Question

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

• A. $0.75\, mol\, min^{-1}$
• B. $1.5\, mol\, min^{-1}$
• C. $2.25\, mol\, min^{-1}$
• D. $3.0\, mol\, min^{-1}$

Answer 1

Answer: (D)

$2H_2O_2 \rightarrow 2H_2O + O_2$
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$^{-1}$
$\therefore \quad$ Rate of formation of H$_{2}O$
$\Rightarrow \quad \frac{1}{2} \frac{d\left[H_{2}O\right]}{dt}=\frac{d\left[O_{2}\right]}{dt}$
$\frac{d\left[H_{2}O\right]}{dt}=2\times\frac{d\left[O_{2}\right]}{dt}=2\times1.5$ = 3 mol min$^{-1}$