In the Classroom
W
Applications of the Gibbs–Duhem Equation P. Nikitas Laboratory of Physical Chemistry, Department of Chemistry, Aristotle University of Thessaloniki, 54006 Thessaloniki, Greece;
[email protected] The Gibbs–Duhem equation (1–5) is usually expressed as r
S dT – V dp + Σ n i dµi = 0 i=1
(1)
where S is the entropy of the system, T is its temperature, V is the volume, p is the pressure, r is the number of constituents of the system, ni is the number of moles of the ith component, and µi is its chemical potential. For a binary mixture under constant pressure and temperature eq 1 reduces to x 1 dµ 1 + x 2 dµ 2 = 0
(2)
where x1, x2 are the mole fractions of components 1 and 2, respectively. The Gibbs–Duhem equation is a limiting case of a more general equation that at constant T and p may be expressed as (1–5) r
x i dX i,m = 0 Σ i=1
(3)
where Xi,m is a partial molar property of i. The Gibbs–Duhem equation is a useful thermodynamic equation because it shows that at constant pressure and temperature the chemical potentials of a mixture cannot change independently. This relationship is also valid for all partial molar properties of the mixture. These interconnections allow the calculation of the activity of a component, say 1, α1, when the activity of component 2, α2, is known. In this case eq 2 yields
ln
α1 = α1*
α2 α 2*
x2 d ln α2 x1
(4)
Analogous equations (see eq 6), can be written for every partial molar quantity of a binary mixture. These are the only applications of the Gibbs–Duhem equation that may be found in some textbooks for undergraduate students. However, the usefulness of the Gibbs–Duhem equation and its extension, eq 3, is much wider. For example, the chemical potentials of the constituents of a solution are equal to the corresponding chemical potentials in the vapor. Moreover, since the chemical potentials in the vapor depend upon the partial vapor pressures, the Gibbs–Duhem equation determines the relation between the partial vapor pressures and therefore the composition of the vapor. In this communication we present two main applications of the Gibbs–Duhem equation. We apply (i) the generalization of the Gibbs–Duhem equation, eq 3, to the direct calculation of partial molar quantities from the composition dependence of the corresponding total quantity, and (ii) the Gibbs–Duhem equation for the vapor of a binary liquid solution to calculate partial vapor pressures from the liquid-phase composition dependence of the total vapor pressure. Other applications, such as the calculation of a partial molar quantity of a component, say 1, from that of component 2, are also discussed. 1070
Integration of the Gibbs–Duhem Equation In a binary mixture eq 3 yields x1 dX1,m + x2 dX2,m = 0
(5)
and therefore x
X 1,m = X 1,m x = 1 –
1–x dX 2,m x x=1
(6)
where x = x1, 1 – x = x2 and X1,m(x = 1) is the value of X1,m when x = 1. If we assume, in agreement with experimental evidence, that X2,m is an analytic function of x, we can write X2,m = X2,m(x = 0) + A1x + A2x 2 + A3x 3 + … + Aq x q
(7)
where A1, A2, etc. are functions of T and p only. Substitution of the above equation into eq 6 and integration of the resulting expression yields X 1,m = X 1,m x = 1 + A – A 1 ln x + A 1 – 2A 2 x + A 2 – A3 –
4A 4 3
x3 + … +
A p–1 –
pA p p –1
3A 3 2
x p–1 + A
x2 +
(8) q
xq
where
A = A 1 + A 2 +
A3 2
+
A4 3
+…+
Aq
(9)
q–1
Equation 8 may be expressed as X1, m = X1, m(x = 1) + A – A1 ln x + a1x + a2x 2 + a3 x 3 + … + aq x q (10)
where
aj = Aj –
j + 1 A j+1
(11)
j
From the above treatment we observe the following. 1. If a partial molar quantity X 2, m of a binary mixture is expressed by a polynomial of q degree, the expression of the other quantity X 1, m is fixed by eq 10. Therefore, if experimental X 2, m versus x data are available, parameters A1, A2, …, Aq can be determined by using a least squares fitting procedure. In this case the dependence of X 1, m on x can be calculated directly from eqs 10 and 11. 2. If we wish the two partial quantities X 1, m and X 2, m to be represented by polynomials, we should put a priori A1 = 0 in order to omit the term A1 ln x from eqs 8 and 10. 3. Consider a set of experimental X versus x data, where X is a thermodynamic quantity that is related to X 1, m and X 2, m by the following equation X = f (X 1, m, X 2, m)
Journal of Chemical Education • Vol. 78 No. 8 August 2001 • JChemEd.chem.wisc.edu
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In the Classroom
16. However, if we wish to increase the degree of polynomials above 10, then the new relationships between ai and Ai must be determined following the above procedure.
