsA model of molecular interactions
We can express the behavior in real solutions as aWe can express the behavior in real solutions as a
deviation from Raoult's law. For example, in the figure
below we have component 1 with a vapor pressure ofbelow we have component 1 with a vapor pressure of
50 torr for the pure component.
That is P * = 50 torr Likewise P * = 100 torrThat is, P1 = 50 torr. Likewise, P2 = 100 torr.
We can consider two cases Deviations from Raoult'sWe can consider two cases. Deviations from Raoult's
law can be positive or negative. Positive deviations (i.e.
t th id l) th t th likgreater vapor pressure than ideal) mean that the unlike
molecules have a repulsive interaction.
Positive deviations from Raoult’s lawPositive deviations from Raoult s law
A mathematical model for the d f d ldeviations from ideality
To obtain the plots we use the following expressions.p g pP1 = x1P1*
P2 = x2P2*
P1 = x1P1*exp(αx22)P2 = x2P2*exp(αx12)where the term exp(αx22) represents a deviation from ideality (α = 0.5 in the plot above). Negative deviations f R lt' l l b d h i th l tfrom Raoult's law are also observed as shown in the plot below (α = ‐2 in the plot below).
Negative deviations from Raoult’s lawNegative deviations from Raoult s law
Limiting behaviorLimiting behavior
For either positive or negative deviations from Raoult's law, if we look at component 1 we can see that as x1 → 1 and x2 → 0 we recover Raoult's law, P1 = x1P1*. In other words when 1 is the
l t h l id l b h i f b t 1pure solvent we have nearly ideal behavior for substance 1. However, at the opposite limit as x1 → 0 (and x2 → 1) we have, P1 = x1P1*exp(α). In this case substance 1 is the solute and it 1 1 1 p( )behaves non‐ideally. This has been expressed as Henry's law. Henry's law states that P1 = x1kH,1 as x1 → 0. Thus, Henry's describes the non‐ideal behavior of a solute in a dilute solution. Note that for the explicit model above we can equate kH 1 = P1*exp(α)kH,1 = P1 exp(α).
Explaining activityThroughout the entire range of concentration we can redefine
activity, a1. Activity is an effective mole fraction. That is we can
define
P1 = a1P1*
for a real solution. Our simple model above shows that the
deviation from ideality can be treated by fitting to expressions of
th f P P * { 2 β 3 }the form P1 = x1P1*exp{αx22 + βx23 + …}The chemical potential for this type of solution is
μ = μ * + RTln x + αRTx 2 + βRTx 3μ1 = μ1 + RTln x1 + αRTx2 + βRTx2We see from the definition of activity as a1 = P1/P1* that we can
express the chemical potential as μ1 = μ1* + RTln a1p p μ1 μ1 1
Thus we can see that a1 = x1exp{αx22 + βx23 + …}.
The activity coefficientThe activity coefficient
We can define an activity coefficient γ1 such thaty γ1γ1 = a1/x1.Therefore γ1 = exp{αx22 + βx23 + …}.Clearly, a1 → x1 and γ 1 → 1 as x1 → 1. Thus, the solution approaches ideal behavior as the composition approaches that
of the pure solution. This definition is based on a Raoult's law
standard state. This is also known as a solvent standard state.
The activities or chemical potentials are meaningless unless weThe activities or chemical potentials are meaningless unless we
know the standard state. The solvent or Raoult's law standard
state holds for substances that are miscible in all proportions.state holds for substances that are miscible in all proportions.
μ1 = μ1* + RTln a1
Raoult’s law standard stateRaoult s law standard state
Up to now we have implicitly discussed only Raoult's lawp p y y
standard states.
Strictly speaking, we should use the words solute and solvent to
describe a binary solution only when one component is sparingly
soluble in the other. In this case that standard state may be
based on Henry's law rather than Raoult's law. Assuming that
component j is the sparingly soluble component we can write
μ = μ * + RTln (P /P *)μ1 = μ1 + RTln (P1/P1 )We calso think of this as the standard state for an ideal solution.
