ChE 101 – Chemical Reaction Engineering
Winter 2014
Homework #6
Due Tuesday, February 25, 2014 in class
Read Schmidt, Chapter 7,
Read Wolfenden and Snider, and Sheldon on Handouts page of the website.
NOTE:
Please list
all of your assumptions
and cite any external sources used when solving
each problem.
1)
Computational Problem of the Week:
In this problem, you are evaluating a system in
which nitrous oxide reacts with carbon monoxide in the presence of a ceria-promoted
rhodium catalyst to form dinitrogen and carbon dioxide. A proposed sequence for this
reaction is shown as follows:
Note that (g) represents a gas phase species, while (s) indicates that a species is bound in a
catalytic surface site.
(a)
Assuming that the surface coverage of oxygen is very small, derive the following rate
expression:
𝑟𝑟
=
𝐾𝐾
1
𝜃𝜃
0
𝑃𝑃
𝑁𝑁
2
𝑂𝑂
1 +
𝐾𝐾
2
𝑃𝑃
𝑁𝑁
2
𝑂𝑂
+
𝐾𝐾
3
𝑃𝑃
𝐶𝐶𝑂𝑂
What are the parameters K
1
, K
2
, and K
3
in your result?

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(b)
Linearize the rate expression derived in part (a) with respect to the reactants. Then use
linear regression on the following data to determine the kinetic parameters found in part
(a).
P
CO
(torr)
P
N2O
(torr)
Turnover rate (s
-1
)
45.6
11.4
0.00566
45.6
22.8
0.00953
45.6
45.6
0.0194
45.6
68.4
0.0242
45.6
91.2
0.0367
11.4
45.6
0.0386
22.8
45.6
0.0255
68.4
45.6
0.0157
91.2
45.6
0.00927
(c)
Use a nonlinear regression to fit the data above. How do the two methods compare?
Under what conditions might each fit be better? It may help to use a graphical argument
in your analysis.
None of the reactions are stated as rate limiting, thus we can only use the pseudo steady state
approximation to derive the rate law. Starting with N
2
O,
0 =
𝑟𝑟
𝑁𝑁
2
𝑂𝑂
(
𝑠𝑠
)
=
𝑘𝑘
1𝑓𝑓
𝑃𝑃
𝑁𝑁2𝑂𝑂
𝜃𝜃
𝑠𝑠
−
(
𝑘𝑘
1𝑏𝑏
+
𝑘𝑘
2
)
𝜃𝜃
𝑁𝑁2𝑂𝑂
Grouping together the constants into a single term,
We have
Looking at oxygen, we have
Finally looking at carbon monoxide,
The rate of product formation is given by
And using a mass balance on the free sites, we have
However, we are told in the problem statement that the surface concentration is negligible, thus
the surface balance becomes:
Making the proper substitutions, we get a rate constant of the form