The Best Proportion of Reactants for Gaseous Reactions Huang Zhenyan Wenzhou Teachers' College, Wenzhou Zhejiang China, 325000 It is well-known that the original proportions of reactants 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 1 - a
n n -?a
0 a
Thus, we can immediately obtain the following equations x,=-
1 +
n x,=-(1 l+n
-Xc)
(6)
-xG)
(7)
If the mole fractions of each component in the equilibrium state are indicated by Xo (shown by X below), we have
initially a t equilibrium
Thus, we get
Now the question has changed: Which value can be adopted for n when X has the maximum value? From eq 8 we can obtain
According to eq 2, for various mole prnportions of reactanttin. both o and the mole fractionsofCH20Hat eauilibrium 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~6nalreacmnts is thv cwfficient proportion of reactants in the stoichiometrir eauation, the e(~uilihriummole fraction of product has the maximum value.
The derivative of eq 9 with respect to n is
General Gaseous Reactions What about the general gaseous reactions? The question is not yet discussed in ordinary physical chemistry textbooks. Consider a reaction involving three gases A, B, and G, all at temperature T, which now will be assumed to behave ideally. When the system reaches a n equilibrium state, we have For the reactions that do not proceed to completion, X t 1,and this inequality leads to
aA+bB=gG
1-XtO
where P is the total pressure of the equilibrium system; and PA,P B , and PG are the partial pressures of the components 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 components A, B, and G. Assume X ~- - n XA
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Journal of Chemical Education
(5)
(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)
Conclusion (13)
(14)
From eq 14 we obtain n = -b a
(15)
As compared with eq 5, we have (16)
The result shows that for the equilibrium state, a s long a s the proportion of mole fractions of reactants A and B is a:b, the mole fraction of product will have the maximum value. Because it is assumed that there are no side reactions and no product in the reactor initially, the proportion of mole fractions of components Aand B can be kept a:b a t equilibrium a s long a s that of reactants Aand B is initially a:h. I t can be proved that the conclusion is true for more reactants 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 equation.
Volume 71 Number 8 August 1994
647
