Applications of Chemical Kinetics in Environmental Systems - Elements of Environmental Engineering: Thermodynamics and Kinetics - page 276

Integrating the above equation with the initial condition (at t = 0) C A = C A0 ,we

obtain

τ d ) C A0 1

e − (k + ( 1 / τ d ))t

1

C A0 e − (k + ( 1 / τ d ))t , C A (t) = − + (6.21)

( 1 + k

where , the above equation reduces to

the steady-state solution obtained earlier. Depending on the functionality of C A0 , the
response of the CSTR will vary. Levenspiel (1999) considered several such input
functions and derived the corresponding response functions.

τ d is as defined earlier. Note that as t →∞

6.1.1.5 Relationship between Steady State and Equilibrium for a CSTR

As discussed in Chapter 2, equilibrium for closed systems is the state of minimum
free energy (or zero entropy production), which is a time-invariant state . For open
systems,theequivalenttime-invariantstateiscalledthe steadystate .Howcloselydoes
a steady state resemble the equilibrium state? This question is important in assessing
the applicability of steady-state models in environmental engineering to depict the
time-invariant behavior of systems.
As an example, consider the reversible reaction

A k f k b B. (6.22)

Consider the reaction in an open system where both species A and B are entering
and leaving the system at a feed rate Q 0 with feed concentrations C A0 and C B0 ,
respectively. The mass balance relations are

V d C A

d t =

Q 0 ( C A0 − C A ) − k f C A + k b C B ,

(6.23)

V d C B

d t = Q 0 (C B0 − C B ) − k f C A + k b C B .

Applying the steady-state approximation to both species, and solving the resulting
equations, we obtain

C A
C B

ss =

1
K eq +

1
k f τ d

, (6.24)

where K eq = k f /k b and τ d = V/Q 0 . Recalling that k f = 0.693 /t 1 / 2 for a first-order

reaction, we have

C A
C B

ss =

1
K eq +

t 1 / 2
0.693

. (6.25)

· τ d

Note that if 1 /K eq . Thus,

if the mean residence time is very large and the reaction is fast, the steady-state ratio
approaches that at equilibrium. For many natural environmental conditions, even
though

τ d is very large and t 1 / 2 is very small, then ( C A /C B ) ss →

τ d may be quite large, t 1 / 2 is also relatively large (since k f is small) and hence
steady state and equilibrium are not identical.