Chapter 6
Elements of Heterogeneous Catalysis
Solutions to Problems
Solution 6.1:
A. The adsorption and desorption isotherms for N2 on bayerite alumina are plotted in
Figure P6.1.1. The data fall on the same curve for relative pressures below 0.4 but
hysteresis is observed at relative pressures above this value. The presence of hysteresis
indicates that there are regions of the solid pore structure that are linked to the ambient
gas by pores that are smaller in diameter than the regions themselves. One interpretation
of this hysteresis is the so-called “ink bottle” effect.
B. The plot of the data in the form of the Langmuir isotherm (equation 6.28 of Hill and
Root) is shown in Figure P6.1.2. The data do not conform to a straight line. Hence the
Langmuir equation does not provide an adequate description (model) of the data. This
result is expected because physical adsorption of nitrogen on alumina involves multilayer adsorption.
C. The plot of the data in the form of the BET equation (Figure P6.1.3) is quite linear
except at very low pressures where the data are least accurate. Hence the BET equation
provides a very good model of physical adsorption of nitrogen on the alumina surface.
Amoount adsorbed (cm^3/g)
250
adsorption
200
desorption
150
100
50
0
0.0
0.2
0.4
0.6
0.8
1.0
Relative pressure P/P0
Figure P6.1.1 Isotherms for adsorption and desorption of N2 on alumina.
Chapter 6 Solutions
6-1
y = -0.0254x2 + 0.0219x + 0.0002
5.0E-03
P/(P0*V)
4.0E-03
3.0E-03
2.0E-03
1.0E-03
0.0E+00
0.0
0.1
0.2
0.3
0.4
0.5
Relative pressure P/P0
Figure P6.1.2 Plot of adsorption data in the form of the Langmuir isotherm to determine
specific surface area.
y = 2.010E-02x + 3.309E-04
9.0E-03
8.0E-03
lhs of BET Eqn
7.0E-03
6.0E-03
5.0E-03
4.0E-03
3.0E-03
2.0E-03
1.0E-03
0.0E+00
0.0
0.1
0.2
0.3
0.4
0.5
Relative Pressure P/P0
Figure P6.1.3 Use of BET plot to determine specific surface area.
From the slope of the best fit straight line in Figure P6.1.3
C "1
= 2.010 #10"2
CVm
and from the intercept
Chapter 6 Solutions
(1)
6-2
1
= 3.309 "10#4
(2)
CVm
From the sum of equations 1 and 2
1
= 2.010 "10#2
Vm
Thus the volume corresponding to monolayer coverage is
Vm = 49.75 cm3 (STP) g
The corresponding BET surface area requires conversion of Vm to an equivalent number
of nitrogen molecules and then multiplication by the area covered per molecule (0.162
nm2). Thus
49.75
2
Sg =
" 6.023 "10 23 "16.2 " 10#20 = 217 m g
22, 400
where we have recognized that 1 mole (6.023 !1023 molecules) of nitrogen occupies
22,400 cm3 at standard temperature and pressure (STP).
Solution 6.2:
For the Langmuir adsorption isotherm
K H PH
" = v vm =
(1)
1+ K H PH
This equation can be manipulated in a variety of ways to obtain forms that might be
employed to test the ability of this equation to fit the adsorption data. Two possible
forms are
!
PH
1
P
=
+ H
(2)
v vmK H vm
(see Figures P6.2.1 and P6.2.1A) and
v
= K H vm " K H v
(3)
PH
!
(see Figures P6.2.2 and P6.2.2A).
y = 1.7403x + 0.507
3
!
P/v (cm Hg/cc)
2.5
2
1.5
1
0.5
0
0
0.5
1
Pressure (cm Hg)
1.5
Figure P6.2.1 Plot of adsorption data for H2 on Cu powder in the form of a Langmuir
isotherm (equation 2).
Chapter 6 Solutions
6-3
y = 1.65x + 0.5802
3
P/v (cm Hg/cc)
2.5
2
1.5
1
0.5
0
0
0.5
1
Pressure (cm Hg)
1.5
Figure P6.2.1A A plot of adsorption data for H2 on Cu powder in the form of a
Langmuir isotherm (equation 2). (Two data points in Figure P6.2.1 were omitted from the
regression analysis).
The data for chemisorption of hydrogen were worked up using a spreadsheet and plotted
in the form of equation 2 in Figures P6.2.1, and P6.2.1A. Inspection of these plots
reveals that in both cases departures from linearity are observed at low pressures. The
points that deviate most significantly from linearity are those that are most susceptible to
error. Equation 2 is more appropriate for use than equation 3, because uncertainties in
the data obtained at the lowest pressures are amplified when the reciprocal of the pressure
is involved.
From the slope of the best straight line fit of the data in Figure P6.2.1A
1
1
vm =
=
= 0.61cm3( STP ) gr
slope 1.65
This value pertains to the data obtained at pressures from 0.19 to 1.195 cm Hg. The
regression analysis omits the two points obtained at the lowest pressures. If these points
are included in the regression analysis, vm = 0.57 cm3( STP ) gr .
If regression of the data is based on equation 3, one obtains the plots shown in Figures
P6.2.2 and P6.2.2A. The fit of the regression line in Figure P6.2.2 is quite poor, because
of the problems with the low pressure points noted earlier. Based on the regression line
that eliminates 4 low pressure points (Figure P6.2.2A), the adsorption equilibrium
constant K H is 2.68 (cm Hg)-1 and K H v m is 1.65cm3/(cm Hg). Thus v m = 1.65 2.68 or
0.62 cm3 (STP). Hence the values of v m based on the data that fall on straight lines are
within roughly 10% of one another and thus can be viewed as internally consistent.
!
!
!
!
Chapter 6 Solutions
6-4
y = -3.921x + 2.1429
1.8
1.6
v/P (cc/cm Hg)
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
Volume (cc)
0.4
0.5
Figure P6.2.2 Alternative test of the linearity of the fit of a Langmuir adsorption
isotherm (equation 3) for H2 on Cu powder.
v/P (cc/cm Hg)
y = -2.6834x + 1.6544
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
Volume (cc)
Figure P6.2.2A Alternative test of the linearity of the fit of a Langmuir adsorption
isotherm (equation 3) for H2 on Cu powder. (Four low pressure points were omitted from
the regression analysis.)
We can now proceed to convert the v m values to specific surface areas.
v " ( Avogadro' s number) " ( area molecule)
Sg = m
volume in cc of a mole
! "10 23 )(14.2 "10#20 ) 22,400 = 2.33m 2 g
= 0.61(6.023
This low value is characteristic of finely divided nonporous materials.
!
Now we shall consider the data for nitrogen on silica gel. These data are plotted in the
form !
of equation 2 in Figure P6.2.3. In this case the data are not consistent with a straight
line. Only the three data points recorded at the lowest pressures are co-linear. These
points, however, are those that are most susceptible to error. Moreover, a plot of the
nitrogen data in the form of equation 3 is meaningless.
Chapter 6 Solutions
6-5
P/v (cm Hg/cc)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
10
20
30
40
50
Pressure (cm Hg)
Figure P6.2.3 Test for linearity of a fit of the Langmuir isotherm to data for N2 on silica
gel.
y = 0.0755x + 0.0002
0.05
0.05
0.04
0.04
P /[v(P 0-P )]
!
The most appropriate form of the BET equation for plotting the data is
$ # "1'$ P '
P
1
(4)
=
+&
)& )
v( P0 " P ) #v m % #v m (% P0 (
where P0 is the saturation pressure at the temperature at which the measurements are
made (76.0 cm Hg in the present case). The data for nitrogen are plotted in the form of
equation 4 in Figure P6.2.4. Note that the BET plot of the data is essentially linear, in
! to the highly nonlinear situation for the Langmuir plot.
contrast
0.03
0.03
0.02
0.02
0.01
0.01
0.00
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Relative Pressure (P/P0)
Figure P6.2.4
BET plot of the data for adsorption of N2.
Chapter 6 Solutions
6-6
The sum of the slope and intercept of the best fit straight line in Figure P6.2.4 is equal to
the reciprocal of v m (see equation 4). Thus
1
vm =
= 13.2cm3( STP )
0.0755 + 0.0002
This value
may
be
converted
to a specific surface area using a procedure analogous to
!
that employed for adsorption of hydrogen
" 13.2 %
23
)20
2
Sg = $
'6.023 (10 (16.2 (10 ) = 57.5 m g
# 22,400 &
This specific surface area is typical of a porous catalyst.
! 6.3:
Solution
We are given data for the amount of hydrogen adsorbed as a function of time and
temperature (see Figure P6.3.1 of Hill and Root). We are asked to interpret these data.
The experimental facts are as follows:
1. From 0 to A there is a gradual increase in the amount adsorbed as time elapses. The
amount adsorbed appears to be approaching an asymptotic value.
2. For the curve segment from A to B the temperature has been increased. Initially there
is a very rapid rise in the amount adsorbed, followed by a slower rate of increase in the
amount adsorbed.
3. For the curve segment from B to C the temperature has again been increased. Initially
there is a very rapid drop in the amount adsorbed followed first by a rapid increase in the
amount adsorbed and subsequently by a slower increase in the amount adsorbed.
4. For the curve segment from C to D the temperature is increased even further. Initially
there is a very rapid decrease in the amount adsorbed, but subsequently there is a slow
increase in the amount of hydrogen adsorbed.
