CP303 Solutions to Set #3 (January-May 2012)

Question Bank 1:
Reaction Kinetics and Isothermal Batch Reactor Design
Solution to Q1:
Design equation for reactant A in an ideal batch reactor is given by
dN
A
dt
= r
A
V (3.1)
a) Since the reaction A+B C is rst-order in A and zero-order in B, we could write the
rate equation as
r
A
= k C
1
A
C
0
B
= k C
A
(3.2)
Combining (3.1) and (3.2), we get
dN
A
dt
= k C
A
V (3.3)
Using N
A
= C
A
V along with the assumption V , the volume of the reacting mixture in
the batch reactor, is a constant for a liquid-phase reaction, (3.3) could be simplied to
dC
A
dt
= k C
A
(3.4)
Upon integrating (3.4), we get
C
A
= C
Ao
exp (k t) (3.5)
where C
Ao
is the initial concentration of A in the batch reactor. Substituting C
A
= 0.5 C
Ao
at t = 10 min in (3.5), we get
0.5 C
Ao
= C
Ao
exp (10 k)
which gives k = ln (0.5)/10 = 0.069 min
1
. Using the numerical value of k in (3.5), we
get
C
A
= C
Ao
exp (0.069 t) (3.6)
Equation (3.6), at t = 20 min, gives
C
A
= C
Ao
exp (0.069 20) = 0.25 C
Ao
from which we know that a quarter of A is left unreacted after 20 minutes of the commence-
ment of the reaction.
1
b) Since the given reaction is rst-order in A and rst-order in B, we could write the rate
equation as
r
A
= k C
1
A
C
1
B
= k C
A
C
B
(3.7)
Combining (3.1) and (3.7), we get
dN
A
dt
= k C
A
C
B
V
which, for constant V , could be simplied to
dC
A
dt
= k C
A
C
B
(3.8)
Equation (3.8) could not be integrated without relating C
B
to either C
A
or t. From the
stoichiometry of the given reaction, we could write
moles of A reacted = moles of B reacted
N
Ao
N
A
= N
Bo
N
B
(3.9)
It is stated in the problem that there are equimolar quantities of A and B at the commence-
ment of the reaction. Therefore, N
Ao
= N
Bo
, which reduces (3.9) to N
B
= N
A
. Dividing it
by the volume of the reacting mixture, we get C
B
= C
A
. Substituting which in (3.8), we get
dC
A
dt
= k C
2
A
(3.10)
which upon integration gives
1
C
A
=
1
C
Ao
+k t (3.11)
Substituting C
A
= 0.5 C
Ao
at t = 10 min in (3.11), we get k = 0.1/C
Ao
min
1
. Using
this in (3.11), we get
1
C
A
=
1 + 0.1 t
C
Ao
(3.12)
Equation (3.12), at t = 20 min, gives
C
A
= C
Ao
/3
from which we know that one third of A is left unreacted after 20 minutes of the commence-
ment of the reaction.
c) Since the given reaction is second-order in A and rst-order in B, the rate equation becomes
r
A
= k C
2
A
C
B
(3.13)
Combining (3.1) and (3.13), we get
dN
A
dt
= k C
2
A
C
B
V
which, for constant V , could be simplied to
dC
A
dt
= k C
2
A
C
B
(3.14)
2
Using C
B
= C
A
, which is proved in part (b), (3.14) is reduced to
dC
A
dt
= k C
3
A
(3.15)
which upon integration gives
1
C
2
A
=
1
C
2
Ao
+ 2 k t (3.16)
Substituting C
A
= 0.5 C
Ao
at t = 10 min in (3.16), we get k = 3/(20 C
2
Ao
) min
1
.
Using this in (3.16), we get
1
C
2
A
=
1 + 0.3 t
C
2
Ao
(3.17)
Equation (3.17), at t = 20 min, gives
C
A
= C
Ao
/

