Question

Decomposition of $H_2{O}_2$ follows a first order reaction. In fifty minutes the concentration of ${\text{H}}_2{\text{O}}_2$ decreases from $0.5$ to $0.125M$ in one such decomposition. When the concentration of $H_2{O}_2$ reaches $0.05M$ , the rate of formation of $O_2$ will be:

• A. $2.66Lmi{n}^{-1}atSTP$
• B. $1.34\times 10^{-2}molmi{n}^{-1}$
• C. $6.93\times 10^{-2}molmi{n}^{-1}$
• D. $6.93\times 10^{-4}molmi{n}^{-1}$

Answer 1

Answer: (D)

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\to {\text{H}}_{\text{2}}\text{O}+\frac{1}{2}{\text{O}}_{\text{2}}$

$-\frac{d[H_2{O}_2]}{dt}=\frac{d[H_{2}O]}{dt}=2\frac{d[O_2]}{dt}$



$t_{\frac{1}{2}}=25$

${\text{t}}_{\frac{\text{1}}{\text{2}}}\text{=}\frac{\text{0}\text{.69314}}{\text{k}}$

$\text{k}\text{=}\frac{\text{0}\text{.69314}}{\text{25}}$

rate of reaction = $k[H_2{O}_2]$

$\therefore \frac{d[O_2]}{dt}=\frac{1}{2}\times k[H_2{O}_2]$

$=\frac{1}{2}\times \frac{0.69314}{25}\times 0.05$

$=6.93\times 10^{-4}molmi{n}^{-1}$