Substitution of eqs 7, 10, 11 into eq 12 makes X depend on the parameters A1, A2, …, Aq. Therefore, by using a least squares fitting these parameters can be determined and lead to the calculation of X 1, m and X 2, m directly from X versus x data. This issue will be discussed in more detail below.
Application to Partial Molar Volumes
An alternative approach to determine the interconnection between X1,m and X2,m implied by the Gibbs–Duhem equation is the following. Let us expand X1, m and X2, m in powers of 1 – x and x, respectively (5). X1, m = X1, m(x = 1) + A1(1 – x) + A2(1 – x)2 + A3(1 – x)3 + A4(1 – x)4 + …
(13)
X2, m = X2, m(x = 0) + a1x + a2x2 + a3x3 + a4x4 + … (14) If we retain the first five terms of eqs 13 and 14 and substitute them into eq 5, we obtain x[A1 + 2A2(1 – x) + 3A3(1 – x)2 + 4A4(1 – x)3] = (15) (1 – x)[a1 + 2a2x + 3a3x2 + 4a4x3] Writing the above equation in terms of powers of x and equating the coefficients of powers of x on both sides of eq 15, we find a4 = A4
a3 =
a2 =
3A 3 + 8A 4 3
(16)
2A 2 + 3A 3 + 4A 4 2 a 1 = A1 = 0
We should point out that the roles of upper case Ai and lower case ai have been exchanged between eqs 7, 10 and eqs 13, 14. We also note that eqs 7, 10 are more flexible than eqs 13, 14, since there is not a general relationship between ai and Ai for eqs 13, 14. Thus, if polynomials of 10th degree are used to represent X1,m and X2,m, eqs 16 are no longer valid for the first four terms. In this case the relationships between ai and Ai are given in the box. Using these relationships, ai can be calculated from Ai for every polynomial with degree lower than 10. For example, if q = 4, then A5 = A6 = A7 = A8 = A9 = A10 = 0 and the relationships of the box are reduced to eqs Relationships Interconnecting Parameters ai and A i of Eqs 13 and 14 a 1 = A1 = 0 .
a 2 = (2A2 + 3A3 + 4A4 + 5A5 + 6A6 + 7A7 + 8A8 + 9A9 + 10A10)/2 .
a 3 = (3A3 + 8A4 + 15A5 + 24A6 + 35A7 + 48A8 + 63A9 + 80A10)/3 .
a 4 = (4A4 + 15A5 + 36A6 + 70A7 + 120A8 + 189A9 + 280A10)/4 .
a 5 = (5A5 + 24A6 + 70A7 + 160A8 + 315A9 + 560A10)/5 .
a 6 = (6A6 + 35A7 + 120A8 + 315A9 + 700A10)/6 .
a 7 = (7A7 + 48A8 + 189A9 + 560A10)/7 .
a 8 = (8A8 + 63A9 + 280A10)/8 .
a 9 = (9A9 + 80A10)/9 .
a 10 = A10 .
