# THE NUMERICAL SOLUTION OF EQUILIBRIUM PROBLEMS

Textbooks of physical chemistry leave us in the dark regarding the best method to follow in hding numerical solutions of equilibrium problems. The fol...
THE NUMERICAL SOLUTION OF EQUILIBRIUM PROBLEMS HORACE G. DRMING, U N ~ R S I TOZ. Y NEBRASKA. LINCOLN. NEBRASKA Textbooks of physical chemistry leave us in the dark regarding the best method to follow in h d i n g numerical solutions of equilibrium problems. The following procedure will apply even to the most complex cases, where the equilibrium proportions are to be calculated from the equilibrium constant. Let r = (a,

+ sl.cjax (a, + hxj"

(a3

+ wj*"

(1)

in which al, az, aa,. . . express the number of moles of each substance originally present in the reaction mixture, while sl, sz, Sa, . . . express the number of moles of each substance that appear in the chemical equation. The values of sl, sr, ss, . . are to be taken as positive if they relate to substances in the right-hand member of the chemical equatibn, and negative in the contrary case. We shall let x represent the number of moles of any resultant that would have been produced by the reaction if one mole of that resultant appeared i n the right-hand manber of the chemical equation. a may be called the mole-product. Lrt

.

in which KO is an expression that becomes identical with the equilibrium constant, expressed in partial pressures, whenever the reacting substances are present in equilibrium proportions; iyhile s = sl s2 sa . . ., namely the sum of the exponents in the mole-product or equilibrium con(a2 s * ~ ) (aa sax) = . ., namely the sum of stant; n = (al six) all the moles in the reaction mixture; and P represents the sum of the partial pressures of the individual components. Now let a reaction take place, which would produce a slight additional amount, dx, of any resultant, if one mole of that resultant appeared in the chemical equation. Then, since dn = s dx,

+ + +

+

+ +

+ +

.

in which Z is the sum of the quantities formed by multiplying the reciprocal of the moles of each substance by the square of the exponent of that substance, in the mole-product or equilibrium constant. For example, in the first approximation below, the value of Z is roughly 4, while the total number of moles is n = 95, and the sum of the exponents is s = - 1 . The term s 2 / n may usually be neglected, since n is usually large in comparison with the denominators of the terms that compose Z. When 581

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the sum of the exponents is zero, or when the reaction is carried out a t constant volume, the small term s2/n vanishes altogether. We may replace Kp in the preceding formula by K,, the equilibrium constant expressed in terms of mole fractions; or by Kc, the equilibrium constant expressed in terms of concentrations. Calculate the composition at equilibrium (480°C. and a constant pressure of 720 mm.) of a mixture originally containing 25 moles of HCl and 75 moles of 02. K, = 11.92, or log K, = 1.07628. The large numerical value, of K* shows that the equilibrium mixture must contain considerable amounts of Clz and HzO. As a first approximation, assume that 20 moles of HC1 react with 5 moles of Oz, to form 10 moles of Clz and 10 moles of HzO. From the concentrations then present we may calculate, by equation (2):

+

lo'

log KI = log log 95 lo' 5& X 70

- log 720/760 = 1.36022

This exceeds the desired result by 1.36022 - 1.07628 = 0.284 unit. To correct i t we must make d log K p = -0.284. We then roughly estimate Z = 4, and 2.3/Z = 0.6, whence by (4) d r , = 0.6 X (-0.284)

=

-0.17

The negative sign indicates that we decrease the substances on the righthand side in such a proportion as to increase the oxygen by 0.17 mole. Thus we arrive a t a second approximation, tabulated below, from which we may calculate log Kz = 1.07835. Thiois too large by about 0.002 unit, hence we may calculate dx* = 0.6 X (-0.002) = -0.001

which enables us to make a slight final adjustment that gives the moles a t equilibrium to the nearest thousandth of a mole. 4HCI

Moles a t start

25 First approx. 5 Second approx. 5.68 5.684 Final result

+

0 3

F?

75 70 70.17 70.171

2Cb

+

2Hz0

s

0 0 10 10 n = 95 9.66 n = 95.17 9.66 9.658 9.658 n = 9 5 . 1 7 1

-

-1

log K2 = 1.36022 log KI = 1.07835 logKz=l.07679

The great advantage of this or any other iteration method is that slight accidental errors or the use of rounded values in approximation do not vitiate the k a l result, but a t worst merely delay the convergence. In the present method it is usually sufficient to make a rough estimate of Z, mentally, or with a slide-rule. Problems in which the equilibrium constant is expressed in terms of the degree of dissociation may be solved in a similar way. If the initial concentrations appearing in a problem may be expressed in round numbers of moles, or if slide-rule accuracy is sufficient, it may be best to calculate and approximate K* rather than log K*.

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In this case the adjustment to be made in the approximate value of K , may be found by a formula derivable from (4), viz., dx

=

1 - dK,,

(5) KZ' in which K is the average of the approximate and true values of KO. In the preceding example, the calculated first approximation for K, is 22.9, as compared with the true value 11.9. Therefore, dKp = - 11, K = 17.4, Z = 4, whence d x = -0.16, in close agreement with the value -0.17, obtained as a first adjustment by the ather method. The second approximation will nearly always give the desired solution within the limits of experimental error.