Chemistry - Chemical Kinetics Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Explanation: }4\mathrm{A}+3\mathrm{B}⟶6\mathrm{C}+9\mathrm{D}\\ & \begin{array}{rl}\text{ Rate of reaction }& =-\frac{1}{4}\frac{d[\mathrm{A}]}{dt}=-\frac{1}{3}\frac{d[\mathrm{B}]}{dt}\\ & =\frac{1}{6}\frac{d[\mathrm{C}]}{dt}=\frac{1}{9}\frac{d[\mathrm{D}]}{dt}\end{array}\end{array}$$ Rate of reaction $=$ Rate of formation of $\mathrm{C}$. $$\begin{array}{rl}\therefore r& =\frac{1}{6}\frac{d[\mathrm{C}]}{dt}=\frac{1}{6}\times 6\times 10^{-2}\\ & =1\times 10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\end{array}$$ Also, rate of reaction $=$ Rate of disappearance of $B$ $$\begin{array}{rl}r& =-\frac{1}{3}\frac{d[\mathrm{B}]}{dt}=\frac{d[\mathrm{B}]}{dt}=3\times r\\ & =3\times 10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\end{array}$$ Consumption of $B$ in 10 seconds $=3\times 10^{-2}\times 10$ $$=30\times 10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$$