5. For the curve segment to the right of D, behavior similar to that of phase 4 was
observed, i.e., a rapid decrease in the amount adsorbed followed by a slow increase as
time evolved.
The above facts may be interpreted as consequences of the occurrence of at least two
different types of adsorption phenomena that take place simultaneously. One type of
adsorption is very rapid and the other occurs at a much slower rate, especially at the
lowest temperatures investigated. The second type of behavior is indicative of some type
of activated chemisorption that takes place in time frames measured in hours. If one waits
for a sufficiently long time after an increase in temperature, additional adsorption occurs
subsequent to an initial very rapid response to the temperature change. This behavior
clearly indicates the presence of some sort of rate-limited adsorption phenomenon. We
can attribute this behavior to an activated chemisorption process. The very rapid initial
decreases in the amount adsorbed following a temperature increase are indicative of an
adsorption process for which equilibrium is established very rapidly. Because all of the
temperatures investigated are well in excess of the critical temperature of hydrogen, we
cannot rationally attribute the decreases to evolution of hydrogen that had been physically
adsorbed on the surface of the oxide. Instead we attribute this observation to the
Chapter 6 Solutions
6-7
occurrence of some form of non-activated (or weakly activated) chemisorption process.
The behavior of the system at 0ºC supports the hypothesis that there is a small but
measurable activation energy associated with this chemisorption process.
Solution 6.4:
Figure P6.4.1 contains plots of the specific surface areas obtained at 90 ºK versus the
porosity of the corresponding sample. The values obtained for plugs 4 and 6 have been
averaged because they correspond to the same porosity.
Specific surface area [(m^2)/g]
BET
Hüttig
1100
1000
900
800
700
0.35
0.45
0.55
0.65
Porosity
0.75
!!
Figure P6.4.1 Specific surface areas of porous pellets formed from carbon powder.
The values obtained at 79 ºK have been omitted from the plots because all of the other
data were obtained at 90 ºK. The results based on an analysis involving a linearized form
of the BET equation are represented by triangles while the results based on the Hüttig
equation are represented by diamonds. Both plots exhibit minima in the vicinity of a
porosity of 0.45. The values of the specific surface area determined using the Hüttig
equation are roughly 3-6 % greater than those based on the traditional BET approach.
These differences are not surprising in light of the analyses on which they are based,
including the different values for the area of surface covered per molecule of adsorbate.
The data for specific surface areas obtained at different temperatures but the same
porosity differ by less than 1% from one another and the difference is not statistically
significant given the uncertainties typically associated with these parameters. However,
the fact that a minimum in Sg exists is clearly evident from inspection of Figure P6.4.1
because major changes are observable. The presence of a minimum reflects the net result
Chapter 6 Solutions
6-8
of the interactions between two factors, one of which acts to increase Sg, and the other of
which acts to decrease Sg.
There are two contributors to the void volume per gram of catalyst pellet – the micropores within
the particles of carbon powder and the void regions between the particles that make up the
pelletized catalysts. At high porosities the large majority of the void volume consists of the void
regions between particles. As compression of the powders is initiated and increased, some of the
void regions shrink or are lost when the particles are forced together. There is some concomitant
loss of surface area that we can attribute to some of the micropores in a given particle being
covered by other particles thereby physically blocking access of gas phase adsorbate molecules to
the volume and surface area contained within the micropores. In addition, there may be a loss of
surface area within the particles by compaction (mechanical deformation) of the individual
particles themselves to close off the micropores contained therein, thereby creating completely
enclosed pores. The gas within such pores has no means of escaping so the surface area
associated therewith cannot interact with the gas external to the pellets. Hence there is a decline
in the area on which gas phase molecules can adsorb and a concomitant decline in Sg. The areas
associated with the completely enclosed pores will not be measured in a conventional adsorption
isotherm experiment. Mechanical deformation continues as the pressure applied to compact the
particles increases and there will be further loss of accessible surface area within the structure of
the micropores. Eventually in the regime of low porosities (say less than 0.40) the forces
necessary to compact the particles of powder become sufficiently great that the mechanical
strength of the particles is not sufficient to retain their physical integrity. When the particles begin
to break apart, there will be a relatively small increase in the surface area available for adsorption
because of either the surface area created by fracture of the particles or opening up of porous
regions that had been inaccessible because of physical blockage or enclosed pore effects.
Solution 6.5:
We are asked to interpret data concerning the reaction of dimethylamine over a
montmorillonite catalyst.
One series of experiments involved measurement of reaction rates for different catalyst
sizes, maintaining all other variables constant. The measured rates were constant for
catalyst particle sizes from - 100, + 120 to - 30, + 40 mesh, a fourfold variation in
average particle diameter. These experiments were carried out at mass velocities
corresponding to the asymptotic portions of the curves in Figure P6.5.1of Hill and Root.
Several other experiments were also carried out at mass velocities corresponding to the
asymptotic portions of the curves shown in Figure P6.5.1. Initial rates were measured as a
function of the dimethylamine pressure for the four temperatures indicated in Figure
P6.5.2 of Hill and Root. These data may be fit by an expression of the form r0 = kD PD0.523 ,
where PD is the partial pressure of dimethylamine.
Chapter 6 Solutions
6-9
The stoichiometry of the transmethylation of dimethylamine is
2 (CH3)2NH (CH3)3N + CH3NH2
There are three sets of data that need to be considered:
A. The fact that the initial rate increases with increasing mass velocity (G) at all three
temperatures (Figure P6.5.1) indicates that external mass transfer limitations on the rate are
present. Even at the lowest operating temperature (slowest rate for a fixed initial
composition) these limitations are observed below a mass velocity of roughly 100 g/(h-cm2).
The asymptotes observed at 278.8 and 292.0 ºC correspond to conditions at which mass
transfer processes between the bulk fluid and the external surface of the catalyst are not ratecontrolling.
B. The fact that the measured rates were independent of particle size indicates that the
effectiveness factors of the catalyst were unity for the conditions studied. That is, intraparticle mass transfer/diffusion phenomena were not rate-limiting.
C. The fact that the rate data could be modeled using a rate law of the form r = k D PD0.523 can
be interpreted in terms of a Hougen-Watson model that involves dissociation of the
dimethylamine on adsorption (see Figure P6.5.2 of Hill and Root). In this model the fraction
of the sites covered by dissociated species can be written as
!D =
K D PD
1+ K D PD
This fraction reduces to K DPD if 1>> K DPD . Under these conditions the coverage of the
surface is sparse. Mechanisms whose rate-controlling step is first-order in the fraction of the
surface covered by the dimethylamine fragments can thus give rise to a rate law that is
essentially half-order in the partial pressure of dimethylamine.
r = k! D0.5 = k K DPD
Several such mechanisms are conceivable.
The activation energy for this reaction can be determined from a semi-log plot of the initial
rate at a fixed dimethylamine pressure (0.7 atm in the present case) versus the reciprocal of
the absolute temperature (see Figure P6.5.3 of this solutions manual).
Chapter 6 Solutions
6 - 10
y = -13357x + 25.425
2
ln(initial rate)
1.5
1
0.5
0
1.7E-03
-0.5
1.8E-03
1.9E-03
2.0E-03
Reciprocal absolute temperature (1/K)
Figure 6.5P.3 Arrhenius plot for determination of activation energy.
The activation energy corresponding to the slope of this plot is 27 kcal/mole. One could also
determine individual rate constants at each temperature for which initial rate data are
available. Moreover, one could also test the rate law reported by these researchers by making
appropriate plots of their data.
Solution 6.6:
There are two routes by which one might proceed to ascertain the order of the reaction
with respect to ammonia. One approach would be to plot the logarithm of the initial rate
versus the logarithm of the initial pressure of ammonia. In this case the order of the
reaction is the slope of the best fit straight line through the data (see Figure P6.6.1). The
slope of the trend line (0.995) indicates that the order of the reaction is unity.
-1.5
1.2
1.7
2.2
-1.7
log (rate)
y = 0.9947x - 3.7739
-1.9
-2.1
-2.3
-2.5
log (Ammonia pressure) for pressure in mm Hg
Figure P6.6.1 Determination of reaction order from initial rate data.
The second approach is to plot the rate versus an assumed order of the reaction in
ammonia, and then check for the linearity of the plot. Figure P6.6.2 is based on the
assumption of first-order kinetics. This approach has the advantage of providing a check
Chapter 6 Solutions
6 - 11
Initial rate (arbitrary units)
on the analysis via the requirement that the best fit straight line should pass through the
origin. In essence one obtains an additional data point based on the expectation that the
rate should be zero if no ammonia is present. The linearity of the trend line in Figure
P6.6.2 and the fact that it essentially passes through the origin confirm the validity of the
assumption of a first-order rate law.
y = 1.67E-04x - 1.50E-04
0.025
0.02
0.015
0.01
0.005
0
0
50
100
150
Initial ammonia pressure (mm Hg)
Figure P6.6.2 Test for the linearity of the fit of a proposed first-order rate law.
Solution 6.7:
The stoichiometry of the reaction of interest is
2NH3 ! N2 + 3H2
We are asked to ascertain whether or not a proposed rate law is consistent with a set of
experimental rate data. We shall accomplish this task by plotting the data in a form that
should yield a straight line if the assumption of this rate law is valid.
