CE 561, Exam 1, October 30, 2009
This exam consists of 4 questions, some with multiple parts.
You should be careful not to get stuck
on one part.
If you do not know how to do a problem, move on and return to it if you have time at
the end.
You may use a calculator and three lettersize sheets (2sided) of notes to aid you on this exam.
You
may not exchange notes with or otherwise consult your fellow students.
If you talk to your fellow
students during the exam, I will assume that you are cheating, you will be asked to leave, and you
will fail the exam.
You will have 2 hours to complete the exam.
Please use a separate blue book for each exam
problem.
Carefully explain any assumptions you make, label what part of what problem you are
working on, and define the symbols that you use.
The point value of each part is indicated – budget
your effort accordingly.
There are 100 points total.
1.
Consider the following reactions:
(1)
C
3
H
8
→
C
3
H
6
+ H
2
with rate
r
1
=
k
1
[C
3
H
8
]
(2)
C
3
H
8
→ C
2
H
4
+ CH
4
with rate
r
2
=
k
2
[C
3
H
8
]
(3)
C
2
H
6
→ C
2
H
4
+ H
2
with rate
r
3
=
k
3
[C
2
H
6
]
(4)
C
2
H
4
→ C
2
H
2
+ H
2
with rate
r
4
=
k
4
[C
2
H
4
]
(a)
Write these reactions in matrix form. (5 points)
In matrix form, these can be written as:
3
8
3
6
2
2
4
4
2
6
2
2
1
1
1
0
0
0
0
1
0
0
1
1
0
0
0
0
0
1
1
0
1
0
0
0
1
1
0
0
1
C H
C H
H
C H
CH
C H
C H
−
−
=
−
−
(b)
Write the rate equations for the concentrations of the seven species in matrix form (for isothermal
reaction at constant volume). (5 points)
The rate equations can be written terms of the vector of concentrations and a matrix of
rate constants as:
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
1
2
3
8
3
8
1
3
6
3
6
1
4
3
2
2
2
4
3
2
4
2
4
2
4
4
3
2
6
2
6
4
2
2
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
k
k
C H
C H
k
C H
C H
k
k
k
H
H
d
k
k
k
C H
C H
dt
k
CH
CH
k
C H
C H
k
C H
C H
−
−
=
−
−
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(c)
Describe briefly how you would solve these equations using matrix methods
(you do not
have to actually solve them)
(5 points).
If
M
is the matrix of rate constants from part (b), and
C
o
is the vector of concentrations at
t
= 0,
then we know that the solution can be written as
( )
exp(
)
o
C t
Mt
C
=
or
1
( )
(
exp(
)
)
o
C t
T
t
T
C
=
Λ
where
Λ
is the diagonal matrix of the eigenvalues of
M
, and
T
is the matrix whose columns
contain the corresponding eigenvectors of
M
.
To write the solution in this form, we find the
eigenvalues of
M
by solving
(
)
det
0
M
I
λ
−
=
for
λ
.
For each solution,
λ
i
, we find the
corresponding eigenvector
x
i
by solving
i
i
i
M x
x
λ
=
.
The eigenvectors and eigenvalues are then
placed in the matrices
T
and
Λ
, respectively.
We then take the inverse of
T
, then multiply the
matrices to get
C
(
t
) according to the equation given above.
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 Fall '09
 Chemical reaction, Rate equation, Reaction rate constant

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