Page 120 - Introduction to chemical reaction engineering and kinetics

102 Chapter 5: Complex Systems

(a) Describe how experiments may be carried out in a constant-volume BR to mea-
sure kAl and kA2, and hence confirm the rate laws indicated (the use of a CSTR is
considered in problem 5-5);
(b) If kAl = 0.001 s-l and kA2 = 0.002 s-l, calculate (i) fA, (ii) the product distribution
(c,, en, etc.), (iii) the yields of B and D, and (iv) the overall fractional yields of B
and D, for reaction carried out for 10 min in a constant-volume BR, with only A
present initially at a concentration CA, = 4 mol L-’ .
(c) Using the data in (b), plot CA, cn and co versus t.

SOLUTION
(a) Since there are two independent reactions, we use two independent material balances
to enable the two rate constants to be determined. We may choose A and B for this pur-
pose.
A material balance for A results in

-dc,/dt = kAlCA + kA$A (5.4-5)

This integrates to

ln CA = ln CA0 - (kAl + kA& (5.4-6)
In other words, if we follow reaction with respect to A, we can obtain the sum of the rate
constants, but not their individual values.
If, in addition, we follow reaction with respect to B, then, from a material balance for
B,

dc,/dt = kAlCA (5.4-7)

From equations 5.4-5 and -7,

-dCA/dC, = ('h + k&&l
which integrates to

cA = cAo - (1 + k,,&d(C, - '& (5.4-8)
From the slopes of the linear relations in equations 5.4-6 and -8, kAl and kA2 can be de-
termined, and the linearity would confirm the forms of the rate laws postulated.
(b) (i) From equation 5.4-6,

CA = 4eXp[-(0.001 + 0.002)10(60)] = 0.661 mol L-’

j-A = (4 - 0.661)/4 = 0.835

(ii) CA is given in (i).
From equation 5.4-8,
cB = cc = (4 - 0.661)/(1 + 0.002/0.001) = 1.113 mol L-’

From an overall material balance,
CD = CB = CA0 - CA - c, = 2.226 Ill01 L-’