Steven Berline and Clark Bricker University of Kansas Lawrence 66044
II
The Law 01 Mas h f i o n
The general theory of equilibria is a topic that is taught in virtually every beginning chemistry course. Students are usually taught to deduce how a chemical system in equilibrium is affected qualitatively by changes in temperature, changes in pressure, and changes in the concentration of reactants and products. When more quantitative aspects of equilihria are approached, the idea of an equilibrium constant is introduced and methods to evaluate these constants are discussed. One of the more common methods used to determine equilibrium constants utilizes kinetic measurements. I n order to apply this method, it is necessary to understand that most chemical reactions do not proceed by the mechanism suggested by the balanced equation for the overall reaction. Instead, several sequential steps may be involved and the balanced equation for the overall reaction is merely the sum of the balanced equations for each of the steps. A reaction consisting of two or more steps cannot proceed any faster than the slowest or rate-determining step. I n the derivation of an equilibrium constant from kinetic data, it is essential to know how the rate determining step is affected by the concentration of reactants and, similarly, how the rate of the reverse reaction is affected by the concentration of products. The Law of Mass Action, proposed by Guldherg and Waage in 1867, states that the rate of the forward reaction is proportional to the product of the concentrations of the reactants, each raised to the power of its coefficient in the balanced equation. Thus, if the rate-determiniug step for a reaction is represented by
According to this equation, the only way one molecule of C can be formed is by the simultaneous collision of two molecules of A with one of B. The experimental evidence eliminates the possibility of the rate-determining step being one A reacting with one B to produce AB and then, AB reacting with another A to produce C. This sequence would give two steps and not the single one which was postulated from the experimental data. If there were two steps in the production of C, the rate-determining step would have to he either the formation of AB or the reaction of AB with A. Now let us assume that we have a unit volume containing one molecule of B and two of A. We would expect the rate of the reaction from the Law of Mass Action to he proportional to [212 [I] or 4. If we now double the number of A molecules and keep the number of B molecules comtant in the same volume, we would expect the rate of the reaction to now be proportional to [412 [I] or 16. This is, of course, four times the rate of the reaction a t the lower concentration of A.
Figure 1. Two rnelsculos of A and one of B in a unit volume.
we would write Rate of forward reaction = u[A]*[B]r
The reason for this relationship is usually left unexplained. Indeed, the student often gains the impression that the parameters in [A]Z[BIu are determined solely by experiment and that it is only by coincidence that they are the same as the coefficients of the reactants in the balanced equation. I n more advanced chemistry courses, the proof of the Law of Mass Action is based on chemical potentials. This approach, however, is not well suited for elementary courses. It is the purpose of this paper to present a derivation which should have meaning and could be utilized a t an elementary level. Let us begin with a simple case. Suppose the ratedetermining step can be represented from the experimental data by 2A+B=C
Figure 2. Twice the number of A rnolec~lefin the same unit volume. There ore now six possible pairs of A insteod of the one pair porriblo in Figure 1.
Figure 1 shows two molecules of A and one of B in a unit volume whereas Figure 2 shows twice the number of A molecules in the same volume. We would expect B = C, to be proporthe rate of the reaction, 2A tional not to the square of the number of A molecules but to the number of ways that the A molecules may be paired because it takes two molecules of A colliding with one B to make a molecule of C. Examination of Figure 1 shows that there is only one pair of A molecules and one B molecule. Thus, the rate of the reaction in this case could be assigned arbitrarily to be unity. I n Figure 2, where there are twice as many A molecules, there are six pairs of A. Thus, the rate of the reaction in this case should he six times unity. The conclusion just deduced concerning the increase in the rate of this reaction when the concentration of A is doubled is not in agreement with what one calculates from the Law
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of Mass Action. Let us examine the origin of this discrepancy to see how the rate of the forward reaction depends on the number of pairs of A molecules. T o do this, a formula which will predict the effect of the concentration of A and B on the rate must be deduced. If we begin with p molecules of A and increase the number n times to np, we must examine how many times the number of pairs of A have increased in order to see B, ought to go. how much faster the reaction, 2A The mathematical symbol for the number of airs that
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(3
can be formed from x things is and the formula for calculating the number of pairs is
Now, therefore, if we start with p molecules of A, we have [ p ( p - 1 ) ] / ( 2 )possible pairs, and when we increase the numher of A molecules to np, we have [np(np - I ) ] / @ ) possible pairs. The number of times we have increased the number of pairs of A is
