J . Phys. Chem. 1990, 94, 3195-3803 surface tension meas~rements.'"'~~'On the other hand, continuum models are valid when only Columbic forces are present. If short-range interactions are important, a good selection is to use the continuum supermolecule approach. Finally, the selection of the cavity's geometrical parameters may start by considering the barycenter of the nuclear charges and the molecular volume.'9-42,43
+
(39) Pierotti, R. A. Chem. Reo. 1976, 76, 717. (40) Kondo, S. J . Chem. Phys. 1956, 25,662. (41) Saylor, J. H.:Battino, R. J . Phys. Chem. 1958, 62, 1334.
3795
In the next related contribution, we shall present a quantummechanical scheme including the models developed here, together with numerical results for comparison with current methodologies.
Acknowledgment. This work has received support from FONDECYT, through Project 11 1 1-88. Also, Austral University of Chile is acknowledged for providing some funds (Project RS88-23). (42) Wada, A. J . Chem. Phys. 1954, 22, 198. (43) Hill, T. L. J . Chem. Phys. 1944, l 2 , 1479.
Exact Solutlon to the One- and Two-Dimensional Models of the Binary Lattice with Nearest-Neighbor Interactions Gholamabbas Pamafar+ Isfahan University of Technology, Isfahan, Iran (Received: April 12, 1989: In Final Form: August 7, 1989)
The "extended sequential construction method", which is newly introduced in this paper, is used for an exact solution to the binary lattice model in the closed form. In section I, this method is applied to a one-dimensional model and a two-dimensional model of a square lattice, for the cases where the fractions of species are not equal (f # OS), and the special case (f=0.5) is being considered in section 11. Application of this method to other two-dimensional, and all three-dimensional, models seems to be straightforward. In the cases wheref # 0.5, no first-order phase transition has been found, which is in agreement with the fact that this transition may occur only whenf= 0.5. However, in the case wheref= 0.5, the results of our calculations show that no first-order phase transition occurs in the case of a one-dimensional model, and the configurational heat capacity (against temperature) goes through a maximum continuously. For the case of a two-dimensionalmodel (of the square lattice), whenf = 0.5, first-order phase-transition phenomenon occurs, and the critical temperature is located roughly at the point x, = 0.41, where x, = exp(t/kT,). A simple exact expression is also obtained for the heat capacity of the one-dimensional model when f = 0.5.
Introduction One may see the terms "binary alloy model", "Ising model", and "lattice gas model" in the literature, all of which may physically be related to the same problem. In these simplified ideal models, the interactions between nearest-neighbors are only considered. These terms are used to explain the nature of different interactions, such as the interactions between different pairs (AA, AB, and BB), in the binary lattice: spins, in Ising model; the atoms that are located on the nearest-neighbor sites, in the lattice gas model. These models have been used in the study of different phenomena in solids, such as adsorption (in the one- and twodimensional (1D and 2D) lattice gas model), absorption (in the three-dimensional (3D) lattice gas model), ferromagnetism and antiferromagnetism, order-disorder, phase transition, and critical temperature. Recently, the sequential construction method, SCM,' and the modified sequential construction method, MSCM,2 were introduced for statistical mechanical studies of special models, specifically Andersen and Mazo's model3 for the feldspar minerals. The main approximation in these methods is that the identical sites of "solid" and "nonsolid" are treated nonidentically, in view of the order of the distribution of atoms among sites. Unlike most other cases, the site energies are more important than the interaction energies in feldspar minerals. The main points of this work are then ( 1 ) to take the interaction energies, instead of site energies, into account and (2) to eliminate the nonidentical treatment of the solid and nonsolid sites when we distribute atoms among sites. The elimination of discriminations among equivalent sites of solid and nonsolid, in SCM and MSCM, leads to the exact sol ut ion. 'The author's last name has been changed from Rajabali to Parsafar.
0022-3654/90/2094-3195~02.50/0
The first exact solution to the one-dimensional problem was given by who used a combinatorial argument. A matrix method was introduced later by Kramer.s The matrix method was used by Onsager6 who, after a very long and sophisticated argument, was able to obtain the solution to a 2D square lattice in closed form, for only the case in whichf= 0.5. In a later effort, Kaufman' succeed in cutting the algebra to a great extent. As far as we know, no exact solution is given for the 3D model. The only available solutions for this model are in the form of a series expansions.* More information on this matter can be obtained from ref 9. Although, the problem of the one-dimensional model has been solved by some approaches, the extended sequential construction method, ESCM, for the one-dimensional model will be considered first because of its simple presentation.
I. General Cases Application of ESCM to One-Dimensional Model. We consider a binary lattice with N sites that are located along a straight line. Each site is only occupied by one atom, A, or B. We also assume an interaction energy between the nearest-neighbor atoms, which is denoted by cAA, CAB, and cBB, respectively for AA, AB, and BB pairs. Now, we show how to calculate the exact com(1) Rajabali, G. A. J . Chem. Sac., Faraday Trans. 2 1986,82,473-476. (2) Rajabali, G.A. A m . Mineral. 1988, 73, 91-96. (3) Andersen, G. R.; Mazo, R. M. J . Chem. Phys. 1979, 71, 1062-65. (4) Ising, E. 2.Phys. 1925, 31, 253. (5) Kramers, H. A,; Wannier, G. H. Phys. Reu. 1941, 60, 252. (6) Onsager, L. Phys. Reu. 1944,65, 117. (7) Kaufman, B. Phys. Reo. 1949, 76, 1232. (8) Rushbrooke, G.S . C.R . REun. Annu. Union In?. Phys. (Paris) 1952, 177. (9) Hill, T. L. Sfatistical Mechanics; Dover: New York, 1987; Chapter 7.
