The rate constant for the oxidation of hydrogen peroxide by $KMnO_4$ is $6.93 \times 10^{-5} \ s^{-1}$. How much time will it take for the volume of a standard $KMnO_4$ solution to decrease from $20 \ mL$ to $8 \ mL$?

  • A
    $1.326 \times 10^4 \ s$
  • B
    $7.3 \times 10^3 \ s$
  • C
    $4.6 \times 10^5 \ s$
  • D
    $3.8 \times 10^3 \ s$

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Similar Questions

If the rate constant is $1.155 \times 10^{-3} \ s^{-1}$,after how many seconds will the concentration of the reactant be reduced to half in a first-order reaction?

The initial concentration of $N_2O_5$ in the first order reaction,$N_2O_5 \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$ was $1.24 \times 10^{-2} \ mol \ L^{-1}$ at $300 \ K$ temperature. The concentration of $N_2O_5$ after $60 \ min$ was $0.20 \times 10^{-2} \ mol \ L^{-1}.$ Calculate the rate constant of the reaction.

Isomerisation of gaseous cyclobutene to butadiene is a first order reaction. At $T \ K$,the rate constant of the reaction is $3.3 \times 10^{-4} \ s^{-1}$. What is the time required (in $min$) to complete $90 \%$ of this reaction at the same temperature? $(\log 2 = 0.3)$

$A$ first order reaction is half completed in $45 \, \text{minutes}$. How long does it need for $99.9 \%$ of the reaction to be completed? (in $hr$)

The following results were obtained during kinetic studies of the reaction $2A + B \to$ products:
Experiment $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ min^{-1})$
$I$ $0.10$ $0.20$ $6.93 \times 10^{-3}$
$II$ $0.10$ $0.25$ $6.93 \times 10^{-3}$
$III$ $0.20$ $0.30$ $1.386 \times 10^{-2}$

The time (in minutes) required to consume half of $A$ is:

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