7 = 0.378 C
Ao
Plot of C
A
/C
Ao
versus time for all the 3 cases analysed above is attached as Set3FigQ1.
Solution to Q2:
Design equation for reactant A in an ideal batch reactor is given by
dN
A
dt
= r
A
V (3.18)
Rate equation is given by
r
A
= k C
A
C
B
(3.19)
Combining (3.18) and (3.19) for a constant volume batch reactor, since the reaction takes
place in liquid phase, we get
dC
A
dt
= k C
A
C
B
(3.20)
a) Equation (3.20) could not be integrated without relating C
B
to either C
A
or t. From the
stoichiometry of the given reaction, one mole of A and one mole of B reacts to give 2 moles
of B. Therefore, we could say that the net reaction uses one mole of A to produce one mole
of B. We could therefore write
moles of A reacted = moles of B produced
N
Ao
N
A
= N
B
N
Bo
which, for a constant volume reactor, gives
C
Ao
C
A
= C
B
C
Bo
Since C
o
= C
Ao
+C
Bo
, the above equation could be rewritten as
C
B
= C
o
C
A
(3.21)
3
Combining (3.20) and (3.21) so as to eliminate C
B
, we get
dC
A
dt
= k C
A
(C
o
C
A
) (3.22)
which upon integration gives
k t =
_
dC
A
C
A
(C
o
C
A
)
=
1
C
o
_ _
1
C
A
+
1
C
o
C
A
_
dC
A
=
1
C
o
_
ln C
A
+
ln (C
o
C
A
)
1
_
+const
=
1
C
o
ln
_
C
o
C
A
C
A
_
+const (3.23)
Substituting C
A
= C
Ao
at t = 0 in (3.23), we get
const =
1
C
o
ln
_
C
o
C
Ao
C
Ao
_
(3.24)
Eliminating the const from (3.23) and (3.24), we get
k C
o
t = ln
_
(C
o
C
A
) C
Ao
C
A
(C
o
C
Ao
)
_
(3.25)
b) When the rate of reaction reaches its maximum,
d|r
A
|
dt
= 0 (3.26)
Using (3.19) and (3.21) in (3.26), we get
d[k C
A
(C
o
C
A
)]
dt
= k
_
C
A
dC
A
dt
+ (C
o
C
A
)
dC
A
dt
_
= k (C
o
2 C
A
)
dC
A
dt
= 0
from which we could conclude that when the reaction rate reaches its maximum,
C
A
= 0.5 C
o
Using the above value of C
A
in (3.25), we could get the time required for |r
A
| to reach its
maximum as
t

at |r
A
| maximum
=
1
k C
o
ln
_
C
Ao
C
Bo
_
since C
o
C
Ao
= C
Bo
.
4
Solution to Q3:
Design equation for reactant A in an ideal batch reactor is given by
dN
A
dt
= r
A
V (3.27)
When the reaction is assumed to be rst-order in A, as stated in the problem, the rate
equation becomes
r
A
= k C
A
(3.28)
Combining (3.27) and (3.28), we get
dN
A
dt
= k C
A
V (3.29)
Note: Since it is a gas-phase reaction, we shall work out the problem in terms of x
A
, con-
version of A, and not in terms of C
A
.
Using C
A
V = N
A
and N
A
= N
Ao
(1 x
A
), (3.29) can be reduced to
dx
A
dt
= k (1 x
A
) (3.30)
a) Upon integration using the initial condition x
A
= 0 at t = 0, (3.30) gives
k =
ln (1 x
A
)
t
(3.31)
We require the value of x
A
at some time during the reaction to be able to determine the
numerical value of k. It is given in the problem statement that there are 80 mole% A and 20
mole% inerts initially at 1 atm, and the total pressure rises by 40% in 3 min.
We could determine the value of x
A
at 3 min from the above information using
V =
P
o
P
T
T
o
V
o
(1 +
A
x
A
) (3.32)
where

A
=
N
Ao
N
To
_
(2 + 1) (1)
1
_
(3.33)
for the reaction A R + 2S.
Since there are 80 mole% A and 20 mole% inerts initially, we get N
Ao
/N
To
= 0.8. There-
fore, (3.33) gives

A
= 0.8 2 = 1.6 (3.34)
Using (3.34) and the data V = V
o
(constant volume reactor), T = T
o
(isothermal condi-
tion), P
o
= 1 atm and P
T
= 1.4 atm (pressure rises by 40% in 3 min) in (3.32), we get
1 =
1
1.4
(1 + 1.6 x
A
)
5
which gives x
A
= 1/4 at t = 3 min. Substituting which in (3.31)
k =
ln (1 1/4)
3 min
=
0.2876
3 min
= 0.0959 min
1
(3.35)
b) In another 3 min, i.e. when t = 6 min, we could determine the value of x
A
from (3.31)
using the numerical value of k from (3.35) as follows:
0.0959 =
ln (1 x
A
)
6
which gives
x
A
= 1 exp (0.0959 6) = 0.4375
Using x
A
= 0.4375, V = V
o
, T = T
o
and P
o
= 1 atm and
A
= 1.6 from (3.34), in (3.32),
we get
P

