Chapter 17 Homogeneous chemical reactions page 17-1
Chapter 17 Homogeneous chemical reactions
1. Accelerating antibiotic adsorption
The adsorption on carbon is apparently slow, probably due to slow (heterogeneous) reaction or
slow diffusion. Assume this, then
k
0 = 6.110-4 cm/s
The adsorption on ion exchanger are much faster. If it is still diffusion controlled, then from
Eq. 16.1-17,
Given D = 9.410-7 cm2/s, we have
By trial and error, = 112.9 s-1
If the carbon adsorption is not diffusion controlled, the answer will be much less.
2. H2S absorption by monoethanol amine solution
The solution of this type of mass transfer is discussed in detail in D. R. Olander, AIChE J., 6,
(a) Reversible case:
For a reversible reaction H2S + RNH2
RNH3
++ HS–
Let 1 = H2S, 2 = RNH2, 3 = RNH3
+, 4 = HS We want to derive an equation similar to Eq.
–
where = 1
D4
2
+ 4
D2c2l
D4
Assuming all diffusion coefficients are equal, and c1l = 0, then becomes
2
and Eq. (1) reduces to
k = 1 + K2 + 4Kc2l/c10 – K
2 (2)
Given K = 275, c2l = 0.1 M, H = 545 atm,
Thus
On the other hand, if c2l is in great excess, such that c2l >> Kc10, then Eq. (2) reduces to
Chapter 17 Homogeneous chemical reactions page 17-2
However this is not the case in this problem.
(b) irreversible case
If the reaction is instantaneous and irreversible, then K , and Eq. (2) reduces to
On the other hand, if the reaction is fast but not instantaneous, then from Eq. 16.3-14,
However, the rate constant is not given in this problem.
3. Hydrocracking of heavy oil
For a 1st order heterogeneous reaction, 2 = ln2/t1/2 = 8.453 s-1,
Assuming 1 2, then the Thiele modulus
By iteration, calculating
4. Reaction rate of CO and H2
5. Staining cells
The amount of dye uptake M is
M = Aj1t
we assume that agar is a semi-infinite slab, thus from Eq. 16.1-21,
Assume the equilibrium constant is
6. Mass transfer in rice
t
Thus the moisture in the rice M varies as
Chapter 17 Homogeneous chemical reactions page 17-3
0.001
0.0024 0.0026 0.0028 0.003 0.0032
1/T
0
0 5 10 15 20
t
1/2
The left figure shows the dependency of M on t, from the slope, we obtained D, which is
the apparent diffusion coefficient. The right figure shows the linear dependence of ln(D) on
1/T. The slope of this curve is E/R = 6560. Thus the activation energy of the diffusion of
moisture in the rice is about 13 kcal/mol. This is higher than the activation energy for
diffusion-only process, which is only a few kcal/mol.
7. Mass transfer with 0° reaction across a thin film
(b) Differential equation:
Boundary conditions:
(c) Mass transfer coefficient
integration yields
8. H2S scrubbing
For fast irreversible reaction, the mass transfer coefficient is from Eq. 16.1-17:
log v
log k
(D)
1/2
(v)
2/3
log
logk
k
0
(D)
1/2
When log(k) is plotted vs log(1), lim
0 k = k, while
0lim
k = D1. Because of stirring, when
log(k) is plotted vs log(v), lim
v0 k = D1, while lim
v k v.
2/3
9. Ammonia stripping
Let 1 = NH3, 2 = NH4
+, 3 = H+
(a) Continuity equation:
By adding Eq. (1) and (2), and integrating twice, we have
l [(c + c)| – (c + c)| ] (3)
(b) Boundary condition at z = l,
(c) Boundary condition at z = 0,
The 2nd condition is replaced with
Thus the flux becomes
l [(c + c c /K) – c] = k[c(1 + c/K) – c] (8)
(d) Overall mass transfer coefficient:
By referring to Eqs. 8.5-8 et seq.,
10. CO2 absorption in carbonates
(a) Mass balance and boundary conditions
Assume pH is high so that no CO2 initially presents, and catalysis gives fast reaction.
11. Carbon’s oxidation
The 1st reaction is heterogeneous, and the 2nd reaction is homogeneous. Thus
dz2 – c
z
= 0
z
=
l
subject to
dz = c
integration gives
c1 = Aexp( 3/D z) + Bexp(- 3/D z)
evaluating the boundary conditions,
Thus,
12. Concentrated film theory with instantaneous reaction
Assume concentrations are mass per volume, and constant density.
When z < zc, c2 = 0, and c1 + c3 = c (= )
At steady state, species 3 is stagnant, thus from Eq. 3.3-11,
zc ln(1 – w) (1)
When z > zc, cv0 = n2 + n3 = 0, and from Eq. 16.2-7,
l – zc (2)
c1ic2l
z
= 0
z
c
z
= l
and from Eq. 16.2-8,
combining these equations to eliminate zc, we find
13. Wine flavor and oxygen
If Pasteur were correct, diffusion in a cylindrical bottle would give
Because the diffusion coefficient of O2 is 2.110-5 cm2/s, the distance that oxygen penetrates
in 1 hr is about
z = 4Dt = 0.55 cm
This is not deep enough: in the absence of free convection, no significant reaction beyond 0.5
cm. This is why decanting wine (or shaking it) makes more sense. (See Cussler and
Moggridge, Chemical Product Design, Cambridge, 2001)