- +44 141 628 6080
- info@tutorsglobe.com

*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})}

**Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)**

Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with an expert at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online chemistry tutoring. Chat with us or submit request at info@tutorsglobe.com

1958926

Questions

Asked

3689

Tutors

1470727

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment©TutorsGlobe All rights reserved 2022-2023.

## Newton Method and Loops

Theory and lecture notes of Newton Method and Loops all along with the key concepts of newton method and loops, Solving equations numerically, Newton iterations, Convergence. Tutorsglobe offers homework help, assignment help and tutor’s assistance on newton method and loops.