Question

In the following reaction $2H_2{O}_2\to 2H_{2}O+O_2$ rate of formation of $O_2$ is 3.6 M min${}^{-1}$.

(a) What is rate of formation of $H_{2}O?$

(b) What is rate of disappearance of $H_2{O}_2$?

• A. (i) $7.2$ mol litre${}^{-1}$ min${}^{-1},$ (ii) $7.2$ mol litre${}^{-1}$ min${}^{-1}$
• B. (i) $3.6$ mol litre${}^{-1}$ min${}^{-1},$ (ii) $3.6$ mol litre${}^{-1}$ min${}^{-1}$
• C. (i) $14.4$ mol litre${}^{-1}$ min${}^{-1},$ (ii) $14.4$ mol litre${}^{-1}$ min${}^{-1}$
• D. None of these

Answer 1

Answer: (A)

(i) $7.2$ mol litre${}^{-1}$ min${}^{-1},$ (ii) $7.2$ mol litre${}^{-1}$ min${}^{-1}$

$\left(a\right)2H_2{O}_2\to 2H_{2}O+O_2$

We know, $\frac{-1}{2}\frac{d\left[H_2{O}_2\right]}{dt}=\frac{-1}{2}\frac{d\left[H_{2}O\right]}{dt}=\frac{d\left[O_2\right]}{dt}$

$\frac{d\left[H_{2}O\right]}{dt}=2\frac{d\left[O_2\right]}{dt}=2\times 3.6Mmi{n}^{-1}$

$\frac{d\left[H_{2}O\right]}{dt}=7.2Mmi{n}^{-1}$

$\left(b\right)\frac{-d\left[H_2{O}_2\right]}{dt}=\frac{2}{2}\frac{d\left[H_{2}O\right]}{dt}=\frac{-d\left[H_2{O}_2\right]}{dt}=7.2Mmi{n}^{-1}$
Option A is correct.