Chemistry - Chemical Kinetics Question with Solution | TestHub

$2{\mathrm{H}}_2{\mathrm{O}}_2⟶2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$ Rate $=-\frac{1}{2}\frac{d\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}{dt}=\frac{1}{2}\frac{d\left[{\mathrm{H}}_2\mathrm{O}\right]}{dt}=\frac{d\left[{\mathrm{O}}_2\right]}{dt}$ Rate of formation of oxygen $=48\mathrm{g}{\min }^{-1}$ $\begin{array}{rl}& =\frac{48}{32}\mathrm{mol}{\min }^{-1}=1.5\mathrm{mol}{\min }^{-1}\\ & \therefore \text{ Rate of formation of }{\mathrm{H}}_2\mathrm{O}\\ & \Rightarrow \frac{1}{2}\frac{d\left[{\mathrm{H}}_2\mathrm{O}\right]}{dt}=\frac{d\left[{\mathrm{O}}_2\right]}{dt}\\ & \frac{d\left[{\mathrm{H}}_2\mathrm{O}\right]}{dt}=2\times \frac{d\left[{\mathrm{O}}_2\right]}{dt}=2\times 1.5=3\mathrm{mol}\min \end{array}$