0 1990 American Chemical Society
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The Journal of Physical Chemistry, Vol. 94, No. 9, 1990
Parsafar
TABLE I: Number of Basic Units with Different Configurations for the One-Dimensional Model" basic unit
"The black circles are solid sites, and the open circles are nonsolid sites bNote that these configurations have a multiplicity of 2 .
binatorial factor and configurational energy of the lattice, when A and B atoms are distributed among sites. The calculation is done in the following way: (1) Sites are divided into two groups, called solid and nonsolid sites, in such a way that the neighboring sites of any solid site are all nonsolid and vice versa. Note that all sites are identical, indeed, and the division of sites into two groups is quite artificial. The set of solid sites that are the nearest-neighbors of a given nonsolid site is called the "basic unit". The basic unit, in this case, of course, includes two solid sites that are located on the sides of a nonsolid site. (2) We consider a basic unit with the configuration of two A atoms on its solid sites. Suppose that the probability of having such a configuration is Q2, the number of basic units with such a configuration will then be equal to Q2(N/2) (note that the total number of basic units in the lattice is equal to N/2). If the probability of having an A atom on the nonsolid site of any of these basic units is represented by P2, then there are as many as Q2(N/2)P2basic units, each with the configuration of AAA, and Q2(N/2)(1 - P 2 ) basic units, each with the configuration of ABA (two A atoms on the solid sites and B on the nonsolid site). There are other configurations for the basic unit, which are all given in Table I along with their numbers (note that there is a multiplicity of 2 for Q, configuration). The subscripts on Qs indicate the number of A atoms on the solid sites of the basic unit. (3) I n order to calculate the combinatorial factor, first we consider those basic units with a Q2 configuration. There are as many as Q2N/2 nonsolid sites at the middle of these units, and these nonsolid sites are occupied by A atoms with the probability of P2 and by B atoms with the probability of 1 - P2. In other words, we have Q2N/2 of such sites that we want to distribute as many as Q2P2N/2A atoms and Q2(1 - P2)N/2 B atoms among them. Such a distribution can of course be done in g2 ways, where (Q2N/2)! g2 = (Q2P2N/2)![Q2(1- P2)N/2]! Similarly, the distribution of Q,P,N/2 A atoms and Q,(l - PI)N/2 B atoms among Q,N/2 nonsolid sites of basic units with Ql configuration can be done in gl ways, where
entropy, by using the Boltzman equation, S = k In g, which leads to the following expression for the entropy:
__
- Q2[P2In P2 + (1 - P 2 ) In(1 - P 2 ) ] + 2Ql[P, In PI + ( 1 - PI)In (1 - PI)]+ Qo[PoIn Po + (1 - Po)In (1 - Po)]
Nk
(4) For the calculation of the number of different pairs and the configurational energy, we have to consider that each basic unit with AAA configuration has two AA pairs and each basic unit with AAB configuration has one AA pair, but other configurations do not have any AA pair. Therefore, the total number of AA pairs, NAA, and the possibility of having a given pair with AA configuration, fAA, are given by NAA = '(2 X Q2P2 + 1 2 = N(Q2P2 + Q I ~ I )
X
2QIPI)
+ QiPi (1) From Table I, it is obvious that each basic unit with ABA, AAB, ABB, and BAB configurations includes 2, 1, 1, and 2 pairs of AB, respectively. Therefore, the total number of AB pairs, N A B , and the fraction of AB pair, fAB, of the lattice are as follows: N NAB = -[2 X Q2( 1 - P2) + 1 X 2QlP1 I X 2Ql(1 - PI) + 2 2 X Qopol = N[Q,(l - P2) + QI + Qopol fAA
=
Q2P2
+
AB = Q2(1 - P 2 ) + QI + QoPo (2) With a similar argument, NEB,the number of BB pairs, andfBB, the fraction of pairs that have BB configuration, are given by NBB= N[Qi(l - PI) + QoCl - Po11 ~ B = B
Qi(1 - PI) + Qo(1 - Po)
(3)
Knowing the number of different pairs, one is able to find the configurational energy, readily. Assuming that there is no energy except the interaction energy between the nearest-neighbors and cAA, CAB, and cBB represent the interaction energy between A and A, A and B, and B and B, on the nearest-neighbor sites, respectively. The configurational energy, E , is then given by E = N(Q2P2 + Q I P I ) ~ A+AIQ2(1 - P2) + QI + Q~PoI~AB + I Q I ( ~ - PI) + Qo(1 - P0)kBBI
in which the power of 2 is due to the multiplicity of 2 for QI configuration. With a similar argument, there are go ways for the distribution of atoms among nonsolid sites of those basic units that are indicated by Qo configuration, where go =
( QoN/ 2) ! (QoPoN/2)![Qo(I - Po)N/21!
