elementary reaction kinetics chapter 35 n...

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Elementary Reaction KineticsChapter 35

Consider the elementary reaction; A → B

Equilibrium reached

Defining the extent of the reaction as ξ .0

i i in n ν ξ= ±

0

A An n ξ= −0

B Bn n ξ= +

For A,

For B,

ksi

in number of moles of i=

stoichimetric coefficients - associated with a sig

for reactants

for products

n

i

i

i

negative

positiveν

ν

ν

0 0 A Bn n

For a general reaction such as;

1

i

d i

dtR

ν=

[ ]

The unique reaction rate R, can be expressed with respect to any component i.

V = reaction volume

Example:

2 52 21 1 1

4 2 2

[ ][ ] [ ] d N Od NO d OR

dt dt dt= − = − =

V = reaction volume, constant

For the generalized reaction;

Reaction rate law is given by;

where,

order w.r.t. A

order w.r.t. B

.

.

overall order of the reaction

= intrinsic reaction rate.

( ...)

k

α

β

α β

=

=

+ + =

The exponents in the rate law of reactions and the stoichiometric coefficients are unrelated in general; the rate law is an experimentally determined relationship.

For non elementary rxn. the order w.r.t. reactant ≠ stoichiometric coefficient.

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Rate for theelementary reaction:

Elementary reaction: One step reaction

[ ]d BR

dt= Rate R, changes with time.

Slope (reactant)

+

-ve

M s-1

M s-1

Elementary (single step)reactions

v

v

For elementary reactions, the order w.r.t. reactant = stoichiometric coefficient.

Overall reaction order and rate-constant units

The sum of the individual orders gives the overall reaction order.

For the unit of rate to come out to be M/s, the units of the rate constant for third-order reactions must be M−2⋅s−1 sinceM/s = (M−2⋅s−1) (M3)

For a second-order reaction, the rate constant has units of M−1⋅s−1 because M/s = (M−1⋅s−1) (M2).

In a first-order reaction, the rate constant has the units s−1

because M/s = (s−1) (M1).

order w.r.t. A

order w.r.t. B

α

β

α β

=

=

+ + = overall reaction or...) der(

Each exponent (α, β,...) of the overall reaction rate law is determined and is followed by substitution to the rate equation ⇒ rate constant.

1111

2222

Rate Law for the step reaction is of the form:

For a one step reaction:

If [A]>>0, [A] variation negligible during reaction, [A] ≅ constant

Rate Law reduces to:

Determining orders –Isolation Method:

Make runs varying one [reactant] but keeping the others in excess.

Plot the appropriate plot. What is it?

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Make runs varying one [reactant] but keeping the others constant.

Find initial reaction rate R for each run;

Determining orders –Initial rates Method:

[B]0 constant

Determining orders –Initial rates Method:

Make two runs varying one [reactant] keeping the others in excess. Generate the appropriate plot;

[B]0 constant

Reaction Mechanisms:

A collection of elementary (one step) reactions that would lead reactants to products. The number of molecules (particles) involved in the elementary reaction is termed as the molecularity of that elementary (single) step.

Examples of elementary (single step) reactions;

1 unimolecular1 unimolecular1 unimolecular1 unimolecular

2 bimolecular2 bimolecular2 bimolecular2 bimolecular

For a given reaction one could propose more than one mechanism.

Usually the simplest mechanism which is consistent with the experimental observations (rate law) is accepted correct, until it is disproven by experimentation.

1

1 2

2

A I

I I

I P

1

2

3

k

k

k

→

→

→

Elementary Reactions and their Rate Laws.

Elementary reactions are one step reactionsinvolving one, two or three molecular specieswhere all species collide (molecularity) to generate the products of that ‘step’.

The initial step of determining the rate law is to follow the reactant(s) concentration over time.

That would be followed by, assuming a rate law (i.e. assuming the orders w.r.t. the reactants) and attempting to fit the experimental data to the integrated rate law.

First order elementary reaction:

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First order elementary reaction:

Differential expression for rate

Molecularity = 1

⇒

Upon integration with limiting conditions of [A]0 at t=0 and [A] at t=t.