at 6 min
= (1 atm) (1 + 1.6 0.4375) = 1.7 atm
Solution to Q4:
Design equation for reactant A in an ideal batch reactor is given by
dN
A
dt
= r
A
V (3.36)
The rate equation is given by
r
A
= k C
A
(3.37)
Combining (3.36) and (3.37), we get
dN
A
dt
= k C
A
V (3.38)
a) Since we have been asked to prove C
A
= C
Ao
exp (k t), we write (3.38) in terms of C
A
using N
A
= C
A
V as
d(C
A
V )
dt
= k C
A
V
which, for a constant volume reactor, becomes
dC
A
dt
= k C
A
Integrating the above with the initial condition C
A
= C
Ao
at t = 0, we get
C
A
= C
Ao
exp (k t) (3.39)
b) We need to verify the assumption that the reaction is rst-order in A using the 6 sets of
C
A
vs t values given. If the data t (3.39) then the assumption is correct. Since (3.39) is
6
an exponential function, we need to rst convert it to get a straight line relationship, which
could be done as follows:
ln C
A
= k t ln C
Ao
(3.40)
Plot of ln C
A
versus t made using the C
A
vs t data is attached as Set3FigQ4. The plot
shows that the ln C
A
versus t data can be tted by the straight line, ln C
A
= 0.0027 t +
4.1352, having R
2
= 99.99%, and therefore the assumption that the reaction in rst-order in
A is a valid one.
The slope of the straight line gives the value of k as 0.0027 min
1
.
c) Applying the ideal gas equation of state to relate P to the other properties of the reacting
mixture at time t, we get
P V = N
T
RT (3.41)
where N
T
= N
A
+ N
P
+ N
Q
in usual notation.
Stoichiometry for the reaction A P +Q gives
N
Ao
N
A
= N
P
N
Po
= N
Q
N
Qo
which becomes
N
Ao
N
A
= N
P
= N
Q
(3.42)
since N
Po
= N
Qo
= 0.
Combining (3.41) and (3.42), we get P V = (2 N
Ao
N
A
) RT, which, for a constant
volume reactor, gives
P =
2 N
Ao
N
A
V
RT = (2 C
Ao
C
A
) RT
Note:
Solutions to Q5, Q6 and Q7 are not provided since they are part of Assignment 1.
Solution to Q8:
Design equation for reactant A in an ideal batch reactor is given by
dN
A
dt
= r
A
V (3.43)
a) The rate equation for the given elementary reversible reaction is
r
A
= k
1
C
A
+k
2
C
B
(3.44)
Combining (3.43) and (3.44) for a constant volume (since it is a liquid-phase reaction)
reactor, we get
dC
A
dt
= k
1
C
A
+k
2
C
B
(3.45)
7
Stoichiometry for the given reversible reaction yields that when one mole of A disappears
in the reaction, all of it would be converted into one mole of B. Therefore, we get N
Ao
N
A
=
N
B
N
Bo
, which, since V is a constant, reduces to C
Ao
C
A
= C
B
C
Bo
It is given that
there are 2 mol/litre of A present in the reactor initially and no B. Therefore, we get
2 C
A
= C
B
(3.46)
Combining (3.45) and (3.46) so as to eliminate C
B
, we get
dC
A
dt
= k
1
C
A
+k
2
(2 C
A
) (3.47)
which could be integrated as follows:
dC
A
dt
+ (k
1
+k
2
) C
A
= 2 k
2
d
dt
{C
A
exp [(k
1
+k
2
)t]} = 2 k
2
exp [(k
1
+k
2
)t]
C
A
exp [(k
1
+k
2
)t] =
2 k
2
k
1
+k
2
exp [(k
1
+k
2
)t] +const (3.48)
Since C
A
= 2 mol/litre at t = 0, (3.48) gives
const = 2
2 k
2
k
1
+k
2
=
2 k
1
k
1
+k
2
(3.49)
Eliminating const from (3.48) and (3.49), we get the following:
C
A
exp [(k
1
+k
2
)t] =
2 k
2
k
1
+k
2
exp [(k
1
+k
2
)t] +
2 k
1
k
1
+k
2
C
A
=
2 k
2
k
1
+k
2
+
2 k
1
k
1
+k
2
exp [(k
1
+k
2
) t] (3.50)
b) The rate of reaction reaches zero value at equilibrium. Therefore, (3.44) combined with
(3.46) gives
r
A