Therefore, the total number of distributions on the nonsolid sites are given by g1I2= ( g 2 X gl X go). Since the solid and nonsolid sites are identical, there are the same number, gl/*, of ways for the distribution of atoms among the solid sites. It means that the total combinatorial factor is given by g: g = (g2 x g, x g0l2 Now, we are in the position of calculating the configurational
(5) In order to make sure that our artificial division of sites into two groups of solid and nonsolid does not introduce any differences between identical sites of solid and nonsolid, appropriate constraints must be introduced. In other words, we have to make sure that in all distributions solid and nonsolid sites have been treated identically. In order to fulfill this condition, those distributions are only accepted in which the number of basic units with similar configuration on solid and nonsolid sites be equal. For instance, the number of units with the configuration of Q2 must be equal on both solid and nonsolid sites. Therefore, in the distribution of atoms, the following constraints have to be imposed: NAA(SS)= NAA(") NAB(SS)= NAB(") N B B W )= NBB(")
The Journal of Physical Chemistry, Vol. 94, No. 9, 1990 3797
Binary Lattice One- and Two-Dimensional Models where NAA(SS) is the number of solid-solid pairs with AA configuration and Nu(") is the number of nonsolid-nonsolid pairs NBB(SS),and with the same configuration. NAB(SS), NAB("), NEB(") represent the number of specific pairs with the appropriate configuration. In order to find the constraints, we need to know the above terms as functions of Ps and Qs. The number o f solid-solid pairs with any configuration are already known (see Table I). However, the number of nonsolid-nonsolid pairs are not known. A way out of this problem is to find the above constraints by using the fact that in the acceptable distributions the total number of pairs in the lattice (with any configuration) must be twice the number of pairs of the corresponding configuration on the solid sites. In other words
TABLE 11: Values for Different Parameters and the Thermodynamic Properties of the One-Dimensional Model at the Equilibrium State for Some Given Values of the Interaction Energy, in the Case Where f = 0.2 -W 0.0000 0.4090 0.8070 1.2009 1.5995 3.0028 5.2873 Q2
Q, Qo p2
PI
PO ~ A A fAB
fBB
0.0457 0.1543 0.6457 0.2578 0.2200 0.1864 0.0457 0.3085 0.6457 0.4992 0.0187 0.5179
0.0400 0.1600 0.6400 0.2000 0.2000 0.2000 0.0400 0.3200 0.6400 0.5004 0.0000 0.5004
0.0529 0.1476 0.6524 0.3245 0.2400 0.1719 0.0524 0.2951 0.6524 0.4951 0.0423 0.5374
0.0605 0.1396 0.6605 0.3998 0.2600 0.1564 0.0605 0.2791 0.6605 0.4870 0.0726 0.5596
0.0703 0.1297 0.6703 0.4829 0.2800 0.1394 0.0703 0.2595 0.6703 0.4732 0.1124 0.5856
0.1 186 0.0814 0.7186 0.7665 0.3400 0.0748 0.1 186 0.1629 0.1786 0.3599 0.3561 0.7159
0.1843 0.0157 0.7843 0.9676 0.3800 0.0124 0.1843 0.0314 0.7843 0.0996 0.9745 1.0740
NAA = ~ N A A ( S S )
(4)
S/Nk -E/NkT
N A B= ~ N A B ( S S )
(5)
-A/NkT
NBB = 2NBB(SS)
(6)
where XI, A,, Xj, and X4 are the undetermined multipliers. The results of these minimizations are
NAA, NAB, N E Bare given in eqs 1, 2, and 3, respectively, and NAA(SS), N A ~ ( S S and ) , NBB(SS) are all given in Table I . By substituting NAA, from eq I , and NAA(SS), from Table I, in eq 4, we will have
z = Po(1 - P1)2/P2(l - Po)2 and
(L)2(L2) )( =
1 - PI
or Q2P2 + Q I P I= Q2
(7)
- P2)
+ QoPo = QI
QI(1 - PI1 = QoPo
(8)
w=
uAA
(9)
QI =
A---=
NkT
E NkT
=
PI[P12 +f(l
(11)
S Nk
where Uij = c,/kT. The free energy, then, has to be minimized, in such a way that the constraints (7), (9). (IO), and (1 1) are satisfied. In order to carry out this minimization, the method of the Lagrange multiplier may be used. The function 4 is defined as
- 3Pd1 1 - 2P1
1 - 2 . - 2P1
Po =
+ SfP, + 31pI2+ P I 3 1 - 2P1
P2 =
(6) In order to find the equilibrium state, we first set up an expression for the reduced Helmholtz free energy, A / N k T , as
+ UBB
1 - 2PI
(10)
-f= QI + Qo
2uAB
+ 3P12 - 3 p 1 )- P13
f(1
Qo =
With a similar argument for B atoms, we get 1
-
Therefore, at the equilibrium state, eqs 12 and 13, along with the constraints, must be satisfied. If one combines eq 13 with the constraints, the following results will be obtained:
Q2
where the factor is used to prevent overcounting of solid sites (each solid site belongs to two basic units). To get the fourth constraints, we then equate NA(S) t o f ( N / 2 ) :
QI
(13)
and
N A ( S ) = f/2(2Q2+ 2 Q i ) ( N / 2 )
+
1-P,
2 = exp(W)
Equations 7-9 are not all independent, however, and we disregard eq 8 as a dependent one. In addition to the above-mentioned constraints, in the acceptable distributions, the number of A atoms on either solid or nonsolid sites must be equal t o f ( N / 2 ) , and in the case of B atoms, this number must be equal to (1 - f l ( N / 2 ) ; f i s the fraction of sites that are occupied by A atoms and N is the total number of sites in the lattice. From Table I, the number of A atoms on solid sites is given by
f = Qz
I-Po
where
With similar substitutions, eq 5 and 6 are reduced to Q2(1
(12)
(14)
(15) (16)
P,Z(l - 3 f + PI) PI'
+ f(1 - 3 p 1 )
( 1 - p ~ ) L f ( l+ 3P12 - 3P1) - P13] 1 - 2f - 2Pl
+ 5fP1 - 3fP12 + P13
(18)
From the above equations, for any given values offand P,,other probabilities can be calculated. These results may then be substituted into eq 12 to get the value of Z (or W).Of course, this calculation can be done for any value off, exceptf = 0.5. Having the values for the probabilities and W, the configurational entropy, the energy and the probabilities for the different pairs can be calculated at the equilibrium state. The results of such calculations are given in Table 11, for the case t h a t f = 0.2. Also the reduced configurational entropy is plotted against the reduced interaction energy for two cases, f = 0.2 and 0.6, in Figure I . Application to Two-Dimensional Model of Square Lattice. Now, we are willing to use ESCM for the square lattice. In this case, a basic unit has four solid sites, which are located on the corners of a square. These solid sites are the nearest-neighbors of a nonsolid site, which is located at the middle of that square. A basic unit may have different configurations, which all have been shown in Table 111, along with their numbers in the lattice.
