Chemistry 433
L t
Lecture
22
Non‐ideal behavior
NC State University
Free energy and entropy of mixing
We can calculate the free energy of mixing for an ideal solution based on the chemical potential. The free energy of mixing is
, , 1,,n2)) ‐ G1*(T,P,n
( , , 1)) ‐ G2*(T,P,n
( , , 2)
ΔmixG = Gsln((T,P,n
ΔmixG = n1μ1sln + n2μ2sln ‐ n1μ1* ‐ n2μ2*
ΔmixG G = nRT(x
nRT(x1lnx1 + xx2lnx2)
The free energy of mixing is always negative because x1
and x2 are always less than one. An ideal solution will and x
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 by
entropy of mixing. The entropy is given by
ΔmixS = ‐(∂ΔmixG/∂T) = ‐ nR(x1lnx1 + x2lnx2)
Definition of ideal behavior
Definition of ideal behavior
This result is exactly the same as the mixing entropy for This
result is exactly the same as the mixing entropy for
an ideal gas. For an ideal solution we find further that ΔmixV = (∂Δ
V = (∂ΔmixG/∂P) = 0
G/∂P) = 0
and
ΔmixH = Δ
H = ΔmixG + T Δ
G + T ΔmixS = 0.
S=0
The volume change and enthalpy of mixing are zero for an ideal solution because the size and shape and
an ideal solution because the size and shape and intermolecular potential of the two species being mixed are nearly the same as the pure species. i d
l th
th
i
A model of molecular interactions
A model of molecular interactions
We can express the behavior in real solutions as a We
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 of
below we have component 1 with a vapor pressure of 50 torr for the pure component. That is P1* = 50 torr. Likewise, P
That is, P
= 50 torr Likewise P2* = 100 torr.
= 100 torr
We can consider two cases. Deviations from Raoult's We
can consider two cases Deviations from Raoult's
law can be positive or negative. Positive deviations (i.e. greater vapor pressure than ideal) mean that the unlike t
th id l)
th t th
lik
molecules have a repulsive interaction. Positive deviations from Raoult’ss law
Positive deviations from Raoult
A mathematical model for the d
deviations from ideality
f
d l
To obtain the plots we use the following expressions.
p
g p
P1 = 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
from Raoult's
R
lt' law are also observed as shown in the plot l
l
b
d
h
i th l t
below (α = ‐2 in the plot below).
Negative deviations from Raoult’ss law
Negative deviations from Raoult
Limiting behavior
Limiting 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 pure solvent we have nearly ideal behavior for substance 1. l t
h
l id l b h i f
bt
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 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 *exp(α). kH,1 =
= P
P
H1
1 exp(α)
Explaining activity
Throughout 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
the form P
P1 = x1P1*exp{αx
{ 22 + βx
β 23 3 + …}}
The chemical potential for this type of solution is
μ1 = μ
= μ1* + RTln x
+ RTln x1 + αRTx
+ αRTx22 + βRTx
+ βRTx23
We see from the definition of activity as a1 = P1/P1* that we can
p
p
μ1 = μ
μ1* + RTln a1
express the chemical potential as μ
Thus we can see that a1 = x1exp{αx22 + βx23 + …}.
The activity coefficient
The activity coefficient
We can define an activity coefficient γ
y
γ1 such that
γ1 = 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 we
The 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’ss law standard state
Raoult
law standard state
Up to now we have implicitly discussed only Raoult's law
p
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
μ1 = μ
= μ1* + RTln (P
+ RTln (P1/P1*)
We calso think of this as the standard state for an ideal solution.
Henry’ss law standard state
Henry
law standard state
For the solute j we use Henry's law P
j
y
j → xjkH,j
H j as xj → 0 where 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 aj = P
Thus, we define a
= Pj/kH,j. The standard state then implies that 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 activity
Standard 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 P
We also need to know P
514.5 and Pdimeth* = 587.7 torr.
587.7 torr.
Two definitions of standard state
Two definitions of standard state
Raoult's Law
a CS2 = PCS2/ P CS2* = 357.2/514.5 = 0.694
a dimeth = Pdimeth/ P dimeth* = 342.2/587.7 = 0.582
γ CS2 = a CS2/x
/ CS2 = 0.694/0.539 = 1.287
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
H CS2 = 357.2/1130 = 0.316
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 energy
Excess 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 + n2μ2sln ‐ n1μ1* ‐ n2μ2*
However, in this case we must use the definition of chemical potential in terms of activity (rather than mole fraction)
μj = μ
= μj* + RTln a
+ RTln aj = μ
= μj* + RTln x
+ RTln xj + RTln γ
+ RTln γj
leading to the expression
ΔmixG G = RT(n
RT(n1ln x
ln x1 + n
n2ln x
ln x2 + n
n1ln γ
ln γ1 + n
n2lnγ2)
The first two terms represent the Gibbs free energy of mixing of an ideal solution To focus on the effect of non ideality we
an ideal solution. To focus on the effect of non‐ideality, we define an excess Gibbs energy of mixing ΔGE.
