Introduction to Chemical Kinetics - Chemical Kinetics Chemistry Notes | EduRev

CHEMICAL KINETICS

“Chemical kinetics involves the study of the rates and mechanism of chemical reactions.” The rates of reactions:

(a) The definition of rate: Consider a reaction of the form
A + 2B  → 3C + D.............(1)
in which the molar concentration of participants are [A], [B], [C] & [D].
The rate of consumption or decomposition of the one of the reactants at a given time is  where R is A or B. The rate of formation of one of the products is  where P is C or D.
The rate of reaction can be expressed with respect to any species in equation (1).
 

Thus, the rate of reaction can be defined with respect to both reactants and products.
For example:
4NO₂ (g) + O₂ (g)  → 2N 2O₂ (g)
find the expression for rate of reaction.
 Sol. 4NO₂ (g) + O₂ (g)  → 2N 2O₂ (g)

(b)  Rate laws and rate constant: The rate of a react ion will generally depends on temperature pressure and concentration of species involving in the reaction.
The rate of reaction is proportional to the molar concentration of reacting species.
i.e. A + B + C + D + ……. → Product
then, rate of reaction = k[A]^a [B]^b [C]^c [D]^d ………..

where [A] is the concentration of reactant A, [B] is the concentration of reactant B and so on. The constant a is known as the reaction order with respect to species A, b the reaction order with respect to species B and so on.
The overall reaction order is equal to the sum of the individual reaction orders (a + b + c + d + ……..). Finally the constant k is rate constant for the reaction.
The rate constant dependent on concentration but also on temperature & pressure.

This relationship is known as a rate law. 

(c) Order of the reaction:

A + B + C + ………  → Product
The rate law = v = k[A]^a [B]^b [C]^c ……..
The order of reaction = a + b + c + ……
For example: if rate law = v = k[A]^1/2 [B]
Then, it is half order in A, first order in B and three half order overall.

Molecularity of a Reaction : The number of reacting species (atoms, ions or molecule) taking parting an elementary reaction, must collide simultaneously, in order bring about a chemical reaction is called molecularity of a reaction.
Relationship between Rate law, order and the rate constant:

Then, rate of react ion = 

The unit of rate or reaction is mol liter^–1 sec^–1 i.e. mol L^–1 s^–1.
where M represent mol L^–1 or mo les per liter & n is order of reaction.
The unit of rate constant (k)
Rate of reaction = k[A]ⁿ
unit of rate of reaction = unit of k × [unit of concentration]ⁿ 
MS^–1 = unit of k × [M]n

 ⇒  unit of k = 

i.e., unit of k     = M^{1 - n} S^-1 = mol^{1 - n} L^{n - 1} S^-1

Problem. Find the order of the reaction if unit of rate constant or the reaction is (dm³)^3/2 mol^–3/2 s^–1.
 Sol. Unit of rate constant = (dm³)^3/2 mol^–3/2 s^–1 (given)
We know that,
Unit of rate constant = M^{1 – n} s^–1
For n^th order
i.e. M^{1 - n} s -1 = (dm³ )^{3 / 2} (mol)^{-3 / 2} s ^-1

∵      1 L = 1 dm³
&  

i.e. it is 5/2 order reaction.
 

Determine Reaction order: 

Using the following data for the reaction, we determine the order of the reaction with respect to A and B, over all order and rate constant for the reaction

Sol. A + B  → Product
rate of react ion = k[A]^a [B]^b
5.25 × 10^–4 = k[2.30 × 10^–4]^a [3.10 × 10^–5]b ...(1)
4.20 × 10^–3 = k[4.60 × 10^–4]^a [6.20 × 10^–5]^b ...(2)
1.70 × 10^–2 = k[9.20 × 10^–4]^a [6.20 × 10^–5]^b ...(3)

Divide equation (2) by equation (3), we get

2.47 × 10^–1 = (0.5)^a
(0.247) = (0.5)^a
(0.5 × 0.5) ≈ (0.5)^a
(0.5)^a ≈ (0.5)^a
or taking log we can find the value of a.
a = 2
Divide equation (1) by equation (2) we get

 
1.25 × 10^–1 = [0.5]^a [0.5]^b = [0.5]² [0.5]^b
= 0.25 [0.5]^b
5 × 10^–1 = [0.5]^b
⇒ 0.5 = [0.5]^b
⇒ b = 1 

Therefore, the reaction is second order in A and first order in B and third order overall.

rate = k[A]² [B]
5.2 × 10^–4 Ms^–1 = k(2.3 × 10^–4 M)² (3.1 × 10^–5]M
⇒ k = 3.17 × 10⁸ M^–2 s^–1
i.e. the over all rate law is 
rate = (3.17 × 10⁸ M^–2 s^–1) [A]² [B]