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### Transcript of Reversible First-Order Reactions - kguo/pchem3/9502/L05_Ch18.pdf2006/03/17 51 Reversible First-Order...

2006/03/17 51

Reversible First-Order Reactions

If a reaction does not go to completion, the rate law is of the form =f b, where is f the rate of the forward reaction and b is the rate of the backward reaction.

A reversible process in kinetics means the backward reaction is significant.

The effect of a product on the reaction rate may be determined by adding it initially.

The products may be inhibitory. A reaction is said to be autocatalytic if a product of the reaction cause it to go faster.

2006/03/17 52

Reversible First-Order Reactions

For a reversible reaction A B the rate law is of the form

where kf is the rate constant of the forward reaction and kb is the rate constant of the backward reaction.

If both A and B are present initially, [A]o = [A] and [B]o = [B] then [B] = [A]o +[B]o [A] as a boundary condition:

The plot of ln (kf[A] -kb[B]) versus time is linear, and the slope is used to calculate (kf + kb):

[ ] [ ] [ ] [ ] [ ]( ) [ ] [ ]( ) ( ) [ ]A kkBAk- A-BAk - Akd

A d- bfo o bo o bf +++=+=t( ) [ ] [ ] [ ]( )

++

+= o o bf

bbf BAkk

k-A kk

[ ] [ ][ ] [ ] ( )tbfo bo f

bf kk- Bk-AkBk-Ak ln +=

[ ] [ ] [ ]Bk-AkdAd- bf== t

2006/03/17 53

Reversible First-Order Reactions

If initially only A is present, [A]o = [A] and [B]o = 0 then [B] = [A]o [A] is a boundary condition,

The expression for [A]eq is obtained as follow:

The equilibrium constant K can be used to eliminate kb.

[ ] [ ] [ ] [ ]( ) [ ] ( ) [ ]A kkAk A-Ak A-kd

A dbfo bo bf +=+=t

( ) [ ] [ ] ( ) [ ] [ ]( )eq b f o b f

b b f AA kk- Akk

k-A kk- +=

++=

[ ][ ]

[ ] [ ][ ] eq

eq o

eq

eq

b

f

AA-A

AB K

kk === [ ] [ ] o

b f

b eq Akk

k A+

=

2006/03/17 54

Reversible First-Order Reactions

Since

The reaction has an initial rate of kf[A] and slows down as B accumulated. When [B]eq/[A]eq = K. The reaction is at equilibrium and the rate is zero.

The integration is obtained as follow:.

[ ] [ ] [ ] [ ] [ ][ ]

=+=K A

B-1 A-kBKkA-k

dA d

ff

ft

[ ] ( ) [ ] [ ]( )eq b f AAkk- dtA d +=

[ ] [ ] ( ) +=t

0 bf[A]

[A]eq

dkk- A-A

[A] do

t

Kkk

b

f =

[ ] [ ][ ] [ ] ( )t kk- A-A

A-Aln b f

eq o

eq +=

[ ] [ ]( ) [ ] [ ]( ) ( )t kk-eq o eq b f e A-AA-A +=

2006/03/17 55

Reversible First-Order Reactions

The concentrations of A and B will be halfway to their equilibrium values in a time of ln(2)/(kf + kb)

The plot of ln([A] - [A]eq) versus time is linear, and the slope is used to calculate (kf + kb).

The integration of [A] and [B] are in exponential forms and the boundary condition are [A] + [B] = [A]o

[ ] [ ] ( )( )

[ ] o bf

kk-fbkk-

b

f

bf

o b Akk

ekkekk1

kkAk A bfbf

++=

+

+=

++

tt

[ ] [ ] ( )[ ]( )

[ ] o bf

kk-ffkk-

bf

o f Akk

ek-ke-1kk

Ak Bbf

bf

+=

+=

++

tt

2006/03/17 56

Reversible First-Order Reactions

Fig. 18.5 Reversible first-order reaction starting with A at concentration [A]o. The values of the rate constants are k1 = 3 s-1 and k2 = 1 s-1 .

B A 1k

2k

2006/03/17 57

Experimental Kinetics and Gas Reactions

6 The approach of concentrations to their equilibrium values for a reaction A B that is first-order in each direction, and for which k = 2k .

[ ]( )

[ ] o '-

A 'e ' Akk

kk tkk+

+=+

B A '

k

k

2006/03/17 58

Consecutive First-Order Reactions

For a consecutive irreversible reaction represented by:

the change of concentrations depend on time, rate law is these simultaneous differential equations:

If initially only A is present, [A]o = [A], [B]o = 0 and [C]o = 0 at t=0, then [A]o = [A]+ [B]+[C] as a boundary condition:

[ ] [ ]A-kd

A d1=t

CBA 21 kk

[ ] [ ] [ ]Bk-Akd

B d21=t

[ ] [ ]Bkd

C d2+=t

2006/03/17 59

Consecutive First-Order Reactions

First, the integration form for A is : Substituted into

Integrated to

Now

[ ] [ ] t -ko 1eAA =[ ] [ ] [ ] [ ] [ ]Bk-eA kB k-A k

dB d

2 k-

o 121 1t

t==

[ ] [ ] [ ]tt 21 k-k-12

1o e-ek-k

k AB

=

[ ] [ ] [ ] [ ]ttt 211 k-k-12

1o 2

k-o 1 e-ek-k

kA k-eA k dtB d

=

[ ] [ ]tt k-2 k-112

1o

21 e k-e k k-k

kA-

=

2006/03/17 60

Consecutive First-Order Reactions

To solve an ODE form like

First substitute with y(t ) = [B], r(t ) = k1[A]o e-k1t , b = k2 , Now as

Use integration factor F = eh = exp( b dt ) = ebt

eh (y + b y) = eh r(t )or y eh = eh r(t ) dty = e-h [ eh r(t ) dt ]