When Xi,m represents the partial molar volume Vi,m, eq 12 may be expressed as Vm = xV1, m + (1 – x)V2, m
(17)
Therefore, from measurements of the total molar volume Vm as a function of the mole fraction x the partial molar volume, Vi,m, can be calculated as follows. V1, m and V2,m are expressed as functions of Ai via eqs 7, 8 or eqs 13, 14. In both cases the total molar volume Vm becomes a function of x and depends on Ai via eq 17. Consequently the determination of Ai using a least squares fitting procedure leads readily to the determination of the partial molar volume, Vi, m. In this communication we use as an example the x – Vm data of methanol (1)–water (2) solutions at 293 K. These data were calculated from data of the density of solution versus the weight percent of methanol taken from ref 6, p 115. The whole treatment was carried out with Microsoft Excel spreadsheets. Microsoft Solver, a powerful routine based on the Levenberg–Marquardt algorithm (7–9), was used for least squares fitting. The use of Solver in chemistry, its simplicity and capabilities, have been pointed out in several articles (10–15). Figure 1 shows a spreadsheet used in the calculations. It contains the experimental x – Vm data in rows 13 through 33 of columns A and B. For simplicity subscript m is omitted in Figure 1. The initial guesses of values for parameters Ai are entered in cells B2–B10. Parameters ai are calculated in cells C2–C10 on the basis of parameters Ai. The quantity A defined by eq 9 is computed in cell G2 by the addition of its terms, which are calculated in cells D2–D10. Columns C, D, and E from row 13 to row 33 contain the analysis of data in terms of eqs 10, 7, and 17, respectively. That is, under the label V1(calc) we calculate the partial molar volume V1, m using eq 10 and the comments at the bottom of Figure 1. Similarly, we calculate the partial molar volume V2,m in column D under the label V2(calc) using eq 7. Finally the total molar volume Vm is calculated by means of eq 17 in the range E13–E33. Note that the values appearing in these columns depend on the initial guesses of values for parameters Ai used in cells B2–B10. Note also that in this application, owing to eq 17, we have V2,m(x = 0) = Vm(x = 0) = 18.0469 and V1, m(x = 1) = V m(x = 1) = 40.4724. These values of V2,m(x = 0) and V1, m(x = 1) are entered in cells G3 and G4, respectively. The next column of values under the label SR contains the square of residuals (SR)—that is, the values [Vm – Vm(calc)]2. The sum of the squares of residuals (SSR) is calculated in cell G6. Solver is used to minimize the content of cell G6 (the sum of squares of residuals) by changing cells B2–B10 (the cells with the parameters Ai). However, before its use Solver must be configured via the command sequence Tools → Solver → Options to use automatic scaling, precision, and convergence of the order of 1012 and central derivatives. In addition, we found it more effective to run Solver in two steps. In the first step, we used as initial estimates for parameters Ai the value 1 for A2, A3, A4 and 0 for A1, A5 –A10 and ran Solver to find the optimum values of A2, A3, and A4 only. In the next step
JChemEd.chem.wisc.edu • Vol. 78 No. 8 August 2001 • Journal of Chemical Education
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1 2 3 4 5 6 7 8 9 10
A j
B Aj
1 2 3 4 5 6 7 8 9
0 10.4852 -105.1965 283.4883 -358.4355 212.5070 -46.5154 0 0
C aj -20.9703 168.2800 -483.1809 731.5326 -613.4438 266.7749 -46.5154 0 0
D Aj/(j-1) 0 10.4852 -52.5983 94.4961 -89.6089 42.5014 -7.7526 0 0
E
F
G
A= V(x=0)= V(x=1)=
-2.4770 18.0469 40.4724
SSR=
0.0002
H
I
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
