Week 2. Chemical Kinetics Analysis of Rate Equation
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8/13/2019 Week 2. Chemical Kinetics Analysis of Rate Equation
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Reaction Engineering
Chemical Kinetics /
Analysis of rate equations
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Chemical Kinetics - outcomes
Understand the significance of kinetics
Aware of concentration dependencies
Understand the difference elementary / non-elementary reactions
Calculate the order of reaction
Aware of temperature dependencies Arrhenius, activation energy
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The Rate Equation / Stoichiometry
aA bB cC dD
dtdN
Vr AA 1 rate of disappearance of A
dt
dN
Vr CC
1 rate of appearance of C
d
r
c
r
b
r
a
r DCBA
Relationship
between rates of
reaction
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Concentration / Temp. dependency terms
.).,( tempconcfr
,...)],()][([ BAA CCfTkr
Reaction rate
(constant)
Concentration
dependent terms
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Concentration-dependent term
,...)],([ BAA CCfkr Almost without exception determined by experimental observation!!!
Most common expression:
DBAA CCkCr ....
Order of reaction: AinorderBinorder
orderoverall...n
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Elementary / Non-elementary Reactions
Elementary reactions
iff stoichiometric coefficients are the same as theindividual reaction order of each species
Non-elementary reactions
stoichiometry does not match the kinetics
DCBA
BAA CkCr
stoichiometric coeff: 1 dcba
1 a
BAA CkCr 2 2
1
BAA CkCr
1 b
A sequence of elementary reactions 6
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Elementary reactions / Molecularity
= the number of molecules involved in a reaction
refers ONLY to elementary reaction!!!
Elementary reactions
PX
PX2
PYX PX3
PYX 2
PZYX
Rate Law
XkC
2
XkC
YXCkC3
XkC
2
YXCkC
ZYX CCkC
Molecularity Comments
unimolecular
bimolecular
bimolecular
termolecular
termolecular
termolecular
X decomposes
X collides with X
X collides with Y
X+X+X collide
X+Y+Y collide
X+Y+Z collide
tymoleculariorder
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Non-elementary reactions
Overall reaction:
)(2)(2)()(4 222 gBrgOHgOgHBr kexp
Sequence of reactionsproposed mechanism:
)()()( 2 gHOOBrgOgHBr
)(2)()( gHOBrgHBrgHOOBr
)()()()( 22 gBrgOHgHBrgHOBr
)()()()( 22 gBrgOHgHBrgHOBr
(1)
(2)
(3)
(4)
k1
k2
k3
k4
elementary
reactions(bimolecular)
(slow)
(fast)
(fast)
(fast)
kexp=func(k1,k2,k3,k4)
kexp=k1 8
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Concentration / Temp. dependency terms
.).,( tempconcfr
,...)],()][([ BAA CCfTkr
Reaction rate
(constant)
Concentration
dependent terms
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Rate constant (k)
,...)],([ BAA CCfkr
Dimensions vary with order of reaction (n):
nionconcentrattime 11 )()(
Reaction order
(mol/m3)
CA
(mol/m3)
(mol/m3)
(mol/m3.s)
(mol/m3.s)
(mol/m3.s)
-rA Rate law
(mol/m3)/skrA
AA kCr
2
AA kCr
1/s
(m3/mol)/s
zero
1st
2nd
k
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Temperature dependency
Activation energy
RTEekk /0
Activation energy (E) calculation:
data for same concentration different temperatures
TR
Ekk 1
lnln 0
TR
Ekk
1
3.2loglog 0or
211
2 11ln
TTR
E
k
k
T1,T2
ln, log
211
2 11
3.2
log
TTR
E
k
k
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Activation energy calculation
example 1
First order reaction: CBA
Data: k (1/s) 0.00043 0.00103 0.0018 0.00355 0.00717T(K) 313 319 323 328 333
211
2 11
3.2log
TTR
E
k
k
21
12
/1/1
)/log()3.2(
TT
kkRE
00303.0/1005.0 11 Tk
00319.0/10005.0 22 Tk
Hint: for 1)/log(1.0 2112 kkkkmolkJE /12013
0
0.5
1
1.5
2
2.5
3
3.5
4
0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325
1/T Vs log k
Log k
1/T
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Activation energy calculation
example 2
To obtain pasteurised milk:
heating at 63C for 30 min
heating at 74C for 15 s
KTt 336min30 11
212
1
1
2 11lnlnTTR
E
t
t
k
k
Reaction rate is inversely
proportional with reaction time
KTt 347min25.0sec15 22
347
1
336
1
314.825.0
30ln
E
molJE /422000
Find the activation energy of the process
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Data on the tenebrionid beetle whose body mass is 3.3 g show that it can
push a 35-g ball of dung at 6.5 cm/s at 27 C, 13 cm/s at 37 C, and 18 cm/s
at 40 C. How fast can it push dung at 41.5 C? [B. Heinrich. The Hot-
Blooded Insects (Cambridge, Mass.: 1993).]