0

[ ]

[ ]

ktAe

A

−=


Linearization

0

[A]ln = -kt

[A]

Linear plots are easier to prove the assumption and extract parameters such as k.

0

0

[ ]; [ ] [ ]

[ ]

kt ktAe A A e

A

− −= =

From the product’s ‘view’;

0

[ ]1

[ ]

ktPe

A

−= −i.e.

Exponential rise of [P]

If [P] is followed and a first order reaction w.r.t. A is assumed, what would be your plot?.

Half life – for first order reactions

0

[A]ln = -kt

[A]

Time t1/2 for [A]0 to reduce to [A]0/2.

Independent of [A]0

Second order elementary reaction Type I:


Molecularity = 2

2effdefine k k=

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00

[A]= 1+[A] kt

[A]

and

0 0

[A] 1 =

[A] 1+[A] kt

Second order elementary reaction Type II:

Molecularity = 2

The relationships between concentrations are:Component concentrationsat time t,

0

using the general solution;

[ ]

[ ]

[ ][ ][ ] [ ]( [ ])

[ ]

[ ]( [ ])

A

A

d Ak A B k A A

dt

d Akdt

A A

= − = − ∆ +

= −∆ +∫


;conservation of matter

What would be the integrated equation for this case if [A]0 = [B]0?

Consecutive first order elementary reactions:

Differential rate equation A, loss only

Differential rate equation I, gain and loss

Differential rate equation P, gain only

⇒

For I;

Solves as:

All concentrations are related as; ⇒

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Concentration profiles of A, I and P for different k values.

Concentration profiles of A, I and P for different k values. Concentration profiles of A, I and P for different k values.

[I]max and time to reach [I]max

⇓

Regardless of the [A]0 value of tmax is a constant.

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Overall rate determining step:

i. kA>>kI

ii. kI>>kA

Depends on the relative values of rate constants. Smaller k generally controls the overall reaction rate.

1st order decay of I

1st order decay of A

1

1

~ 1st order decay of IkA>>kI

rds = step 2

For kA > 20 kIrls approximationvalid.

Smaller k generally controls the rate.How small should it be?

rls =rds rate limiting or rate determiningstep

~ 1st order decay of AkI >> kA

rds = step 1

For kI > 25 kArls approximationvalid.

Steady state approximation:

Differential rate Equations:

consecutive.reactions.xls

Steady state approximation (SSA) - visualization:

[I] very small and do not changemuch. Therefore approximatedas,

Applying SSA to the intermediates:

and

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For P (i.e. [P]ss = [P] assuming SSA)

~ 1st order decay of A !!

Validity of SSA

2

0

1

2

1 0

2

1 0

0

0

small makes, 1

A

A

A

ss

k t

A

k t

A

k t

A

A

d I

dt

k A e

k

i e k k A e

k e

then k k A

−

−

−

=

− =

>>

→

>>

[ ]

[ ]

. . [ ]

[ ]

Using

At SS,

kA (0.02/s) < k1(0.2/s) = k2

SSA predicts a considerablydifferent concentration profile!! if kA is not very low,as is here.

Parallel Reactions:

Differential rate equations:

Solves as:

Also;

0

0

0 0

( )

[ ] [ ]

( )

[ ][ ] [ ]

[ ] [ ]

B C

B C

k k t

B B

B B

k k t

B

d Bk A k A e

dt

d B k A e dt

− +

− +

= =

=∫ ∫ B Ck k>

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Product (fractional) yield, parallel reactions:

Yield of component i;

where,

For the parallel reaction;

e.g.

assume kB =2kC

Temperature dependence of k:

Much faster rise of k than √T.

k

>Ea

<Ea

Ea

Low T

High T

Reaction progress: Energy diagram

Part of Ea – energy required to overcome repulsive forces amongelectron clouds of reacting molecules.

Reaction at EquilibriumOne step elementary reactions in both directions

Reaction coordinatediagram

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Differential rate equations:

;conservation of matter

;substituting for [B].

⇓

and

Once the equilibrium is reached at t >>0;

Division:

KC is dependent only on the ki values (ratio of, in this case).