Let f represent the mole fraction of the ammonia initially present that has reacted by time
t, and let n0 represent the moles of ammonia present initially. We can then construct the
following mole table:
Species
Moles at time zero
Moles at time t
NH3
n0
n0(1-f)
H2
0
3 n0f/2
N2
0
n0f/2
Total moles
n0
n0(1+f)
Total pressure
P0
P0(1+f)
Because P = P0(1+f) then
P - P0
P0
where P is the pressure measured at time t.
f=
Chapter 6 Solutions
(1)
6 - 12
The entries in the mole table and the definitions of partial pressures indicate that
PNH 3 y NH 3 P n NH 3 nTotal n0 (1 - f ) 2(1 - f)
(2)
=
=
=
=
PH 2
y H 2 P nH 2 / nTotal 3n0 f / 2
3f
From the basic definition of the reaction rate
1 d! - 1 dnNH 3 n 0 df
=
=
V dt 2V dt
2V dt
(3)
Combination of this definition of the rate with the proposed rate law gives
n 0 df
2(1 - f )
=k
2V dt
3f
Separation of variables and integration gives
f
4Vkt
fdf
=!
= - f - ln(1 - f )
3n 0 0 (1 - f )
(4)
Combination of equations 1 and 4 and the ideal gas law gives
4 RTkt
P - P0
P - P0
(5)
= -[
] - ln(1 - [
]) = [1 - (P P0 )] - ln[2 - (P P0 )]
3P0
P0
P0
Consequently if the proposed rate law is valid, a plot of the data in the form of the right
side of equation 5 versus time should be linear. The data were worked up using a
spreadsheet and plotted in the indicated form in Figure P6.7.1.
0.35
right side of
equation 5
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
200
400
Time (ksec)
600
800
Figure P6.7.1 Test of the linearity of the fit of a proposed rate law.
Visual inspection of the plot indicates marked curvature. Hence the assumed rate law
does not provide a good representation of the data.
One Hougen-Watson mechanism that would lead to the indicated rate expression
involves a situation in which (a) the rate is proportional to the fraction of the surface
covered by ammonia; and (b) the large majority of the surface sites are occupied by
molecular hydrogen species. In its most complete form where one allows all species to be
adsorbed (with hydrogen in a molecular form), the rate expression is
Chapter 6 Solutions
6 - 13
k ' K NH 3 PNH 3
'
r = k ! NH 3 =
(6)
1+ K NH 3 PNH 3 + K H 2 PH 2 + K N 2 PN 2
If neither N2 nor NH3 is adsorbed to a significant degree and if there are very few vacant
sites, equation 6 reduces to
k ' K NH 3 PNH 3
PNH 3
'
(7)
r = k ! NH 3 =
=k
K H 2 PH 2
PH 2
Solution 6.8:
A. In terms of the proposed mechanism, equilibrium for the first reaction requires that
k1! V CA PH 2 = k 2 ! AH 2
or
! AH 2 = k1 CA PH2 / k2 ! V
(1)
(
)
Because ! Z " 0
1 = !V + ! AH 2
Combination of equations 1 and 2 gives
(k1 / k 2 )CA PH2
! AH 2 =
1 + (k1 / k2 )CA PH2
The observed rate will equal to that of the rate limiting step.
( k1 k3 / k2 )CA PH2
r = k3 ! AH 2 =
1 + (k1 / k2 )CA PH 2
(2)
(3)
B. Figure P6.8.1 is a plot of the total rate of product formation versus mass of suspended
catalyst for reaction at 130 °C, an initial benzoic acid concentration of 0.4 M and a
hydrogen pressure of 50 atm.
O
6
O
O
O
5
O
4
Rate
(mole/hr)
O
3
O
2
O
1
0
1
2
mass of catalyst (g)
Figure P6.8.1 Dependence of reaction rate on the mass of catalyst present.
Inspection of Figure P6.8.1 indicates that up to a catalyst mass about 1.1 grams, the total
rate of production of species B is proportional to the amount of catalyst present. Beyond
1.1 grams the rate becomes independent of the quantity of catalyst present. This region
must be one in which neither surface reaction phenomena nor mass transfer of either
reactant to the catalyst surface is rate controlling. The most plausible explanation for the
Chapter 6 Solutions
6 - 14
presence of the asymptote is that in this region the rate becomes controlled by the rate of
dissolution of gaseous hydrogen in the butanol solvent.
rate in mmoles/(g-h)
Figure P6.8.2 is a plot of the reaction rate per unit mass of catalyst versus the product
CAPH2. The data correspond to reaction at 130°C at a catalyst loading of 0.5 gr.
16
14
12
10
8
6
4
2
0
C=0.4
0
10
20
30
C(PH2)
(M-atm)
PH2 =50
40
50
Figure P6.8.2 Plot of rate versus presumed dependence of the rate law on hydrogen
pressure and acid concentration.
To determine whether or not the proposed mechanism and the kinetic data are consistent,
we can note that the general shape of the curve in Figure P6.8.2 is vaguely similar to that
expected for the mathematical form of equation 3. A better test for consistency, however,
involves the rearrangement of equation 3 to a form against which the data can be tested
for a linear fit. Rearrangement of equation 3 gives
1
k2
1
(4)
=
+
r k1k3 PH2 CA k 3
Both sets of data are worked up together using a spread sheet and plotted in the form of
equation 4 in Figure P6.8.3.
Chapter 6 Solutions
6 - 15
1/rate [(g-h)/mmole]
y = 1.259x + 0.0547
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0.0
0.1
0.2
0.3
1/(CPH2) (M-atm)^(-1)
Figure P6.8.3 Test for linearity of the fit of a proposed rate law.
While there is considerable scatter of the data, it is evident that the data do approximate a
straight line. From the intercept it is evident that
k3 = (1/0.0547)=18.3 hr gr catalyst/mmole. From the slope of the plot
k2
= 1.26
k1 k3
Thus
k2
= 1.26(18.3) = 23.1 atm M hr gr catalyst/mmole
k1
or
k1 / k 2 = 0.043 mmole/atm M hr g catalyst
Solution 6.9:
For the single site mechanism that has the surface reaction as the rate determining step
(1)
r = k f C - k r BPP
Because species C and species B are in adsorptive equilibrium and because it is assumed
that propylene is not adsorbed.
1
(2)
!V =
1+ K BPB + K C PC
K C PC
(3)
!C =
1+ K BPB + K C PC
K BPB
(4)
!B =
1+ K BPB + K C PC
Combination of equations 1-4 leads to the predicted rate expression
k K P -k K P P
(5)
r= f C C r B B P
1+ K BPB + K C PC
Chapter 6 Solutions
6 - 16
The initial rate will correspond to the situation in which both PB and PP are zero. Thus
equation 5 gives
k K P
(6)
r0 = f C C
1+ K C PC
This rate expression can be rearranged to give a form that can be used in the analysis of
the data shown in Figure P6.9.1 of Hill and Root..
PC
1
P
(7)
=
+ C
r0 k f K C k f
The straight line in Figure P6.9.1 of Hill and Root depicts the same mathematical form as
equation 7. Hence this mechanism is consistent with the experimental rate data.
For the dual-site mechanism for which the rate determining step is reaction on the surface
of the catalyst
(8)
r = k f !C!V - k r!B!P
Because all species are assumed to be in adsorptive equilibrium
1
(9)
!V =
1+ K BPB + K C PC + K P PP
K C PC
(10)
!C = K C PC!V =
1+ K BPB + K C PC + K P PP
K BPB
(11)
! B = K BPB!V =
1+ K BPB + K C PC + K P PP
K P PP
(12)
! P = K P PP!V =
1+ K BPB + K C PC + K P PP
Combination of equations 8 and 10-12 gives the corresponding rate law.
k f K C PC - k r K B PB K P PP
(13)
r=
[1+ K BPB + K C PC + K P PP ]2
The initial rate then becomes
k f K C PC
r0 =
[1+ K C PC ]2
Algebraic manipulation then gives
PC
1
2P K P 2
(14)
=
+ C+ C C
r0 k f K C k f
kf
Equation 14 is not consistent with the experimental data shown in Figure P6.9.1.
Solution 6.10:
The rate data were worked up using a spreadsheet in order to test several simple forms of
the rate expression by plotting the rate versus some function of the partial pressures of
SO2, CO2, and CO. Inspection of the various plots indicated that the rate law was neither
simple first- or second-order in SO2. Plots of the rate versus the product of the partial
pressure of SO2 and the ratio of the partial pressure of CO to the partial pressure of CO2
Chapter 6 Solutions
6 - 17
as well as the product of the partial pressure of SO2 and the reciprocal of the indicated
ratio were prepared. In addition, plots of the rate versus the product of these ratios raised
to the second power and the square of the partial pressure of SO2 were also prepared. Of
the various figures the only plot that provided a fit of the data that was anywhere close to
reasonable is that shown in Figure P6.10.1. All of the other fits were extremely
unreasonable. The value of the rate constant corresponding to the best fit straight line in
Figure P6.10.1 is 6.13 mole m 2 ! s ! atm , a value that is consistent with the rate
(
)
constant reported by these authors, namely 6.4, in the same units.
y = 6.127x
R2 = 0.7785
Rate [(mole/(sec-m^2)]
0.025
0.020
0.015
0.010
0.005
0.000
0.000
0.001
0.002
0.003
0.004
(PSO2)(PCO)/(PCO2) atm
Figure P6.10.1 Test of proposed rate law that is first-order in both CO and SO2
and of order minus one in CO2.