x 0.00000 0.02874 0.05880 0.09026 0.12324 0.15783 0.19417 0.23239 0.27263 0.31507 0.35989 0.40729 0.45751 0.51080 0.56745 0.62780
V 18.0469 18.6117 19.1949 19.7958 20.4259 21.0893 21.7957 22.5535 23.3699 24.2483 25.1891 26.2131 27.3111 28.5035 29.8045 31.2163
V1(calc) 37.9954 37.5207 37.2542 37.1632 37.2146 37.3757 37.6152 37.9041 38.2167 38.5317 38.8331 39.1104 39.3591 39.5798 39.7768 39.9555
V2(calc) 18.0469 18.0533 18.0650 18.0718 18.0652 18.0385 17.9871 17.9086 17.8029 17.6719 17.5185 17.3461 17.1570 16.9501 16.7198 16.4544
V(calc) 18.0469 18.6128 19.1932 19.7950 20.4251 21.0905 21.7983 22.5553 23.3684 24.2442 25.1894 26.2105 27.3146 28.5093 29.8035 31.2083
SR 0 1.19E-06 2.64E-06 5.98E-07 5.47E-07 1.41E-06 6.57E-06 3.28E-06 2.48E-06 1.66E-05 6.27E-08 6.47E-06 1.27E-05 3.39E-05 9.71E-07 6.47E-05
V' 19.8620 19.4414 19.1956 19.1096 19.1660 19.3453 19.6262 19.9861 20.4016 20.8502 21.3115 21.7691 22.2126 22.6403 23.0609 23.4937
V1 37.9089 37.4944 37.2618 37.1805 37.2299 37.3814 37.6111 37.8950 38.2094 38.5292 38.8309 39.1158 39.3612 39.5792 39.7795 39.9608
V2 18.0469 18.0529 18.0662 18.0709 18.0639 18.0360 17.9849 17.9090 17.8078 17.6790 17.5193 17.3467 17.1486 16.9389 16.7186 16.4671
0.69221 0.76111 0.83499 0.91440 1.00000
32.7365 34.4017 36.2440 38.2552 40.4724
40.1188 40.2639 40.3800 40.4519 40.4724
16.1372 15.7513 15.2926 14.7961 14.3800
32.7374 34.4080 36.2402 38.2558 40.4724
7.83E-07 3.9E-05 1.43E-05 3.62E-07 2.02E-28
23.9661 24.5027 25.1010 25.6809 25.9844
40.1132 40.2552 40.3860 40.4534 40.4724
16.1471 15.7526 15.2850 14.7725 14.4880
34 3 5 Comments: 3 6 Contents of cell C2: 3 7 Contents of cell C3: 3 8 Contents of cell D3: 3 9 Contents of cell G2: 4 0 Contents of cell G6: 4 1 Contents of cell C13: 4 2 Contents of cell D13: 4 3 Contents of cell E13: 4 4 Contents of cell F13: 4 5 Contents of cell G13: 46 4 7 Contents of cell H13: 4 8 Contents of cell I13:
= = = = = = = = = =
B2-A3*B3/A2 B3-A4*B4/A3 B3/A2 SUM(D2:D9) SUM(F13:F33) $G$4+$G$2+$C$2*A13+$C$3*(A13^2)+…+$C$10*(A13^9) $G$3+$B$3*(A13^2)+$B$4*(A13^3)+...+$B$10*(A13^9) A13*C13+(1-A13)*D13 (B13-E13)^2 19.862-6*15.808*(A13^5)+5*53.548*(A13^4)-4*70.347*(A13^3) + 3*44.281*(A13^2)-2*9.1123*A13 = B13+G13*(1-A13) = B13-G13*A13
Figure 1. Spreadsheet showing the calculation of partial molar volumes from data for total molar volume of the system methanol (1)–water (2) at 293 K. The comments describe the specific equations used.
we used Solver to find the best values of all parameters from A2 to A7. That is, we fitted a 7th degree polynomial to the experimental data. The results are shown in Figure 1. It is seen that from the experimental Vm versus x data the partial molar volumes V1, m and V2,m can be obtained. A plot of V1, m versus x is given in Figure 2. If the degree of the polynomial is increased by including parameters A8, A9, the results do not change, indicating that the selected degree of the polynomials q = 7 is correct. This 1072
way of choosing the degree of the polynomials is very simple but empirical. However, there are several techniques in the literature that can be used for the proper selection of q (16, 17). The simplest of them is the s criterion, which is based on the calculation of the standard error of estimate N
s=
Σ i=1
V m x i – V m(calc) x i N –q
Journal of Chemical Education • Vol. 78 No. 8 August 2001 • JChemEd.chem.wisc.edu
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(18)
In the Classroom
V1,m / (cm3 mol1)
40
39
38
37 0.0
0.2
0.4
0.6
0.8
1.0
x Figure 2. Plot of the methanol partial molar volume as a function of its mole fraction x. Points (o) were calculated from the method of intercepts, points (ⴛ) from the slope method, and the line from the method proposed in text.
