Complex Reaction, Chemistry tutorial

Introduction to Complex Reactions:

Not all the chemical reactions carry on to a phase at which the concentration of the reactions becomes vanishingly small. Such a reaction proceeds through a more complex reaction procedure or mechanism. Most of the industrial chemical reactions, the kind probable to be encountered in a chemical laboratory or plant, comprise multiple steps between the reaction and products.  

These reactions are known as complex reactions.

Parallel Reaction:

This is not uncommon for a reaction to generate or formed more than one product, and the reaction is frequently kinetic and thermodynamic. The series is:

97_Parallel Reaction.jpg

Fig: Parallel Reaction

Rate = d[A]/dt - K1[A] + K2[A]

[A] = ae-(K1+K2)t

The rate of formation of the products is represented as:

d[B]/dt = K1[A]

            = K1ae-(K1+K2)t

And d[C]/dt = K2[A]

                   = K2ae-(K1+K2)t

Integrating the equation provides:

[B] = [K1a/(K1+K2)][(1 - e-(K1+K2)t]

And [C] = [K2a/(K1+K2)][(1 - e-(K1+K2)t]

The rate of products formed is proportional to their rate constant.

[B]/[C] = K1/K2

Reactions Approaching Equilibrium:

Let's take a reaction in which both the forward and reverse reaction are first order as represented by the scheme below.

A ↔ B

The rate of change of [A] consists of two contributions. This is depleted by the forward reaction at a rate K[A] however is replenished via the reverse reaction at a rate K1[B]. The total rate of change is thus,

d[A]/dt = - K[A] + K1[B]

Whenever the critical concentration of A and [B]o is [A]o and there is no B present initially, at all times [A] + [B] = [A]o, and therefore

d[A]/dt = - K[A] + K1([A]o - [A])

           = - (K + K1)[A] + K1[A]o

[A]t = [A]o {K1 + Ke-(K+K1)t}/(K + K1)

If time't' approaches infinity the concentrations reach their equilibrium values

[A]α = K1[A]/(K1 + K1) and

[B]α = [A]o - [A] = K[A]o/(K + K1)

The ratio of such equilibrium concentrations that is the equilibrium constant is:

KC = [B]α/[A]α = K/K1

Other types of Equilibria:

In case of a reaction which is bimolecular and second-order in both directions as illustrated by the plan below:

A + B ↔ C + D

The rate of change of concentration of A, as an outcome of the forward and reverse reactions is:

A + B → C + D             uA = - K [A] [B]

And C + D → A = B      uA = K1 [C] [D]

At equilibrium, the total rate of change is zero. Therefore, at equilibrium:

- K [A] [B] + K1 [C] [D] = 0

As well, KC = {[C][D]/[A][B]}eq = K/K1

In case of a reaction that carries on by a sequence of simple reactions, like:

A + B ↔ C + D        uA forward = - Ka [A][B]

                                uA reverse = K1a [C][D]

C ↔ E + F                uC forward = - Kb [C]

                                uC reverse = K1b [E][F]

At equilibrium, each and every reaction is individually at equilibrium, in such a way that:

{[C][D]/[A][B]}eq = Ka/K1a and {[E][F]/[C]}eq = Kb/K1b

The total reaction equilibrium is:

A + B ↔ D + E + F

K = {[D][E][F]/[A][B]}eq = {[C][D][E][F]/[A][B][C]}eq

  = {[C][D]/[A][B]}eq {[E][F]/[C]}eq

  = Ka Kb/K1a K1b

Whenever the total reaction is the sum of a sequence of steps

K = Ka Kb/ K1a K1b

Consecutive Reactions:

Some of the reactions carry on through the formation of an intermediate as in radioactive decay:

239U → 239Np → 239Pu

Let's take a first-order consecutive reaction as represented below:

A → (K1) → B → (K2) → C

The rate of disappearance of A is:

d[A]/dt = - K[A]

And rate of formations of B and C are:

d[B]/dt = K1[A] - K2[B]

d[C]/dt = K2[B]

At initial time t = 0, concentration of [A] = [A]o and those of [B] = 0 and [C] = 0