[ ]

[ ]

eq Aeq

eq B

B kK

A k= =

apparent decay constant = (kA + kB) Single step reactions

Energy hump to overcome as the reaction moves toward products.

Involves bringing AB and C closer together (collision); breaking A-B and making B-C bonds and separation of A and BC molecular entities.

i.e. energy demand followed by an energy loss.

k

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Energetics of elementary reactions:

AB + C A + BC

Involves breaking A-B and making B-Cbonds.

← →→ ←

The reaction path follows an energy profile, but it is not necessarily a simple linear one.

Transition state. The transition state (‡)))) of a chemical reaction is a precise atomicconfiguration along the reaction coordinate, located at the highest potential energy position along this reaction coordinate. Transition state has an equal probability of forming either the reactants or the products of the given reaction.

The activated complex differs from the transition state. The activated complex refers to all the configurations that the atoms acquire in the transformation to products. The activated complex refers to any point along the energy profile along reaction coordinate of a reaction in progress.

‡

Consider the bond formation of AB and BC.

In reactions - Formation of bonds and break up of bonds occur at the same time on way to and away from the transition state.Energy plots therefore cannot be 2 dimensional but are threedimensional. Morse curve.

Morse curve.

B + C→ ←

A + B→ ←

Potential energy surface

Position of the transition state (‡) on the PE surface –saddle point ( ).

Reaction coordinate= lowest energy path.

Reaction coordinateReaction coordinateReaction coordinateReaction coordinate is a ‘one-dimensional coordinate’.It represents the reaction progress along a reaction pathway.

Saddle point

Stationary point on the PE surface.

Minimum on blue curve.Maximum on the red curve.Blue and red curves are in orthogonal planes.

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Contour plot showing equi-potential heights.

Reaction coordinate d d d d →→→→ cccc = lowest energypath.

Reaction coordinate diagram. Activated Complex (Transition State ) Theory; elementary bimolecular reaction.

Assumptions:

1.

2.

ABK

A B

≠≠ =

[ ]

[ ][ ];from thermodynamics

1k

1k−

2k

Therefore the reaction rate R;

Assumption 1 ⇒

For P;‘K’ dimensionless.

1 1

[ ][ ][ ] [ ]

d Ak A B k AB

dt

≠−= − +For A; k2

ν vibrational frequency of the complex of breaking bond –no restoration.

Frequency of vibration = rate of breaking one bond.

One of two bonds can break ( ), one leading to the product, other back to the reactants = transmission coefficient (≈ 1); probability of the ‘complex’ formed dissociating into products.

κ

k-1

[A][B]R k=

κ

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[A][B]kR =

Using thermodynamic functions to replace the partition function based , where ∆G‡, ∆H‡ and ∆S‡ are ‘quantities’ of activation.

cKν ≠

we get

i.e.

Pre-exponent = Arrhenius’ A

Arrhenius’ EEEEaaaa

Besides the energy requirement (T) to overcome the activation energy (Ea), the proper orientation of reactants (entropy, ∆S‡ ), the phase of molecular vibrations (transmission factor) are the factors determining the intrinsic reaction rates (k) of elementary reactions.

Expressions for A and Ea depend on the molecularity and the physical phase of the reaction.

Diffusion Controlled Reactions:

Condensed phase – ‘reactants’

collide with solvent molecules and therefore has a lesser chance to collide with the reactant. ‘Reactants’ diffuses thro’ the solvent.

In both phases the average molecular velocities are the same.

Gas phase – reactants collide with leasthindrance with the molecules present in the reaction mixture.

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kd

kr

Reactants collide after diffusion formingan intermediate AB.

Intermediate can generate products.

Intermediate can revert to reactants.

i.e.

Applying SSA to the intermediate AB;

Reaction rate:

Substituting for [AB]

kp>>kr kp<<kr

Diffusion controlledlimit (reaction).

Activation controlledlimit (reaction).

for limitingcases

Also kinetic theory gives;

; diffusion coefficients of A and BAB A B

D D D= +where

and

Diffusion controlled reactions are dependent on the viscosityof the medium.

Ionic reactions are faster in solution because of the additionalattractive impetus due to Coulombic attractions

elementary reaction kinetics chapter 35 n...

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