The Best Proportion of Reactants for Gaseous Reactions Huang Zhenyan Wenzhou Teachers' College, Wenzhou Zhejiang China, 325000

I t is well-known that the original proportions of reac- tants that enter the reactor are different, and so are the proportions of each component in the equilibrium state. By way of a n example, we study the synthetic reaction of CH30H from Hz and CO. The equilibrium constant of this reaction, a t 663.2K and 300 atm, is 3.3 x Assuming that n is the initial mole proportion of reactants Hz and CO and that a represents the conversion rate of CO, we get the following

1 n 0 initially 1 - a n -?a a at equilibrium

Thus, we get

According to eq 2, for various mole prnportions of reac- tanttin. both o and the mole fractionsofCH20H at eauilib- rium can be calculated. (See the table.) 1 t i s easili seen that when n = 2, in other words, when the mole mo~ortion . . for ori~6nal reacmnts is thv cwfficient proportion of reac- tants in the stoichiometrir eauation, the e(~uilihrium mole fraction of product has the maximum value.

General Gaseous Reactions What about the general gaseous reactions? The question

is not yet discussed in ordinary physical chemistry text- books. Consider a reaction involving three gases A, B, and G, all a t temperature T, which now will be assumed to be- have ideally. When the system reaches a n equilibrium state, we have

a A + b B = g G

where P is the total pressure of the equilibrium system; and PA, P B , and PG are the partial pressures of the compo- nents A, B, and G. Under the given conditions, Kp and Kx are constants.

Suppose there are no side reactions. In other words, only the components A, B, and G exist in the reactor. Thus,

X , + X , + X , = l (4)

where XA, XB, and Xo stand for the mole fractions of com- ponents A, B, and G.

Assume

- - X ~ - n XA (5 )

Thus, we can immediately obtain the following equa- tions

1 x , = - + - X c )

(6 )

n x,=-(1 -xG)

l + n ( 7 )

If the mole fractions of each component in the equilib- rium state are indicated by Xo (shown by X below), we have

Now the question has changed: Which value can be adopted for n when X has the maximum value? From eq 8 we can obtain

The derivative of eq 9 with respect to n is

For the reactions that do not proceed to completion, X t 1, and this inequality leads to

1 - X t O (12)

Thus, in order to get dxldn = 0, the following equation must be obtained

Values of or and Mole Fractions of CHjOH at Equilibrium for Various n (at 683.2 K and 300 atm)

646 Journal of Chemical Education

From eq 14 we obtain

b n = - a

As compared with eq 5, we have

Conclusion

(13) The result shows that for the equilibrium state, as long as the proportion of mole fractions of reactants A and B is a:b, the mole fraction of product will have the maximum

(14) value. Because it is assumed that there are no side reac- tions and no product in the reactor initially, the proportion of mole fractions of components Aand B can be kept a:b a t equilibrium as long as that of reactants Aand B is initially a:h.

(15) I t can be proved that the conclusion is true for more re- actants and products. For general gaseous reactions the best proportion of reactants to place in the reactor is the coefficient proportion of reactants in the stoichiometric

(16) equation.

Volume 71 Number 8 August 1994 647