as a function of q. This quantity decreases with increasing q and the optimum value of q is that value after which no significant decrease occurs in s (16, 17). Application of the s criterion to our data confirmed that q should be equal to 7. It is useful to point out that taking into account the above precautions Solver always converged to the optimum fitting. However, we should have in mind that Solver, as well as every other nonlinear least squares routine may be trapped into a local minimum instead of finding the global one. This defect can be overcome by using a Monte Carlo search for initial estimates as suggested in ref 15. Up to now the standard method for determination of partial molar quantities has been the method of intercepts (2–5). According to this method, the molar quantity Xm of a binary mixture is plotted as a function of x, and at the value of x of interest the tangent to the experimental Xm – x curve is drawn. The tangent intercepts the ordinate at the value X2, m when x = 0 and at the value X1, m when x = 1. To compare the results obtained by the method we propose (i.e., the calculated values of the partial molar volumes in columns C and D of Fig. 1) with those of the method of intercepts, the three last columns of Figure 1 under the labels V ′, V1, and V2 were added. In fact the following variation of the method of intercepts was adopted. Fitting a 6th-degree polynomial to the x – Vm data by using the command Add Trendline, we obtained the equation Vm = 15.808x 6 + 53.548x5 – 70.347x4 + 44.281x3 – 9.1123x2 + 19.862x + 18.048. On the basis of this equation, the derivative V ′ was calculated in column G. If the derivative of a function y = f (x) is known, the equation for the tangent line at the point x0 is given by y – f (x0) = f ′(x0)(x – x0). Therefore, the two partial molar volumes, V1, m and V2, m, may be calculated by means of the equations V1, m = Vm – Vm′x and V2, m = Vm + Vm′ (1 – x) (19) These equations were used in columns H and I for the calculation of the partial molar volumes by means of the method of intercepts. It is seen from Figure 1 that the two methods give comparable results. Thus, the sum of squares of the differences between the V1, m values calculated from the method of intercepts and our method is 0.0128, and the corresponding quantity between V2, m values is 0.0092.
At this point it is worth noting the following. The increasing use of computers in science has led some authors to suggest the calculation of partial molar quantities directly from the slope of the X versus n1 curve at the points of interest (slope method). Thus, the method of intercepts is described in the 3rd edition of Atkins’s Physical Chemistry, but in the 5th edition it is replaced by the direct calculation of the slope of the X versus x curve using a computer curve-fitting program (1). However, the slope of the X versus x curve does not give the partial molar quantity X1, m but the difference X1,m – X2,m. This is clearly depicted in Figure 1, where the value of the slope of the Vm versus x curve in column G is equal to the difference V1, m – V2, m. Thus we must be very careful in using the slope method. By definition X1, m is equal to the partial derivative of X with respect to n1 when T, p, n2 are kept constant. Therefore, if we wish to compute the partial molar volumes by the slope method, we should first find the dependence of V upon n1 when n 2 is constant. This can be achieved as follows. From the known composition of the binary solution, n1 and n2 can be readily calculated. Then the quantity n1* = n1/n2 gives the number of moles of component 1 in the solution when the number of moles of component 2 is kept constant and equal to 1. In this case V is given by the equation V = Vm(1 + n1*). Therefore, the slope of the V versus n1* curve at a certain value of n1* gives the partial molar volume V1, m at this value of n1*. We applied the above procedure to the data of Figure 1 and found the following. The V versus n1* data cannot be represented adequately by a single polynomial, irrespective of its degree, because these data are not uniformly distributed over the range of the n1* values. For this reason we split the data into two subsets. The first set is from n1* = 0 to n1* = 0.8433 (x = 0.4575), and the second one is from n1* = 0.8433 to n1* = 10.682 (x = 0.9144). The first set of data may be described by the equation V = 10.201(n1*)5 – 24.137(n1*)4 + 19.872(n1*)3 – 5.0248(n1*)2 + 37.715n1* + 18.049 and the second one by V = 0.0088(n1*)3 + 0.1718(n1*)2 + 39.413n1* + 16.951. Therefore, V1, m = 51.005(n1*)4 – 96.548(n1*)3 + 59.616(n1*)2 – 10.0496n1* + 37.715 when x varies from 0 to 0.4575 and V1, m = 0.0264(n1*)2 + 0.3436n1* + 39.413 when x varies from 0.4575 to 0.9144. The values of V1, m calculated from the above equations are shown in Figure 2 by points ⴛ. There is a scatter of the partial molar volumes of methanol calculated by the slope method, which shows the limited accuracy of this method. In general, this method is less accurate than the previous methods because its results depend on both the degree of the fitting polynomials and the point of separation of the original data set into two (or more) subsets. The method proposed above to calculate partial molar volumes from experimental data for the total molar volume of a binary mixture can be readily extended to the calculation of the partial molar volume of, say, component 1 if the dependence of V1, m upon x is known. In this case eq 7 with V1, m in place of X2, m is fitted to the experimental V1, m – x data and the partial volume of the other component, V2, m, is calculated from eq 8. As an example, let us assume that for the system methanol (1)–water (2) only the dependence of V2, m upon x is known and that this dependence is the one determined by the method of intercepts. To find the dependence of V1, m upon
JChemEd.chem.wisc.edu • Vol. 78 No. 8 August 2001 • Journal of Chemical Education
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In the Classroom 50
40
p2
p / kPa
30
20
10
p1
0 0.0
0.2
0.4
0.6
0.8
1.0
0.8
1.0
x 8
p / kPa
6
p1
4 p2
2
0 0.0
0.2
0.4
0.6
x Figure 3. Plots of the partial vapor pressures of (top) chloroform (1)–acetone (2) and (bottom) ethyl alcohol (1)–water (2) mixtures. Points are experimental data; lines (—) were calculated by the method proposed in text.
