Question

In the following reaction
$2H_2{O}_2⟶2H_{2}O+O_2$
rate of formation of $O_2$ is 36g/min, what is the rate of disappearance $H_2{O}_2$?

• A. 2.25
• B. 3.5
• C. 1.5
• D. 2.75

Answer 1

Answer: (A)

2.25

The given equation is

$2H_2{O}_{2\left(eq\right)}\stackrel{}{\to }2H_{2}O\left(g\right)+O_{2\left(g\right)}$

Also given

Rate of formation of $O_2=36g/min$

$\Rightarrow \frac{36}{32}mol{min}^{-1}$ $=1.12mol{min}^{-1}$

Now we can calculate of rate,

$Rate=\frac{d\left[O_2\right]}{dt}=-\frac{1}{2}\frac{d\left[H_2{O}_2\right]}{dt}$

$\frac{d\left[H_2{O}_2\right]}{dt}=2\times \frac{d\left[O_2\right]}{dt}$

$\Rightarrow 2\times 1.12mol{min}^{-1}$

$\Rightarrow 2.24mol{min}^{-1}$

Hence,

This is required solution.