equations of Kinetics 3

7/30/2019 equations of Kinetics 3

1/29

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


2/29

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

=


3/29

)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 =


4/29


5/29

What if

Can we treat it like in the example?

4.0

B

6.0

AA

CkCdt

dC

=


6/29

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

==


7/29

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

==


8/29


9/29

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


10/29

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


11/29


12/29

Irreversible reactions in parallel

Irreversible reactions in series

First-order reversible reactions

Second-order reversible reactions


13/29

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.


14/29

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 ==


15/29

Nonlinear-curve

fitting can be used


16/29

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.


17/29

Variable-volume batch reactor

dt

)VC(d

V

1

dt

dN

V

1r iii ==

+

= dtdVCVdC

V

1r

iii

dtdV

VC

dtdCr iii +=


18/29

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!


19/29

Thus

or

AA

A

A

A

x1

x1

C

C

0+

=

0

0

A

AA

A

A

A

C

C1

C

C1

x

+

=


20/29

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


21/29

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


22/29

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

+


23/29

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


24/29

ktV

Vln

C)x1ln(

C

)x1(

dxC

0A

Ax

0 AAA

A

AA

A

A

0

A

0

0

=

=+

=+


25/29

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

+

=


26/29

=

==

Ax

0 0AA

A

A ktV

V1ln)x1ln(

x1

dx


27/29

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

+

=


28/29

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 =+

+=

+


29/29

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(

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equations of Kinetics 3