First equation is the first order rate law, and therefore [A] = [A]o e-K1t

The rate of second equation is:

d[B]/dt = K1[A]o e- K1t - K2[B]

d[B]/dt = K2[B] = K1[A]o e-K1t

By integrating the equation, the solution is:

[B] = (K1/K2-K1) [Ao] {1 + (K1e-K1t - K2e-K2t)/K2 - K1}


In this case, a consecutive reaction in which the intermediate reaches equilibrium by the reactions before making a product, as represented in the scheme below:

A + B ↔ (K1/K2) ↔ (AB) → C

As we assume that A, B and (AB) are in equilibrium, we can represent:

K = {[(AB)]/[A][B]}eq

Having, K = K1/K2

By overlooking the fact that [AB] is gradually leaking away as it forms C. The rate of formation of C might now be representing as:

d[C]/dt = K2 [(AB)]

           = K2 K1 [A][B]

           = K [A][B]

Here, K = K1K2/K2

Enzyme Reaction:

The other illustration of a pre-equilibrium reaction is the Michaelis-Menten method of enzyme action. The proposed method is:

E + S ↔ (ES) → P + E

d(p)/dt + K3[(ES)]

(ES) represents a bound state of the enzyme 'E' and its substrates 'S'. In order to associate [(es)] to the enzyme concentration we represent its rate law and then impose the steady-state approximation,

d[(ES)]/dt = K1[E][S] - K2[(ES)] - K3[(ES)] = 0

This reorganizes to:

[(ES)] = {K1/(K2 + K3)}[E][S]

[E] and [S] are the concentration of enzyme and substrate and [E]o in total concentration of enzyme

[E] + [(ES)] = [E]o, a constant

As only few E is added, we can ignore the fact that [S] varies slightly from [S] total.


[(ES)] = {K1/(K2 + K3)} {[E]o - [(ES)]}[S]

That can as well reorganize to:

[(ES)] = {K1[E]o[S]}/{K3 + K2 + K1[S]}

It follows that the rate of formation of products is:

d[P]/dt = {K3K1[E]o[S]}/{K3 + K2 + K1[S]}

           = {K3 [E]o[S]}/{Km + [S]}

Here, Km is the Michael constant and is:

Km = (K3 + K2)/K1

Unimolecular Reaction:  

The number of gas phase reactions obeys first-order kinesis and are supposed to carry on through a Unimolecular rate-determining phase. These are termed as Unimolecular reactions. In the Lindemann-Heinshelwood method it is assumed that a reactant molecule A collides with the other M, a diluents gas molecule and becomes energetically excited at the expense of M's translational kinetic energy,

A + M → A* + M             d[A*]/dt = K1[A][M]

And the energized molecule might lose its surplus energy by colliding with the other

A + M → A + M               d[A*]/dt = K2[A*][M]

Or the excited molecule might shake itself apart and form the product

A* → P                             d[b]/dt = K3[A*]

                                         d[A*]/dt = - K3[A*]

By applying the steady state approximation to the total rate of formation of A*,

d[A*]/dt = K1[A][M] - K2[A*][M] - K3[A*] = 0

This resolves to:

[A*] = K1[A][M]/(K3 + K2[M])

And therefore the rate law for the formation of P is:

d[P]/dt = K3[A] = {K1K3[A][M]}/{K3 + K2[M]}

Whenever the rate of deactivation by A*, M collisions is much more than the rate of Unimolecular decay, in such a way that,

K2[A*][M] >> K3[A*] or K2 [M] >> K3

Then we ignore K3 in the denominator and get

d[P]/dt = {K1K3[A][M]}/K2[M] = {K1K3/K2}[A]

a first-order rate law, as we set out to show.

The Lindemann-Hinshelwood method can be tested as it predicts that as the concentration of M is decreased, the reaction must switch to overall second-order kinetics. This is due to reason if K2[M] << K3, the rate is around:

d[P]/dt = {K1K3[M][A]}/K3 = K1[A][M]

The physical cause for the change of order is that at low pressure the rate-determining step is the bimolecular formation of [A*], if we write the full rate law as:

d[P]/dt = Keff[A]

Keff = {K1K3[M]}/{K3 + K2[M]}

Then, the expression for the effective rate constant can be reorganized to,

1/Keff = 1/{Ki[M] + (K2/K1K3)}

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