x we may work as follows. In the spreadsheet of Figure 1 the column with the Vm values is replaced by that with the V2, m values. That is, the content of cells I13–I33 is inserted in cells B13–B33. In addition, the column with the SR values is corrected from [Vm – Vm(calc)]2 to [V1, m – V1, m(calc)]2 and then Solver is used to find the optimum fitting. Now the column V1(calc) gives the dependence of V1, m upon x. A comparison of the calculated V1, m values with those obtained by the method of intercepts is given in Figure 2. It is seen that the calculated values are almost identical to the reported ones. Application to the Vapor Pressure of a Liquid Mixture When a binary solution of volatile liquids is at equilibrium with its vapor, the chemical potentials of the constituents of the solution are equal to the corresponding chemical potentials in the vapor. Assuming that the gaseous solution over the liquid is ideal, we have µi = µi* + RT ln( pi /pi*)
i = 1,2
(20)
where pi is the partial vapor pressure of i, pi* is the vapor pressure of pure i, and µi* is the chemical potential of i in its standard state (i.e., when pi = pi*). Substitution of eq 20 into eq 2 yields x d ln p1 + (1 – x)d ln p2 = 0
(21)
Thus, in the case of vapor pressures we have Xi,m = ln pi. In addition, Dalton’s law states that p = p1 + p2 1074
(22)
We observe again that p1 and p2 can be expressed as functions of Ai via eqs 7, 8 or 13, 14, which means that the total pressure p becomes function of x and the fitted parameters Ai. Therefore, by fitting this function of p to the corresponding experimental x versus p data, the adjustable parameters Ai can be determined, leading to the determination of the partial vapor pressures p1 and p2. The literature contains much tabulated x versus p1 and p2 data, which can be used to test either the validity of the proposed procedure or the thermodynamic consistency of these data. Here we give two tests of our method using vapor pressures of mixtures of (i) chloroform (1) and acetone (2) (6, p 286), which exhibit negative deviations from Raoult’s law, and (ii) ethyl alcohol (1) and water (2) (18, 19), which show positive deviations from Raoult’s law. The original data were transformed to kPa units from mmHg and the calculations were carried out using spreadsheets similar to the one in Figure 1. The analysis of chloroform– acetone system was based on eqs 7, 8, and the ethyl alcohol– water system was analyzed by means of eqs 13, 14. The results are shown in Figure 3, where the calculated and the experimental data for vapor pressures are plotted as a function of x. The spreadsheets used for these two cases, as well as the spreadsheet in Figure 1, are available as supplemental material.W It is seen that our method can determine the partial vapor pressures from the total vapor pressure. The accuracy of the calculated partial vapor pressure is not as good as that of the partial molar volumes in our first example. The average accuracy in the vapor pressures examples is ±2.4%, whereas that in the partial volumes is better than 0.1%. This difference is attributed to the lower accuracy of the experimental data for the total vapor pressure. For example, in the chloroform– acetone system the error in the total vapor pressure is ±0.21 kPa, leading to an average percent error of 2.1%, comparable to the accuracy of the calculated partial vapor pressures. Discussion This treatment shows that the Gibbs–Duhem equation is a very useful thermodynamic equation, which can be used for the calculation of (i) a partial molar quantity of a binary mixture from measurements of the corresponding total molar quantity, (ii) the partial molar quantity of a component, say 1, from measurements of the corresponding partial molar quantity of component 2, and (iii) the partial vapor pressures from measurements of the total vapor pressure. The first application can replace the method of intercepts for the determination of partial molar quantities. Its main advantage is that it does not use derivatives, which may amplify experimental errors. The accuracy of the results depends upon the accuracy of the fitting of the total molar quantity with a polynomial of a certain degree. Since the fitting procedure, especially using Solver, may be as accurate as the experimental error allows, the determination of the partial molar quantities is achieved with the highest possible accuracy. However, the most interesting application of the Gibbs– Duhem equation is the last one, which allows several other applications. From measurements of the total vapor pressure of a binary mixture of volatile liquids we can calculate (i) the partial vapor pressures and examine positive, negative, or more complex deviations from Raoult’s law, (ii) the composition of the