Henry’s law standard stateHenry s law standard state
For the solute j we use Henry's law Pj → xjkH j as xj → 0 where j y j j H,j j
kH,j is the Henry's law constant for component j. Thus we have.
μj = μj* + RTln (xjkH,j/Pj*) = μj
* + RTln (kH,j/Pj*) + RTln (xj) (xj → 0)We define the activity of component j by
μj = μj* + RTln (kH,j/Pj*) + RTln aj.
Just as above for the Raoult's law standard state
aj → xj as xj → 0. Thus we define a = P /k The standard state then implies thatThus, we define aj = Pj/kH,j. The standard state then implies that
kH,j = Pj*. This standard state may not exist in practice, so it is
called a hypothetical standard state.called a hypothetical standard state.
Standard state for activityStandard state for activity
The numerical value of the activity depends on the standard y p
state. This is best demonstrated using an example. We consider
a solution of CS2 and CH3OCH2OCH3. The Henry's law constants
for this solution are kH,CS2 = 1130 torr and kH,dimeth = 1500 torr.
Based on vapor pressure data we can calculate the activity and
(the activity coefficient based on each standard state (Raoult's
Law and Henry's Law).
For x CS2 = 0.5393 we have P CS2 = 357.2 and Pdimeth = 342.2 torr.
We also need to know P CS2* = 514.5 and Pdimeth
* = 587.7 torr.We also need to know P CS2 514.5 and Pdimeth 587.7 torr.
Two definitions of standard stateTwo definitions of standard stateRaoult's Law
a CS2 = PCS2/ P CS2* = 357.2/514.5 = 0.694
a dimeth = Pdimeth/ P dimeth* = 342.2/587.7 = 0.582
/ 0 694/0 539 1 287γ CS2 = a CS2/x CS2 = 0.694/0.539 = 1.287
γ dimeth = a dumeth/x dimeth = 0.582/(1‐0.539) = 1.262
Henry's Law
a CS2 = PCS2/ kH CS2 = 357.2/1130 = 0.316CS2 CS2 H,CS2
a dimeth = Pdimeth/ kH,dimeth = 342.2/1500 = 0.228
γ CS2 = a CS2/x CS2 = 0.316/0.539 = 0.586
γ dimeth = a dumeth/x dimeth = 0.228/(1‐0.539) = 0.494
Excess free energyExcess free energy
We can calculate the Gibbs energy of mixing of binary solutions gy g y
in terms of the activity coefficients to obtain a free energy of
mixing for non‐ideal solutions. As above for ideal solutions we
can define
ΔmixG = n1μ1sln + n2μ2
sln ‐ n1μ1* ‐ n2μ2
*
However, in this case we must use the definition of chemical
potential in terms of activity (rather than mole fraction)
μ = μ * + RTln a = μ * + RTln x + RTln γμj = μj + RTln aj = μj + RTln xj + RTln γjleading to the expression
ΔmixG = RT(n1ln x1 + n2ln x2 + n1ln γ1 + n2lnγ2)ΔmixG RT(n1ln x1 n2ln x2 n1ln γ1 n2lnγ2)
The first two terms represent the Gibbs free energy of mixing of
an ideal solution To focus on the effect of non ideality wean ideal solution. To focus on the effect of non‐ideality, we
define an excess Gibbs energy of mixing ΔGE.
ΔGE = Δ i G ‐ Δ i GidΔG ΔmixG ΔmixG
Thus, ΔGE = RT(n1ln γ1 + n2ln γ2)If we divide by the total number of moles n1 + n2, we obtain the1 2
molar excess Gibbs energy of mixing ΔGE.
ΔGE = RT(x1ln γ1 + x2ln γ2)Using the definition of the activity coefficient for the solution
example above we can calculate the excess free energy for the
d l l ti bmodel solution above.
γ1 = exp{αx22}andand
γ2 = exp{αx12}.