at equilibrium
= k
1
C
A,eqm
+k
2
(2 C
A,eqm
) = 0 (3.51)
where C
A,eqm
stands for the equilibrium concentration of A. Since C
A,eqm
is given as 0.33
mol/litre, (3.51) gives
k
1
k
2
=
2 C
A,eqm
C
A,eqm
=
2 0.33
0.33
= 5 (3.52)
c) The concentration of A is said to be reduced to 1 mol/litre in 0.9 min. Substituting the
above in (3.50), we get
1 =
2 k
2
k
1
+k
2
+
2 k
1
k
1
+k
2
exp [(k
1
+k
2
) 0.9]
Since k
1
= 5 k
2
from (3.52), substituting which in the above we get
1 =
2
6
+
2 5
6
exp [6 k
2
0.9]
8
which reduces to
k
2
= ln (0.4)/5.4 = 0.17 min
1
and therefore, we get
k
1
= 5 k
2
= 5 0.17 min
1
= 0.85 min
1
Solution to Q9:
Design equation for reactant A in an ideal batch reactor is given by
dN
A
dt
= r
A
V (3.53)
a) The rate equation for the given reversible reaction is given by
r
A
= k
f
C
A
C
B
+k
b
C
P
(3.54)
Stoichiometry for the given reversible reaction yields that one mole of A combines with one
mole of B to produce one mole of P. Therefore, we get N
Ao
N
A
= N
Bo
N
B
= N
P
N
Po
,
which, since V is a constant, reduces to C
Ao
C
A
= C
Bo
C
B
= C
P
C
Po
. Initially, there
are 1 kgmol/m
3
each of A and B present in the reactor and no P. Therefore, we get
1 C
A
= 1 C
B
= C
P
(3.55)
Combining (3.54) and (3.55) so as to eliminate C
B
and C
P
, we get
r
A
= k
f
C
2
A
+k
b
(1 C
A
) (3.56)
b) To determine k
f
and k
b
, (3.56) should be rewritten as follows:
r
A
C
2
A
= k
f
k
b
_
1 C
A
C
2
A
_
(3.57)
We shall then construct the data set y = r
A
/C
2
A
versus x = (1 C
A
)/C
2
A
from the C
A
versus (r
A
) data set provided. The slope of the best straight line t to the y versus x data
points, as per (3.57), gives k
b
and the intercept of which gives k
f
.
Plot of y = r
A
/C
2
A
versus x = (1 C
A
)/C
2
A
is attached as Set3FigQ9. The plot shows
that the data can be tted by the straight line, y = -0.2933 X + 1.5031, having R
2
= 98.83%.
From which, we get k
b
= 0.2933 0.3 per min and k
f
= 1.5031 1.5 m
3
per (kgmol.min).
c) Using N
A
= C
A
V along with the expression for r
A
from (3.56) in (3.53) and utilizing the
fact V may be assumed to be a constant for a liquid-phase reaction, we get
dC
A
dt
= k
f
C
2
A
+k
b
(1 C
A
)
Substituting the numerical values of k
f
and k
b
in the above and integrating it from C
A
=
1 kgmol/m
3
at t = 0 to C
A
= 0.5 kgmol/m
3
at t = t
f
, we get
t
f
=
_
0.5
1
dC
A
1.5 C
2
A
0.3 C
A
+ 0.3
=
1
1.5
_
1
0.5
1
C
2
A
+ 0.2 C
A
0.2
dC
A
By carrying out the above integration we could get t
f
.
9
Solution to Q10:
a) The numerical values of the forward reaction rate constant k
1
and the backward reaction
rate constant k
2
are to be determined. We therefore require two independent equations in k
1
and k
2
. One equation could be obtained from the information the concentration of A drops
from 0.8 mol/litre to 0.6 mol/litre in one minute, provided we rst nd the dependence of
C
A
on time. The other equation could be obtained from the information that the equilibrium
concentration of A is 0.52 mol/litre.
To determine the dependence of C
A
on time, let us start with the design equation for
reactant A in an ideal batch reactor, which is
dN
A
dt
= r
A
V (3.58)
and the rate equation for the given rst-order reversible reaction, which is
r
A
= k
1
C
A
+k
2
C
R
(3.59)
Stoichiometry for the given reversible reaction yields that when one mole of A disappears
in the reaction, all of it would be converted into one mole of R. Therefore, we get N
Ao
N
A
=
N
R
N
Ro
, which, since V is a constant, reduces to C
Ao
C
A
= C
R
C
Ro
. It is given that
there are 0.8 mol/litre of A and 0.5 mol/litre of R present in the reactor initially. Therefore,
we get
0.8 C
A
= C
R
0.5 (3.60)
Combining (3.59) and (3.60) so as to eliminate C
R
, we get
r
A
= k
1
C
A
+k
2
(1.3 C
A
) (3.61)
Combining (3.58) and (3.61) so as to eliminate r
A
, we get
dC
A
dt
= k
1
C
A
+k
2
(1.3 C
A
) (3.62)
since N
A
= C
A
V and since V could be taken as a constant for a liquid-phase reaction.
Equation (3.62) could be integrated as follows:
dC
A
dt
+ (k
1
+k
2
) C
A
= 1.3 k
2
d
dt
{C
A
exp [(k
1
+k
2
)t]} = 1.3 k
2
exp [(k
1
+k
2
)t]
C
A
exp [(k
1
+k
2
)t] =
1.3 k
2
k
1
+k
2
exp [(k
1
+k
2
)t] +const (3.63)
Since C
A
= 0.8 mol/litre at t = 0, (3.63) gives
const = 0.8
1.3 k
2
k
1
+k
2
=
0.8 k
1
0.5 k
2
k
1
+k
2
(3.64)
10
Eliminating const from (3.63) and (3.64), we get the following:
C
A
exp [(k
1
+k
2
)t] =
1.3 k
2
k
1
+k
2
exp [(k
1
+k
2
)t] +
0.8 k
1
0.5 k
2
k
1
+k
2
C
A
=
1.3 k
2
k
1
+k
2
+
0.8 k
1
0.5 k
2
k
1
+k
2
exp [(k
1
+k
2
) t] (3.65)
Since C
A
= 0.6 mol/litre at t = 1 min, (3.65) reduces to
0.6 =
1.3 k
2
k
1
+k
2
+
0.8 k
1
0.5 k
2
k
1
+k
2
exp [(k
1
+k
2
)] (3.66)
which gives one equation relating k
1
and k
2
.
It is also given in the problem statement that at equilibrium, the concentration of A is
0.52 mol/litre. Since r
A
= 0 at equilibrium, (3.61) gives
r
A