Journal of Chemical Education • Vol. 78 No. 8 August 2001 • JChemEd.chem.wisc.edu
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gaseous phase, since the mole fraction of component i in the vapor is given by yi = pi /p, (iii) the activities of the constituents of the liquid mixture, since we have ai = pi /pi*, and (iv) the dependence of p upon xi and yi allowing the construction of the plots of p versus the overall mole fraction of component 1 in the system, which are useful for evaluating distillation. Note that the calculation of partial vapor pressures directly from measurements of total pressure was a subject of scientific work some years ago. Measurement of partial pressures requires the determination of the composition of both phases at equilibrium. Thus, it is much easier to determine the partial vapor pressures from the total pressure. For this reason the majority of textbooks on experimental physical chemistry describe methods for determining the total vapor pressure but not those for partial vapor pressures (20–22). Therefore, the applications of the Gibbs–Duhem equation described in this communication may considerably enrich laboratory experiments concerning vapor pressures of binary liquid mixtures. Alternatively, using tabulated VLE data concerning the variation of the total and the partial vapor pressures with the composition of the solution, a student can test the consistency of either the experimental data or the Gibbs– Duhem equation itself. Since this equation depends directly on the first two laws of thermodynamics, a test of the Gibbs– Duhem equation is in fact a strong test of the principles of thermodynamics. W
Supplemental Material
The spreadsheet in Figure 1 and the spreadsheets used for Figure 3 are available as supplemental material in this issue of JCE Online. Literature Cited 1. Atkins, P. W. Physical Chemistry; Oxford University Press: Oxford, 1996.
2. Moore, W. J. Physical Chemistry; Longman: London, 1996. 3. Alberty, R. A.; Silbey, R. J. Physical Chemistry; Wiley: New York, 1977. 4. Rock, P. A. Chemical Thermodynamics; Macmillan: London, 1969. 5. Kirkwood, J. G; Oppenheim, I. Chemical Thermodynamics; McGraw-Hill: New York, 1961. 6. International Critical Tables, Vol 3; McGraw-Hill: New York, 1928. 7. Levenberg, K. Q. Appl. Math. 1944, 2, 164. 8. Marquardt, D. W. J. SIAM 1963, 11, 41. 9. Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; Vetterling, W. T. Numerical Recipes in Pascal; Cambridge University Press: New York, 1986. 10. Walsh, S.; Diamond, D. Talanta 1995, 42, 561. 11. Machuca-Herrera, J. O. J. Chem. Educ. 1997, 74, 448. 12. Billo, E. J. Excel for Chemists, A Comprehensive Guide; WileyVCH: New York, 1997. 13. Harris, D. C. J. Chem. Educ. 1998, 75, 119. 14. DeLevie, R. J. Chem. Educ. 1999, 76, 1594. 15. Nikitas, P.; Pappa-Louisi, A. Chromatographia 2000, 52, 477. 16. Palston, A.; Rabinowitz, Ph. A First Course in Numerical Analysis; McGraw-Hill; New York, 1988. 17. Draper, N. R; Smith, H. Applied Regression Analysis; Wiley: New York, 1981. 18. Guggenheim, E. A. Thermodynamics, An Advanced Treatment for Chemists and Physicists; North-Holland: Amsterdam, 1950. 19. Guggenheim, E. A.; Adam, J. Proc. R. Soc. London, Ser. A 1933, 139, 218. 20. Shoemaker, D. P.; Garland, C. W.; Nibler, J. W. Experiments in Physical Chemistry; McGraw-Hill: New York, 1989. 21. Levitt, B. P. Findlay’s Practical Physical Chemistry; Longman: London, 1973. 22. James, A. M. Practical Physical Chemistry; Longman: London, 1974.
JChemEd.chem.wisc.edu • Vol. 78 No. 8 August 2001 • Journal of Chemical Education
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