Question

A first order reaction is 75% completed in 60 minutes. Find the half-life of the reaction.

Answer 1

Hint: Let us first understand what is meant by order of a reaction. The order of a reaction is a number that relates the rate of a reaction with the concentration of the reactants or products.

Complete solution:
A reaction in which the rate of the chemical reaction is directly proportional to the concentration of one of the reactants is called First order chemical reaction. Half life of a reaction is defined as the time taken by the reactant to reduce to one-half of its initial concentration
For a first order reaction we can use
 $
 k_1 = \left( {\dfrac{{2.303}}{t}} \right)\log \left( {\dfrac{a}{{a - x}}} \right) \\ \\
$ to find the value of $k_1$
Where $k_1$ is the reaction rate constant.
Substituting the values in the above equation we get
$
k_1 = \left( {\dfrac{{2.303}}{{60}}} \right)\log \left( {\dfrac{1}{{1 - 0.75}}} \right) \\
 k_1 = 0.0231 $
To calculate half-life, we can use \[t_{\frac{1}{2}} = {\text{ }}\dfrac{{0.693}}{k} = \dfrac{{0.693}}{{0.0231}} = 29.98\min \]
Upon substitution of values in above equation we get,
Half-life, $t_{\frac{1}{2}} = 29.98\min $

Therefore,$t_{\frac{1}{2}}\simeq\,30\min $

Additional information:
When a second order reaction is intentionally made to behave like a first order reaction, then it is called a Pseudo first order reaction. This reaction takes place when one of the reacting components is kept constant or present in great extent compared to the other.

Note:
Remember that in first order reaction, the rate of reaction is influenced by all the reactants but in pseudo first order reaction, the rate of the reaction depends only on the isolated reactant.