CE 561, Exam 1, November 7, 2008
This exam consists of 3
questions, each 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 two reactions:
(1)
A
→
B with rate
r
1
=
k
1
C
A
(2)
B
↔
C with rate
r
2
=
k
2
C
B
–
k
3
C
C
(a)
Write these reactions in matrix form. (5 points)
In matrix form, these can be written as:
11 0
0 11
0
01
1
A
B
C
−
−=
−
If the forward and reverse directions of the second reaction are treated as separate
reactions, or as
0
01 1
A
B
C
−
=
−
If the second reaction is treated as a single reaction
(b)
Write the rate equations for the concentrations of the three species in matrix form. (5 points)
The rate equations can be written terms of the vector of concentrations and a matrix of
rate constants as:
1
1
23
00
0
AA
BB
CC
Ck
C
d
C
k
kkC
dt
C
k
kC
−
=
−
−
(c)
Explain how you would solve these equations to obtain expressions for the concentrations of A,
B, and C as functions of time, for initial conditions of
C
A
(
t
= 0) =
C
Ao
,
C
B
(
t
= 0) =
C
C
(
t
= 0) = 0.
Take your solution as far as you can in the time available.
(10 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
Ct
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
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View Full Documenteigenvalues of
M
by solving
( )
det
0
MI
λ
−=
for
.
For each solution,
i
, we find the
corresponding eigenvector
x
i
by solving
i
ii
Mx
x
=
.
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.
( ) ( ) ( )( )
( )
( ) ( )
( )
( )( )
1
1
2
3
1
2
3
23
2
1
2
3
1
2
3
00
det
det
0
0
0
k
M
I
k
k
k
k
k
k
kk
k
k
k
k
k
k
λλ
−−
− ±
±
−− −− −− −
±
+ +
+ −
=
=
So, the three eigenvalues are
λ
1
= 
k
1
,
λ
2
= (
k
2
+
k
3
), and
λ
3
= 0.
Corresponding eigenvectors are
obtained by substituting the eigenvalues into
i
x
=
.
For
λ
1
= 
k
1
this gives
( )
( )
( )
( )
( )
1,1
1
2
1
3
1,2
2
3
1
1,3
1 1,1
1
2
3 1,3
2 1,2
1
3
2
31
2
21
3
2
2
1
3
1
12
13
1
1
1
13 1
0
0
0
0
0
x
k
k
x
k
kkx
kx
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 Fall '09
 k2, Rate equation, Reaction rate constant, K3

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