Page 179 - Modeling of Chemical Kinetics and Reactor Design

Reaction Rate Expression 149

At equilibrium with concentrations of A and B as C , C , respectively,
Ae
Be
then k C = k C . Since dC /dt = 0 at equilibrium
2 Be
1 Ae
A
k 1 = K = C Be
k C (3-168)
2 Ae
Substituting Equation 3-168 into Equation 3-167 gives

  C Be  
  +1   •C A − (C AO −C BO )
 = (k
−ln    C Ae  1 + k 2 )t (3-169)
  C Be •C + C  
  Ae AO BO   
 C


A plot of

 (  C C + )•C − (C −C )

1
−ln  Be Ae A AO ) Bo 
 ( ( C Be C Ae )•C AO + C BO 


versus time t gives a straight line with the slope (k + k ). From the
1
2
slope and the equilibrium constant K, k and k can be determined.
1
2
Figure 3-14 shows a plot of the reversible first order reaction.

Figure 3-14. First order reversible reaction.