B. For the proposed mechanism, the reaction rate is approximately that of the rate
controlling step.
r = r3 = k3! SO
(1)
where ! SO refers to the concentration of adsorbed SO species (an intermediate).
From equations 1, 2, 7, 8, and 9 of the problem statement
! SO = K2! SO2 !O
! SO2 = K1PSO2
! O2 = K7PO2
! O = K8! O2
PO2 1 / 2 =
(2)
(3)
(4)
(5)
PCO2
(6)
K9 PCO
From equations 1-6 of this solution
r = k3K2 K1PSO2 K8!1 2K7!1 2 K9PCO PCO2
Chapter 6 Solutions
(7)
6 - 18
This equation is of the same mathematical form as the experimental rate expression
determined in Part A. Hence the rate expression obtained from the proposed mechanism
is consistent with the data.
Solution 6.11:
A. Consideration of adsorption equilibria on type 1 sites indicates that:
" H 2 S,1 = K H 2 S,1PH 2 S"V ,1
" H 2O,1 = K H 2O,1PH 2O"V ,1
"V ,1 = "V ,1
Because
! the fractional coverages must sum to unity
1
"V ,1 =
!
1+ K H 2O,1PH 2O + K H 2 S,1PH 2 S
!
and
K H 2 S,1PH 2 S
" H 2 S,1 =
1+ K H 2O,1PH 2O + K H 2 S,1PH 2 S
!
Consideration of the adsorption equilibria on type 2 sites indicates that
"O,2 2 = KO2,2 PO2 "V ,2 2
!
or
"O,2 = KO2,2 PO2 "V ,2
" H 2O,2 = K H 2O PH 2O"V ,2
"V ,2 = "V ,2
Addition
! of equations 7-9 gives
1 &
#
! "O,2 + " H 2O,2 + "V ,2 = 1 = "V ,2 %1+ K H 2O PH 2O + KO2,2 PO2 2 (
$
'
!
or
1
"V ,2 =
1
1+ K H 2O PH 2O + KO2,2 PO2 2
!
(
(4)
(5)
(6)
(7)
(8)
(9)
!
(
(1)
(2)
(3)
)
(10)
)
and
!
"O,2 =
(
KO2,2 PO2
)
1
2
(
1+ K H 2O PH 2O + KO2,2 PO2
)
1
(11)
2
The observed rate of reaction will be that for the rate-limiting step
r = k1" H 2 S,1"O,2
!
1
k1K H 2 S,1PH 2 S KO2,2 PO2 2
=
1 %
"
2
1+
K
P
+
K
P
1+
K
P
+
K
P
[
]
H 2O,1 H 2O
H 2 S,1 H 2 S $
H 2O,2 H 2O
O2,2 O2
'&
!
#
(
)
(
Chapter 6 Solutions
!
(12)
)
6 - 19
B. The data in the figure indicate that the reaction is inhibited by the presence of H 2O .
This result is consistent with the implications of equation 12 derived from the indicated
mechanism.
! independent
C. The data in Table P6.11.1 indicate that the observed rate is substantially
of the linear velocity of the feed gas through the catalyst bed. Hence, one can conclude
that external mass transfer effects are probably not significant. The data in Table P6.11.2
indicate that there is little or no effect of particle size on the observed rate. Hence, one
may conclude that intra-particle diffusional limitations are absent. Note that at the highest
temperature (where one would be most likely to observe such limitations) the observed
rate is independent of particle size.
Solution 6.12:
A. The rate of formation of CO2 is the same as that for reaction 2.
d (CO2 )
r=
= k2! CO[O ]
dt
If the equilibrium is established with respect to the adsorption step
(1)
k1PCO!V = k"1!CO
or
!CO = (k1 / k"1 )PCO!V = KCOPCO!V
(2)
Because CO is the only adsorbed species
! CO + !V = 1
(3)
From equations 2 and 3
(K COPCO + 1)!V = 1
or
! V = 1/ (1+ K COPCO )
(4)
From equations 2 and 4
KCO PCO
(5)
1 + KCOPCO
From equations 1 and 5
k K COPCO [O]
(6)
r=
+ K COPCO
This rate expression is consistent with the experimental findings, provided that the
concentration of lattice oxygen atoms is independent of the partial pressures of CO and
O2. Because the lattice oxygen atoms are present in large excess, this is a valid
assumption.
! CO =
B. To test the data for consistency, it is convenient to rewrite equation 6 in the form
1
1
1
1
=[
•
]+
r
k2 KCO [O] PCO k2 [O]
The data are worked up below and plotted in Figure P6.12.1.
Chapter 6 Solutions
6 - 20
PCO
(atm)
0.035
0.109
0.194
0.250
0.301
r ! 104
mol/(hr-g catalyst)
8.2
19.6
27.3
31.6
35.2
1/r
(hr-g catalyst /mole)
1220
510
366
316
284
y = 36.806x + 169.59
1400
1/r [(h-g catalyst)/mole]
1/PCO
(atm-1)
28.57
9.17
5.15
4.00
3.32
1200
1000
800
600
400
200
0
0
5
10
15
20
25
30
1/PCO (atm)^(-1)
Figure P6.12.1 Test for the linearity of the fit of a proposed rate law.
The linearity of the plot indicates that the data are consistent with the rate expression
associated with the proposed mechanism. Thus this mechanism is consistent with the
kinetic data.
From the intercept of the best-fit straight line in Figure P6.12.1
k2[O] = kapparent = 1/170 = 5.58!10-3 mole/(h-g catalyst)
From the slope
1
= 36.8 (h-g catalyst-atm)/mole
k 2 KCO[O]
Thus
KCO = 1/[5.58!10-3 (36.8)] =4.87 atm-1
Solution 6.13:
The reaction rate may be taken as that corresponding to the slow step in the reaction, i.e.,
step 3
r = k3! A,1! H ,0 " k"3! v,0!C ,1
where ! v,0 represents the fraction of the sites of type zero that are vacant and
Chapter 6 Solutions
6 - 21
(1)
! C ,1 represents the sites of type 1 that are occupied by the half hydrogenated
intermediate.
Because adsorption equilibrium prevails for reactions 1, 2, and 5
!
K 1 = A,1
P A !v,1
! 2H ,0
K2 =
or
(2)
P H 2 ! 2v,0
!
K 5 = B ,0
P B ! v,0
(3)
! A,1 = K 1P A ! v,1
(5)
(4)
K 2P H 2 !v,0
! H ,0 =
(6)
! B ,0 = K 5P B! v,0
(7)
Because the fractional coverages of sites of type zero must sum to unity, use of equations
6 and 7 gives
1= ! v,0 1+ K 2P H 2 + K 5P B
[
]
or
! v,0 =
1
(9)
1+ K 2P H 2 + K 5P B
Use of the Bodenstein steady state approximation for the intermediate
! C ,1 gives
d!C ,1
= k3! A ,1! H ,0 " k"3! v,0! C ,1 " k4!C ,1! H ,0 + k"4! B ,0!v,1 # 0
dt
Solution for ! C ,1 gives
k! !
+k ! !
! C ,1 = 3 A ,1 H ,0 "4 B ,0 v,1
k"3!v,0 + k4! H ,0
Combination of equations 5, 6, 7 and 11 gives
k3K A P A K 2P H 2 + k"4K 5P B !v,1!v,0
! C ,1 =
! "# v,1
k"3 + k4 K 2P H 2 ! v,0
[
(
(8)
)
(10)
(11)
]
(12)
Because the fractional coverages of sites of type 1 must sum to unity, combination of
equations 5 and 12 gives
1= ! v,1[ 1+ K 1P A + " ]
or
1
! v,1 =
1+ K 1P A + "
Chapter 6 Solutions
6 - 22
(13)
Substitution of equations 5, 6, 9, 12, and 13 into equation 1 gives
r=
k3K 1P A K 2P H 2 ! k!3"
( 1+ K P
H2
H 2 + K 5P B
)( 1+ K P + " )
1 A
To compare the proposed rate expression with the experimental data we may assume that
for the initial rate the k!3 term disappears and that
k3K A P A K 2P H 2 + k"4K 5P B
!=
k4 K 2P H 2
If
K 2P H 2 !! (1+ K 5P B ) and k3K A P A K 2P H 2 !! k4K 5P B the rate is independent of
P H 2 . The rate will also increase with increasing acetone pressure and decrease with
increasing alcohol pressure. Note that this situation implies that the sites of type zero are
essentially completely occupied by hydrogen atoms and that the sites of type 1 are not
fully occupied.
Solution 6.14:
For the various reaction mechanisms, the reaction rate will be identical with that for the
rate-limiting step. Each mechanism will be considered in turn.
Mechanism 1
In terms of the rate limiting step
r1 = k 4 !N 2 O
(1)
From the adsorption equilibrium relations
! CO = KCOPCO! V
(2)
!N O = K N OPN O!V
(3)
! V = !V
(4)
and
2
2
2
However,
The fractions of the surface, which are vacant or covered by a particular species, must
sum to unity.
