MATHEMATICAL PROBLEM PAGE

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MATHEMATICAL PROBLEM PAGE Directed by PAUL C. CROSS Gates Chemical Laboratory, California Institute of Te&ology,

SOLUTIONS of the following problems will be given in the January issue. 46. Show that (dxz/x2)/(dxl/xl) = or, where or is a constant, if the rate of change of each substance xl and x2follows the first order law, &/dt = -kx. (k, the specific reaction rate, is a constant characteristic of the particular system.) 47. Assuming or = (dxz/xz)/(dx~/xJ to express the ratio of the rates of electrolysis of HZand H1 from water in which the amounts present are xz and XI, respectively, show that where xz0and x10are the amounts initially present. 48. Let b be defined as

b = (XI

+ xd/(xl0 + .a0).

b is the fraction of the total hydrogen (or water) which remains nnelectrolyzed. Derive the relation

from the relation %/xzO = ( x I / x I ~ ) ~ .

+ x2)

49. Let NI = XI/(XI

Pasadena, California

(The N's represent the atomic fractions of the isotopes in the total hydrogen.) Show that Eq. (1) of problem 48 may be reduced to 61-m

=

(Nz0/Nz)(NI/NIO)".

50. Suppose that 1 part in 5000 of the hydrogen in ordinary water is H z (Nz = 0.0002), and that a has the value 0.2 under the conditions with which we are concerned. (a) What is the proportion of H2 in the hydrogen in a liter of water which remains from the electrolysis of 100 liters? (Simplify the calculation by using the approximation N,/Nlo = 1.) (h) Suppose the reduction of the 100 liters had been carried out as follows: 50 liters electrolyzed until the residue was 1 liter then 50 more liters added and the electrolysis continued until the residue was 1 liter. What proportion of the hydrogen in the final residue is H a ? [Use the same approximation as in part ( 4 . I (c) Let 5 cc. of water in which 60% of the hydrogen is Hz be reduced to 1 cc. by electrolysis. Show by numerical calculation that over 99% of the hydrogen in the residue is Hz.

SOLUTIONS O F PROBLEMS 4 1 4 5

[J. CEEX.Eouc., 10, 706 (1933)l 44.

0.0539 AR(308.1 - 298.1). log ---- = 0.0180 4.576.308.1298.1

Clearing of fractions. PoN pan - p N pn = $on poN = p ( N n). 9 = PoN/(N n) = P& 42. '/.mua = Ee = Ve/300. o = (Ve/150 m)'/'. A = h/m(Ve/l50 m)'h an. A = (150)'/~~h.lO~/(rncV)'/* Angstr6rns.

+

+

-

+

45. Molar refraction of GHsI is (2.2.501)

+ (5.1.051) + 14.12 = 24.377