Research: Science and Education
Examples for Non-Ideal Solution Thermodynamics Study Carl W. David Department of Chemistry, University of Connecticut, Storrs, CT 06269-3060;
[email protected] Theoretical chemistry holds out the promise that the physical and chemical properties of species can be determined solely from fundamental constants such as the charge on the electron, the masses of nuclei, Planck’s and Boltzmann’s constants, etc. All one need know is the identity of the substance in question, and given enough human energy and computational effort, one could calculate, say, the vapor pressure of said substance at any desired temperature. This tantalizing promise, in effect, weaves a chemical tapestry using quantum mechanics and statistical thermodynamics as its threads. Since all of practical chemistry takes place in mixtures, rarely in pure states, it seems reasonable to strengthen the ideas concerning the thermodynamics of solutions (mixtures), as noted recently (1), in constructing our tapestry. We offer a mathematical model of a non-ideal solution, and show where and how this non-ideality manifests itself in the standard thermodynamics tableau. The one-dimensional hard line model, originally introduced by Rayleigh (2) and extended to hard line mixtures (3), was used to explicitly discuss solution thermodynamics (4). Unfortunately, the hard line mixture (suitably extended to three dimensions) is an ideal solution, and as a result, part of the value of using it as an example is lost. Starting with a partition function for a “one-dimensional hard sphere” mixture of substances A and B which can be obtained exactly (see ref 3): Z =
3 NA 2
( L − NA SA − NB SB ) NA + N B
ΛA
NA! NB!
3 NB 2
ΛB
and then ad hoc converting to three dimensions by writing L → V, and then artificially constructing a non-ideal system gives us a mathematical form for a partition function which allows us to follow the thread from the partition function through to thermodynamics. This example exemplifies all but the initial step in the tapestry, that is, we can not obtain this new partition function from an intermolecular potential energy form itself derived from quantum mechanics. For all but the simplest problems, complicated approximation schemes are required to pass from the potential energy functions between particles to the appropriate partition function. This inability is tied intimately to the many-body problem that plagues chemistry in other contexts.
− NA SA − NB SB + xA xB ( NA + NB ) SAB )
NA + NB
NA ! NB ! 3 NA 2
× Λ A
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3 NB 2
2π mi kT h
2
where ‘i ’ refers to either substance A or B. The plus sign in eq 1 before the term xA xB ( NA + NB ) SAB =
NA N B SAB NA + NB
is arbitrary (i.e., we can define the constant SAB as either positive or negative in the non-ideal term). The intention in choosing a positive sign was to include some form of “attraction” in this non-ideal term, while a negative value would have been analogous to the SA and SB terms in being exclusion volumes. However, there is no real system whose potential energy of dissimilar interaction would yield this term in the partition function, and the following discussion therefore solely concerns the mathematical explication of various differentiation and integration techniques in the hope that the exercise will reinforce mastery of elements of calculus in a non-calculus setting. Gibbs Free Energy of an Artificially Non-Ideal Mixture From this partition function we obtain the Helmholtz free energy A = − ( NA + NB ) ln V − NA SA − NB SB kT + xA x B ( NA + NB ) SAB 3NA 3N B ln Λ A − ln ΛB + ln( NA !) + ln( NB ! ) 2 2
(2)
by following the standard prescription. Taking the partial derivative of A with respect to V we generate the equation of state:
Given the partition function:
(V
Λi =
−
The Partition Function of an Artificially Non-Ideal Mixture
Z =
where NA is the number of molecules of A in the system, NB is the number of molecules of B in the system, SA is the molecular hard sphere diameter of A molecules, SB is the molecular hard sphere diameter of B molecules, and SAB is a non-additive molecular interaction between A and B molecules (xA and xB are the mole fractions of A and B respectively). This form makes NASA, NBSB, and (NA + NB)SAB all have volume units, with the latter term modulated by the product of mole fractions which disappears when the solution becomes either pure A or pure B, as it must. The translational partition functions are given by
