Chemistry - Chemical Kinetics Question with Solution | TestHub

For the given reaction $\text{A}+3\text{B}⟶2\text{C}$, the rate of reaction can be expressed as:

$\text{Rate}=-\frac{d[\text{A}]}{dt}=-\frac{1}{3}\frac{d[\text{B}]}{dt}=\frac{1}{2}\frac{d[\text{C}]}{dt}$

Given that the rate of consumption of A is $-\frac{d[\text{A}]}{dt}=1.4{\text{ mol dm}}^{-3}{\text{ sec}}^{-1}$.

From the rate expression, we have:

$\frac{1}{2}\frac{d[\text{C}]}{dt}=-\frac{d[\text{A}]}{dt}$

$\frac{d[\text{C}]}{dt}=2\times \left(-\frac{d[\text{A}]}{dt}\right)$

$\frac{d[\text{C}]}{dt}=2\times 1.4=2.8{\text{ mol dm}}^{-3}{\text{ sec}}^{-1}$

The rate of formation of C is $2.8{\text{ mol dm}}^{-3}{\text{ sec}}^{-1}$.

Answer: $2.8{\mathrm{moldm}}^{-3}{\sec }^{-1}$