On the Chemically Impossible 'Other' Roots in Equilibrium Problems Oliver G. Ludwig Villanova University, Villanova, PA 19085 When doing equilibrium problems for a class of freshmen I have often been asked: "How do you know that the other root(s) of the equilihrium expression are physically impossible?" (i.e., giving negative or too large concentrations). Of course, one possible answer is that nature is never in doubt as to the position of an equilibrium; this appeal to a "higher authority," however, is somewhat unsatisfactory. I have found that, a t least for quadratics, a more convincing oroof is easilv constructed once a basic theorem is shown. ...
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~ ~ ~ e n d~ ih xe .i properties e say that the sum of the roots is the negative of the coefficient of the linear term divided by the coefficient of the quadratic term (i.e., that X I + x~ = -bla for the quadratic a x 2 bx + c = 0). Also, the product of the roots is the constant term divided by the coefficient of the
+
or xZ
( K - 1) - x ( p
+ y ) K t Kpq = 0.
Using property (11)we have x~.x?=K.p.ql(K-I!
Now for the case K < 1we see that the product of the roots must be negative, but since XI is positive, the second root must be negative and therefore spur~ous.For the case K > 1, the ratio KI(K - I ) is > 1, so the product of the roots must exceed p q . Since X I is less than p, the second root must be greater than q , a clearly impossible result. The above techniques are extendahle to more complicated equilibria, albeit with a lessening of impact upon the student. For cubic and higher equations the techniques required are heyond the scope of a first-year course in chemistry. Appendix Let 21 and xn be the two roots of the quadratic equation
volume for simplicity.)~ Let x be the additional amount of Q produced to reach equilibrium; manifestly then, -q