The origin of excess free energy is the enthalpy
S b tit ti th ti it ffi i t i t thSubstituting these activity coefficients into the
expression for the excess free energy we have
ΔGE = RT(α x x 2 + α x x 2) = α RTx xΔG = RT(α x1x2 + α x2x1 ) = α RTx1x2using the fact that x2 = 1 ‐ x1.
If components 1 and 2 are distributed randomly throughout the p y g
solution then the entropy of the ideal solution and the non‐ideal
solution will be the same. Such a solution is known as a regular
solution. In a regular solution ΔHE ≠ 0 and ΔSE = 0. Thus, the difference in the Gibbs energy is due to an interaction energy
( b h h l f l )term (a contribution to the enthalpy of solution).
Molecular model for d l lnon‐ideal solutions
We can express the potential energy of the solution in the formp p gy
U = N11e11 + N12e12 + N22e22where Nij is the number of neighboring pairs of molecules of
type i and j and where eij is the interaction energy of a pair of
molecules of type i and j when they are next to each other. We
assume a coordination number z where z is between 6 and 10.
There are N1 component 1 molecules in solution so the number
of 1 1 neighboring pairs is N = zN x /2 where the factor of usedof 1‐1 neighboring pairs is N11 = zN1x1/2 where the factor of used
to avoid counting each 1‐1 pair twice. Similarly, for component 2
we have N22 = zN2x2/2. The same value of z is used because wewe have N22 zN2x2/2. The same value of z is used because we
assume that molecular sizes are about the same.
Aspects of the microscopic model
e11Interaction energies
self interaction
e22 self interaction
e12 cross term
z is the solvation number (here it is 6)
Molecular model for d l lnon‐ideal solutions
The number of 1‐2 neighboring pairs is given by g g p g y
N12 = zN1x2 = zN2x1.
The total interaction energy in the solution is
U = zN1x1e11/2 + zN2x2e22/2 + zN1x2e12Using the definitions x1 = N1/(N1+N2) and x2 = N2/(N1+N2)
we can reexpress the interaction energy as
U = (N12e11/2 + N2
2e22/2 + N1N2e12)z/(N1+N2)
We can focus on the non ideality of the solution by introducingWe can focus on the non‐ideality of the solution by introducing
the quantity
w = U = e11 + e22 ‐ 2e12w U e11 e22 2e12
Comparing ideal and real energiesFor an ideal solution e11 = e22 = e12 and so w = 0. However, for a
non ideal solution e ≠ e ≠ enon‐ideal solution e11 ≠ e22 ≠ e12. Substituting e12 = (e11 + e22 ‐ w)/2 we have
U = (N12e11/2 + N2
2e22/2 + N1N2(e11 + e22 ‐ w)/2)z/(N1+N2)U (N1 e11/2 N2 e22/2 N1N2(e11 e22 w)/2)z/(N1 N2)
U = zN1e11/2 + zN2e22/2 ‐ zwN1N2/2(N1+N2)
The last term represents the non‐ideality in the solution.
Therefore, we can express the Gibbs energy of the solution as
Gsoln = Gideal ‐ zwN1N2/2(N1+N2)
or units of moles
Gsoln = Gideal ‐ zwNAn1n2/2(n1+n2)
The chemical potential of component 1 is given by
μ ∂G ∂Gid zwN A ∂n1n2/(n1 + n2)
Note that the chemical potential of an ideal solution is given by
μ1 = ∂G∂n1
= ∂n1– A
21 2 ( 1 2)
∂n1
μ1 = μ1* + RT lnx1
The derivative is
This leads to the expression
∂n1n2/(n1 + n2)∂n1
= n2n1 + n2
– n1n2
n1 + n22 = x2 – x1x2 = x2(1 – x1) = x2
2
This leads to the expression
μ1 = μ1* + RT lnx1 ‐zwNAx22/2
μ1 = μ1* + RT ln(x1e ‐zwx22/2kT)μ1 μ1 ( 1 )
where we have used the fact that k = R/NA.
Our simple model of a solution has led directly to an expression
for the activity in terms of the interaction strength w,
coordination number z, mole fraction x2, and thermal energy kT.
