MATHEMATICAL PROBLEM PAGE Directed by PAUL C. CROSS Gates Chemical Laboratory, California Institute of Technology. Pasadena, California
OLUTIONS of the following problems will be gwen m the February issue. 51. The unimolecular rate constant is given by -xz k = -1-.I an- X- I -- 1 In a. . . etc. tz-tl a-xz t3-& a-x3 where a is the initial concentration and x,. -. x -.~xr, are the amounts which have reacted after the times tl, tz, ts respectively. Find the value of a in terms of the x's when the measurements are taken a t equal time intervals (tz - tl = ta - h). 52. From the results of problem 51 show that the rate constant can be determined from measurements of the change in concentration over two equal successive time intervals without data on 1 the total concentration. (Derive k = - In
54. For a himolecular reaction
dxldt
=
k(a
- x.) (.b - x )
where a and b are the initial molar concentrations of the two reacting substances and x is the amount of each transformed a t the time t. Integrate by oartial fractions between the limits x = 0 to x = x and t = 0 to t = t to obtain the relation ~
~
k =
-in b(a - x ) .
t ( a - b)
a(b-x)
55. Given a reversible reaction A+B=C+D, let kl be the rate constant for the forward reaction and kz be that for the reverse reaction. The equilibrium constant K is defined as kl/kz. The temperature variation of the reaction rate may be represented by
T
where T = t* - tl = ts - tr.) x3 - xz 53. In a unimolecular decomposition, if measurements of the concentrations, C., are taken a t times, ti, for a series of equal time intervals, ti+, - ti = T , a set of values for the rate constant may be obtained,
k
= se - E / R T
where T is the absolute temperature, R the molar gas constant, E a constant known as the energy of activation, and s is relatea to the number of collisions between the reacting molecules. The variation of the equilibrium constant with temperature is given by the relation
Show that the arithmetical average of the ( n - 1) values of k obtained from n such measurements is equal to the value of ko,,: i: e., show that
Show that A H should be equal to El (s,/s, may be taken as independent of T.)
- Ez.
SOLUTIONS 03 PROBLEMS 46-50
[J. Crr~aa.E ~ u c .10, , 755 (Dec.. 1933)l From the definition of b XI xs = b(xtO xz0). Substituting in the right side of Eq. ( 1 ) xd(x1 n) = [x#/b(x~* xzO)l(rdxiO)". (2) 49. Multiplying numerator and denominator by ( X I x,) (.%lo x 8 and substituting, x , h Q = bN,/N>O. Substituting N's in Eq. (2) of the answer to problem 48 N* = ( N n @ / b(bN,/NlO)". ) b = ( N s o / N 2 )( N I / N I O ) = ~ = . 61-a = ( N z 0 / N d ( N L I N I O ) ~ .
46. dz,/dl = - k , x ~ . dxx/xl = -krdl. Similarly, d r 9 / x 2 = -kzdt. ( d m / x J / ( d x l / x r )= k d k ~= a. u is a constant since k1 and k2 are constants. 47. dx2/xs = a d x z l x ~ . Integrating between the limits m0 to and xlQ to X L , respectively.
+
+
+
+
In(xdxtO) = u l n ( x r / r ~ ~ ) .
50.
57
~
+
+
( a ) 61-0: = (Na0/Np)( N d N l o ) " .
-
Substituting numerical values and making the approximation (c) (b)
(0.02)0.8= 1/5000 NI Nz = 1/(5000.0.0436) = 1/218 = 0.0046. After the addition of 50 liters of ordinary water Nz = [0.0046 (50.0.0002)1/51 = 0.000286 = Nno for the second step in the process. (0.0196)0.8= O.O00286/Nt
+
Nr = 0.00028(i/0.04:1 = 0 GOGG5 1, 150. Note that thc d m c t clrctroly~isac in (3) is t h e more rfticient method of concentrarinq H' (0.2)0.8= (0.6/Nn) (NJO.4)"' Ns = 0.6.N,Q.1/(0.276.0.833) = 2.61 NIO.~. Substituting NI = 0.99 and N, = l-Nz = 0.01, 0.99 < 2.61,0.398(= 1.04). Thus NI = 0.99 nearly satisfies the equation, and since the right-hand term becomes smaller as N2 increases. i t is obvious that the value of Na which satisfies the equation is greater than 0.99. (Actually about 0.992.)