(1)
V =
ΛB
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( NA + NB ) kT p
+ NA SA + NB SB
− xA xB ( NA + NB ) SAB
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Compare this to the “ideal solution” equation of state (SAB = 0): V =
( N A + N B ) kT p
which is
=
+ NA SA + NB SB
from ref 4. Since G = A + pV
∂ x A x B ( nA + n B ) RT + σA − σAB ∂ nA p
that is =
pV = ( NA + NB ) kT
∂ xB RT + σA − σAB xA (nA + n B ) p ∂ nA
+ p NA SA + NB S B − x A x B (NA + NB ) S AB + x B ( nA + n B )
we obtain
3 NB 3NA ln ΛB − ln Λ A 2 2 3 NB − ln ΛB + ln ( NA !) + ln (NB ! ) 2
−
(4)
p
+ nA σA + nB σ B − xA x B ( nA + n B ) σAB
(5)
We obtain the partial molar volumes vi by explicit partial differentiation. We have
∂ V ∂ nA
=
+ xA x B T, p, n B
xB RT + σA + σAB xA ( nA + nB ) p nA + nB − xB − xA xB nA + nB
that is RT + σA − x B2 σAB p
vA =
Our final result then is vA =
RT + σA + xB x A σAB − x B σAB p
Assume that we have experimental information on the molar volume of a solution as a function of the mole fraction of A (or B). If V is the total volume, and v is the volume per mole of solution, that is, if v =
p
V nA + n B
then it follows that T, p, nB
+
∂ ( nA σA ∂ nA
∂ V ∂ nA
) T, p, nB
≡ vA = ( n A + nB ) nB
∂ v + v ∂ nA
and using the chain rule one obtains −
1656
(6)
with a similar term for vB . Later, we show that these expressions can be used to verify that V = nA vA + nB v B (Euler’s Theorem).
( nA + n B ) RT ∂ nA
T, p, n B
∂ xA ∂ nA
Constructed Determination of the Partial Molar Volume
The Partial Molar Volumes
∂
=
+ xB (nA + nB )
where we have written G as G(T, p) explicitly, that is, every mention of V has been obliterated! (It may be appropriate to mention here that the Legendre transformations that relate E → H → A → G carry with them definite rules [5] about which variables are appropriate, and which are not.) Converting eq 3 to a “per mole” basis gives us (using lowercase letters for “per mole” quantities and uppercase letters referring to “per particle” quantities, and NAvogadrok = R, NiSi → niσi)
(nA + nB ) RT
T, p, n B
(see Appendix 1 for help with the partial derivatives) which becomes
( NA + NB ) RT G 3NA ln Λ A = − (NA + N B ) ln − kT p 2
V =
T, p, nB
∂ xA x B ( nA + n B ) σAB ∂ nA
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T, p, n B
•
v A = ( nA + nB )
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∂ v ∂ xA
∂ x A ∂ nA
+ v
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which yields (again, see Appendix 1 for help with partial derivatives) vA = v + x B
∂ v ∂ xA
(7)
nB
the partial of G (eq 4) with respect to nA yields, ∂ G ∂ nA
= RT − ln T,, p, n B
which phrases the calculation of the partial molar volume of A in terms of the experimental molar volume and mole fraction. This result is a direct recovery of the standard textbook form for graphically obtaining the partial molar volume. We can verify that this form holds in our case. From eq 5 we have: v =
V RT = + x A σA + x B σB − x A x B σAB (8) nA + n B p
But ∂ v ∂ xA
∂ x B ∂ xA
= σA + nB
σB − x B σA B − x A vB
∂ x B ∂ x A
σAB nB
(nA + n B )RT 3 + ln nA − ln λA p 2
+ p σA − xA xB σAB − ( nA + nB ) x B
∂ xA σAB ∂ nA
− ( nA + n B ) x A
∂ x B σAB ∂ nA
Employing the results in Appendix 1 we obtain ∂ G (n + n B )RT 3 = RT − ln A + ln nA − ln λA ∂ nA p 2
which yields = σA − σB − xB σAB + xA σAB
∂ G (n + nB ) R T 3 = RT − ln A + ln nA − ln λ A ∂ nA p 2
nB
so that eq 7, using eq 8, becomes: vA = v + x B =
∂ v ∂ xA
(
+ p σA − x B2 σAB
RT + xA σA + x B σB − x A x B σAB p
( σA − x B2 σAB ) p
(9)
−3 p µA = ln xA λA 2 e RT RT
RT + xA σ A + x B σB − σAB x A x B + x B σA p − xB σB − x B2 σAB + xA x B σAB
an equation that succumbs to impressive cancellation yielding:
p −3 2 µB = ln x B λB e RT RT
(12)
( σB − xA2 σAB ) p RT
(13)
In the limit of pure A (xB → 0) the chemical potential of A properly approaches that of pure A, that is,
RT + σA − x B σAB + x A xB σAB p
p −3 2 µA pure λA e = ln RT RT
which recovers eq 6.
