**differential rate low shows the dependence of the rate with the concentration of the reacting species.**

*Chemical kinetics*- But the integrated rate law of the chemical kinetics provides the concentration of these species at any time from the start of the reaction.

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Zero-order kinetics questions

- The rate of the zero-order chemical kinetics reaction does not depend on the concentration of the reactants.

Zero-order chemical kinetics |

- The rate constant of a chemical reaction is 5× 10⁻⁸ mol lit⁻¹sec⁻¹. What is the order of this reaction? How many secs need to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹?

In chemical kinetics unit of the rate constant in nth order reaction

= (unit of concentration)

Given unit of the rate constant = mol lit⁻¹sec⁻¹

= (unit of concentration)(unit of time)⁻¹

= (unit of concentration)

^{1-n}(unit of time)⁻¹Given unit of the rate constant = mol lit⁻¹sec⁻¹

= (unit of concentration)(unit of time)⁻¹

Compare the above two equation

We have 1 -n = 1

or, n = 0

Thus the reaction is a zero-order reaction

And the integration rate equation at two times

(x₂ - x₁) = k (t₂ - t₁)

Here at the time t₂, x₂ = 2 × 10⁻² moles lit⁻¹ and at time t₁, x₁ = 4 × 10⁻⁴

Hence the time required to change the above concentration

(t₂ - t₁) = (x₂ - x₁)/t

= (2 × 10⁻² - 4 × 10⁻⁴)/5× 10⁻⁸ sec

= 3.92 × 10⁵ Sec

QuestionWe have 1 -n = 1

or, n = 0

Thus the reaction is a zero-order reaction

And the integration rate equation at two times

(x₂ - x₁) = k (t₂ - t₁)

Here at the time t₂, x₂ = 2 × 10⁻² moles lit⁻¹ and at time t₁, x₁ = 4 × 10⁻⁴

Hence the time required to change the above concentration

(t₂ - t₁) = (x₂ - x₁)/t

= (2 × 10⁻² - 4 × 10⁻⁴)/5× 10⁻⁸ sec

= 3.92 × 10⁵ Sec

- The half-life of a zero-order reaction is x and the reaction is completed on t₁ time. What is the relation between x and t₁?

Form the Zero-order kinetics half-life(t½) = [A]₀/2k

or, x = [A]₀/2k

or, [A]₀ = 2kx

or, x = [A]₀/2k

or, [A]₀ = 2kx

- Again for zero order chemical kinetics, [A]₀ - [A] = kt when the reaction completed concentration of [A] = 0.

Thus [A]₀ = kt₁

Compare the above two equation we have,

kt₁ = 2kx

or, t₁ = 2x

QuestionCompare the above two equation we have,

kt₁ = 2kx

or, t₁ = 2x

- When the rate of the reaction is equal to the rate constant. What is the order of the reaction?

- For zero-order chemical kinetics, the rate of the reaction is proportional to the zero power of the reactant.

That means r ∝ [A]⁰

or, r = k

Thus the reaction is a zero-order reaction.

or, r = k

Thus the reaction is a zero-order reaction.

Question

- For a reaction, N₂ + 3 H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴ mol lit⁻¹sec⁻¹, What is the order and the value of - d[H₂]/dt of this reaction?

- From the unit of the rate constant, we can easily find out the order of this reaction. Here the unit of the rate constant is mol lit⁻¹sec⁻¹, this is the unit of zero-order reaction thus the reaction is zero-order chemical kinetics.

Rate of reaction of zero-order chemical kinetics is

- d[N₂]/dt = - ⅓ d[H₂]/dt

= ½ d[NH₃]/dt

Thus form the above equation, - ⅓ d[H₂]/dt = ½ d[NH₃]/dt

or, - d[H₂]/dt = (3/2) × d[NH₃]

Given d[NH₃]/dt = 2 × 10⁻⁴ mol lit⁻¹sec⁻¹

-d[H₂]/dt = (3 × 2 × 10⁻⁴ mol lit⁻¹sec⁻¹)/2

= 3 × 10⁻⁴ mol lit⁻¹sec⁻¹

Question- d[N₂]/dt = - ⅓ d[H₂]/dt

= ½ d[NH₃]/dt

Thus form the above equation, - ⅓ d[H₂]/dt = ½ d[NH₃]/dt

or, - d[H₂]/dt = (3/2) × d[NH₃]

Given d[NH₃]/dt = 2 × 10⁻⁴ mol lit⁻¹sec⁻¹

-d[H₂]/dt = (3 × 2 × 10⁻⁴ mol lit⁻¹sec⁻¹)/2

= 3 × 10⁻⁴ mol lit⁻¹sec⁻¹

- For a zero-order reaction N₂O₅ → 2NO₂ + ½ O₂, the rate of disappearance of N₂O₅ is 6.25 × 10⁻³ mol lit⁻¹sec⁻¹, what is the rate of formation of NO₂ and O₂ respectively?

