Question

For a reaction, $2A+B\to 3Z$, if the rate of consumption of $A$ is $2\times {10}^{-4}{\text{mol dm}}^{-3}{\text{s}}^{-1}$ the rate of formation of $Z$ $\left({\text{in mol dm}}^{-3}{\text{s}}^{-1}\right)$ will be:

• A. $3\times {10}^{-4}$
• B. $2\times {10}^{-4}$
• C. $\frac{4}{3}\times {10}^{-4}$
• D. $4\times {10}^{-4}$

Answer 1

The correct option is A $3\times {10}^{-4}$

$2A+B\to 3Z$
The rate of comsumption of $A$ is $=-\frac{d\left[A\right]}{dt}=2\times {10}^{-4}{\text{mol dm}}^{-3}{\text{s}}^{-1}$.
The rate of the reaction is
$=-\frac{1}{2}\frac{d\left[A\right]}{dt}=\frac{1}{3}\frac{d\left[Z\right]}{dt}\Rightarrow \frac{d\left[Z\right]}{dt}=-\frac{3}{2}\frac{d\left[A\right]}{dt}\Rightarrow \frac{d\left[Z\right]}{dt}=\frac{3}{2}\times 2\times {10}^{-4}{\text{mol dm}}^{-3}{\text{s}}^{-1}=3\times {10}^{-4}{\text{mol dm}}^{-3}{\text{s}}^{-1}$