### Chemical kinetics half-life derivation

In **chemical kinetics** differential rate low shows the dependence of the rate with the concentration of the reacting species and **half-life**.

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

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

#### Zero-order kinetics example problem

Question

The rate constant of a **chemical reaction** is 5× 10^{-8} mol lit^{-1} sec^{-1}. What is the order of this reaction?

How many seconds need to change concentration from 4 × 10^{-4} moles lit^{-1} to 2 × 10^{-2} moles lit^{-1}?

Answer

Unit of the rate constant for the nth-order reaction

= (unit of concentration)^{1-n} (unit of time)^{-1}

But from the above question unit of the rate constant

(mol lit^{-1}) (sec^{-1}) = (unit of concentration)^{1-n} (unit of time)^{-1}

Compare the concentration term

1 -n = 1

or, n = 0

Thus the chemical reaction is a zero-order chemical reaction.

From the integration rate equation

∴ (x_{2} – x_{1}) = k(t_{2} – t_{1})

Here at the time t_{2}, x_{2} = 2 × 10^{-2} moles lit^{-1} and at time t_{1}, x_{1} = 4 × 10^{-4}.

Hence the time required to change the above concentration

(t_{2} – t_{1}) = (x_{2} – x_{1})/t

= 3.92 × 10^{5} sec

#### The half-life period of a chemical reaction

Question

The half-life of a zero-order reaction x but the reaction completed on t_{1} time. What is the relation between x and t_{1}?

Answer

Form the half-life of zero order chemical kinetics

t_{½} = [A]_{0}/2k

∴ x = [A]_{0}/2k

where t_{½} = x

Thus [A]_{0} = 2kx

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

Thus [A]_{0} = kt_{1}

Compare the above two equation we have,

kt_{1} = 2kx

or, t_{1} = 2x

Question

For a reaction,

N_{2} + 3H_{2} → 2NH_{3} and

Find out the order of the reaction an the value of

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^{-1} sec^{-1}, but this is the unit of zero-order reaction, thus the reaction is zero-order chemical kinetics.

Rate of reaction of a zero-order reaction

Form the above equation,

= 3 × 10^{-4} mol lit^{-1} sec^{-1}

Question

For a zero-order reaction

If the rate of disappearance of N_{2}O_{5} = 6.25 × 10^{-3} mol lit^{-1} sec^{-1}, what is the rate of formation of NO_{2} and O_{2} respectively?

Answer

Rate of reaction in zero-order chemical kinetics

Rate of disappearance of N_{2}O_{5}

Thus the rate of formation of NO_{2}

= (2 × 6.25 × 10^{-3} mol lit^{-1} sec^{-1})

= 1.25 × 10^{-2} mol lit^{-1} sec^{-1}

But the rate of formation of O_{2}

= (6.25 × 10^{-3} mol lit^{-1} sec^{-1})/2

= 3.125 × 10⁻² mol lit^{-1} sec^{-1}

#### How to determine the order of reaction?

Question

When the rate of the reaction equal to the rate constant, what is the order of the reaction?

Answer

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

That means r ∝ [A]⁰

or, r = k

Thus the reaction is a zero-order reaction.

Question

For the reaction H_{2} + Cl_{2} → 2HCl, on sunlight and taking place on the water. What is the order of the reaction?

Answer

This is the example of a zero-order chemical reaction.

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 this reaction with respect to B?

Answer

Let the order of the reaction, in terms of A = ɑ and in terms of B = β.

Thus the rate of the reaction,

r = k [A]^{ɑ} [B]^{β}

where k = rate constant of the reaction.

If the initial concentration of A = [A]_{₀} and B = [B]_{₀}

Then the initial rate of the reaction

r_{₀} = k [A]_{₀}^{ɑ} [B]_{₀}^{β}

When the concentration of B doubled then the rate of the reaction is one-fourth of the initial rate.

∴ r_{₀}/4 = k [A]_{₀}^{ɑ} [2B]_{₀}^{β}

Compare these two equations,

r_{₀}/(r_{₀}/4) = (k [A]_{₀}^{ɑ} [B]_{₀}^{β})/(k [A]_{₀}^{ɑ} [2B]_{₀}^{β})

or, 4 = 2 – β

or, β = – 2

#### Kinetics chemistry half-life

Question

In a chemical reaction, the rate constant of this reaction 2.5 × 10^{-3} mol lit^{-1} sec^{-1} and 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^{-1} sec^{-1}, but this is the unit of zero-order kinetics reaction thus the reaction is zero-order kinetics reaction.

Thus for the Zero-order kinetics half-life

t_{½} = [A]_{0}/2k

∴ t_{½} = 1/(2.5 × 10^{-3}) sec

= 0.4 × 10^{3} sec

#### First-order reaction example problems

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

Unit of the rate constant for an nth-order reaction

= (unit of concentration)^{1-n}(unit of time)^{-1}

Given the unit of the rate constant, = sec^{-1}

Thus sec^{-1} = (unit of concentration)^{0}(unit of time)^{-1}

Compare the above two-equation,

1 -n = 0

or, n = 1

Thus the reaction is the first-order chemical reaction.