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The Journal of Physical Chemistry, Vol. 94, No. 9, 1990
TABLE 111:
Parsafar
Number of Basic Units with Different Configurations for the Sauare Lattice Model
basic unit configuration
no.
&--4 A
configuration
I
configuration
no.
no.
U A A
A
2Q,’ P2’NI 2
I B
e
B
B
B
B
The number of different pairs and the configurational energy can be calculated exactly in the same way as in the case of the one-dimensional model. For example, the number of AA pairs can be calculated as follows: There are Q4P4N/2 basic units with A atoms on their nonsolid sites (see Table 111); the total number of AA pairs of these basic units will then be equal to 4Q4P4N/2. There are 4Q3P3N/2 basic units with A atoms on their nonsolid sites, which include as many as 3(4Q3P3N/2)AA pairs. Similarly, there are 4Q2P2N/2basic units, each with two AA pairs and a basic units, total number of 2(4Q2P2N/2) AA pairs, 2Q,’P,”/2 AA each with two AA pairs and a total number of 2(2Q,’P,”/2) pairs, and 4QIP1N/2 basic units, each with one AA pair and a total number of 1(4Q,PlN/2) AA pairs. When these terms are added up, the total number of AA pairs of the lattice is obtained as NAA= 2NQ4P4 + 3Q3P3 + 2Q2P2 + QiP,’
+ QiPi)
2
0.50 0.r) 0. P 0.40
(19)
With a similar argument, NABand NBBcan be calculated. The results are Ngg = 2N[Q3( 1 - Ps) + 2Q2(1 - P2) + Q i ( 1 - P2’) + 3Q1(1 - PI) + Qotl - Po11 (20) and
With the number of different pairs, the configurational energy can be calculated by making use of the following equation: E = N A A ~ A+ANABCAB + NBB~BB The combinatorial factor for the two-dimensional model (square lattice) can also be calculated with a similar argument as in the case of the one-dimensional model. For instance, there are Q P / 2 basic units where each one has four A atoms on its solid sites.
B
/
F.O.2
0 x)
0.m
10 .
2.0
M
LO
50
Figure 1. Reduced configurational entropy of the one-dimensional model against the reduced interaction energy for the cases wheref = 0.2 (the lower curve) a n d f = 0.6 (the upper curve).
The nonsolid sites of these basic units are occupied by A atom, with the probability of P4, and by B atom, with the probability of 1 - P.,. The distribution of atoms on nonsolid sites of these units can be done in g4 ways, where
The Journal of Physical Chemistry, Vol. 94, No. 9, 1990 3799
Binary Lattice One- and Two-Dimensional Models
g4 =
(Q4N/ 2)!
QO
(Q4P4N/z)![Q4(1 - P4)N/21!
Similarly, the number of ways for the distribution of atoms among nonsolid sites of those basic units where each one has three A and one B atoms is given by g3:
(4Q3N/2)! = (4Q3P3N/2)![4Q3(1- P,)N/2]! Also, the number of distributions on the nonsolid sites of those units where each one has two A atoms on one side of basic unit is given by g2, where
(4Q2N/ 2) ! g2 = (4Q2P2N/2)![4Q2(1 - P2)N/2]! The number of distributions on the nonsolid sites for the other configurations are given in a similar way, that is, g i for the case of two A atoms on the diagonal of the basic unit, gl for one A atom, and go for the case where the basic unit has no A atom on its solid sites, where
(2QiN/2)! g’ = (2Q,’P,”/2)![2Q,l(l - P,’)N/2]!
+~
Q I+
2Q2 + Qz’ +
= 1
Q3
-f
(24)
Other constraints will be obtained by requiring equal numbers for NSS triangles (each with one nonsolid and two solid sites) and SSS triangles (each with three solid sites) with any given configuration. The number of NSS triangles with AAA configuration can be calculated by referring to Table 111. There are Q4P4N/2basic units each with five A atoms, four on its solid sites and one on its nonsolid site. Each one of these basic units has four NSS triangles with AAA configuration. Therefore, these basic units have a total number of 4(Q4P4N/2)of such triangles. There are also 4Q3P3N/2basic units that have as many as 2(4Q3P3N/2) of such triangles. Finally, there are 4Q2P2N/2basic units that have the same number of such triangles. The total number of NSS triangles with AAA configuration is, then, given by
4(Q4P4N/2) + 2(4Q3P3N/2) + (4Q2P2N/2) From Table 111, the number of SSS triangles with AAA configuration can also be calculated. There are as many as Q4N/2 basic units, each with four of such triangles, and 4Q3N/2 basic units, each with one of such triangles. Therefore, the number of SSS triangles with AAA configuration is
4(Q4N/2) + (4Q3N/2) By a similar argument, the number of NSS and SSS triangles with the other configuration can be calculated. The results of such calculations are given in Table IV. Since the number of NSS and SSS triangles with the same configurations has to be equal, then the following constraints have to be satisfied (see Table IV):
(~QIN/~Y g1 = (4QlPlN/2)![4Ql(1 - Pl)N/2]! (QoN/ 2)! go = (QoPdv/2)![Qo(1 - Po)N/21!
