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**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:**

4NO_{2} (g) + O_{2} (g) → 2N 2O_{2} (g)**find the expression for rate of reaction. Sol. **4NO

**(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 (dm ^{3})^{3/2} mol^{–3/2} s^{–1}.**Unit of rate constant = (dm

Sol.

We know that,

Unit of rate constant = M

For n

i.e. M

∵ 1 L = 1 dm^{3}

&

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 × 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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