Question

For the reaction $3A\left(g\right)\stackrel{k}{\to }B\left(g\right)+C\left(g\right),k\text{is}{10}^{-14}\text{L/mol~min}$
if $\left[A\right]=0.5M$, then the value of $-\frac{d\left[A\right]}{dt}\left(inM{s}^{-1}\right)$ is :

• A.
  $7.5\times {10}^{-5}$
• B.
  $3\times {10}^{-4}$
• C.
  $2.5\times {10}^{-5}$
• D. $1.25\times {10}^{-6}M{s}^{-1}$

Answer 1

The correct option is D $1.25\times {10}^{-6}M{s}^{-1}$

Given, $\text{rate constant}k={10}^{-4}\text{L/mol.min}$
So it is a second order reaction.
$\therefore -\frac{1}{3}\frac{d\left[A\right]}{dt}=k[A{]}^2\Rightarrow -\frac{d\left[A\right]}{dt}={10}^{-4}\times 3\times (0.5{)}^2=7.5\times {10}^{-5}{\text{Mmin}}^{-1}=\frac{7.5\times {10}^{-5}}{60}=1.25\times {10}^{-6}M{s}^{-1}$