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*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:

Fig: Parallel Reaction

Rate = d[A]/dt - K_{1}[A] + K_{2}[A]

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

The rate of formation of the products is represented as:

d[B]/dt = K_{1}[A]

= K_{1}ae^{-(K1+K2)t}

And d[C]/dt = K_{2}[A]

= K_{2}ae^{-(K1+K2)t}

Integrating the equation provides:

[B] = [K_{1}a/(K_{1}+K_{2})][(1 - e^{-(K1+K2)t}]

And [C] = [K_{2}a/(K_{1}+K_{2})][(1 - e^{-(K1+K2)t}]

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

[B]/[C] = K_{1}/K_{2}

*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 K_{1}[B]. The total rate of change is thus,

d[A]/dt = - K[A] + K_{1}[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] + K^{1}([A]_{o} - [A])

= - (K + K^{1})[A] + K^{1}[A]_{o}

[A]_{t} = [A]_{o} {K^{1} + Ke^{-(K+K1)t}}/(K + K^{1})

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

[A]_{α} = K^{1}[A]/(K_{1} + K^{1}) and

[B]_{α} = [A]_{o} - [A] = K[A]_{o}/(K + K_{1})

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

K_{C} = [B]_{α}/[A]_{α }= K/K^{1}

*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 = K^{1} [C] [D]

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

- K [A] [B] + K^{1} [C] [D] = 0

As well, K_{C} = {[C][D]/[A][B]}_{eq} = K/K^{1}

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 = K^{1}a [C][D]

C ↔ E + F uC forward = - Kb [C]

uC reverse = K^{1}b [E][F]

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

{[C][D]/[A][B]}_{eq} = Ka/K^{1}a and {[E][F]/[C]}_{eq }= Kb/K^{1}b

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/K^{1}a K^{1}b

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

K = Ka Kb/ K^{1}a K^{1}b

*Consecutive Reactions:*

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

^{239}U → ^{239}Np → ^{239}Pu

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

A → (K_{1}) → B → (K_{2}) → 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 = K_{1}[A] - K_{2}[B]

d[C]/dt = K_{2}[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 = K_{1}[A]_{o} e^{- K1t} - K_{2}[B]

d[B]/dt = K_{2}[B] = K_{1}[A]_{o} e^{-K1t}

By integrating the equation, the solution is:

[B] = (K_{1}/K_{2}-K_{1}) [A_{o}] {1 + (K_{1}e^{-K1t} - K_{2}e^{-K2t})/K_{2} - K_{1}}

*Pre-Equilibria:*

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 ↔ (K_{1}/K_{2}) ↔ (AB) → C

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

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

Having, K = K_{1}/K_{2}

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 = K_{2 }[(AB)]

= K_{2 }K_{1 }[A][B]

= K [A][B]

Here, K = K_{1}K_{2}/K_{2}

*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 + K_{3}[(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 = K_{1}[E][S] - K_{2}[(ES)] - K_{3}[(ES)] = 0

This reorganizes to:

[(ES)] = {K_{1}/(K_{2} + K_{3})}[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.

Thus,

[(ES)] = {K_{1}/(K_{2} + K_{3})} {[E]_{o} - [(ES)]}[S]

That can as well reorganize to:

[(ES)] = {K_{1}[E]_{o}[S]}/{K_{3} + K_{2} + K_{1}[S]}

It follows that the rate of formation of products is:

d[P]/dt = {K_{3}K_{1}[E]_{o}[S]}/{K_{3} + K_{2} + K_{1}[S]}

= {K_{3} [E]_{o}[S]}/{Km + [S]}

Here, K_{m} is the Michael constant and is:

K_{m} = (K_{3} + K_{2})/K_{1}

*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 = K_{1}[A][M]

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

A + M → A + M d[A*]/dt = K_{2}[A*][M]

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

A* → P d[b]/dt = K_{3}[A*]

d[A*]/dt = - K_{3}[A*]

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

d[A*]/dt = K_{1}[A][M] - K_{2}[A*][M] - K_{3}[A*] = 0

This resolves to:

[A*] = K_{1}[A][M]/(K_{3} + K_{2}[M])

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

d[P]/dt = K_{3}[A] = {K_{1}K_{3}[A][M]}/{K_{3 }+ K_{2}[M]}

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

K_{2}[A*][M] >> K_{3}[A*] or K_{2} [M] >> K_{3}

Then we ignore K_{3} in the denominator and get

d[P]/dt = {K_{1}K_{3}[A][M]}/K_{2}[M] = {K_{1}K_{3}/K_{2}}[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 K_{2}[M] << K_{3}, the rate is around:

d[P]/dt = {K_{1}K_{3}[M][A]}/K_{3} = K_{1}[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 = K_{eff}[A]

K_{eff} = {K_{1}K_{3}[M]}/{K_{3} + K_{2}[M]}

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

1/K_{eff} = 1/{K_{i}[M] + (K_{2}/K_{1}K_{3})}

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## Imperfection in Solids

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