1 = ! V + !CO + ! N 2 O
(5)
Combination of equations 2-5 gives
1
!v =
1+ K COPCO + K N2 O PN 2 O
Combination of equations 1, 3, and 6 gives
[
]
Chapter 6 Solutions
(6)
6 - 23
k 4 K N 2 O PN 2 O
(7)
1+ K N 2 O PN2 O + K COPCO
or, in linearized form
PN 2 O 1+ KN 2 O PN 2 O + KCOPCO
(8)
=
r1
k 4 KN 2 O
To test the data, one should plot the left side of equation 8 versus PN 2 O for data taken at a
constant value of PCO. One should also plot the data obtained at constant values of PN 2 O in
the form of the left side of equation 8 versus PCO.
r1 =
Mechanism 2
In terms of the rate limiting step
r2 = k 6! N 2O! V
(9)
From the adsorption equilibrium relations
! CO = KCOPCO! V
(10)
!N O = K N OPN O!V
(11)
and
2
2
2
From equations 4, 5, 10, and 11
[
1 = !V 1 + K COPCO + K N 2OPN 2O
]
(12)
From equations 9, 11, and 12
k 6 KN 2 O PN 2 O
(13)
r2 =
2
1+ KCO PCO + K N2 OPN 2 O
In linearized form
1
! PN 2 O $ 2 1+ KCOPCO + KN 2 O PN2 O
#
& =
(14)
" r2 %
k 6 K N2 O
Hence, the data obtained at constant PCO should be tested by plotting the left side of
equation 14 versus PN 2 O and data obtained at constant PN 2 O should be tested by plotting
the left side of equation 14 versus PCO.
[
]
Mechanism 3
In terms of the rate controlling step
r3 = k 7! N 2 O! CO
(15)
From equations 2, 3, and 5
!V = 1
1 + KCO PCO + KN 2 O PN 2 O
[
]
(16)
From equations 2, 3, 15, and 16
Chapter 6 Solutions
6 - 24
r3 =
k 7KN 2 O PN 2 OK COPCO
[1 + K P
+ KCO PCO
N 2O N 2O
(17)
2
]
In linearized form
1
!PN 2 O PCO $ 2 1+ K N 2 OPN 2 O + KCO PCO
=
1
#" r3 &%
k 7 KN 2 O KCO 2
[
]
(18)
Thus data obtained at constant PN 2 O should be tested by plotting the left side of equation
18 versus PCO. Data obtained at constant PCO should be tested by plotting the left side of
equation 18 versus PN 2 O.
Mechanism 4
For the rate controlling step
r4 = k7 PN 2O!V
(19)
For adsorption equilibrium of CO
!CO = KCO PCO!V
(20)
If we make the steady state approximation for the adsorbed oxygen atoms
d! O
" k 8 PN2 O!V # k9 !CO! O $ 0
dt
From equation 21
k8 PN2 O! V
!0 =
k 9! CO
From the fact that the various fractional coverages must sum to unity
(21)
(22)
1 = !V + !O + ! CO
(23)
From equations 20, 22, and 23
k8 PN 2 O
"
%
!V = 1/ $1 +
+ KCO PCO '
# kK P
&
(24)
9
CO CO
Combination of equations 19 and 24 gives
k7 PN2 O
(25)
r4 =
k8 PN 2 O
!
$
1+
+ KCO PCO
#" k9 K COPCO
&%
It is not possible to completely linearize this expression, but a potentially useful form is
PN 2 O 1 !
k8PN2 O
$
(26)
=
1+
+ KCO PCO
&%
r
k #" k K P
4
7
9
CO CO
Data obtained at constant PCO can be tested by plotting the left side of equation 26
versus PN 2 O.
Mechanism 5
Chapter 6 Solutions
6 - 25
In terms of the rate controlling step
r5 = k10PN2O !CO
(27)
Equation 20 is again applicable to the adsorption equilibrium for CO. Because the
fractional converges must sum to unity
(28)
1 = !V + !CO = !V + K COPCO! V
or
(29)
!V = 1/[1 + KCOPCO ]
Thus combination of equations 20, 27 and 29 gives
k10 K COPCO PN2 O
(30)
r5 =
1+ KCO PCO
In linearized form
PCOPN2 O 1 + K COPCO
(31)
=
r5
k10 KCO
B. In order to determine if any of the proposed mechanisms can be ruled out, it is helpful
to consider the various rate expressions [equations 7, 13, 17, 25, and 31]. Examination of
these rate expressions to determine if they are consistent with the experimental
observation that the order in CO is negative reveals that Mechanism 5 can be eliminated
from consideration. Mechanism 4 can also be eliminated because it is not consistent
with the observation that the order in N2 O is ca 0.66.
Discrimination among the remaining mechanisms would require more information than is
provided in the problem statement.
Solution 6.15:
From the equation describing the rate controlling step
r = k4! ABPB
(1)
If the adsorption steps are at quasi-equilibrium
k1PA! V = k1! A
(2)
Thus
!A =
k1
PA!V # K1 PA! V
k"1
(3)
Similarly
k3 PB! V
# K3 PB! V
k "3
If the second reaction is at equilibrium
or
!B =
(4)
k2!APB = k"2! AB
(5)
k2
! A PB # K 2! A PB
k "2
Combination of equations 1-3, and 6 gives
! AB =
(6)
r = k4 K1K2PAPB2! V
(7)
Chapter 6 Solutions
6 - 26
Now
!V = !V
(8)
and from equations 3 and 6
! AB = K2PBK1PA!V
(9)
Addition of equations 3, 4, 8, and 9 and recognition that the fractional coverages must
sum to unity gives
(10)
1 = ! A + ! B + ! AB + ! V = !V [K1PA + K3PB + K1K2PAPB +1]
or
1
(11)
!V =
1+ K1 PA + K3 PB + K1 K2 PA PB
Combination of equations 7 and 11 gives
k4 K1 K2 PA PB 2
r=
1 + K1PA + K3 PB + K1 K2 PA PB
(12)
This general form is consistent with the observed rate expression when
(K1PA + K3PB) >> (1 + K1K 2PAPB) , i.e., when the number of vacant sites is very small
and when the fraction of the surface covered by (AB )! is small.
To further test the mechanism one could ascertain what happens when one alters the
partial pressures of species A and B. For example if the amount of isobutane in the feed
is decreased further, one would expect to enter a regime where K1PA >> K3PB and the
rate expression will then become zero order in PA and second order in PB . At higher
values of PA one would also expect to observe the same result. By contrast, if one
increases PB to a sufficient degree, one could enter a region in which K3PB >> K1PA and
the rate expression should then become first order in both A and B. A similar result
should be observed if PA is decreased to a sufficient degree.
Solution 6.16:
If we let !1i represent the fraction of the type 1 sites covered by species i and !2i
represent the fraction of the type 2 sites covered by species i, the following relations may
be derived for the adsorption equilibrium steps.
(1)
"1K = K K PK "1#
(2)
" 2A = K A PA " 2#
2
"1H
= K H PH 2 "1#
2
1
or
"1H = (K H PH 2 ) 2 "1#
(3)
!
(4)
" 2P = K P PP " 2#
!
(5)
"1W = KW PW "1#
where!subscript K refers to ketone, A to ammonia, H2 and H to molecular and atomic
!
hydrogen respectively, P to the product, W to water and " to a vacant site.
!
!
Chapter 6 Solutions
6 - 27
If the two surface reactions are at equilibrium:
" "
K 4 = 1W 2D
"1K " 2A
" "
K 5 = 1# 2C
"1H " 2D
where!D refers to R1CNHR2 and C to R1CNH2R2.
From equations 1, 2, 5 and 6
! " =K K P K P "
2D
4
K
K
A
A
2#
(6)
(7)
(8)
KW PW
From equations 3, 7 and 8
1
" 2C = K 5 (K H PH 2 ) 2 " 2D
!
!
1
= K 5 (K H PH 2 ) 2 K 4 K K PK K A PA " 2# KW PW
(9)
Because the fractional coverages must sum to unity
"1K + "1H + "1W + "1# = 1
or using equations 1, 3, 5 and the identity "1! = "1!
!
(10)
[
]
"1! K K PK + K H PH 2 + KW PW + 1 = 1
! Hence
[
(11)
]
"1# = 1 K K PK + K H PH 2 + KW PW + 1
!
!
!
(12)
Similarly
(13)
" 2A + " 2P + " 2C + " 2D + " 2# = 1
! Combination of equations 2, 4, 8 and 13 gives
1
*
$
'
" 2# ,K A PA + K P PP + %K 5 (K H PH 2 ) 2 K 4 K K K A PK PA KW PW ( + (K 4 K K K A PA PK KW PW ) + 1/ = 1
&
)
+
.
!
(14)
or
1 '
0
*
-3
$
" 2# = 1 21+ K A PA + K P PP + ,K 4 K K K A PA PK %1+ K 5 (K H PH 2 ) 2 ( KW PW /5
(15)
&
)
+
.4
1
Now the observed rate of reaction will be that of the rate determining step.
(16)
r = k6"1H " 2C
From equations 3, 9, 11, 15, and 16:
1
1
k 6 (K H PH 2 ) 2 K 5 (K H PH 2 ) 2 K 4 K K K A PK PA KW PW
r=
(17)
1 %"
1 +
!"
.
1%
(
2
2
1+ KW PW + K K PK + (K H PH 2 ) '$1+ K A PA + K P PP + 0K 4 K K K A PA PK )1+ K 5 (K H PH 2 ) , KW PW 3'
#$
&#
*
/
2&
In order to reduce this equation to the form of the experimental rate expression, several criteria must be
satisfied:
1. The large majority of the type 2 sites must be occupied by imine-type species,
12
i.e., " 2C # 1, or K 5 (K H PH 2 )
>> 1 and
Chapter 6 Solutions
!
!