at equilibrium
= k
1
(0.52) +k
2
(1.3 0.52) = 0
which gives
k
1
k
2
=
0.78
0.52
= 1.5 (3.67)
Using k
1
= 1.5 k
2
from (3.67) in (3.66), we get
0.6 =
1.3
2.5
+
0.8 1.5 0.5
2.5
exp (2.5 k
2
)
which gives k
2
= 0.5 min
1
. Substituting it in (3.67), we get k
1
= 0.75 min
1
.
The equilibrium constant K
eq
is dened by K
eq
= k
1
/k
2
, which as per (3.67) is 1.5. And,
of course, K
eq
has no unit in this case.
b) Since K
eq
= 1.5 at 720 K, we could write
exp
_
H
720 R
_
1.5 (3.68)
It is given in part (b) of the problem statement that C
eqm
= 0.520/2 mol/liter at 630 K.
Using the C
eqm
in (3.61), we get
r
A

at equilibrium
= k
1
(0.52/2) +k
2
(1.3 0.52/2) = 0
which gives
K
eq
=
k
1
k
2
=
1.3 0.26
0.26
= 4
Since K
eq
= 4 at 630 K, we could write
exp
_
H
630 R
_
4 (3.69)
11
Dividing (3.68) by (3.69), we get
exp
_
H
R
_
1
720

1
630
__
=
1.5
4
which gives
H = 4943.4 K R = 41, 100 kJ/kgmol
when using R = 8.314 kJ/(kgmol.K) in the above.
Solution to Q11:
a) Design equation for reactant A in an ideal batch reactor with constant volume is given by
dC
A
dt
= r
A
(3.70)
The rate equation for the rst-order reaction given by A B is
r
A
= k
1
C
A
(3.71)
Combining (3.70) and (3.71) and integrating it with the initial condition C
A
= C
Ao
at t
= 0, we get
C
A
= C
Ao
exp (k
1
t) (3.72)
To write the design equation for B, which is a product in A B and a reactant in
B C, let us start with the mass balance for B over the volume of the ideal batch reactor
as follows:
mass of B entering the reactor during time dt
+ mass of B being generated in the reaction A B during time dt
= mass of B leaving the reactor during time dt
+ mass of B accumulated within the reactor during time dt
+ mass of B disappearing in the reaction B C during time dt
which becomes
0 + (r
B
)