= e-bt [ ebt r(t ) dt ]= e-bt ebt r(t ) dt

[ ] [ ] [ ] [ ] [ ]Bk-eAkBk-Akd

B d2

tk-o 121 1==t

( ) ( )tr e dey d h h =t

( ) ( ) )y( b -r d

)y(dy' ttttt ==

2006/03/17 61

Consecutive First-Order Reactions

Now y(t ) = e-bt ebt r(t ) dtsubstitute with b = k2, r(t ) = k1 [A]o e-k1t

B(t ) = e-k2t e k2t k1 [A]o e-k1t dt= e-k2t e (k2 - k1)t k1 [A]o dt

Since [B]o = B(0) = 0

[ ] ( ) e c eA k-k

k e k- k-k o 12

1 k- 2122 ttt +

=

[ ] e c eA k-k

k e 0 0 o 12

10 +

=

( ) [ ] ( ) [ ]

[ ] ( )tt

tttt

k- k-

12

1o

k- o

12

1 k-k o

12

1 k-

21

2122

e-e k-k

kA

eA k-k

k- eA k-k

k e B

=

=

2006/03/17 62

Consecutive First-Order Reactions

The concentration of [C] is:

[ ] [ ] [ ] [ ] [ ] [ ] [ ] ( )ttt k- k-12

1o

k-o o o 211 e-e k-k

kA-eA-A B-A-A C

==

[ ] ( )

[ ] [ ] [ ] [ ]tt

tt

k-

12

1o

k-

12

2o

k-1

k-2

12o

21

21

e-1 k-k

kA - e-1 k-k

kA

ek-ekk-k

1-1A

=

=

2006/03/17 63

Consecutive First-Order Reactions

Fig. 18.6 Plots of concentrations of reactants in A1 A2 A3 when k1 = 1 s-1 and k2 is (a) 1, (b) 5, and (c) 25 s-1. These plots could be calculated using equations but they were actually calculated by solving the three simultaneous differential equations that describe the system.

2006/03/17 64

Consecutive First-Order Reactions

The depletion of A does not depend upon B and C, it can be solved exactly:

Accumulation of B approached its maximum in the induction period.

If the production of C is faster than the rate of accumulation of the intermediate B, i.e., k2 k1. The concentration of the intermediate does not have sufficient time to build up. Consequently, its concentration remains small throughout the reaction.

[ ] [ ] t-ko 1eAA =

2006/03/17 65

Consecutive First-Order Reactions

In the steady state, the rate of change of [B] is zero at t = tm, so

The steady state of [B] is: . Also

So, the time for [B]ss is

[ ] [ ] [ ] [ ] [ ] 0B k-eA kB k-A kd

B d2

k-o 121 1 === tt

[ ] [ ] t k-o 2

1SS 1eA k

kB

=

[ ] [ ] [ ] [ ]

[ ] [ ]tt

tttt

k-2

k-1

12

1o

k- k-

12

1o 2

k-o 1

21

211

e k-e k k-k kA -

e-e k-k k Ak-eA k 0

dB d

=

==

m -k

2 -k

1 at e ke k 21 tttt == m22m11 k-k lnk-kln tt =

=

2

1

21m k

k lnk-k

1t

2006/03/17 66

Consecutive First-Order Reactions

At the steady state, the concentration of [A] at tm is:

And further more, we have the

[ ] [ ] [ ]

==

2

1

21

1o m1o SS k

k lnk-k k-A lnk-A lnA ln t

[ ] [ ] [ ] kkA A or

kkA ln

12

1

12

1k-k

k

2

1o SS

k-kk

2

1o

=

=

[ ] [ ]

=

+=

12

1k-k

k

2

1o

2

1

12

1o k

kA lnkk ln

k-kk A ln

[ ] [ ] m1 -ko SS eAA t=

=

2

1

21m k

klnk-k

1t

2006/03/17 67

Consecutive First-Order Reactions

At the steady state, the concentrations of [B] at tm is

And further more, we have the

[ ] [ ] ( )m2m1 k- k-12

1o SS e-e k-k

k AB tt

=

[ ] m1 k-2

1o e k

kA t

=[ ]

= m1m1 k-

2

1 k-

12

1o ek

k-e k-k

kA tt

[ ] [ ] [ ]

+=

=

2

1

21

1

2

1o m1

2

1o SS k

k ln k-k

k-kk ln A ln k-

kkA lnB ln t

[ ] [ ] [ ] kkA B or

kkA ln

12

2

12

2k-k

k

2

1o SS

k-kk

2

1o

=

=

[ ] [ ]

=

=

21

2k-k

-k

2

1o

2

1

21

2o k

kA lnkk ln

k-kk- A ln

2006/03/17 68

Consecutive First-Order Reactions

At the steady state, the concentration of [C] at tm is:

And further more, we have the

[ ] [ ] [ ] [ ] B-A-AC o SS =

[ ] [ ]

=

21

2

21

1k-k k

2

1k-k k

2

1o SS k

kkk A C 1

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