Table of contents 
Chemical Kinetics 
Rate of Reaction 
Rate Law and Rate Constant 
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“Chemical Kinetics involves the study of the rates and mechanism of chemical reactions.”
The rate of reaction refers to the speed at which the products are formed from the reactants in a chemical reaction.
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 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:
4NO_{2} (g) + O_{2} (g) → 2N 2O_{2} (g)
Q.1. Find the expression for the Rate of Reaction.
4NO_{2} (g) + O_{2} (g) → 2N 2O_{2} (g)
Sol:
The rate of a reaction will generally depend on temperature pressure and the 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 the rate constant for the reaction.
The rate constant is not only dependent on concentration but also on temperature & pressure.
This relationship is known as a rate law.
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, firstorder in B, and three half order overall.
The number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction, must collide simultaneously, in order to bring about a chemical reaction is called the molecularity of a reaction.
Then, rate of reaction =
The unit of rate of reaction is mol liter^{–1} sec^{–1} i.e. mol L^{–1} s^{–1}.
where M represents mol L^{–1} or moles per liter & n is the 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}
Q.2. Find the order of the reaction if unit of rate constant or the reaction is (dm^{3})^{3/2} mol^{–3/2} s^{–1}.
Sol. Unit of rate constant = (dm^{3})^{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 = (dm^{3} )^{3 / 2} (mol)^{3 / 2} s ^{1}
∵ 1 L = 1 dm^{3}
&
i.e. it is 5/2 order reaction.
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 thirdorder 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 × 10^{8} M^{–2} s^{–1}
i.e. the over all rate law is
rate = (3.17 × 10^{8} M^{–2} s^{–1}) [A]^{2 }[B]
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