p σA RT
with an equivalent expression for pure B. At the standard pressure (1 bar), one would have
The Chemical Potentials Although interest in partial molar volumes may not be intense, interest in partial molar quantities in the area of the partial molar free energy, nominally the chemical potential, is. The mixture under discussion here lends itself gracefully to exercising our manipulative skills in this area also. Taking
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RT
with an equivalent expression for B:
Selectively expanding xB we obtain:
vA =
)
so we finally obtain
nB
+ x B ( σA − σB − x B σAB + x A σAB )
vA =
(11)
xA n B σAB nA + n B
+ p σA + − xA x B − x B2 +
which becomes ∂ v ∂ xA
(10)
•
pure
µA° RT
= ln
p0 − 3 2 λA e RT
p0 σA RT
σ
= ln
A −3 1 λ A 2 e RT RT
again, with an equivalent expression for B.
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The Gibbs–Duhem Equation
having employed the chain rule (see Appendix 1) to obtain
One consequence of the chemical potential being a homogeneous function of order zero is the Gibbs–Duhem equation, which allows us to obtain the chemical potential of a solute from that of a solvent. Since the Gibbs–Duhem equation nA
∂ ( µ A / RT ) ∂ nA
∂ ( µA / RT ) ∂ n B
+ nB nB
= 0 nA
(14)
x B + 2 x A x B2
We finally obtain ∂ µB RT − 2 x A p σAB = − 1 − xA ∂ x A
which allows us to write
can be rewritten dµB = −
∂ µA ∂ nA
nA
∂ G ∂ nA
∂ + nB
RT d x A − 2 p σAB xA d xA 1 − xA
yielding
nB
= 0
∂ n B
nB
p σAB x 2 ∂ µB = − B RT RT ∂ x A
µ B = RT ln x B − p σAB x A2 + C
nA
(C is a constant of integration), which can be rewritten and since the order of differentiation is irrelevant, that is, 2
∂ µA ∂ nB
∂ G = ∂ nB ∂ nA
= nA
∂ µB ∂ nA
∂ µA ∂ nA
+ nB nB
∂ µ A ∂ nB
nB
= 0 nA
giving us nA
∂ µA ∂ nA
+ nB nB
∂ µB ∂ nA
∂ µA ∂ nA
= nB
1 xA
∂ xA ∂ nA
+ ln C ′
Euler’s Theorem Holds for V
= 0
It is important to show that Euler’s theorem holds properly for the partial molar volumes, that is,
nB
Since we have (from eq 12) 1 RT
− xA2 p σAB
This expression recovers the mole fraction dependence in eq 13. We have therefore obtained the form of the chemical potential of B(solute?) from that of A(solvent?) which is a direct example of how the Gibbs–Duhem equation is used in practice. If the chemical potential of one component has been measured, and is therefore known as a function of concentration, then this integration would be performed either graphically or numerically (6) to obtain the chemical potential of the other component.
