**, the rate of these reactions does not depend on the concentration of the reactants.**

*Zero-order kinetics*###
**Mathematical derivation of zero-order kinetics**

- Let us take a reaction represented as

- A → Product

- Let the initial concentration of the reactant a and product is zero. After the time interval t, the concentration of the reactant is (a-x) and concentration of the product is x. Thus x is decreases of concentration in zero-order reaction.

*Mathematical derivation of zero-order kinetics in terms of product.*

- Thus the mathematical equation of

**in terms of product,**

*zero-order kinetics*
dx/dt = k₀

Where k₀ is the rate constant of the zero-order reaction.

or, dx = k₀dt

Where k₀ is the rate constant of the zero-order reaction.

or, dx = k₀dt

Integrating the above reaction,

∫dx = k₀ ∫dt

or, x = k₀t + c

where c is the integration constant of the reaction.

∫dx = k₀ ∫dt

or, x = k₀t + c

where c is the integration constant of the reaction.

When t = o, x is also zero thus, C = o Thus the above equation is,

x = k₀ t |

- This is the relationship between decreases of concentration of the reactant(x) within time(t).

*Mathematical derivation of zero-order kinetics in terms of reactant.*

Rate equation in terms of reactant,

-d[A]/dt = k₀ [A]⁰ = k₀

Where [A] is the concentration of the reactant at the time t.

or, - d[A] = k₀dt

-d[A]/dt = k₀ [A]⁰ = k₀

Where [A] is the concentration of the reactant at the time t.

or, - d[A] = k₀dt

Integrating the above equation,

We have - ∫d[A] = k₀ ∫ dt

or, - [A] = k₀t + c

where c is the integration constant of the reaction.

We have - ∫d[A] = k₀ ∫ dt

or, - [A] = k₀t + c

where c is the integration constant of the reaction.

If initial at the time t = 0 concentration of the reactant [A]₀ Then from the above equation,

- [A]₀ = 0 + c

or, c = -[A]₀

- [A]₀ = 0 + c

or, c = -[A]₀

- Putting the value on the above equation,

- [A] = kt - [A]₀ |

- This is another form of the rate equation in

**.**

*zero-order kinetics*###
*The half-life of zero-order kinetics*

*The half-life of zero-order kinetics*

- The time required for half of the reaction to be completed is known as the half-life of the zero-order reaction. It means 50% of reactants disappear in that time interval.

####
**Half-life in zero-order kinetics**

- If in a chemical reaction initial concentration is [A]₀ and after t time interval the concentration of the reactant is [A].

- Then, [A]₀ - [A] = kt

- Thus when t = t

_{½}, that is the half-life of the reaction, the concentration of the reactant [A] = [A]₀/2. Putting the value on the above equation,

We have [A]₀ - [A]₀/2 = k t

or, k t

_{½}or, k t

_{½}= [A]₀/2t_{½} = [A]₀/2k |

- Thus for the

*the half-life of the reaction proportional to its initial concentration.*

**zero-order kinetics**### Examples of the zero-order kinetics

- The only heterogeneous catalyzed reactions may have

**.**

*zero-order kinetics*Examples of zero-order kinetics |

###
**Characteristics of zero-order kinetics**

- The rate of the reaction is independent of concentration.
- Half-life is proportional to the initial concentration of the reactant.
- The rate of the reaction is always equal to the rate constant of the reaction at all concentration.

###
**Unit of the rate constant in zero-order kinetics**

- The rate equation in terms of product for the nth-order reaction is,

d[A]/dt = k [A]

or, k = (d[A]/dt) × (1/[A]

^{n}or, k = (d[A]/dt) × (1/[A]

^{n})- Thus the unit of rate constant(k) = (unit of concentration)/{unit of time × (unit of concentration)

^{n}}

- = (unit of concentration)

^{1-n}/unit of time

- Thus if

**the concentration is expressed in lit mole⁻¹ and time in sec**

*zero-order kinetics*
Then the rate constant = (lit mol⁻¹)/sec

= mol lit⁻¹sec⁻¹

= mol lit⁻¹sec⁻¹

###
*Questions and Answers of zero-order kinetics*

*Questions and Answers of zero-order kinetics*

*Question*

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

*Answer*

- The reaction is a zero-order reaction and 3.92 × 10⁵ Sec take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹.

*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*

- 2x = t₁

*Question*

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

*Answer*

- Zero-order reaction.

*Question*

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

*Answer*

- Zero-order reaction and the value of d[H₂]/dt = 3 × 10⁻⁴ mol lit⁻¹sec⁻¹.

*Question*

- For the 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*

- 1.25 × 10⁻² and 3.125 × 10⁻³ 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.