Definition of a critical pointDefinition of a critical point
The activity is a1 = P1/P1* = x1exp(‐zwx22/2kT)y 1 1/ 1 1 p( 2 / )
The activity coefficient is γ1 = exp(‐zwx22/2kT)The expression derived is exactly analogous to the expressions
used above for deviations from Raoult's law provided
α = ‐zw/2kT.
This simple model not only allows us to calculate activities from
molecular properties, but it also includes the possibility of phase
separation A critical point in a two component phase diagramseparation. A critical point in a two component phase diagram
indicates that there is a region of temperature or pressure
beyond which the solution separates into two phases.beyond which the solution separates into two phases.
Phase separationPhase separation
In the case of the model liquid discussed here phase separation q p p
will occur if zw/2kT > 2 or in other words if the parameter α < ‐2. We first show the relationship between the activities of two
species in a non‐ideal binary solution and then use all of the
information to discuss the free energy of mixing. This leads γ
naturally to the idea of limited solubility even with this very
simple model.
Application of the bb hGibbs‐Duhem equation
At this point, you might well ask whether the activity coefficient p , y g y
for component 1 has any bearing on the magnitude of the
activity coefficient for component 2. In fact, they are related as
we now prove using the Gibbs‐Duhem relation. The Gibbs‐
Duhem equation states that
n1dμ1 + n2dμ2 = 0.Dividing through by n the Gibbs‐Duhem equation is
x dμ + x dμ = 0x1dμ1 + x2dμ2 = 0.
dμ2 = ‐ x1dμ1 / x2with μ1 = μ1
* + RT ln(x1e ‐αx22) where α = ‐zw/2kT,with μ1 μ1 RT ln(x1e ) where α zw/2kT,
Application of the bb hGibbs‐Duhem equation
dμ1 = ‐ RTdx1/x1 ‐ 2α(1‐ x1)dx1 and dμ2 = ‐ RTdx1/x2 ‐ 2αx1dx1μ1 1/ 1 ( 1) 1 μ2 1/ 2 1 1
Change variables from dx1 to ‐ dx2.
dμ2 = RTdx2/x2 + 2α(1‐x2)dx2Now we integrate,
μ2 ‐ μ2* = RTlnx2 + α(1‐x2)2
and finally
μ2 = μ2* + RT ln(x2e –αx12)
which shows that the activity of coefficient of component 1which shows that the activity of coefficient of component 1
implies the magnitude of the activity coefficient for
component2.component2.
γ1 = exp(‐zwx22/2kT) implies γ2 = exp(‐zwx12/2kT)
Parameter for non‐ideal solutionsParameter for non ideal solutionsAs a result of our model for the interaction energy of particles
in a non‐ideal solution we can calculate ΔmixG. For a two component solution
Δ G n RTln x + n RTln x zwN /2(n x 2 + n x 2)ΔmixG = n1RTln x1 + n2RTln x2 ‐ zwNA/2(n2x12 + n1x22)
ΔmixG = n1RTln x1 + n2RTln x2 ‐ zwNA(n1 + n2)x1x2/2
Our final expression for Δ i G isOur final expression for ΔmixG is
ΔmixG = nRT(x1ln x1 + x2ln x2 ‐ zwx1x2/2kT)which is clearly separable into an ideal part and the excess free
energy discussed previously. We can plot ΔmixG/nRT for various values of zw/2kT as shown below.
Parameter for non‐ideal solutionsParameter for non ideal solutions
Phase separation and critical pointPhase separation and critical point
Note that for α < ‐2 the curvature is negative in the central region (around a mole fraction of x1 = 0.5). This corresponds to a region where mixing is not spontaneous. I th d th ill b h ti i thiIn other words, there will be a phase separation in thisregion.By assumption, the entropy of mixing is still equal to theBy assumption, the entropy of mixing is still equal to the entropy of mixing of an ideal solution. Thus,ΔmixS = ‐nR(x1ln x1 + x2ln x2)and therefore the non‐ideality appears in the enthalpyΔmixH = ‐ zwNAx1x2/2.
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