AB
M
B
V dt = 0 +d(N
B
M
B
) + (r
B
)

BC
M
B
V dt (3.73)
Removing the molar mass M
B
from (3.73) and rearranging it, we get the design equation
for B in an ideal batch reactor as follows:
dN
B
dt
=
_
(r
B
)

AB
(r
B
)

BC
_
V (3.74)
Since (r
B
)

AB
= (r
A
)

AB
= k
1
C
A
and (r
B
)

BC
= k
2
C
B
, (3.74)could be re-
duced to
dN
B
dt
= (k
1
C
A
k
2
C
B
) V
12
which, for a constant volumes reactor, becomes
dC
B
dt
= k
1
C
A
k
2
C
B
(3.75)
The stoichiometry of a series reaction could NOT be used to relate A and B. And,
therefore, we shall use C
A
from (3.72) in (3.75) to get a dierential equation in C
B
as follows:
dC
B
dt
= k
1
C
Ao
exp (k
1
t) k
2
C
B
(3.76)
which could be integrated as follows:
dC
B
dt
+k
2
C
B
= k
1
C
Ao
exp (k
1
t)
d
dt
{C
B
exp (k
2
t)} = k
1
C
Ao
exp (k
1
t) exp (k
2
t)
C
B
exp (k
2
t) =
k
1
C
Ao
k
1
+k
2
exp [(k
1
+k
2
)t] +const (3.77)
Since C
B
= 0 at t = 0, (3.77) gives
const =
k
1
C
Ao
k
1
+k
2
(3.78)
Eliminating const from (3.77) and (3.78), we get the following:
C
B
exp (k
2
t) =
k
1
C
Ao
k
1
+k
2
exp [(k
1
+k
2
)t]
k
1
C
Ao
k
1
+k
2
C
B
=
k
1
C
Ao
k
1
+k
2
(exp (k
1
t) exp (k
2
t)) (3.79)
b) When C
B
is at its maximum, dC
B
/dt = 0. Therefore, dierentiating (3.79) and equating
it to zero gives
k
1
exp (k
1
t
opt
) = k
2
exp (k
2
t
opt
) (3.80)
where t
opt
is the time at which C
B
reaches its maximum value.
Rearranging (3.80), we get
t
opt
=
ln (k
2
/k
1
)
exp (k
1
+k
2
)
= 4.5 hr
since k
1
= 0.35 hr
1
and k
2
= 0.13 hr
1
. Substituting the values of t
opt
, k
1
and k
2
in (3.79),
along with C
Ao
= 4 mol/m
3
and C
Bo
= C
Co
= 0, we get
C
B

max
=
0.35 4
0.35 + 0.13
(exp (0.35 4.5) exp (0.13 4.5)) = 2.23 mol/m
3
c) C
A
and C
B
as functions of time are given by (3.72) and (3.79), respectively. We need to
determine C
C
as a function of time, which can be done as follows for a series reaction in which
stoichimetry cannot be used.
13
The rate of change of C
A
in the reactor can be obtained by combining (3.70) and (3.71)
as
dC
A
dt
= k
1
C
A
(3.81)
The rate of change of C
B
in the reactor is given by (3.75).
The rate of change of C
C
in the reactor can be obtained by combing the design equation
for product C in an ideal batch reactor with constant volume, given as
dC
C
dt
= r
C
,
and the rate equation for the rst-order reaction B C, given as
r
C
= k
2
C
B
as
dC
C
dt
= k
2
C
B
(3.82)
Adding (3.81), (3.75) and (3.82) gives the following:
d (C
A
+C
B
+C
C
)
dt
= k
1
C
A
+k
1
C
A
k
2
C
B
+k
2
C
B
= 0
which upon integration yields
C
A
+C
B
+C
C
= constant .
Intitial conditions gives constant as 4 mol/m
3
. Therefore,
C
C
= (4 C
A
C
B
) . (3.83)
Plot of C
A
, C
B
and C
C
as functions of time is attached as Set3FigQ11.
Observe in the plot that the concentration of the reactant, C
A
, keeps decreasing with
time and the concentration of the product, C
C
, keeps increasing with time. Whereas, the
concentration of the intermediate component, C
B
, experiences a maximum.
Note:
Solutions to Q12 and Q13 are not provided since they are part of Assignment 1.
14