Q4P4 + 2Q3P3 + Q2Pz =
The product of the above terms (go-g4) gives the total number of distributions of atoms among nonsolid sites. The number of distributions among all sites, then, is given by g = (goglgzg2’g3g4)2
Q4(1
- P4)
+ 2Q3(1 - P3) + Q2(1
Q3P3 + Q2P2 + Q2’P2’ + Q,(l
- P3) + Q2(1
- P2)
+ Q3
Q4
- P2) =
QiPi
=
Q3
Q3
+ Qi
+ Q2
+ Q2’(1 - P2’) + Q I ( ~- PI) =
Therefore, the reduced configurational entropy is as follows:
+ +
-S/Nk = Q4[P4In P4 + (1 - P4) In ( 1 - P4)] 4Q3[P3In P3 + (1 - P3)In (1 - P,)]+ 4Q2[P2 In P2 (1 - P 2 ) In (1 - P2)] 2Qi[Pz’ In P2’ + (1 - Pi) In ( 1 - P2’)] 4Q,[P, In PI + (1 - P I ) In (1 - P I ) ] + Qo[PoIn Po + (1 -Po)In (1 -Po)] (22)
+
+
Now that the configurational energy and entropy have been calculated, we have to obtain the constraints that guarantee that solid and nonsolid sites are being treated equally. Those distributions are only accepted in which the number of basic units with similar configurations on solid and nonsolid sites are equal. In addition, the normalization condition must be taken into account. The constraints may be obtained as follows, however. All different configurations for a basic unit are given in Table 111. The configurations from the top to the bottom of this table have 4, 3, 2, 2, 1 , and 0 A atoms on its solid sites, respectively. Therefore, on one hand, there are as many as
A atoms on the solid sites (the factor is used to prevent the overcounting of solid sites). On the other hand, half of the A atoms have to be on the solid sites. Hence,
1/4(4Q4+ 12Q3
=t
+ 8Q2 + 4 Q i +
or Q4
+ 3Q3 + 2Q2 + Qz’+
QI = f
(23)
Using a similar argument for the number of B atoms on solid and nonsolid sites leads to
Q2
Q2(1
Q2P2 + 2QiPi + QoPo =
Q2’
+ QI
+ 2Q1(1 - PI) + Qo(1
-
Po) =
- P2)
QI
+ QI
+ Qo
The above equations are not all independent. We take the following ones as the independent equations: Q2
=
(25)
Q2‘
Q4P4 + 2Q3P3 + Q2P2 =
Q4
Q3P3 + Q2P2 + Q2’P2’ + QIPI= Q2P2 + 2QiPi + QoPo =
+ Q3 Q3
+ Q2
Q2’ + QI
(26) (27)
(28)
Now that the six constraints (eqs 23-28) are found, one can set up the reduced free energy, by using the expressions for the configurational energy and entropy that have already been obtained. The minimization of the free energy is done in such a way that all the constraints are being satisfied. The result of this minimization is to get the equilibrium state. Again, as in the case of the one-dimensional model, this straightforward method can be applied whenf # 0.5. The minimization is done numerically by using a software program so called “gino”. One sample of such calculation is given in Table V, for the case wheref= 0.2. In these calculations, cAB and egg are being set equal to zero and U is defined to be c,/kT. With the equilibrium values of Ps and Qs, the fraction of pairs with different configurations can be calculated by making use of eqs 19-21. Some of the values of such calculations are given in Table VI. The configurational entropy and energy are plotted against the reduced interaction energy, U = cAA/kT,in Figures 2 and 3, respectively. 11. Special Case
Application of ESCM to the One-Dimensional Model When f = 0.5. The probabilities for the basic units, Qs, and for the
3800 The Journal of Physical Chemistry, Vol. 94, No. 9, 1990
Parsafar
TABLE IV: Number of NSS and SSS Triangles with Different Configurations NSS triangle configuration no.
SSS triangle configuration
no.
A
r
E
E ,-
6 B
J"\, B
B
e
VI
1.00
-n .7n .0.50
nonsolid sites, when occupied by A atoms, Ps,are given in section 1 (eqs 14-18). W h e n f = 0.5, these equations reduce to 1 - 2 P , P,(1 - P,) Q 2 = 2
Po=--
010
- 2 P , PI - I
1 - 2P, P , - 1
1 - 2P,
Figure 2. Reduced configurational entropy of two-dimensional model of the square lattice against the reduced interaction energy for f = 0.2 and 0.6.
