Chapter 7 – Reaction Mechanisms, Pathways,

Bioreactions, and BioreactorsPart 1 by Akawat Sirisuk

Objectives

• Discuss the pseudo-steady-state-hypothesis and explain how it can be used to solve reaction engineering problems.

• Explain what an enzyme is and how it acts as a catalyst.• Describe the Michaelis-Menten enzyme kinetics and rate law along

with its temperature dependence.• Discuss how to distinguish the different types of enzyme inhibition.

Sections 7.1.4 and 7.4 are omitted

Companion website at umich.edu/~elements/5e

Elementary VS. Non-elementary reactions

• Rate law describes the rate of chemical reaction as a function of various concentrations of species (Ci) in the system. It is determined from experimental data only.

• Elementary reaction occurs as it is written and evolves a single step such as the bimolecular reaction between oxygen and methanol

• In this case, the stoichiometric coefficients in this reaction are identical to the powers in the power rate law

3 3O CH OH CH O OH+ → +

3O O CH OHr kC C− =

Elementary VS. Non-elementary reactions (cont.)• For non-elementary reactions, the powers in the power rate law are

not the same as the stoichiometric coefficients. It can be a simple integer of 0, 1, or 2 corresponding to a zero-, first-, and second-order reaction.

• The power can be non-integer such as the decomposition of acetaldehyde at 500 ̊C

3 4CH CHO CH CO→ +3 3

3/2CH CHO CH CHOr kC− =

Non-elementary reactions (cont.)

• The rate law for non-elementary reactions can be of a form where there are concentration terms in both the numerator and denominator such as the formation of HBr from H2 and Br2

• Rate law of this form usually involve a number of elementary reactions in sequence and at least one active intermediate

2 2 2H Br HBr+ → 2 2

2

3/21

2

H BrHBr

HBr Br

k C Cr

C k C=

+

Active intermediate

• An active intermediate is a molecule that is in a highly energetic and reactive state. It is short lived as it disappears virtually as fast as it is formed. That is, the net rate of reaction of an active intermediate, A*, is zero.

• Active intermediate, A*, can be formed by collision or interaction with other molecules.

• Other types of intermediates include free radicals (with one or more unpaired electrons), ionic intermediates (e.g. carbonium ion), and enzyme-substrate complex.

*A M A M+ → +

Active intermediate (cont.)

• The work of Ahmed Zewail, who received the Nobel Prize in 1999 for femtosecond spectroscopy, proves the existence of active intermediate

• His work on cyclobutane showed the reaction to form two ethylene molecules did not proceed directly but the active intermediates were formed residing in the small trough at the top of energy reaction coordinate diagram

Active intermediate (cont.)

Pseudo-Steady-State Hypothesis (PSSH)

• Because an active intermediate reacts virtually as fast as it is formed, the net rate of formation of an active intermediate is zero

• This condition is referred to as the Pseudo-Steady-State Hypothesis (PSSH)

* , *1

0n

A i Ai

r r=

= =∑

Application of Pseudo-Steady-State Hypothesis• Consider the gas-phase decomposition of azomethane, A, to give

ethane (B) and nitrogen (C)

• From experimental observations,

262223 NHCN)(CH +⇒

2N AZOP 1atm; r C> ∝

AZO2

N CrmmHg;50P2∝<

Application of Pseudo-Steady-State Hypothesis (cont.)• The following reaction mechanism consisting of three elementary

reactions is proposed

1 * (1)kA A A A+ → +

2* (2)kA A A A+ → +

3* (3)kA B C→ +

21A* 1 Ar k C=

2A* 2 A A*r k C C= −

3A* 3 A*r k C= −

Application of Pseudo-Steady-State Hypothesis (cont.)• Overall reaction rate can be written as

• To find the concentration of the active intermediate, we employ PSSH

• Solve for CA*

3 *B C Ar r k C= =

* 1 * 2 * 3 *2

* 1 2 * 3 *

0

0A A A A

A A A A A

r r r r

r k C k C C k C

= + + =

= − − =

21

*2 3

AA

A

k CCk C k

=+

Application of Pseudo-Steady-State Hypothesis (cont.)• Substituting CA*

• At high CA, k2CA >> k3

• At low CA, k2CA << k3

21 3

2 3

AB

A

k k Crk C k

=+

21B Ar k C=

1 3

2

'B A Ak kr C k Ck

= =

Application of Pseudo-Steady-State Hypothesis (cont.)• Comments on application of PSSH

• PSSH can be applied to only active intermediates, not reactants or products

• Each reaction step in the mechanism is an elementary reaction• Net rate of intermediate formation is zero

• Overall reaction is non-elementary• The overall reaction rate should contain only the concentrations of

reactant and product. No intermediate concentration should appear in the final rate expression.

