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 - 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}
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 ↔ (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.
Thus,
[(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)}
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 [email protected]
types of processes tutorial all along with the key concepts of work, heat and heat capacity, first law of thermodynamics, internal energy, mathematical forms of first law of thermodynamics, isothermal expansion and isothermal irreversible-reversible process
Fermi Surfaces tutorial all along with the key concepts of Brillouin zones, Effect of a crystal potential, Alkali metals, The noble metals, cubic divalent metals and Trivalent metals
density tutorial all along with the key concepts of Concept of density, Mass of the Body, Volume of a Body, Volume of Regular Objects, Volume of Irregular Objects, Density of Mixtures
www.tutorsglobe.com assignment help - systems have been classified in different ways, physical or abstract, open or closed, man-made information systems
the chemical industry-an overview tutorial all along with the key concepts of complex characteristics of the chemical industry, what does the chemical industry produce and economic aspects
tutorsglobe.com gaseous exchange in the alveoli assignment help-homework help by online respiration tutors
tutorsglobe.com harmful effects of algae assignment help-homework help by online algae tutors
Energy effects in Chemical reactions tutorial all along with the key concepts of Procedure of energy effects, Results of energy effects
tutorsglobe.com microscopy assignment help-homework help by online general microbiology tutors
types of vacuum cleaner: cylinder vacuum cleaner (generally more versatile, and are easier to store.), upright vacuum cleaner, wet and dry vacuum cleaner.
tutorsglobe.com flowers fruits and seeds assignment help-homework help by online plant morphology tutors
seeking for top-rated atomic physics assignment help for a++ grades? avail it from the qualified and skilled tutors!
Theory and lecture notes of Unemployment and measuring the Macroeconomy all along with the key concepts of unemployment, unemployment rate, Economists, measuring the Macroeconomy. Tutorsglobe offers homework help, assignment help and tutor’s assistance on unemployment.
composite video signal contains - video signal, horizontal blanking pulses, vertical blanking pulses, horizontal synchronizing pulses, vertical synchronizing pulses
tutorsglobe.com essentiality of a mineral element assignment help-homework help by online mineral nutrition tutors
1941015
Questions Asked
3689
Tutors
1450313
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!