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5.60 Thermodynamics & Kinetics
Spring 2008

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5.60 Spring 2008

Lecture #32

page 1

Complex Reactions and Mechanisms (continued)
III) Reversible Reactions

A

k

1 -1

k

B

K

eq

=

[A]eq

[B]eq

If 1st order,

R R

forward backwa

= Rf = k1[A] rd = Rb = k-1[B]

At Equilibrium,

Rf = R

b



k1[A]

eq

= k-1[B]

eq

K

eq

=

k1 k-1

a) 1st order reversible reactions

A

k

1 -1

k

B

-

d[A] = k1 [A] - k -1 [B] dt

[B] = [B]o + ([A]o ­ [A]) So... - d[A] = k1 [A] - k dt

-1

([B]o + [A]o - [A])


5.60 Spring 2008

Lecture #32

page 2

At Equilibrium,

d[A] =0 dt k-1 k1 + k



[A]eq =

-1


([B]o
+ [A]o )

-

d [A] - [A]eq dt

(

)

=-

d([A]) = (k1 + k dt

-1

)([A] - [A]eq
- (k1 +k-1 )t

)

[A] - [A]eq = ([A]o - [A]eq )e

Conc [A]0 > [A] [A]eq [A]0 < [A]
eq

ln (|[A] - [A]eq|)

Slope = -(k1+k-1)

eq

time

time

Can measure:

K

eq

=

k1 k -1

and

k1 + k

-1

k

obs

And extract

k1 and k-1


5.60 Spring 2008

Lecture #32

page 3

b) Higher order reactions
k
2 -1

e.g.

A+B

k

C

2nd order forward, 1st order backward

-

d[A] = k2 [A][B] - k -1 [C] , dt

K=

[C ]eq [A]eq [B]eq

, K=

k2 k -1

After much calculation, get...

A mess!

We must begin simplifying from the beginning!

Use Flooding in this case: Then -

[B]o >> [A]o,[C]

o

k1 k2[B]o k2[B] d[A] = k1 [A] - k -1 [C] dt

This is now pseudo 1st order in A Looks the same as in part a) Measure: K = k2
-1

k

,

kobs k1 + k-1 = k2[B]o + k

-1 -1

By changing [B]o over a few experiments, can extract k2 and k


5.60 Spring 2008

Lecture #32

page 4
st

IV)
k

Series Reversible Reactions (1
1 -1

order)

A

k

B

k

2

C

-

d[A] = k1 [A] - k -1 [B] dt d[C] = k2 [B] dt

d [ B] = k1[ A] - k-1[ B] - k2 [ B] dt

-

Can solve this, but it is an even bigger mess than in part IIIb)!! And here Flooding, as an approximation, is not going to do much for us. We need to find new approximations for more complicated mechanisms!


5.60 Spring 2008

Lecture #32

page 5

IV) Steady State and Equilibrium Approximations

a) Steady State Approximation
k
1 -1

A

k

B

k

2

C

Assume that [B] is small and slowly varying e.g. d[B] 0 and (k2 + k-1) >> k dt
1

[B] reaches a steady state concentration [B] remains there d[B] = k1 [A] - k-1 [B]SS - k2 [B]SS 0 dt
Steady State approximation Solving... [B]SS = k1 [A]
k-1 + k2

SS

and







So

-

d[A] = k1 [A] - k-1 [B]SS dt - d[A] k1k2 [A] = dt k-1 + k2

k k [A] d[C] d[A] = k2 [B]SS = 1 2 =- dt k -1 + k2 dt


5.60 Spring 2008

Lecture #32

page 6

Looks like

A

k'

C

(first order) with k' =

k1k2 k -1 + k2

**Necessary Condition for use of Steady State Approximation** Data must be taken after B has built up to a steady i) state value. (k2 + k-1) >> k1 [B]SS is small ii)

______________________________________ b) Equilibrium Approximation
k
1 -1

A

k

B

k

2

C

Assume

k2 << k-1 and k1
k

That is...

B

2

C

is the rate limiting step.

Then... A and B quickly come into equilibrium, while C slowly builds up. K = k1 [B] k -1 [A]

eq

[B]

=

k1 [A] = K k -1

eq

[A]

Equilibrium approximation kk d[C] = k2 [B] = k2Keq [A] = 1 2 [A] So... dt k-1


5.60 Spring 2008

Lecture #32

page 7

Or,




kk d[C ] = 1 2 [A ] dt k -1

Looks like

A

k'

C

(first order) with k' =

k1k2 k-1

In general, for a mechanism with multiple pre-equilibria...

e.g.

A A A A



I1 I2 I3 I4

K1 K2 K3 K4

In B

(Rate Determining)

n d[B] =k n [In ] = kn Ki [A] dt i=1

Examples:

A)

Apparent Termolecular Reactions (Reaction Chaperones)


5.60 Spring 2008

Lecture #32

page 8

I+I+M

k

I2 + M

M is a rare gas molecule or the wall of the reaction vessel Mechanism: I+I I2* + M (M
*


k

1 -1

k

I2
k
2

*

I2 + M*




M) that is the Steady State approximation!

where (k2 + k-1) >> k1 ,

So

* d[I2 ]

dt

* * = k1 [I] 2 - k -1 [I2 ]SS - k2 [I2 ]SS [M] 0

Steady State approximation

Solving...

[

* I2 ]SS

k1 [I] 2 = k -1 + k2 [M]

And ...

d[I2 ] k1 [I] 2 * = k2 [I2 ]SS [M] = k2 [M] dt k-1 + k2 [M]

Limiting Cases


5.60 Spring 2008

Lecture #32

page 9

i)

k2[M] >> k

-1

th e n

d[I2 ] =k 1 [I] dt

2

(high pressure) ii) k2[M] << k th e n

second order d[I2 ] k 1 k2 = [M][I] dt k -1 third order
2

-1

(low pressure)

B)

Gas decomposition (Lindemann Mechanism)

A(g) products Mechanism: A+M A*
k
2

k

1 -1

k

A* + M products (B + ...)

M is a rare gas molecule and/or A,
k
1

is fast,

k

-1

is very fast,

k

2

is slow

So... (k2 + k-1) >> k1 ,

Steady State approximation again.

d[A * ] = k1 [A][M] - k -1 [A * ]SS [M] - k2 [A * ]SS 0 dt


5.60 Spring 2008

Lecture #32

page10

Steady State approximation [A * ]SS = - k1 [A][M] k -1 [M] + k2

k k [A][M] d[A] d[B] = = k2 [A * ]SS = 1 2 dt dt k-1 [M] + k2

Limiting Cases

i)

High pressure (1 bar) -

k-1[M] >> k

2

d[A] k1k2 = [A] = k [A] dt k -1

(1st order)

ii)

Low pressure (~10-4 bar)

k-1[M] << k2

-

d[A] = k1 [A][M] dt

(if MA, then 2nd order in A)