Page 62 - The engineering of chemical reactions

46 Reaction Rates, the Batch Reactor, and the Real World

- CA$ = -kc; = kc&(1 - X)2
so that
dX
- = kCA,(l - X)’
dt
This equation can be integrated to give X(t) = kCA,t/(l + kc&) and substituted back
into CA = CB = c~,(l - X) and CC = CA,X to find CA(~), C,(t), and C,(t). However,
for most simple reaction systems it is just as convenient (and there is less chance of a
mathematical error) to work directly in concentration units.

Reversible reactions

The next more complex rate expression is the reversible reaction
A 2 B, r = kfCA - k&B
IfCA=CAoandCs=Oatt=O,wehave
CA t
dC.4
z- dt = -t
s kfCA - kb(CAo - CA> s
CA0 0

= &MWA - kb(Ch - CA)] - ln[kfCAd
Rearranging and solving for CA(~), we obtain

We can proceed to examples of increasing complexity. For example,
A2 2B, Y = kfCA - k&i
can be written as a function of CA alone from the relation Ce = z(C.4, - CA). Substitution
gives the integral

CA
t=- dCA
s k&A - 4kb(CAo - CA)’
CA0
which must be solved by factoring the denominator of the integrand and solving by partial
fractions.

“n Example 24 Solve the preceding example by partial fractions to find CA (t).
“L’ n”
z; “; You have seen partial fraction solutions to integral equations in math courses. We
,^,a >I i 4 :
‘da: will find many situations where rate equations can only be integrated by partial
‘ij8
z&J
I/(
,& fractions; so it is worth reviewing this procedure. The solutions are simple but they
i@jl require some algebraic manipulation. The integrand of the previous integral can
j’.$i I* be written as