equations of Kinetics 3
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Transcript of equations of Kinetics 3
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7/30/2019 equations of Kinetics 3
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AA
A kCdt
dCr ==
=t
0
C
C A
A dtkC
dCA
0A
ktC
Cln
0A
A =
Irreversible unimolecular-type first-order
reaction
A Products
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For convenience let
0
0
A
AA
AN
NNx
=
Fractionalconversion
)x1(CV
x1N
V
NC
AA
A
A
A
A 00=
==
AAAA xCCC 00 =
AAA dxC0dC 0=
AAA dxCdC0
=
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)xCC(kdt
dxCAAA
AA
00
0 =
)x1(kdt
dxA
A =
Substitute dCA and CA into the rate equation
=
t
0
x
0 A
A dtkx1
dxA
kt)x1ln( A =
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What if
Can we treat it like in the example?
4.0
B
6.0
AA
CkCdt
dC
=
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Irreversible bimolecular-type second-order
reaction
A + B Products
A and B have reacted equally
Then
BABA
A CkCdt
dC
dt
dCr ===
BBAA xCxC 00 =
)xCC)(xCC(kdt
dxCr AABAAA
AAA 00000
==
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Let0
0
A
B
C
CM =
)xM)(x1(kC
dt
dxCr AA
2
AA
AA 00==
=
A
0
x
0
t
0
A
AA
A dtkC
)xM)(x1(
dx
A
B
AB
AB
A
A
A
B
MC
Cln
CC
CCln
)x1(M
xMln
x1
x1ln
0
0 ==
=
1M,kt)CC(kt)1M(C000 ABA
==
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Overall order of irreversible reactions from
half-life ( )
Consider the irreversible reaction
At any time
Therefore
21t
ProductsBA ++ Lb
B
a
AA
A CkC
dt
dCr ==
=A
B
C
C
L
LL
++
=
== baA
bb
A
a
AA
A CkCkCdt
dCr
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nAA Ckdt
dC=
Integrating for n 1
Defining the half-life as the time needed for the
reduction of concentration of reactant for one-half of
the original value then we get
t)1n(kCC n1An1
A 0=
n1
A
1n
021 C
)1n(k
12t
=
kk
b
=
L
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Irreversible reactions in parallel
Irreversible reactions in series
First-order reversible reactions
Second-order reversible reactions
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Differential method of analysis of data
Guess the rate equation
Find (1/V)(dN/dt)
Plot of the data
Good fit Unfit
Rate equation
is satisfied by
the data.
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General procedure
1. Hypothesize a mechanism.
2. Plot concentration vs. time from
experimental data3. Draw a smooth curve through the data.
4. Determine the slope at each concentration.
)C,k(fdt
dC
r
A
A ==
)C(kf
dt
dCr AA ==
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Nonlinear-curve
fitting can be used
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5. Evaluate f(C) for each concentration
If we obtain straight line through the
origin, our mechanism is consistent to the
data. If not hypothesize another one.
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Variable-volume batch reactor
dt
)VC(d
V
1
dt
dN
V
1r iii ==
+
= dtdVCVdC
V
1r
iii
dtdV
VC
dtdCr iii +=
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LetA = fractional change in volume
)x1(VV AA0 +=
0x
0x1x
A
AA
VVV
=
== =
Complete
conversion
Noconversion
)x1(NN AAA 0 =
)x1(
)x1(C
)x1(V
)x1(N
V
NC
AA
AA
AA0
AAAA 0
0
+
=
+
==
Therefore
Linear change of
volume with
concentration!
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Thus
or
AA
A
A
A
x1
x1
C
C
0+
=
0
0
A
AA
A
A
A
C
C1
C
C1
x
+
=
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Substitute V and NA into the rate equation
or
IfA = 0
dt
dN
V
1r AA =
dt
)x1(dN
)x1(V
1rAA
AA0
A0
+=
dt
dx
)x1(
Cr A
AA
A
A0
+=
dt
dxCr AAA 0= Constant-volume
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Integral method of analysis
The analysis requires again the integration of
the rate expression to be tested.
For T, P = constant
dt
dx
x1
C
dt
dN
V
1r A
AA
AAA
0
+
==
tdt)r)(x1(
dx
C
A
0
x
0
t
0AAA
AA ==+
We need to know the exact form of rA in
order to integrate this equation
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Differential method of analysis
The procedure is the same as for constant-
volume batch reactor analysis except that we must
replace
dt
VlndC
dt
dCby
dt
dCA
AA +
dtdx
x1Cor A
AA
A0
+
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Zero-order reactions
For homogeneous zero-order reaction (-rA f(C))
with and
we obtain
dt
dN
V
1r AA
=
kdt
dx
x1
C
rA
AA
A
A0
=+=
)x1(VV AA0 += =+
A
0
x
0
t
0AAA
AA dt)r)(x1( dxC
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ktV
Vln
C)x1ln(
C
)x1(
dxC
0A
Ax
0 AAA
A
AA
A
A
0
A
0
0
=
=+
=+
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First-order reactions
For unimolecular-type first-order reaction, P = const
Substitute xA and -rA
A
A
A
kCdt
dN
V
1r ==
AA
AAA
AA
A
Ax1
)x1(kC
dt
dx
x1
Cr 00
+=
+=
dt
dx
)x1(
Cr A
AA
A
A0
+=
0
0
A
AA
A
A
A
C
C1
CC1
x
+
=
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=
==
Ax
0 0AA
A
A ktV
V1ln)x1ln(
x1
dx
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Second-order reactions
For bimolecular-type second-order reaction
2A Products
or
A + B Products,
The rate is
Substitute xA and -rA
00 BACC =
2
AA kCr =
dt
dx
)x1(
Cr A
AA
A
A0
+=
0
0
A
AA
A
A
A
C
C1
C
C1
x
+
=
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2
AA
A2
AA
AA
A
A
x1
x1kC
dt
dx
x1
Cr
0
0
+
=
+
=
tkC)x1ln(
x1
x)1(dx
)x1(
x10
A
AAA
A
AAA
x
0
2
A
AA =+
+=
+
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nth-order reactions
From
We obtain
n
AA
An
A
n
AAx1
x1kCkCr
0
+
==
tdt
)r)(x1(
dxC
A
0
x
0
t
0AAA
AA ==
+
=+A
0
x
0
1n
AAn
A
1n
AA ktCdx)x1(
)x1(