Question

A reaction is 50% complete in 2 hours and 75% complete in 4 hours .The order of the reaction is:
A) 0
B) 1
C) 2
D) 3

Answer 1

Recall the relation of half life with the initial concentration of the reactant for reactions if different order. For 1st order reactions, half life is independent of the initial concentration of reactant that means if for 1st half reaction, half life is x, then for the other half it will also be x.

Complete step by step answer:
Given that, 50% reaction completes in 2 hours. It means half life given is 2 hours. Half life of a reaction is the time at which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as ${t_{1/2}}$.
Let initial concentration of reactant is ${A_o}$ .Then, for 50% completion of reaction , the remaining amount of reactant is $\dfrac{{{A_o}}}{2}$ .
$\Rightarrow {A_o}\xrightarrow[{2hours}]{{{t_{1/2}}}}\dfrac{{{A_o}}}{2}$

Now, also given that 75% reaction complete in 4 hours. 75% reaction completion means 75 % of the initial concentration of reactant is used , then the remaining amount of reactant is 25%(100-75=25) of the initial concentration. 25% means $\dfrac{{{A_o}}}{4}$ of ${A_o}$.

Therefore,${A_0}\xrightarrow{{4hours}}\dfrac{{{A_o}}}{4}$
Also, we know that $\dfrac{{{A_o}}}{4}$ is half of $\dfrac{{{A_o}}}{2}$ . Thus, it will be 2 hours from$\dfrac{{{A_o}}}{2}$ to $\dfrac{{{A_o}}}{4}$ .
$\Rightarrow {A_o}\xrightarrow[{2hours}]{{{t_{1/2}}}}\dfrac{{{A_o}}}{2}$
$\dfrac{{{A_o}}}{2}\xrightarrow{{2hours}}\dfrac{{{A_o}}}{4}$

It is clear that first half life from ${A_o}$ to $\dfrac{{{A_o}}}{2}$ is 2 hours and other half life from $\dfrac{{{A_o}}}{2}$ to $\dfrac{{{A_o}}}{4}$ is again 2 hours . It means that ${t_{1/2}}$ is constant for 2 hours. It is independent of ${A_o}$.
Let me here give you a simple formula to find the relation of ${t_{1/2}}$ with initial concentration of reactant${A_o}$.
For, any order of reaction ,
${t_{1/2}}\alpha \dfrac{1}{{{{[{A_o}]}^{1 - n}}}}$ ,where n= order of the reaction.
Now, if n=1 ,
${t_{1/2}}\alpha \dfrac{1}{{{{[{A_o}]}^{1 - 1}}}}$

So, for 1st order ${t_{1/2}}$ is independent of ${A_o}$
Thus, the answer of the given question will be 1st order reaction because half life is constant that is 2
So, the correct answer is “Option B”.

Note: Remember that –
-Half life is directly proportional to initial concentration of reactant for zero order of reaction.
-Half life is independent of initial concentration of reactant for 1st order.
-Half life is inversely proportional to initial concentration of reactant for 2nd order.