Question

During the decomposition of $H_2{O}_2$ to give oxygen, $48g{O}_2$ is formed per minute at a certain point of time. The rate of formation of water at this point is

• A. $0.75molmi{n}^{-1}$
• B. $1.5molmi{n}^{-1}$
• C. $2.25molmi{n}^{-1}$
• D. $3.0molmi{n}^{-1}$

Answer 1

Answer: (D)

$2H_2{O}_2\to 2H_{2}O+O_2$
Rate $=-\frac{1}{2}\frac{d\left[H_2{O}_2\right]}{dt}=\frac{1}{2}\frac{d\left[H_{2}O\right]}{dt}=\frac{d\left[O_2\right]}{dt}$
Rate of formation of oxygen = 48 g min${}^{-1}$
$=\frac{48}{32}$ mol min${}^{-1}$ = 1.5 mol min${}^{-1}$
$\therefore$ Rate of formation of H${}_{2}O$
$\Rightarrow \frac{1}{2}\frac{d\left[H_{2}O\right]}{dt}=\frac{d\left[O_2\right]}{dt}$
$\frac{d\left[H_{2}O\right]}{dt}=2\times \frac{d\left[O_2\right]}{dt}=2\times 1.5$ = 3 mol min${}^{-1}$