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  • Decomposition of  follow a first order reaction. In fifty m

Decomposition of \mathrm{H}_2 \mathrm{O}_2 follow a first order reaction. In fifty minutes the concentration of \mathrm{H}_2 \mathrm{O}_2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of \mathrm{H}_2 \mathrm{O}_2  reaches  0.09 M, the rate of formation of \mathrm{O}_2 will be

Option: 1

\mathrm{11.6 \times 10^{-3} \mathrm{~mol} \mathrm{~min}^{-1}}


Option: 2

12.47 \times 10^{-4} \mathrm{mol \ min}^{-1}


Option: 3

\text { 2. } 2 \times 10^{-5} \text { molmin }^{-1}


Option: 4

1.4 \times 10^{-6} \mathrm{~mol} \mathrm{~min}^{-1}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}+\frac{1}{2} \mathrm{O}_2 \\ & -\frac{d\left[\mathrm{H}_2 \mathrm{O}_2\right]}{d t}=\frac{d\left[\mathrm{H}_2 \mathrm{O}\right]}{d t}=2 \frac{d\left[\mathrm{O}_2\right]}{d t} \end{aligned}}

\mathrm{\begin{aligned} & t_{1 / 2}=25 \\ & t_{\frac{1}{2}}=\frac{0.69314}{k} \\ & k=\frac{0.69314}{25} \\ & \frac{d\left[\mathrm{O}_2\right]}{d t}=\frac{1}{2} \times k\left[\mathrm{H}_2 \mathrm{O}_2\right] \\ & =\frac{1}{2} \times \frac{0.69314}{25} \times 0.09 \\ & =12.47 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1} \\ & \end{aligned}}

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