Chemical kinetics and half-life

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.

Chemical kinetics equation
Chemical kinetics

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

\frac{dx}{dt}=k

∴ (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

=\frac{\left ( 2\times \times 10^{-2} \right )-\left (4\times 10^{-4} \right )}{5\times 10^{-8}}\, sec

= 3.92 × 105 sec

The half-life period of a chemical reaction

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

Answer
Form the half-life of zero order chemical kinetics

t½ = [A]0/2k

∴ x = [A]0/2k
where t½ = 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 = kt1

Compare the above two equation we have,
kt1 = 2kx
or, t1 = 2x

Question
For a reaction,

N2 + 3H2 → 2NH3 and

\frac{d\left ( NH_{3} \right )}{dt}=2\times 10^{-4}\, mol\, lit^{-1}\, sec^{-1}

Find out the order of the reaction an the value of

-\frac{d\left ( H_{2} \right )}{dt}

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

-\frac{d\left ( N_{2} \right )}{dt}=-\frac{1}{3}\frac{d\left ( H_{2} \right )}{dt}=\frac{1}{2}\frac{d\left ( NH_{3} \right )}{dt}

Form the above equation,

-\frac{1}{3}\frac{d\left ( H_{2} \right )}{dt}=\frac{1}{2}\frac{d\left ( NH_{3} \right )}{dt}

or,-\frac{d\left ( H_{2} \right )}{dt}=\frac{3}{2}\frac{d\left ( NH_{3} \right )}{dt}

Given\, \frac{d\left ( NH_{3} \right )}{dt}=2\times 10^{-4}\, mol\, lit^{-1}\, sec^{-1}

\therefore -\frac{d\left ( H_{2} \right )}{dt}=\frac{3}{2}\times 2\times 10^{-4} \, mol\, lit^{-1}\, sec^{-1}

= 3 × 10-4 mol lit-1 sec-1

Question
For a zero-order reaction

N_{2}O_{5}\rightarrow 2NO_{2}+\frac{1}{2}O_{2}

If the rate of disappearance of N2O5 = 6.25 × 10-3 mol lit-1 sec-1, what is the rate of formation of NO2 and O2 respectively?

Answer
Rate of reaction in zero-order chemical kinetics

-\frac{d\left ( N_{2}O_{5} \right )}{dt}=\frac{1}{2}\, \frac{d\left ( NO_{2} \right )}{dt}=2\, \frac{d\left ( O_{2} \right )}{dt}

Rate of disappearance of N2O5

-\frac{d\left ( N_{2}O_{5} \right )}{dt}=6.25\times 10^{-3}\, mol\, lit^{-1}\, sec^{-1}

Thus the rate of formation of NO2

= (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 O2

= (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 H2 + Cl2 → 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] and B = [B]

Then the initial rate of the reaction
r = k [A]ɑ [B]β

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

∴ r/4 = k [A]ɑ [2B]β

Compare these two equations,
r/(r/4) = (k [A]ɑ [B]β)/(k [A]ɑ [2B]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

t½ = [A]0/2k

∴ t½ = 1/(2.5 × 10-3) sec

= 0.4 × 103 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.