Question

A reaction is $50$% completed in $2$ hours and $75$% completed in $4$ hours. What is the order of the reaction?

Answer 1

We should know the half-life formula to determine the answer. From the half-life formula we know that half-life is inversely proportional to the concentration of the reactant. So, we can compare the half-life and concentration relation at two different conditions to determine the answer.

Complete solution:
The relation between half-life and concentration of reactant is as follows:
${{\text{t}}_{{\text{1/2}}}}\, = \,\dfrac{1}{{({\text{n}} - 1){\text{k}}}}\dfrac{{{{\text{2}}^{{\text{n}} - 1}} - 1}}{{{\text{A}}_{\text{o}}^{{\text{n}} - 1}}}$
Where,
${{\text{t}}_{{\text{1/2}}}}$is the half-life.
${\text{n}}$is the order of reaction.
${\text{k}}$is the rate constant.
${\text{A}}$is the concentration of reactant.
So, the half-life time is inversely proportional to the concentration of the reactant.
${{\text{t}}_{{\text{1/2}}}}\, \propto \,{\text{1/}}{{\text{A}}^{\text{n}}}$
At two different life time we can write the above relation as follows:
${{\text{t}}_{\text{x}}}\, \propto \,{\text{1/A}}_{\text{x}}^{\text{n}}$….$(1)$
${{\text{t}}_{\text{y}}}\, \propto \,{\text{1/A}}_{\text{y}}^{\text{n}}$….$(2)$
On comparing the equation $(1)$and $(2)$ we get,
$\dfrac{{{{\text{t}}_{\text{x}}}\,}}{{{{\text{t}}_{\text{y}}}\,}}{\text{ = }}\,\,\dfrac{{{\text{1/A}}_{\text{x}}^{\text{n}}}}{{{\text{1/A}}_{\text{y}}^{\text{n}}}}$
$\dfrac{{{{\text{t}}_{\text{x}}}\,}}{{{{\text{t}}_{\text{y}}}\,}}{\text{ = }}\,\,\dfrac{{{\text{A}}_{\text{y}}^{\text{n}}}}{{{\text{A}}_{\text{x}}^{\text{n}}}}$
$\dfrac{{{{\text{t}}_{\text{x}}}\,}}{{{{\text{t}}_{\text{y}}}\,}}{\text{ = }}\,\,{\left( {\dfrac{{{{\text{A}}_{\text{y}}}}}{{{{\text{A}}_{\text{x}}}}}} \right)^{\text{n}}}$….$(3)$
Where,
${{\text{t}}_{\text{x}}}$ is the lifetime when the concentration of the reactant is x.
${{\text{t}}_{\text{y}}}$ is the lifetime when concentration of the reactant is y.
We assume that the initial concentration of the reactant is $100$. After the $50$% completion of the reaction in $2$hours the concentration of the reactant will be $50$% and after the $75$% completion of the reaction in $4$hours the concentration of the reactant will be $25$%.
On substituting $4$ for ${{\text{t}}_{\text{x}}}$, $2$ for${{\text{t}}_{\text{y}}}$, $25$ for ${{\text{A}}_{\text{x}}}$ and $50$ for ${{\text{A}}_{\text{y}}}$,
$\dfrac{{\text{4}}}{{\text{2}}}\,\,{\text{ = }}\,{\left( {\dfrac{{{\text{50}}}}{{{\text{25}}}}} \right)^{\text{n}}}$
${\text{2}}\,\,{\text{ = }}\,{\left( {\text{2}} \right)^{\text{n}}}$
${\text{n}}\,{\text{ = }}\,{\text{1}}$

Therefore, the order of the reaction is ${\text{1}}$.

Note: For the first order reaction, half-life is independent from the initial concentration of the reactant, so half-life will remain the same for whatever the concentration of reactant. Here, as in two hour the concentration of reactant decreases from$100$to $50$. Then again in two hour the concentration of reactant decreases form $50$to $25$ so, all over the $75$ % reaction goes completed. It means a reaction goes $50$ completed in every two hours. It means the half-life is two which is independent from the initial concentration of the reactant so, it is a first order reaction.