* , *1

0n

A i Ai

r r=

= =∑

Chain reactions

• Chain reactions are complex reaction mechanism, consisting of the three following sequences

• Initiation – formation of an active intermediate from collision or decomposition of reactant molecules

• Propagation or chain transfer – interaction of an active intermediate with the reactant or product to produce another active intermediate

• Termination – deactivation of the active intermediates to form products

Thermal cracking of ethane

• The thermal decomposition of ethane to ethylene, methane, butane, and hydrogen is believed to proceed in the following sequence

• Initiation

• Propagation

• Termination

12 6 3(1) 2kC H CH→

2

3

4

3 2 6 4 2 5

2 5 2 4

2 6 2 5 2

(2)

(3)

(4)

k

k

k

CH C H CH C H

C H C H H

H C H C H H

+ → +

→ +

+ → +

52 5 4 10(5) 2 kC H C H→

[ ]1 1 2 6r k C H=

[ ][ ][ ][ ][ ]

2 2 3 2 6

3 3 2 5

4 4 2 6

r k CH C H

r k C H

r k H C H

=

=

=

[ ]25 5 2 5r k C H=

Thermal cracking of ethane (cont.)

• The overall reaction rate is determined from the rate of formation of ethylene (the major product)

• For the three active intermediates (CH3•, C2H5•, and H•), the net rates of formation are zero according to PSSH

[ ]2 4 3 3 2 5(6) C Hr r k C H= =

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

[ ] [ ][ ]

2 5

3

2 3 4 5

22 3 2 6 3 2 5 4 2 6 5 2 5

3 4 3 2 5 4 2 6

1 2 1 2 6 2 3 2 6

(7) 2 0

2 0

(8) 0

(9) 2 2 0

= + − + − =

= + − + − =

= + − = + − =

= + − = + − =

C H

H

CH

r r r r r

k CH C H k C H k H C H k C H

r r r k C H k H C H

r r r k C H k CH C H

Thermal cracking of ethane (cont.)

• From (9)

• Adding (7)+(8) yields

• Solve for [C2H5•]

[ ] [ ][ ] [ ] 11 2 6 2 3 2 6 3

2

22 0 kk C H k CH C H CHk

+ − = ⇒ =

[ ][ ] [ ]22 3 2 6 5 2 52 0+ − = k CH C H k C H

[ ] [ ][ ] [ ] [ ]1/21/2 1/2

2 2 1 12 5 3 2 6 2 6 2 6

5 5 2 5

22 2

= = =

k k k kC H CH C H C H C Hk k k k

Thermal cracking of ethane (cont.)

• Substitution of [C2H5•] in (6) yields the rate of formation of ethylene

• So the overall reaction is half order with respect to ethane. The reaction rate constant and the corresponding apparent activation energy of the reaction are

[ ] [ ]2 4

1/21/21

3 2 5 3 2 65

= =

C H

kr k C H k C Hk

( )1/2

13 3 1 5

5

12

= ⇒ = + −

app

kk k E E E Ek

Thermal cracking of ethane (cont.)

• To determine the rate of disappearance of ethane

• Solve for [H•] from (8)

• Substitution gives

[ ] [ ][ ] [ ][ ]2 6 1 2 4 1 2 6 2 3 2 6 4 2 6C Hr r r r k C H k CH C H k H C H− = + + + = + + +

[ ] [ ][ ] [ ] [ ]1/2

1/23 13 2 5 4 2 6 2 6

4 5

0 − + − = ⇒ =

k kk C H k H C H H C Hk k

( )[ ] [ ] [ ] [ ]2 6

1/2 1/21/2 1/21 1

1 1 2 6 3 2 6 1 2 6 3 2 65 5

2 3

− = + + = +

C Hk kr k k C H k C H k C H k C Hk k