Question

The rate constant for the reaction $A\to B$ is $2\times {10}^{-4}Lmo{l}^{-1}mi{n}^{-1}$. The concentration of $A$ at which rate of the reaction is $\left(\frac{1}{12}\right)\times {10}^{-5}Mmi{n}^{-1}$ is:

• A. $\frac{0.25}{\sqrt{40}}M$
• B. $\left(\frac{1}{20}\right)\sqrt{\frac{5}{3}}M$
• C. $\frac{0.25}{\sqrt{15}}M$
• D. None of these

Answer 1

The correct option is C $\frac{0.25}{\sqrt{15}}M$

Given,
$\text{Rate of the reaction}\left(r\right)=\frac{1}{12}\times {10}^{-5}Mmi{n}^{-1}$
$\text{Rate constant}\left(k\right)=2\times {10}^{-4}Lmo{l}^{-1}mi{n}^{-1}$

Since the unit of rate constant is $Lmo{l}^{-1}mi{n}^{-1}$, thus it is a second order reaction.
So, rate would be $r=k[A{]}^2$
$\therefore \frac{1}{12}\times {10}^{-5}=2\times {10}^{-4}\times [A{]}^2$
$\Rightarrow [A{]}^2=\frac{1}{240}{M}^2$
$\Rightarrow \left[A\right]=\frac{1}{4\sqrt{15}}M$