Activation energy calculation
example 3
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The behavior of the beetle can be modeled by the Arrhenius equation.
The beetle's speed (k) increases exponentially with temperature.
Arrhenius equation:
To calculate how fast the beetle can push the ball determine:
the activation energy (E)
the Arrhenius coefficient (k0) of the beetle
Activation energy calculation
example 3
RTEekk /0
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k1 = 6.5 T1 = 300K
k2=13 T2 =310 K
k3= 18 T3=313 K
k4 = ? T4 = 314.5 K
ln(k) for T=314.5 is equal to ~2.95
This corresponds to a k4value of 19.1 cm/s.
Activation energy calculation
example 3
Plot ln(k) vs. 1/T
Predict ln(k) for T=314.5 K
Rate (cm/s) Temp (K)
How would you solve the
problem analytically?17
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Analysis of rate equations - outcomes
Be able to use experimental data to get empirical rate equations
Necessity of a rate equation for design purposes
Able to analyse different types of reactions
Expand the analysis to multiple reactions
Chain reactions and their interpretation & analysis
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Analysis of rate equations general
issues
dt
dN
Vr ii
1The rate equation:
The form might be suggested by:Theoretical considerations
Empirical curve-fitting procedure
The value of equation constants experimental only!
2 step-procedure:
Find concentration dependency at Temp=const.
Find temperature dependency of the rate constants
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dt
dN
Vr ii
1The rate equation:
Equipment used for getting the empirical data
batch reactor container that holds the reactants usually for
homogeneous reactions
flow reactor continuous flow in / flow out - usually for
heterogeneous reactions
Measurements:
concentration of a component
change in physical properties (conductivity, refractive index)
change in total pressure when V=constant.
change in volume when P=constant.
Analysis of rate equations general
issues (contd.)
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dt
dN
Vr ii
1The rate equation:
Procedures for analysing the kinetic data:
integral method:
guess a form of rate equation
integration & mathematical manipulation
predict the plot of C=f (time) as a straight line
plot the data and if good fit accept the equation
differential method:
find (1/V)(dN/dt) from the data
test the fit directly (no integration)
Analysis of rate equations general
issues (contd.)
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0
0
A
AAA
N
NNX
Conversion of A:
Constant volume batch reactor
conversion
= the fraction of A
reacted away
0000
0 1/
/11
A
A
A
A
A
A
A
AAA
C
C
VN
VN
N
N
N
NNX
The volume of reaction mixture = constant
= constant density reaction system dt
dCr AA
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A. Integral method- irreversible, unimolecular, first order rxs. -
sProductA
A
A
A kCdt
dC
r
Test the first-order rate equation form:
using concentration
using conversion
0
1A
AA
CCX
0A
AA
C
dCdX
AAAA
A kCdt
dXC
dt
dCr 0
0A
AA
C
Ck
dt
dX
)1( AA
A Xkdt
dXr
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A. Integral method- irreversible, unimolecular, first order rxs. -
tX
A
A dtk
X
dXA
00
1
Separate & integrate:
tC
CA
A dtkC
dCA
A 00
ktC
C
A
A 0
ln
using concentration
using conversion
ktXA )1ln(
AA kC
dt
dC
)1( AA Xkdt
dX
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A. Integral method- irreversible, unimolecular, first order rxs. -
Plot the experimental data & the proposed equation line:
-ln(CA
/CA0
)orl
n(1-XA
)
t
If the data fit Accept first-order rate equation
kslope
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t
A
X
AA
A dtkCXMX
dXA
00
0 1
Separate & integrate:
A
B
AB
AB
A
A
A
B
MC
C
CC
CC
XM
XM
X
Xlnln
)1(ln
1
1ln
0
0
B1. Integral method- irreversible, bimolecular, second order rxs. -
ktCCktMC ABA )()1( 000
1MCondition:00 BA
CC or
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B2. Integral method- irreversible, bimolecular, second order rxs. -
sProductA2
22
0
2
)1( AAAA
A XkCkCdt
dCr
The second-order rate equation form:
ktX
X
CCC A
A
AAA 1
111
00
t
AC
1
kslope
0
1
ACerceptint
t
kCslope A0
A
A
X
X
1
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B3. Integral method- irreversible, bimolecular, second order rxs. -
sProductBA 2
)2)(1(20 AAABAAA XMXkCCkCdtdCr
The second-order rate equation form:
kt
X
X
CCC A
A
AAA
2
1
111
00
ktMCXM
XM
CC
CCA
A
A
AB
AB )2()1(
2lnln 0
0
0
2M
2M31
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