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Chemistry 433
Lecture 22
Non‐ideal
Non
ideal behavior
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
ΔmixG = Gsln(T,P,n1,n2) ‐ G1*(T,P,n1) ‐ G2*(T,P,n2)
ΔmixG = n1μ1sln + n2μ2sln ‐ n1μ1* ‐ n2μ2*
ΔmixG = nRT(x1lnx1 + x2lnx2)
The free energy of mixing is always negative because x1
and 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 by
ΔmixS = ‐(∂ΔmixG/∂T) = ‐ nR(x1lnx1 + x2lnx2)
Definition of ideal behavior
A model of molecular interactions
This result is exactly the same as the mixing entropy for an ideal gas. For an ideal solution we find further that ΔmixV = (∂ΔmixG/∂P) = 0
and
ΔmixH = ΔmixG + T ΔmixS = 0.
The volume change and enthalpy of mixing are zero for 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. 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 50 torr for the pure component
50 torr for the pure component. That is, P1* = 50 torr. Likewise, P2* = 100 torr.
Positive deviations from Raoult’s law
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 molecules have a repulsive interaction. A mathematical model for the deviations from ideality
To obtain the plots we use the following expressions.
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 from Raoult's law are also observed as shown in the plot below (α = ‐2 in the plot below).
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Negative deviations from Raoult’s law
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. However at the opposite limit as x1 → 0 (and x
However, at the opposite limit as x
0 (and x2 → 1) we have, 1) we have
P1 = x1P1*exp(α). In this case substance 1 is the solute and it 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). 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 the form P1 = x1P1*exp{αx22 + βx23 + …}
The chemical potential for this type of solution is
μ1 = μ1* + RTln x1 + αRTx22 + βRTx23
We see from the definition of activity as a1 = P1/P1* that we can
express the chemical potential as μ1 = μ1* + RTln a1
Thus we can see that a1 = x1exp{αx22 + βx23 + …}.
The activity coefficient
We can define an activity coefficient 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 know the standard state. The solvent or Raoult's law standard state holds for substances that are miscible in all proportions.
μ1 = μ1* + RTln a1
Raoult’s law standard state
Henry’s law standard state
Up to now we have implicitly discussed only Raoult's law
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 = μ1* + RTln (P1/P1*)
We calso think of this as the standard state for an ideal solution.
For the solute j we use Henry's law Pj → xjkH,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 = 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.
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Standard state for activity
The numerical value of the activity depends on the standard 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.
Excess free energy
We can calculate the Gibbs energy of mixing of binary solutions 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 = μj* + RTln aj = μj* + RTln xj + RTln γj
leading to the expression
ΔmixG = RT(n1ln x1 + n2ln x2 + n1ln γ1 + n2lnγ2)
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
γ dimeth = a
= a dumeth/x dimeth = 0.582/(1‐0.539) = 1.262
= 0 582/(1 0 539) = 1 262
Henry's Law
a CS2 = PCS2/ kH,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
The first two terms represent the Gibbs free energy of mixing of an ideal solution. To focus on the effect of non‐ideality, we define an excess Gibbs energy of mixing ΔGE.
ΔGE = ΔmixG ‐ ΔmixGid
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.
γ1 = exp{αx22}
and γ2 = exp{αx12}.
The origin of excess free energy is the enthalpy
Molecular model for non‐ideal solutions
Substituting these activity coefficients into the expression for the excess free energy we have
ΔGE = RT(α x1x22 + α x2x12) = α RTx1x2
using the fact that x2 2 = 1 using the fact that x
1 ‐ x1. .
If components 1 and 2 are distributed randomly throughout the 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).
We can express the potential energy of the solution in the form
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 = 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 we assume that molecular sizes are about the same. 3
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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)
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. Substituting e12 = (e11 + e22 ‐ w)/2 we have
U = (N12e11/2 + N22e22/2 + N1N2(e11 + e22 ‐ w)/2)z/(N1+N2)
( 1+N2)
U = zN1e11/2 + zN2e22/2 ‐ zwN1N2/2(N
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)
Molecular model for non‐ideal solutions
The number of 1‐2 neighboring pairs is given by 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 the quantity
w = U = e11 + e22 ‐ 2e12
The chemical potential of component 1 is given by
id
zwN A ∂n 1n 2/(n 1 + n 2)
μ 1 = ∂G = ∂G –
∂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
∂n1n2/(n1 + n2)
n1n2
n
= n +2 n –
∂n1
1
2
n1 + n2
2
= 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)
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
Phase separation
The activity is a1 = P1/P1* = x1exp(‐zwx22/2kT)
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 indicates that there is a region of temperature or pressure
beyond which the solution separates into two phases.
In the case of the model liquid discussed here phase separation 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.
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Application of the Gibbs‐Duhem equation
At this point, you might well ask whether the activity coefficient 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 + x2dμ2 = 0.
dμ2 = ‐ x1dμ1 / x2
with μ1 = μ1* + RT ln(x1e ‐αx22) where α = ‐zw/2kT,
Parameter for non‐ideal solutions
Application of the Gibbs‐Duhem equation
dμ1 = ‐ RTdx1/x1 ‐ 2α(1‐ x1)dx1 and dμ2 = ‐ 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 implies the magnitude of the activity coefficient for component2.
γ1 = exp(‐zwx22/2kT) implies γ2 = exp(‐zwx12/2kT)
Parameter for non‐ideal 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 = n1RTln x1 + n2RTln x2 ‐ zwNA/2(n2x12 + n1x22)
ΔmixG = n
G = n1RTln x
RTln x1 + n
+ n2RTln x
RTln x2 ‐ zwNA(n1 + n
+ n2)x1x2/2
Our 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.
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. In other words, there will be a phase separation in this
region.
region
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.
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