PI
- 2 P , P,2 - P, +
Q,=,_ZP,
-v
+1
1 - 2 P , PI - 1 PI2 - PI 1
0.30
1 - 2 P , P,2 -1
B
$1
TABLE V: Values of Different Parameters at the Equilibrium State for the Square Lattice, for Some Given Values of the Interaction Energies and in the Case Where f = 0.2 -U 0.0000 0.2000 0.4000 0.7000 1.0000 2.0000 3.0000 4.0000 Q4 0.0000 0.0097 0.0098 0.0000 O.OOO0 0.1457 0.1953 0.1996 Ql 0.0078 0.0000 0.0000 0.0179 0.0300 0.0106 0.0007 0.0000 Q2 0.0241 0.0330 0.0331 0.0176 0.0063 0.0000 0.0001 0.0000 Ql 0.1040 0.0912 0.0911 0.0934 0.0910 0.0224 0.0024 0.0003 Qo 0.4076 0.4274 0.4277 0.4490 0.4780 0.7222 0.7901 0.7990 P4 0.0240 0.2608 0.2609 0.2816 0.2816 0.9593 0.9966 0.9996 PI 0.1973 0.3865 0.3865 0.3649 0.4679 0.7834 0.9235 0.9235 P2 0.2003 0.2209 0.2213 0.2791 0.3231 0.3271 0.3369 0.3383 P,' 0.2012 0.2215 0.2217 0.2733 0.3073 0.3087 0.3114 0.3125 PI 0.2003 0.2028 0.2030 0.2066 0.2016 0.1041 0.0327 0.0335 Po 0.2005 0.1868 0.1868 0.1502 0.1225 0.0247 0.0028 0.0004
p2 =
+ Qo)
1
The common solution to the above equations is PI = 0.5 (if PI < 0). This value for P I reduces the other probabilities to O/O, which are ambiguous. Now the question that we have to answer is whether or not the
2
# 0.5, then eq 29 gives P2 = P12/(PI- 1 )
PI(] - PI)
Qo=,_ZP,
2
TABLE VI: Probabilities for the Different Pairs at the Equilibrium State of the Two-Dimensional Model of a Square Lattice for Some Given Values of f and U f = 0.2 f = 0.6 f = 0.8
-U
fAA
fAB
fBB
fAA
JAB
fBB
fAA
fAB
fee
0.00 0.05 0.10 0.20 0.30 0.40
0.0400 0.0407 0.041 3 0.0429 0.0430 0.0430 0.0665 0.1671 0.1967 0.1996
0.3203 0.3 188 0.3 175 0.3143 0.3142 0.3142 0.267 1 0.0660 0.0064 0.0008
0.6397 0.6405 0.6412 0.6428 0.6428 0.6428 0.6663 0.7669 0.7950 0.7996
0.3600 0.3615 0.3630 0.3664 0.3703 0.3747 0.4179 0.5637 0.5966 0.5997
0.4800 0.4770 0.4739 0.4671 0.4594 0.4506 0.3643 0.0726 0.0068 0.0006
0.1600 0.1615 0.1630 0.1665 0. I703 0.1747 0.2178 0.3637 0.3966 0.3997
0.6399 0.6406 0.641 3 0.6426 0.6441 0.6462 0.6654 0.7668 0.7968 0.7996
0.3201 0.3188 0.3173 0.3147 0.3117 0.3074 0.2691 0.0666 0.0064 0.0009
0.0400 0.0406 0.0414 0.0427 0.0442 0.0464 0.0655 0.1666 0.1968 0.1995
1 .oo
2.00 3.00 4.00
The Journal of Physical Chemistry, Vol. 94, No. 9, 1990 3801
Binary Lattice One- and Two-Dimensional Models
q
d(A/NkT)/dQ, = -W- In 4 - 3 In P2 + In (1 - Pz)
/
where W is defined as
w =U A A -
5.0
2uAB
+ UB,
Hence, for the equilibrium state 40
(33)
30
where Z is defined as exp( W). Now, we can investigate the possibility of the phase transition. For that, we consider the lattice gas model, in which
w = UAA= €AA/kT
20
There are as many as QzN/2 basic units with AA configuration. The number of AA pairs in the lattice must be equal to 2(Q2N/2). Therefore, the configurational energy is
10
E = ~(Q~N/~)EAA 0
1.0
PO
2.0
4.0
50
+ 60
Figure 3. Reduced configurational energy against the reduced interaction energy of the two-dimensional model of the square lattice for some given values o f f .
value of 0.5 for PI is meaningful at the equilibrium state, when f = 0.5. Remember that P I is the probability of having an A atom on the nonsolid site of the basic units with Q l configuration. A basic unit with Ql configuration has one A atom and one B atom on its solid sites as 0
A
6--0-@
The configurational heat capacity is given by
c = -d E
dT
If we eliminate QI between eqs 30 and 3 1, we will obtain Q2 in terms of Pz as
X
Qz =
3-2pz
and the derivative dQ2/dPz is then given by
In this case,f= 0.5, A and B have equal abundance, and putting either A or B atom on the nonsolid site of this basic unit will also make no difference, energetically, because the energy of the basic
+
unit in both cases is the same and equal to cAA eAB. Hence, the probability of having an A atom on this site (PI) is equal to that for having a B atom (1 - P I ) on it; then
PI = I-PI
dQz/dPz = 1/(3 - 2PZ)’
(35)
The other derivative, dPz/dT, can be calculated by making use of eq 33; the result is
The derivatives can be substituted from eqs 35 and 36 into eq 34; we get
or PI = 0.5 Therefore, the value of 0.5 for PI at the equilibrium state, when f = 0.5, is quite understandable. To get out of this ambiguity, first we substitute PI = 0.5 into the equations expressing constraints as well as into the equation for the reduced free energy and then will carry out the minimization, in order to obtain the equilibrium state of the system. The constraints will then reduce to
P z = l--
(37) The reduced heat capacity, after some rearrangement, is given by
The heat capacity is plotted against the reduced temperature, 1/lw = kT/lCAA(, ill Figure 4. Application of ESCM to the Two-Dimensional Model of a Square Lattice When f = 0.5. A similar argument that led to P, = 0.5, for the equilibrium state of the one-dimensional model, will lead to the following constraints for the two-dimensional model:
