Question

A reaction is \[50\% \] complete in two hours and \[75\% \] complete in \[4{\text{ }}hours\]. What is the order of the reaction? Give reason.

Answer 1

Hint: The order of a reaction refers to the relationship between the rate of a chemical reaction and the concentration of the reactants. It can be defined as the power dependence of rate on the concentration of all reactants. Here, in this question, we have been given a hint about the half-life of the reaction. So, we will try to find out the order of the reaction by its half-life.

Complete solution:
We have been told that a reaction completes \[50\% \] in \[2{\text{ }}hours\]. So from this given data, we can conclude that the half-life of the reaction \[{t_{1/2}}\] is \[2{\text{ }}hours\].
Let us suppose that the initial concentration of the reactant was A. So, in \[2{\text{ }}hours\], the concentration of the reactant would remain $\dfrac{A}{2}$.
We have also been given that for \[75\% \] completion, it takes \[4{\text{ }}hours\]. From this data, we can conclude that it consumes two half-lives to reach \[75\% \] completion. This means that for going from \[\dfrac{A}{2}\]to\[\dfrac{A}{4}\], it again takes \[2{\text{ }}hours\].It means that whatever be the concentration of the reactants, the value of half-life is constant. Hence, half-life is not dependent on the concentration of the reactant.
So, now, we need to figure out in which order of reaction is the half-life of the reaction independent of the concentration of the reactant.
We know, that half-life of a reaction is related with the concentration of reactant as per the following relation:
${t_{1/2}}\propto {A^{1 - n}}$,
where A in the concentration of the reactant and n is the order of the reaction. So from this relation, it is clear that for a first order reaction, the half-life is independent of the concentration of the reaction.

So, we can say that the reaction follows first order kinetics.

Note:
Students often get confused between the order of a reaction and its molecularity. Always make sure that you do not mix up these two terms and interpret them separately.