we have nA
µB = RT ln x B + ln e
nA v A + nB v B = V
− nB
2 p x B σAB RT
∂ xB ∂ nA
nB
(15)
by substituting eq 6 (and its analog for vB ) into this last equation to obtain V =
substituting (see Appendix 1), we obtain
( nA + nB ) RT p
+ nA σA + nB σB − nA x B σAB
+ n A x A xB σAB − nB xA σAB + n B x A xB σAB nA RT
∂ µ A ∂ nA
= x B + 2 xA x B2 nB
p σAB n = − B RT RT
∂ µB ∂ nA
nB
This equation is almost in integrable form. We may rewrite it as
which simplifies to V =
( n A + n B ) RT p
+ nAσ A + nB σB
− 2 x A x B ( n A + n B ) σAB + x A x B ( n A + nB ) σAB nA RT
∂ µ A ∂ nA
= x B + 2 xA x B2 nB
= −
nB RT
= −nB
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p σAB RT
∂ µ B ∂ nA ∂ µ B ∂ x A
∂ xA ∂ nA
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which is exactly correct (i.e., simplifies into eq 5). In the context of this paper Euler’s theorem demands that any valid expression for a partial molar volume must not only be a homogeneous function of the numbers of moles, but that it must be of order zero! This is because the volume itself is homogeneous of order one, so that removing one order of mole number brings the homogeneity to zero.
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Euler’s Theorem Holds for G
Discussion
It is worthwhile verifying that G = nA µ A + nB µ B
that is, regenerates eq 4 (this would be Euler’s theorem for G ). Using eqs 12 and 13 we obtain
p −3 2 nA RT ln x A λA e RT
( σA − xB2 σAB ) p RT
( σB − xA2 σAB ) p
p −3 2 + nB RT ln x B λB e RT
?
= G
RT
(where the question mark indicates that we are asking, does the right hand side equal the left hand side? ). Focusing our attention solely on the logarithm of the exponential we then have
( x B2 nA
p nA σA + nB σB −
)
+ x A2 nB σAB
With σAB negative, we have an approximation of a repulsive intermolecular potential operating between dissimilar constituents of a solution, in analogy with σA and σB. Unfortunately that does not mean that if σAB were positive, we would have the converse. Attraction is much more difficult to imagine as operating between molecules. Instead of using a term xAxBSAB we could have used a term such as xA2xBSAB or any other polynomial form that vanished appropriately at the boundaries of the mole fraction domain. We could have chosen a temperature dependent interaction term also, that is, SAB(T ). In fact, we could even use a series of such terms, leading to a more accurate representation of possible non-ideal behavior. The fact that the calculus of generalized solution thermodynamics can be applied to a solution which is non-ideal (i.e., holds for any solution), allows us to follow the entire thread, from partition function to thermodynamics. Come the day we can evaluate partition functions for realistic potential energy functions, we can be assured that we have the tools to generate the thermodynamic relations from those partition function in a straightforward and interesting way. Acknowledgments
which, upon algebraic manipulation, generates the correct value for the Gibbs free energy of the mixture (eq 4).
The author thanks two superb referees for their help in making this manuscript more accurate and informative.
Chemical Potential Is Homogeneous of Order Zero
Literature Cited
Since the chemical potential is homogeneous of order zero (one less than the Gibbs free energy itself ), eq 14 should hold, which can now be verified. Substituting eq 12 and its nB equivalent into eq 14, we obtain p ∂ x 1 ∂ x A + −2 x B B σAB ∂ nA xA ∂ nA RT
nA RT +
p ∂ x 1 ∂ xA −2 x B B σAB + ∂ nB xA ∂ nB RT
nB RT
?
