Decomposition of H_2O_2 follows a first order | vedclass

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(D) The decomposition reaction is: $H_2O_{2(aq)} \rightarrow H_2O_{(l)} + \frac{1}{2} O_{2(g)}$For a first order reaction,the rate constant $k$ is giv

Vedclass · Accepted Answer

$6.93 	imes 10^{-4} \ mol \ min^{-1}$

Vedclass · Answer

(D) The decomposition reaction is: $H_2O_{2(aq)} ightarrow H_2O_{(l)} + \frac{1}{2} O_{2(g)}$For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given $[A]_0 = 0.5 \ M$,$[A]_t = 0.125 \ M$,and $t = 50 \ min$:$k = \frac{2.303}{50} \log \frac{0.5}{0.125} = \frac{2.303}{50} \log(4) = \frac{2.303 	imes 0.602}{50} \approx 0.0277 \ min^{-1}$When $[H_2O_2] = 0.05 \ M$,the rate of disappearance of $H_2O_2$ is:$Rate_{H_2O_2} = k[H_2O_2] = 0.0277 	imes 0.05 = 1.385 	imes 10^{-3} \ mol \ L^{-1} \ min^{-1}$From the stoichiometry,$-\frac{d[H_2O_2]}{dt} = 2 \frac{d[O_2]}{dt}$,so the rate of formation of $O_2$ is:$\frac{d[O_2]}{dt} = \frac{1}{2} 	imes Rate_{H_2O_2} = \frac{1.385 	imes 10^{-3}}{2} = 6.925 	imes 10^{-4} \ mol \ L^{-1} \ min^{-1} \approx 6.93 	imes 10^{-4} \ mol \ min^{-1}$

Decomposition of $H_2O_2$ follows a first order reaction. In $50 \ min$,the concentration of $H_2O_2$ decreases from $0.5 \ M$ to $0.125 \ M$. For such decomposition,when the concentration of $H_2O_2$ reaches $0.05 \ M$,the rate of formation of $O_2$ will be:

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