6 - 28
12
K 4 K K K A PA PK K 5 (K H PH 2 )
KW PW >> (1+ K A PA + K P PP )
In this event the rate expression 17 reduces to
12
!
r=
k 6 (K H PH 2 )
(18)
12
1+ KW PW + K K PK + (K H PH 2 )
2. The occupied type 1 sites are occupied primarily by water molecules, i.e.,
1 2%
"
KW PW >> $K K PK + (K H PH 2 ) '
#
&
! In this case equation 17 becomes
12
!
r=
k6 (K H PH 2 )
1+ KW PW
This expression is consistent with the experimentally observed kinetics.
! Solution 6.17:
A. For Mechanism I the rate is that for the rate-controlling step:
r = k f ! A! H2 " kr! #! V
where the various fractional coverages can be determined from the adsorption
equilibrium relations.
! H2 = K H2•" PH2 ! V
! A = K A•" PA! V
! " = K"•# P"! V
The fractional coverages must sum to unity.
! H2 + ! A + ! " + ! V = 1
Combination of equations 2-5 gives
! V K H2 •" PH2 + K A•" PA + K #•" P# + 1 = 1
[
]
(1)
(2)
(3)
(4)
(5)
(6)
Combination of equations 1-4 and 6 gives
k f K A •! PA K H2 •! PH2 " kr K #•! P#
r=
2
1+ K H2 •! PH2 + K A•! PA + K#•! P#
(7)
In its initial rate form, equation 7 becomes
k f K A•! PA K H2 •! PH2
r0 =
2
1+ K H2 •! PH2 + K A•! PA + K "•! P"
(8)
[
[
]
]
For Mechanism II, the corresponding rate of the rate controlling step is
r = k f !" A# H 2 $ k r! " %# V
(9)
where the ! ' s refer to the fractional coverages of the ! sites and the !' s refer to the
fractional coverages of the Z sites. The pseudo equilibrium coverages of the sites can
be expressed in terms of the adsorption equilibrium constants
! A = K A•" PA! V
(10)
Chapter 6 Solutions
6 - 29
! " = K"•# P"! V
(11)
!V = !V
The fractional coverages must sum to unity
! A + !" + !V = 1
Combination of equations 10-13 gives
! V [ K A•" PA + K #•" P# + 1] = 1
(12)
In addition
(13)
(14)
A similar analysis for the type Z sites gives
! H2 = K H2 •Z PH2 ! V
and
!V K H2 • Z PH2 + 1 = 1
[
(15)
]
(16)
Combination of equations 9-11, and 14-16 gives
k f ! K A•" PA K H2 • Z PH 2 # k r! K$•" P$
r=
[1+ K A •" PA + K $•" P$ ] 1+ K H2 •Z PH2
[
(17)
]
The corresponding form of the initial rate expression is
K H2 • Z PH2
k f ! K A• " PA
r0 =
•
[1 + KA•" PA + K #• " P# ] 1 + K H2• Z PH2
[
(18)
]
B. Equations 8 and 18 may be used to ascertain whether either of the proposed
mechanisms is consistent with the kinetic data.
Mechanism I is not consistent with the two experimental observations that when the
partial pressure of one reactant is held constant, the initial rate depends on the partial
pressure of the other reactant in a manner that produces an asymptotic limit. Instead,
equation 8 indicates that the initial rate should pass through a maximum when PA is
increased at constant PH2 (and also that the initial rate will pass through a maximum as
PH2 increases at constant PA). By contrast, Mechanism II predicts the existence of two
asymptotic limits, as well as inhibition of the rate by the alcohol.
C. It would not be reasonable to assume that dissociative adsorption of hydrogen is
involved in the reaction mechanism. In this case, the initial rate expression
corresponding to equation 8 is:
k f K A •! PA K H2 •! PH2
r0 =
3
1+ K H2 •! PH2 + K A•! PA + K "•! P"
[
]
This rate expression is not consistent with the two reported asymptotic limits.
Similarly, the corresponding form of equation 18 is
k f ! K A •" PA
K H2 •Z PH2
r0 =
•
[1+ K A•" PA + K#•" P# ] 1+ KH 2• Z PH2 2
[
Chapter 6 Solutions
]
6 - 30
This mechanism is consistent with the experimental data. This expression will approach
an asymptotic limit as PH2 is increased at constant PA and PB.
Solution 6.18:
For mechanism A with reaction 2 as the rate limiting step, the rate is given by
r = k f " B "V # k r " H "C 4 H 7
From the adsorption equilibria
" B = K B PB "V
" D = K D PD "V
!
" H 2 = K H 2 PH 2 "V
!
If reactions
3 and 5 are at equilibrium
!
" "
"C 4 H 7 = H D
!
K 3"V
and
12
" H = [K 5" H 2 "V ]
[1]
[2]
[3]
[4]
[5]
[6]
!
Combination
of equations 4 and 6 gives
12
" H = [K 5K H 2 PH 2 ] "V
[7]
Combination
of equations 3, 5, and 7 gives
!
K D PD
"V
[8]
K3
Because the fractional coverages must sum to unity, equations 2-4, 7 and 8 indicate that
)
1 2#
K P &,
[9]
1 = "V *1+ K B PB + K D PD + K H 2 PH 2 + [K 5K H 2 PH 2 ] %1+ D D (K 3 '.
$
+
!
Combination of equations 1 and 7-9 then gives
k f K B PB " k rK 5K H 2 PH 2 K D PD K 3
[10]
r=
2
)
,
#
&
1
2
K P
!
*1+ K B PB + K D PD + K H 2 PH 2 + [K 5K H 2 PH 2 ] %1+ D D (K 3 '.
$
+
12
!
"C 4 H 7 = [K 5K H 2 PH 2 ]
For mechanism B the reaction rate is given by
r = k f " B "V # k r "C 4 H 7 " H
!
The adsorption equilibria given by equations 2 and 3 remain valid.
If reactions 3 and 5 are at equilibrium
" "
"C 4 H 7 = D H
!
K 3"V
12
" H = [K H 2 PH 2 ] "V
!
!
[11]
[12]
[13]
Chapter 6 Solutions
6 - 31
Combination of equations 3, 12, and 13 gives
12
"C 4 H 7 =
K D PD [K H 2 PH 2 ] "V
K3
Since the fractional coverage must sum to unity, Equations 2, 3, 13, and 14 give
{
12
1 = "V 1+ K B PB + K D PD + [K H 2 PH 2 ]
[1+ (K D PD K 3 )]}
[14]
[15]
!
Combination
of equations 2, 11, and 13-15 gives the following expression for the
reaction rate
k f K B PB " k rK D PD K H 2 PH 2 K 3
[16]
r=
!
2
12
1+ K B PB + K D PD + [K H 2 PH 2 ] [1+ (K D PD K 3 )]
{
}
For mechanism C the reaction rate is given by
r = k f " B "V # k r " D " H 2
[17]
!
Equations 2 and 3 again apply to the adsorption equilibria for butene and butadiene. For
molecular hydrogen
" H 2 = K H 2 PH 2 "V
[18]
!
Since the fractional coverages must sum to unity, Equations 2, 3, and 18 indicate that
1 = "V [1+ K H 2 PH 2 + K D PD + K B PB ]
[19]
Combination
of equations 2, 3, and 17-19 gives
!
k f K B PB " k rK D PD K H 2 PH 2
r=
[20]
2
1+
K
P
+
K
P
+
K
P
[
]
!
H2 H2
D D
B B
Initial rate measurements can be used to discriminate between various proposed
mechanisms, If one carries out a series of initial rate measurements for which the initial
concentration of butadiene (D) is zero, mechanisms A, B, and C lead to the following
!
simplified
forms of equations 10, 16, and 20
k f K B PB
rA =
[21]
1 2 %2
"
$#1+ K B PB + K H 2 PH 2 + (K 5K H 2 PH 2 ) '&
k f K B PB
rB =
[22]
1 2 %2
"
1+ K B PB + (K H 2 PH 2 ) '
#$
&
!
k f K B PB
rC =
[23]
2
1+
K
P
+
K
P
[ H2 H2 B B ]
!
Modified reciprocal plots can be used to assess the validity of the several mechanisms.
Equations 21-23 can be rearranged in the following manner:
!
" PB %1 2
1 2+
1 (
[21a]
1+ K B PB + K H 2 PH 2 + (K 5K H 2 PH 2 ) $ ' =
*
,
kK )
# rA &
f
!
B
Chapter 6 Solutions
6 - 32
" PB %1 2
$ ' =
# rB &
1 2+
1 (
1+ K B PB + (K H 2 PH 2 ) *
,
k f KB )
[22a]
" PB %1 2
$ ' =
# rB &
1
[1+ K B PB + K H 2 PH 2 ]
k f KB
[23a]
!
For initial rate data taken at constant PB and variable PH 2 , a different dependence of the
left side
! of equations [21a]-[23a] on PH 2 will be observed for the various mechanisms.
The various functional forms of the rate laws can be used as a basis for discrimination.
!
!
!
Solution 6.19
We are asked to consider Hougen-Watson models for the reaction between CO2 and H2
and to ascertain what form of this general model is appropriate for consistency with the
available experimental data.
CO2 + H2 ! CO + H2O
There are four basic facts with which any proposed model must be compatible:
1. At constant PCO2 the rate law is first order in hydrogen.
2. At constant PH 2 the rate goes through a maximum as PCO2 is varied.
3. CO has no retarding influence on the rate.
4. H2O cannot appear in the rate expression because it is continuously removed in an
ice trap.