ΔGE = Δ
Δmixi G G ‐ Δmixi Gid
Thus, ΔGE = RT(n1ln γ1 + n2ln γ2)
If we divide by the total number of moles n1 + n2, we obtain the
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 model solution above.
d l l ti
b
γ1 = exp{αx22}
and and
γ2 = exp{αx12}.
The origin of excess free energy is the enthalpy
SSubstituting these activity coefficients into the b tit ti th
ti it
ffi i t i t th
expression for the excess free energy we have
ΔGE = RT(α
= RT(α x1x22 + α
+ α x2x12) = α
) = α RTx1x2
using 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 term (a contribution to the enthalpy of solution).
(
b
h
h l
f l
)
Molecular model for non‐ideal solutions
d l l
We can express the potential energy of the solution in the form
p
p
gy
U = N11e11 + N12e12 + N22e22
where 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 N11 = zN
of 1‐1 neighboring pairs is N
= zN1x1/2 where the factor of used /2 where the factor of used
to avoid counting each 1‐1 pair twice. Similarly, for component 2 we have N22 = zN
we have N
zN2x2/2. The same value of z is used because we /2. The same value of z is used because we
assume that molecular sizes are about the same. Aspects of the microscopic model
e11
Interaction energies
self interaction
e22
self interaction
e12
cross term
z is the solvation number (here it is 6)
Molecular model for non‐ideal solutions
d l l
The number of 1‐2 neighboring pairs is given by g
gp
g
y
N12 = zN1x2 = zN2x1. The total interaction energy in the solution is
U = zN1x1e11/2 + zN2x2e22/2 + zN1x2e12
Using the definitions x1 = N1/(N1+N2) and x2 = N2/(N1+N2) we can reexpress the interaction energy as
U = (N12e11/2 + N22e22/2 + N1N2e12)z/(N1+N2)
We can focus on the non ideality of the solution by introducing
We can focus on the non‐ideality of the solution by introducing the quantity
w = U w U = ee11 + ee22 ‐ 2e12
Comparing ideal and real energies
For an ideal solution e11 = e22 = e12 and so w = 0. However, for a non ideal solution e11 ≠ e22 ≠ e12. non‐ideal solution e
Substituting e12 = (e11 + e22 ‐ w)/2 we have
U = (N
U (N12e11/2 /2 + N
N22e22/2 /2 + N
N1N2(e11 + ee22 ‐ w)/2)z/(N1+N
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
id
zwN A ∂n 1n 2/(n
( 1 + n 2)
∂G
∂G
μ1 =
=
–
∂n 1
∂n 1
2
∂n 1
Note that the chemical potential of an ideal solution is given by
μ1 = μ1* + RT lnx1
The derivative is
n
n1n2
∂n1n2/(n1 + n2)
= n +2 n –
∂n1
1
2
n1 + n2
2
= x2 – x1x2 = x2(1 – x1) = x2 2
This leads to the expression
This
leads to the expression
μ1 = μ1* + RT lnx1 ‐zwNAx22/2
μ1 = μ
μ1* + RT ln(x
( 1e ‐zwx22/2kT)
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 point
Definition of a critical point
2/2kT))
The activity is a
y
/P1* = x1exp(‐zwx
p(
1 = P1/
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 diagram
separation. 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 separation
Phase 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 γ
information to discuss the free energy of mixing. This leads naturally to the idea of limited solubility even with this very simple model.
Application of the Gibbs‐Duhem
bb
h
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
x1dμ1 + x
+ x2dμ2 = 0.
=0
dμ2 = ‐ x1dμ1 / x2
with μ1 = μ
with μ
μ1* + RT ln(x
RT ln(x1e ‐αx22) where α
) where α = ‐zw/2kT,
zw/2kT,
Application of the Gibbs‐Duhem
bb
h
equation
dμ
μ1 = ‐ RTdx1//x1 ‐ 2α(1‐
( x1))dx1 and dμ
μ2 = ‐ RTdx1//x2 ‐ 2αx1dx1
Change variables from dx1 to ‐ dx2.
dμ2 = RTdx2/x2 + 2α(1‐x2)dx2
Now 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 1
which shows that the activity of coefficient of component 1 implies the magnitude of the activity coefficient for component2.
γ1 = exp(‐zwx22/2kT) implies γ2 = exp(‐zwx12/2kT)
Parameter for non‐ideal
Parameter for non
ideal solutions
solutions
As 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
ΔmixG = n
G n1RTln x
RTln x1 + n
+ n2RTln x
RTln x2 ‐ zwNA/2(n2x12 + n
+ n1x22)
ΔmixG = n1RTln x1 + n2RTln x2 ‐ zwNA(n1 + n2)x1x2/2
Our final expression for Δmixi G is
Our final expression for Δ
G 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
Parameter for non
ideal solutions
solutions
Phase separation and critical point
Phase 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
In other words, there will be a phase separation in this
d th
ill b
h
ti i thi
region.
By assumption, the entropy of mixing is still equal to the
By 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 enthalpy
ΔmixH = ‐ zwNAx1x2/2.