Rate of reaction in zero-order chemical kinetics is

- d[N₂O₅]/dt = ½ d[NO₂]/dt

= 2d[NH₃]/dt

Rate of disappearance of N₂O₅ is,

6.25 × 10⁻³ mol lit⁻¹sec⁻¹ = - d[N₂O₅]/dt

Thus the rate of formation of NO₂

= (2 × 6.25 × 10⁻³ mol lit⁻¹sec⁻¹)

= 1.25 × 10⁻² mol lit⁻¹sec⁻¹

Thus the rate of formation of O₂

= (6.25 × 10⁻³ mol lit⁻¹sec⁻¹)/2

= 3.125 × 10⁻² mol lit⁻¹sec⁻¹

Question- d[N₂O₅]/dt = ½ d[NO₂]/dt

= 2d[NH₃]/dt

Rate of disappearance of N₂O₅ is,

6.25 × 10⁻³ mol lit⁻¹sec⁻¹ = - d[N₂O₅]/dt

Thus the rate of formation of NO₂

= (2 × 6.25 × 10⁻³ mol lit⁻¹sec⁻¹)

= 1.25 × 10⁻² mol lit⁻¹sec⁻¹

Thus the rate of formation of O₂

= (6.25 × 10⁻³ mol lit⁻¹sec⁻¹)/2

= 3.125 × 10⁻² mol lit⁻¹sec⁻¹

- For the reaction H₂ + Cl₂ → 2HCl on sunlight and taking place on the water. What is the order of the reaction?

- This is a zero-order reaction in chemical kinetics.

- A reaction carried out within A and B. When the concentration of B doubled then the rate of the reaction is one-fourth of the initial rate. What is the order of B in this reaction?

- Let the order of the reaction in terms of A is É‘ and in terms of B is Î².

Thus the rate of the reaction(r) = k [A]

where k is the rate constant of the reaction.

The initial concentration of A = [A]₀ and B = [B]₀

Thus the initial rate of the reaction(r₀) = k [A]₀

^{É‘}[B]^{Î²}where k is the rate constant of the reaction.

The initial concentration of A = [A]₀ and B = [B]₀

Thus the initial rate of the reaction(r₀) = k [A]₀

^{É‘}[B]₀^{Î²}- When the concentration of B doubled then the rate of the reaction is one-fourth of the initial rate.

Thus, (r₀/4) = k [A]₀

Compare these two equations we have, r₀/(r₀/4) = (k [A]₀

or, 4 = 2

or, Î² = -2

^{É‘}[2B]₀^{Î²}Compare these two equations we have, r₀/(r₀/4) = (k [A]₀

^{É‘}[B]₀^{Î²})/(k [A]₀^{É‘}[2B]₀^{Î²})or, 4 = 2

^{-Î²}or, Î² = -2

#### Half-life in chemical kinetics

Question- In a chemical reaction, the rate constant of this reaction is 2.5 × 10⁻³ mol lit⁻¹sec⁻¹. If the initial concentration of the reactant is one Find out the half-life this reaction?

- From the unit of the rate constant, we can easily find out the order of this reaction. Here the unit of the rate constant is mol lit⁻¹sec⁻¹, this is the unit of

**thus the reaction is zero-order chemical kinetics.**

*zero-order reaction*Thus for the Zero-order kinetics half-life(t½)

= [A]₀/2k

or, t½ = 1/(2.5 × 10⁻³) sec

= 0.4 × 10³ sec

= [A]₀/2k

or, t½ = 1/(2.5 × 10⁻³) sec

= 0.4 × 10³ sec

### First-order chemical kinetics

Question- In a radioactive reaction, the rate constant of this reaction is 2.5 × 10⁻³ sec⁻¹. What is the order of this reaction?

In chemical Kinetics unit of the rate constant in nth order reaction

= (unit of concentration)

Given unit of the rate constant, = sec⁻¹

= (unit of concentration)⁰(unit of time)⁻¹

Compare the above two-equation,

we have 1 -n = 0

or, n = 1

= (unit of concentration)

^{1-n}(unit of time)⁻¹Given unit of the rate constant, = sec⁻¹

= (unit of concentration)⁰(unit of time)⁻¹

Compare the above two-equation,

we have 1 -n = 0

or, n = 1

- Thus the reaction is the first-order reaction.