There are several simple Hougen-Watson mechanisms that might be proposed to explain
the data. We shall consider models based on the following rate-controlling steps.
A. A simple bimolecular gas phase collision
r = kPCO2 PH 2
B. First-order surface reaction involving ! CO2
r = k! CO2
C. First-order surface reaction involving ! H 2
r = k! H 2
D. Second-order surface reaction involving ! H 2 and ! CO2
r = k! H 2 ! CO2
E. Rideal mechanism involving ! CO and gaseous H2
r = kPH 2 ! CO2
F. Rideal mechanism involving ! H and gaseous CO2
r = kPCO2 ! H 2
2
2
In the general Hougen-Watson model the surface coverages can be expressed as
K H 2 PH 2
(1)
!H 2 =
1+ K H 2 PH 2 + KCO2 PCO2 + KCO PCO + K H 2O PH 2O
and
!CO =
2
KCO2 PCO2
1+ K H 2 PH 2 + KCO2 PCO2 + KCO PCO + K H 2O PH 2O
Chapter 6 Solutions
(2)
6 - 33
By virtue of the facts that water is removed in a dry ice trap and that CO does not inhibit
the reaction we may simplify the expressions for the fractional coverages (equations 1
and 2) to
K H 2 PH 2
(3)
! H2 =
1+ K H 2 PH 2 + K CO2 PCO2
and
!CO =
2
KCO2 PCO2
(4)
1+ K H 2 PH 2 + KCO2 PCO2
Substitution of equations 3 and 4 into the rate laws for mechanisms A-F gives the
following forms of the rate laws.
Mechanism
A
Rate Expression
r = kPCO2 PH 2
B
r = kK CO2 PCO2 (1+ K H 2 PH 2 + KCO2 PCO2 )
C
r = kK H 2 PH 2 (1+ K H 2 PH 2 + KCO 2 PCO 2 )
D
r = kK H 2 PH 2 KCO2 PCO2 (1+ K H 2 PH 2 + KCO2 PCO2 )2
E
r = kPH 2 KCO 2 PCO 2 (1+ K H 2 PH 2 + KCO 2 PCO 2 )
F
r = kPCO2 K H 2 PH 2 (1+ K H 2 PH 2 + KCO2 PCO2 )
Of the six rate expressions in this table, only that for Mechanism D can give a maximum
in the rate as a function of PCO2 at constant PH 2 . The fact that the rate is first-order in H2
also implies that the term K H 2 PH 2 is negligible compared to 1. Thus the working form of
the rate law reduces to
r = kK H 2 PH 2 KCO2 PCO2 (1+ KCO2 PCO2 )2
This rate expression is consistent with the observation that at constant PH 2 the rate goes
through a maximum as PCO2 is varied. When K CO 2 PCO 2 << 1, the rate starts out linear in
PCO 2 , but when K CO 2 PCO 2 >>1, the rate must fall off as 1/ PCO 2 . In physical terms when
K CO 2 PCO 2 << 1 only a very small fraction of the surface sites are occupied by CO2
molecules. The fractional coverage increases with increasing PCO 2 . However, as the CO2
coverage increases, some of the adsorbed hydrogen molecules are displaced until one
eventually reaches a condition in which the increase in ! CO2 with increasing PCO 2 is not
sufficient to offset the decline in ! H 2 that must occur to create sites on which CO2 can
adsorb.
Chapter 6 Solutions
6 - 34
Solution 6.20:
This problem calls for the derivation of a variety of Hougen-Watson rate expressions and
for tests for consistency of these expressions against some kinetic data.
I. The first four mechanisms involve the following sequence of molecular events:
k1
A. SO2 + ! !SO2 • !
Adsorption equilibrium constant = K SO2
B. O2 + 2! ! 2[O• ! ]
Adsorption equilibrium constant = K O2
k -1
k2
k -2
k3
C. SO2 • ! + O• ! ! SO3 • ! + !
Surface reaction equilibrium constant = K r
k -3
k4
D. SO3• ! ! SO3 + !
Desorption equilibrium constant = 1/ K SO3
k -4
where ! denotes a vacant surface site.
1. In this mechanism adsorption of oxygen is rate-limiting.
r1 = k2 PO2 !V2 - k - 2!O2
(1)
where !V and ! O refer to the fractions of the sites that are vacant and occupied by an
oxygen atom, respectively. Equilibrium relations for mechanistic equations A, C, and D
then give
(2)
! SO2 = K SO2 PSO2 !V
!SO3 = KSO PSO !V
3
!O =
(3)
3
! SO !V K SO PSO !V
=
! SO K r K SO PSO K r
3
3
3
2
2
2
(4)
The sum of these three fractional coverages and the fraction of the sites that are vacant
must be unity.
K SO3 PSO3
(5)
1= !V { 1+ K SO2 PSO2 + K SO3 PSO3 +
}
K SO2 PSO2 K r
Combination of equations 1-5 yields
K SO3 PSO3 2
k2 PO2 - k - 2 [
]
K SO2 PSO2 K r
r1 =
K SO3 PSO3
{ 1+ K SO2 PSO2 + K SO3 PSO3 +
}2
K SO2 PSO2 K r
(6)
If we denote the total pressure by ! , the initial rate in the absence of product species
{ PSO3 = 0} when PSO2 = PO2 = ! 2 is given by
Chapter 6 Solutions
6 - 35
r1,0 =
k 2 (! / 2)
{
(7)
}
{ 1+ K SO2 (! / 2) }2
The initial rate will pass through a maximum as the total pressure is increased.
2. For this mechanism adsorption of SO2 is rate-limiting.
r2 = k1PSO2!V - k -1!SO2
(8)
Equilibrium relations for mechanistic equations B- D then give
!O = KO2 PO2 !V
(9)
! SO = K SO PSO !V
3
3
! SO =
2
(10)
3
! SO !V K SO PSO !V
=
!O K r
KO PO K r
3
3
(11)
3
2
2
The sum of these three fractional coverages and the fraction of the sites that are vacant
must be unity.
1= !V { 1+ KO2 PO2 + K SO3 PSO3 +
K SO3 PSO3
KO2 PO2 K r
}
(12)
Combination of equations 8-12 then gives
K SO3 PSO3
k1PSO2 - k -1
KO2 PO2 K r
r2 =
K SO3 PSO3
1+ KO2 PO2 + K SO3 PSO3 +
KO2 PO2 K r
(13)
In this case the expression for the initial rate becomes
k1PSO2
r2 =
1+ K O2 PO2
(14
In terms of the total pressure, equation 14 becomes
k1 (! 2)
r2,0 =
1+ K O2 (! 2)
(15)
{
}
{
}
For this type of rate law the initial rate will increase continuously with increasing
pressure.
3. For this mechanism the rate-limiting step is desorption of SO3.
r3 = k4! SO3 - k -4 PSO3!V
(16)
Equilibrium relations for mechanistic equations A-C then give
! SO2 = K SO2 PSO2 !V
(17)
Chapter 6 Solutions
6 - 36
!O = KO PO !V
K r! SO !O
! SO =
= K r K SO PSO
!V
2
(18)
2
2
3
2
2
KO2 PO2 !V
(19)
The sum of these three fractional coverages and the fraction of the sites that are vacant
must be unity.
(20)
1= !V 1+ KSO2 PSO2 + KO2 PO2 + Kr KSO2 PSO2 KO2 PO2
{
}
Combination of equations 16-20 then gives
k4 K r K SO2 PSO2 KO2 PO2 - k - 4 PSO3
r3 =
1+ K SO2 PSO2 + KO2 PO2 + K r K SO2 PSO2 KO2 PO2
In this case the expression for the initial rate becomes
k4 K r K SO2 PSO2 KO2 PO2
r3 =
1+ K SO2 PSO2 + KO2 PO2 + K r K SO2 PSO2 KO2 PO2
In terms of the total pressure, equation 22 becomes
k4 K r K SO2 KO2 (! 2)3 2
r3,0 =
1+ K SO2 (! 2) + KO2 (! 2)1 2 + K r K SO2 KO2 (! 2)3 2
(21)
(22)
(23)
For this type of rate law at low pressures the initial rate will first increase as the pressure
increases, but at sufficiently high pressures, the initial rate will approach an asymptotic
limit.
4. For this mechanism the rate-limiting step is the surface reaction
r4 = k3!SO2!O - k -3!SO3!V
(24)
Adsorption equilibrium relations for the various species then give
! SO2 = K SO2 PSO2 !V
(25)
!O = KO PO !V
! SO = K SO PSO !V
2
3
(26)
2
3
(27)
3
The sum of these three fractional coverages and the fraction of the sites that are vacant
must be unity.
(28)
1= !V 1+ KSO2 PSO2 + KO2 PO2 + KSO3 PSO3
{
}
Combination of equations 24-28 then gives
k3 K SO2 PSO2 KO2 PO2 - k -3 K SO3 PSO3
r4 =
[1+ K SO2 PSO2 + KO2 PO2 + K SO3 PSO3 ]2
In this case the expression for the initial rate becomes
k3 K SO2 PSO2 KO2 PO2
r4,0 =
[1+ K SO2 PSO2 + KO2 PO2 ]2
(29)
(30)
In terms of the total pressure, equation 30 becomes
Chapter 6 Solutions
6 - 37
r4,0 =
k3 K SO2 KO2 (! / 2)3 2
[1+ K SO2 (! / 2) + KO2 (! / 2) ]2
(31)
For this type of rate law the initial rate will pass through a maximum as the pressure
increases.
