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## 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:

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 therefored[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] = 0As 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}bThe 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}bWhenever the total reaction is the sum of a sequence of steps

K = Ka Kb/ K

^{1}a K^{1}bConsecutive Reactions:Some of the reactions carry on through the formation of an intermediate as in radioactive decay:

^{239}U →^{239}Np →^{239}PuLet's take a first-order consecutive reaction as represented below:

A → (K

_{1}) → B → (K_{2}) → CThe 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] = 0First 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) → CAs 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)] = 0This 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 constantAs 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*] = 0This 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 getd[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})}Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

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