= 0
which is nA RT
p σAB 1 xB x B2 − 2 xA n A + n B n A + nB RT
Appendix 1. Useful Partial Derivatives
1 − xA − x A x B p σAB − 2 nA + n B RT x A nA + n B
n + B RT
1. Nieto, R.; Gonzales, M. C.; Herrero, F. Am. J. Phys. 1999, 67, 1096 . 2. Rayleigh, Lord. Nature (London) 1891, 45, 80; Tonks, E. Phys. Rev. 1936, 50, 955; Gursey, F. Proc. Cambridge Phil. Soc. 1950, 46, 182. 3. Kikuchi, K. J. Chem. Phys. 1955, 23, 23–27; David, C. W. J. Chem. Phys. 1964, 40, 2318. 4. David, C. W. J. Chem. Educ. 1987, 64, 484 5. David, C. W. Int’l J. Math. Ed. in Science and Technology 1986, 17, 201; David, C. W. J. Chem. Educ. 1988, 65, 876; David, C.W. J. Chem. Educ. 1991, 68, 893. 6. Adamson, A. W. A Textbook of Physical Chemistry; Academic Press: New York, 1979, p 321.
?
= 0
We start with the definition of the mole fraction of component A (the other is B), xA =
which finally is 2 p σAB 1 x − xA x B2 RT B RT
(
+
1 RT
− xB −
(16)
where n refers to number of moles and x to mole fraction. By explicit differentiation we obtain
)
2 p σ AB − x A x B2 RT
(
)
!
= 0
∂ xA ∂ nA
i.e., we have shown that eq 14 holds in this case. www.JCE.DivCHED.org
nA nA + n B
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∂ = nB
nA nA + n B ∂ nA nB
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which is
and we had in eq 12 ∂ xA ∂ nA
= nB
1 nA − nA + n B ( nA + n B )2
which, bringing the right hand side to a common denominator, simplifies into ∂ xA ∂ nA
= nB
nB
( nA + nB )2
=
p −3 2 µA λA e = ln x A RT RT
RT
and pure
µA° RT
xB nA + nB
= ln
p0 σA RT
p0 − 3 2 λA e RT
so,
It then follows that ∂ x B ∂ nB
pure
xA nA + n B
= nA
µA RT
However, ∂ x A ∂ n B
( σA − x B2 σAB ) p
pure
µA° RT
−
=
= − nA
nA
( n A + n B)
2
= −
solvent compartment
ln xA
xA n A + nB
+
p + π −3 2 λA RT
(σA
− x B2 σAB
= − nB
) ( p + π)
RT
with the symmetric result ∂ x B ∂ nA
(17)
xB nA + n B
solution compartment
− ln
p0 − 3/ 2 λA RT
−
p0 σA RT
which translates into Appendix 2. The Osmotic Pressure Assume we have a solution of A and B in equilibrium with pure A (call that the solvent). Assume the two are separated by the mythic semi-permeable membrane, which keeps the B molecules in the solution compartment from infecting the pure solvent (A) in the solvent compartment side. Since there is no possibility of B molecules existing on the pure A side, A molecules must migrate across the semipermeable membrane attempting to dilute the solution infinitely, whereupon “there would be vanishingly small number of B molecules there also”. This migration is balanced by an external pressure (π) called the osmotic pressure. At equilibrium, the chemical potential of A (solvent) must be equal on the two sides of the semipermeable membrane, so pure
µA
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solvent compartment
=
µAin solution
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solution compartment
•
µA° + RT ln
p ( e p0
p − p0 ) σA RT
= µA° + RT ln xA
p + π e p0
( p + π)
( σA − x B2 σAB ) − p0 σA RT
which simplifies to pe
p0 ) σA RT
( p + π)
(p −
= xA ( p + π ) e
( σA − x B2 σAB ) − p0 σA RT
or
(
)
2 2 π σA − x B σAB − p x B σAB
p = xA ( p + π ) e
RT
which is, as in the case of “ideal solutions”, a very complicated transcendental equation for the osmotic pressure.
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