II. The next three mechanisms involve the following sequence of mechanistic events.
k1
E. SO2 + ! !SO2 • !
Adsorption equilibrium constant = K SO2
k -1
F. SO2 • ! +
k3
1
O2 ( gas) ! SO3• !
k -3
2
k4
G. SO3• ! ! SO3 + !
Surface reaction equilibrium constant = K r
Desorption equilibrium constant = 1/ K SO3
k -4
5. In this mechanism the rate-controlling step is adsorption of SO2
r5 = k1PSO2!V - k -1!SO2
(32)
Equilibrium relations for mechanistic steps F and G then give
! SO3 = K SO3 PSO3!V
(34)
! SO =
2
! SO
3
PO122 K r
=
K SO3 PSO3!V
(35)
PO122 K r
The sum of these two fractional coverages and the fraction of the sites that are vacant
must be unity.
1= !V { 1+ K SO3 PSO3 +
K SO3 PSO3
PO122 K r
}
Combination of equations 32-36 then gives
K SO PSO
k1PSO2 - k -1 1 32 3
PO2 K r
r5 =
K SO PSO
1+ K SO3 PSO3 + 1 32 3
PO2 K r
(36)
(37)
The initial rate is then
(38)
r5,0 = k1PSO2
In terms of the total pressure and the specified initial composition
(39)
r5,0 = k1 (! 2)
For this mechanism the initial rate increases continuously with increasing pressure.
6. For this mechanism the rate-limiting step is desorption of SO3.
r6 = k4!SO3 - k -4 PSO3!V
Chapter 6 Solutions
(40)
6 - 38
Equilibrium relations for mechanistic equations A-C then give
! SO2 = K SO2 PSO2 !V
!SO = Kr!SO PO1 2 = Kr KSO PSO PO1 2!V
3
2
2
2
2
(41)
(42)
2
The sum of these two fractional coverages and the fraction of the sites that are vacant
must be unity.
(43)
1= !V 1+ KSO2 PSO2 + Kr KSO2 PSO2 PO1 2
{
2
}
Combination of equations 40-43 then gives
k4 K r K SO2 PSO2 PO122 - k - 4 PSO3
r6 =
1+ K SO2 PSO2 + K r K SO2 PSO2 PO122
In this case the expression for the initial rate becomes
k4 K r K SO2 PSO2 PO122
r6,0 =
1+ K SO2 PSO2 + K r K SO2 PSO2 PO122
(44)
(45)
In terms of the total pressure, equation 45 becomes
k4 K r K SO2 (! 2)3 2
(46)
r6,0 =
1+ K SO2 (! 2) + K r K SO2 (! 2)3 2
For this type of rate law at low pressures the initial rate will first increase as the pressure
increases, but at sufficiently high pressures, the initial rate will approach an asymptotic
limit.
7. The rate-limiting step for this mechanism is an Ely-Rideal reaction involving a gas
phase species reacting with a chemisorbed species.
(47)
r7 = k2!SO2 PO1 2 - k-2!SO3
2
Adsorption equilibrium relations for the various species then give
! SO2 = K SO2 PSO2 !V
! SO = K SO PSO !V
3
3
(48)
(49)
3
The sum of these two fractional coverages and the fraction of the sites that are vacant
must be unity.
(50)
1= !V 1+ KSO2 PSO2 + KSO3 PSO3
{
}
Combination of equations 47-50 then gives
k2 K SO2 PSO2 PO1 22 - k - 2 K SO3 PSO3
r7 =
1+ K SO2 PSO2 + K SO3 PSO3
In this case the expression for the initial rate becomes
k2 K SO2 PSO2 PO1 22
r7,0 =
1+ K SO2 PSO2 + K SO3 PSO3
(51)
(52)
In terms of the total pressure, equation 52 becomes
Chapter 6 Solutions
6 - 39
r7,0 =
k2 K SO2 (! / 2)3 2
(53)
1+ K SO2 (! / 2)
For this mechanism, the initial rate will increase continuously as the pressure increases.
III. The final three mechanisms involve the following sequence of molecular events:
k1
H. SO2 + ! !SO2 • !
Adsorption equilibrium constant = K SO2
k -1
I.
k2
1
SO2 • ! + ! + O2 ( gas) ! SO3• ! + !
k -2
2
J.
SO3 • ! ! SO3 + !
k3
Surface reaction constant = K r
Desorption equilibrium constant = 1/ K SO3
k -3
8. For this mechanism, the rate-controlling step is adsorption of SO2
r8 = k1PSO2!V - k-1!SO2
(54)
Equilibrium relations for mechanistic steps I and J then give
! SO3 = K SO3 PSO3!V
(55)
! SO =
2
! SO !V
3
K r PO122!V
=
K SO3 PSO3!V
(56)
K r PO122
The sum of these two fractional coverages and the fraction of the sites that are vacant
must be unity.
1= !V { 1+ K SO3 PSO3 +
K SO3 PSO3
K r PO122
}
Combination of equations 54-57 then gives
K SO PSO
k1PSO2 - k -1 1 32 3
PO2 K r
r5 =
K SO PSO
1+ K SO3 PSO3 + 1 32 3
PO2 K r
(57)
(58)
The initial rate is then
(59)
r5,0 = k1PSO2
In terms of the total pressure
(60)
r5,0 = k1 (! / 2)
For this mechanism, the initial rate would increase continuously with increasing pressure.
9. For this mechanism, the rate-controlling step is desorption of SO3
r9 = k3!SO3 - k -3PSO3!V
(61)
Equilibrium relations for mechanistic steps H and I then give
Chapter 6 Solutions
6 - 40
! SO = K SO PSO !V
2
2
! SO =
(62)
2
K r! SO2 PO122!V
!V
3
= K r K SO2 PSO2 PO122!V
(63)
The sum of these two fractional coverages and the fraction of the sites that are vacant
must be unity.
(64)
1= !V 1+ KSO2 PSO2 + Kr KSO2 PSO2 PO1 2
{
2
}
Combination of equations 61-64 then gives
k3 K r K SO2 PSO2 PO122 - k -3 PSO3
r9 =
1+ K SO2 PSO2 + K r K SO2 PSO2 PO122
}
(65)
{1+ K P + K K P P }
(66)
{
The initial rate is then
r9,0 =
k3 K r K SO2 PSO2 PO122
SO2 SO2
r
12
SO2 SO2 O2
In terms of the total pressure
k3 K r K SO2 (! / 2)3 2
(67)
r9,0 =
1+ K SO2 (! / 2) + K r K SO2 (! / 2)3 2
For this mechanism at very low pressures the initial rate would increase with increasing
pressure, but at higher pressures would approach an asymptotic limit.
{
}
10. For this mechanism, the rate-controlling step is the surface reaction
r10 = k2!SO2!VPO1 22 - k-2!SO3!V
(68)
Equilibrium relations for mechanistic steps H and J then give
! SO2 = K SO2 PSO2 !V
(69)
! SO = K SO PSO !V
3
3
(70)
3
The sum of these two fractional coverages and the fraction of the sites that are vacant
must be unity.
(71)
1= !V 1+ K SO2 PSO2 + K SO3 PSO3
{
}
Combination of equations 68-71 then gives
k2 K SO2 PSO2 PO1 22 - k - 2 K SO3 PSO3
r10 =
1+ K SO2 PSO2 + K SO3 PSO3 2
{
}
The initial rate is then
k2 K SO2 PSO2 PO1 22
r10,0 =
1+ K SO2 PSO2 2
{
(73)
}
In terms of the total pressure
k2 K SO2 (! / 2)3 2
r10,0 =
1+ K SO2 (! / 2) 2
{
(72)
}
Chapter 6 Solutions
(74)
6 - 41
For this mechanism the initial rate will pass through a maximum as the pressure is
increased.
Of the ten mechanisms, three (1, 4, and 10) lead to rate laws that indicate a maximum in a
plot of initial rate versus total pressure. Four mechanisms (2, 5, 7, and 8) yield rate
expressions that predict the initial rate will continuously increase as the total pressure is
increased. Another three mechanisms (3, 6, and 9) indicate that the initial rate will first
increase with increasing total pressure, but then reach an asymptotic limit.
B. Plots of the data for the three temperature levels of interest are shown in Figures
P6.20.1 and P6.20.2.
701 F
649F
0.12
Initial rate
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
Total pressure (atm)
10
Figure P6.20.1 Tests of initial rate behavior of various Hougen-Watson models for
reactions at 649 and 701 ºF.
752 F
0.12
Initial rate
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
Total pressure (atm)
Figure P6.20.2 Test of initial rate behavior of various Hougen-Watson models for
reaction at 752 ºF.
Chapter 6 Solutions
6 - 42
Maxima are observed in the plots for reaction at 649 ºF and 701 ºF. Thus only
mechanisms 1, 4, and 10 will be consistent with these data. For reaction at 752 ºF the plot
does not reveal either a maximum or an asymptotic limit. Consequently, there may be a
shift in the mechanism at high temperatures to one with a different rate-controlling step.
Because neither a maximum nor an asymptotic limit is observed, the remaining candidate
models are numbers 2, 5, 7, and 8.
Last visited 11 May, 2014
Chapter 6 Solutions
6 - 43