QI 1 -2Qi
and the reduced free energy becomes A / N k T = ( X Q I + QZPZIUAA + (Qz(1 - Pz) + QI + QOPOIUAB + %QI + Qo(1 - Po)IUBB+ QdP2 In PZ+ (1 P2) In ( 1 - P2))- 2Ql In 2 Q o P o In Po + ( 1 - Po) In (1 - Po))
+
All the variables in A/NkT are dependable, except one. We take QI as an independent variable; then, in order to get the equilibrium state, the derivative d(A/NkT)/dQ, must be set equal to zero. The result is
Pz = P,’ = 0.5
(39)
Po = 1 - P4
(40)
PI = 1 - P,
(41)
If we make use of these equations in those five constraints that were introduced to guarantee that the solid and nonsolid sites are being treated equivalently, the following results will be obtained: Q4
= Qo
(42)
Q3
=
QI
(43)
3802 The Journal of Physical Chemistry, Vol. 94, No. 9, I990
Parsafar TABLE VII: Values of F and C for Some Given Values of Z F Z Z G 1.000 000 0.500 000 1.000000 0.500 000 0.899 502 0.527 000 0.899 503 0.527 000 0.558 000 0.558000 0.800 205 0.800219 0.595 000 0.595 000 0.699 705 0.699 193 0.600076 0.549 262 0.450 964 0.350 590 0.249675 0.150 353 0.099 453 0.000 0
0.638 999 0.664 999 0.723 999 0.798 998 0.889 998 0.972 997 0.993 997 1.ooo 0
0.600 440 0.549 947 0.450 174 0.350 446 0.249 779 0.234 846 0.594 688
0.638 999 0.664 999 0.725 999 0.804998 0.9 19 998 0.969 998 0.998 997
Figure 4. Heat capacity versus the reduced temperature for the onedimensional model.
The only two independent constraints that will then remain are the following: g(1) =
+ 4Q3 + 3Q2 - 1/2 = 0
Q4
g(2) = Q4(p4 - 1)
+ Q3(2P3 - 1) + Y2Q2
(44) =0
(45)
By substituting variables Po,PI, P2,Pi, Q l , and Qo (from eqs 39-43) into the expression for the reduced free energy, it will reduce to A/NkT = 2[@84 + 3Q3P3 + 72Q2 + Q3(1 - P ~ U A +A 12Q4(1 - p4) + 4Q3 - 2Q3P3+ 3Q2 + 2Q3(1 - P W A B+ k?s(l - P3) + Y2Q2 + 3Q3P3 + Q s p d U ~ ~ +l 2QdP4 In p4+ (1 - p4) In (1 P4)} 8Q3(P3In P3 + (1 - P3) In (1 - P 3 ) )- 6Q2 In 2
+
Gm
I
Now we want to minimize the free energy in such a way that constraints 44 and 45 are satisfied. We use the method of Lagrange multipliers for this purpose. Let us define the function 4 as
Figure 5. (a) Functions F and G versus Z . (b) Functions F and G plotted, partly, in a larger scale by computer.
where A, and A2 are the undetermined multipliers, and minimize it with respect to all variables. The results of such minimization are
(solid curve) has a similar behavior as the F function and in fact it lies almost on F when Z > 0.41. The other segment (shown by broken lines) corresponds to a repulsion between the nearest occupied sites (because P4or the order of the lattice increases by raising the temperature). Since we are interested in the case of attraction between the nearest-neighbors, we disregard the upper segment of G. The point at which the two segments merge can be calculated. If we take the logarithm of both sides of eq 48 and then take the derivative, we find
2
p4
2 4
Z=
1 - P4 =64P47
(47)
1 64P35(1 - P3)
In principle, on one hand, eq 47 can be solved for P4 as a function of Z. Let us denote this function by F, Le.,
F = P4(Z) (49) On the other hand, we may first solve eq 46 for P3 in terms of P4 and then substitute that, P3(P4),into eq 48, and finally the resulted equation can be solved for P4 as a function of Z . Let us denote this function by G; then, G = P4(Z) (50) The values of F and G are given in Table VII, for some given values of Z,and these functions are also plotted in Figure 5. As we can see from Figure 5, the function F reduces continuously with 2 (or temperature), which means that the lattice becomes more disordered when temperature increases. The other function (G) has two different segments, however; the lower one
0.00
0,2(0
0.9
z-
d I-n Z - 6 p 3 - 5 dP3 Pd1 - P3) At the merging point, d In Z/dP3 = 0, or
5 25 P, = -, P4 = -,
36 Z = - = 0.233 6 26 55 The functions F and G, in Figure 5, denote the probabilities for P4of the two phases that are at the equilibrium. Constraints 44 and 45 are applicable to both phases, cy (its P4 is denoted by F) and @ (its P4 is denoted by C). Then, + 4@ + 3@ - '/z = 0 (51)
me-1) +@(ZIT-
1)
+ )/2@ = 0
(52)
Similar equations can be written for the other phase, @. In addition to these four equations, two phases must have the same value of the molar free energy (at any given temperature), when two phases are at equilibrium. Therefore, there are five constraints among Ps and Qs. The values for Ps,however, can be calculated by using eqs 47, 46,41, and 40 for the a-phase and eqs 48,46, 41, and 40 for the @-phase. Therefore, the values for Ps can be
J . Phys. Chem. 1990, 94, 3803-3806 calculated at equilibrium for both phases at any given temperature; however, in order to calculate the values for Qs of two phases, one more equation is needed. It is not clear how to obtain this equation, here.
