Q.During the decomposition of $H_2O_2$ to give dioxygen, 48 g $O_2$ is formed per minute at certain point of time. The rate of formation of water at this point is :

Using the balanced equation $2H_2O_2\rightarrow 2H_2O+O_2$, the rate of $O_2$ formation is found from the given mass, and the rate of water formation is twice that (since $H_2O$ and $H_2O_2$ share the coefficient 2, while $O_2$ has coefficient 1).

For the reaction $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$, the rate of reaction is expressed uniformly across all species using their stoichiometric coefficients: $\text{Rate} = -\dfrac{1}{2}\dfrac{d[H_2O_2]}{dt} = \dfrac{1}{2}\dfrac{d[H_2O]}{dt} = \dfrac{d[O_2]}{dt}$

Step 1 — rate of $O_2$ formation: $48\ g$ of $O_2$ forms per minute. Molar mass of $O_2 = 32\ g\,mol^{-1}$, so $\dfrac{d[O_2]}{dt} = \dfrac{48}{32} = 1.5\ mol\,min^{-1}$

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