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

*Chemical kinetics***provides the concentration of these species at any time from the start of the reaction.**

*chemical kinetics*###
*Zero-order kinetics questions*

*Zero-order kinetics questions*

- Rate of the zero-order

*chemical kinetics*reaction does not depend on the concentration of the reactants.

Rate laws of zero-order chemical kinetics |

*Question*

- The rate constant of a chemical reaction is 5× 10⁻⁸ mol lit⁻¹sec⁻¹. Find out 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?

*Answer*

- In

**unit of the rate constant in nth order reaction = (unit of concentration)**

*chemical Kinetics*^{1-n}(unit of time)⁻¹

- Given 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 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

*Questions*

- 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?

*Answer*

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

- Again for zero order

*, [A]₀ - [A] = kt when the reaction completed concentration of [A] = 0.*

**chemical kinetics**- Thus [A]₀ = kt₁
Compare the above two equation
we have, kt₁ = 2kx
or, t₁ = 2x

*Question*

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

*Answer*

- For zero order

**, the rate of the reaction is proportional to zero power of the reactant.**

*chemical kinetics*- That means r ∝ [A]⁰
or, r = k
Thus the reaction is 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?

*Answer*

- 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

**is - d[N₂]/dt = - ⅓ d[H₂]/dt = ½ d[NH₃]/dt**

*chemical kinetics*- 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*

- 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?

*Answer*

- Rate of reaction in zero order

**is - d[N₂O₅]/dt = ½ d[NO₂]/dt = 2d[NH₃]/dt**

*chemical kinetics*- 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*

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

*Answer*

- This is a zero order reaction in

**.**

*chemical kinetics**Question*

- 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?

*Answer*

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

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

^{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

^{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

###
*The Half-life of zero order chemical kinetics*

*The Half-life of zero order 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?

*Answer*

- 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*- Thus for the Zero order kinetics half-life(t½) = [A]₀/2k
or, t½ = 1/(2.5 × 10⁻³) sec
= 0.4 × 10³ sec

###
*First order chemical kinetics questions*

*First order chemical kinetics questions*

*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?

*Answer*

- In

**unit of the rate constant in nth order reaction = (unit of concentration)**

*chemical Kinetics*^{1-n}(unit of time)⁻¹

- 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

- Thus the reaction is the first-order reaction.