Discussion ESCM has been applied to solve the nearest-neighbor problems exactly. The calculation that is presented in section I is valid when f # 0.5, however. Remember that f = 0.5 corresponds to the critical density (see ref 9). It may be because of the possibility of the phase-transition phenomenon that the straightforward method (section I) cannot be applied for the case wheref= 0.5; for that, it has been handled specifically in section 11. There are a few main points that may be deduced from our results. On the one hand, in the absence of interaction energy (or at very high temperatures), the equilibrium state corresponds to the random distribution of atoms on sites. This point may be seen in Table I1 (for the 1D model) when W = 0 and in Table VI (for the 2D model) when U = 0. In these cases, W = 0 and U = 0, the probability for any given ij pair,fij, is
fi
where fi and are the fraction of i and j atoms in the lattice, respectively, and the lattice has complete disordered configurations. On the other hand, when the interaction energy is very high (or at very low temperatures), fAA - f , f A B 0, a n d & , 1 -f, and the lattice has a complete ordered configuration; such types of behavior are expected. Another point is related to those thermodynamic properties of the lattice that are the first derivatives of the free energy, namely, the configurational energy and entropy (and volume in the 3D model). The configurational entropies are plotted against the interaction energy in Figure 1, for the 1D model, and Figure 2 , for the 2D case, for some values off (for the cases wheref # 0.5). The configurational energy is also plotted in Figure 3 for the square lattice. These figures do not show any step (discon-+
-
3803
tinuity) at all. Such a result is expected, by the fact that the interaction energy between the nearest-neighbors does not govern any first-order phase transition, except when f = 0.5. The configurational heat capacity is given by eq 38 and is plotted against the reduced temperature in Figure 4, for the one-dimensional model, when f = 0.5. The heat capacity goes through a limited maximum, continuously at about kT/Jtl= 0.25. Therefore, the nearest-neighbor interaction does not govern any first-order phase-transition phenomenon in the I D model, at all. If there was any phase transition in this case, the heat capacity function would have gone through a step at the maximum or at some other points. Similar curves have been presented by the other exact calculated methods (for instance see ref 9). For the 2D model (square lattice), however, different ordering behavior has been observed whenf= 0.5. The ordering behavior of the lattice can be related to the value of P4, in such a way that a larger value for P4 corresponds to more ordering in the lattice. The probability P4 is plotted against Z in Figure 5. This figure shows that at high temperatures there is only one value for P4 of the lattice; however, when the temperature is decreased, the curve splits into two curves ( F and G ) roughly at Z = 0.41. Therefore, at this splitting point the system begins to become as two phases, C (more ordered phase) and F (more disordered phase). This point corresponds to the critical temperature, the exact value of Z , = 2'/* - 1 = 0.414 has been obtained by the other exact calculation^.^ The system coexists at the equilibrium between two phases until the temperature becomes so low as to give Z = exp(t/kT) = 0.233. Therefore, the system has two phases in equilibrium when 0.233 Q 2 < 0.41. If the temperature becomes very low, in such a way that Z < 0.233, the lattice will exist as a one-phase system, which seems to be logical. Although ESCM has been used here for the one-dimensional and two-dimensional models (of square lattice), as the special cases, it is applicable to other 2D and 3D models as well; for instance, in a simple cubic lattice, a nonsolid site along with its six nearest-neighbor solid sites should be considered the basic unit.
Thermodynamic Properties of Benzyl Radicals: Enthalpy of Formation from Toluene, Benzyl Iodide, and Dibenzyl Dissociation Equilibria H. Hippler and J. Troe* Institut f u r Physikalische Chemie der Universitat Gottingen, Tammannstrasse 6, 0-3400 Gottingen. West Germany (Received: April 20, 1989; In Final Form: November 9, 1989)
Measurements of the dissociation and reverse recombination rates of toluene, benzyl iodide, and dibenzyl in shock waves have provided a direct access to the dissociation equilibria. A third-law analysis of these data is made, providing a new value = 210.5 f 4 kJ mol-' which is higher than that given by earlier evaluations. of the enthalpy of formation of benzyl of On the basis of this value, thermodynamic properties of benzyl are tabulated over the range 0-3000 K. Revised values of the bond energies of toluene, benzyl iodide, dibenzyl, and ethylbenzene of = 378.4, 190.0,277.2, and 328.8 kJ mol-', respectively, are also given.
1. Introduction
There is considerable interest in the precise enthalpy of formation of benzyl radicals since "there is probably no other single radical upon which so many bond energy values depend".l A variety of methods have been employed to derive this quantity. Among these are studies of iodination kinetics,'-" comparative (1) McMillen, D. F.; Golden, D. M. Annu. Reu. Phys. Chem. 1982, 33,
493. (2) Rossi, M.; Golden, D. M. J . Am. Chem. SOC.1979, 101, 1230. (3) Golden, D. M.; Benson, S. W. Chem. Reu. 1969, 69, 125.
0022-3654/90/2094-3803$02.50/0
single-pulse shock tube dissociation experiment^,^ and the ion cyclotron resonance bracketing technique.6 Over the years the has moved up from3 188 f 4 to4" 201 f 9 kJ value of AHfo298 mol-I. The applied methods have their merits but also their problems. Recent activities in this field (see, e.g., refs 7-14 and (4) Walsh, R.; Golden, D. M.; Benson, S.W. J . Am. Chem. Soc. 1966,88, 650. ( 5 ) Tsang, W. I n t . J . Chem. Kinet. 1969, I , 245; Ibid. 1978, 10, 41. (6) DeFrees, D. J.; Mclver, R. T.; Hehre, W. J. J . Am. Chem. SOC.1980, 102, 3334.
0 1990 American Chemical Society