Therefore, if the concentration of A is increased n times, we would expect the reaction, 2 A B, to go [n(np - 1)1/[(p- I ) ] times faster. Let us now calculate from the above equation what the effect on the rate of the reaction would be when the number of A molecules is increased three times from 10 to 30 but the number of B molecules and the volume are held constant. We would expect that the reaction would be [3(3 X 10 - 1 ) ] / ( 1 0- 1) = (3 X 29)/(9) = 9.67 times as fast. The Law of Mass Action would predict 9 times faster. Similarly, if we started with only 2 molecules of A and increased to 4 , the Law of Mass Action would lead us to expect a ratc 4 times greater whereas equ. (2) predicts a rate 6 times as fast. The reason for this apparent discrepancy is that the Law of Mass Action, as we know it, is applicable ant/ when p is large. For example, let us calculate the increase in the ratc of the reaction when we have a mole of A molecules and when we have twice this number of A molecules. Applying eqn. ( 2 ) to find the increase in the rate of the reaction, 2A B, we obtain [2(2 X G X loz3- 1 ) ] / ( 6X loz3- 1) times faster. Since 6 X loz3and 12 X lo2' are very large numbers, it is virtually impossible to tell the difference between G X loz3and G X loz3- 1 or between 12 X loz3and 12 X lo2' - 1. Thus, the equation [2(2 X 6 X loz3- 1 ) ] / ( 6 X loz3 - 1 ) becomes [2(2 X 6 X 1023)]/(6X loz3)which equals 4. Thus, eqn. (2)predicts that when avery large number of A molecules is doubled, the rate of the reaction is increased four times. This is exactly the result given by the Law of Mass Action. It is necessary, therefore, to dcrive the Law of Mass Action for 2A B as an approximation from eqn. ( 2 ) when p is large. I n other words, we must consider
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lim " ' 9
4 %- ~ 1) p -1
As we have already seen, when n is large. n - 1 is approximately p and np - 1 is approximately n,p. Therefore - 1) = lim n 2 - nip lim n(np -p - 1 pt.. 1 - 1/p
" ' P
-
nP
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Journol of Chemicol Education
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G).
If we start with p molecules of A, the numher of mof A that could collide with a molecule of R
erouos .
(i).
would be If we hold the concentration of B ~, constant and iucrease the number of molecules of A to n.p, the numher of m-groups of A would increase to
(tf). Since the rate of reaction is proportional to
the number of m-groups of A, the increase, R, in the rate of reaction should he /nv\
It is essential to recall that, for the iricrease in the rate p must he a of reaction to equal the fraction in eqn. (3, large number. Therefore R
=
.--
("2) (K)
lim -
(6)
If we use eqn. (4) to evaluate the numher of nz-groups in eqn. ( 6 ) ,we obtain
(7)
The factorial quantities can he written np(np - 1) . . . (np - m l)(np R = lim (np m)!p(p 1 ) . . . (p -m pt-
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- m)!(p
- mi,!
+ l ) ( p - m)!
and after cancelling R
= lim pm
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np(np - 1) . . . (np - m 1) p(p - 1) . . . (p - m 1)
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(8)
The numerator and denominator of eqn. (8) can be expanded into polynomials and the terms in the numerator arrangcd accordiug to descending powers of n p and the denominator according to descending powers of p. The highest power of n.p in the numerator will he obtained when all the first terms of t,he binomials, np, np - 1, np - 2, . . ., n p - (m 1) are multiplied. Since therc are m factors in the numerator, the highest term will he nmpm. The next term would he kin-Ipm-', where l i ~is some coefficient which will not hc explicitly cvaluated. Similarly, the highest term in the denominator will he p", thc next, t,erm llpm-I, et,c. Thus, eqn. (8) becomes
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(3)
Equation (3) is the Law of R1lassAction as applied to the 500
concentration of A molecules in determining the rate for the reaction, 2A B. Now, in the more general case, mA B = C, we would expect the rate of reaction to he proportional to the number of groups of m molecules that could be formed from the molecules of A. (A group of m molecules of A will he referred to as an m-group of A.) The mathematical symbol for the number of groups of y objects that can be formed from a total of x objects is Equation (1) was a special case of this symbol. The formula is
Si~iceit has been showu it1 t,he more spccific cnse that.
thc Law ot Mass Action is applicable only when the number of molecules is very large, p must also be a very large number in this general case. If p is approximately of a mole of molecules, p = loz3. Thus, p" equals and the first term of the denominator is approximately loz3times greater than the second term. Therefore, when p is a very large number, the only term in the denominator that is significant is the first term. By similar reasoning, all but the first term of the numerator is insignificant. Equation (9) can now he written R = lim nmpm - = n"
-
P"
concentration of A is increased n times and the concentration of B is held constant. Consider briefly the most general case mlA
+ m2B = C
The rate of the forward reaction must depend on the number of ml groups of A and mr groups of B. Every m, group of A can combine with (,")i3 groups of B, ,...*, and vice versa. Thus, if there are p, molecules of A and pz of B, and increase each n1 and n2 times, respectively, the increase in the rate of the forward reaction ,-\
will he
This equation is, of course, in agreement with the Law of Mass Action which states that the rate of the forward reaction for mA B = C will increase nm times if the
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