For the reaction $2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g)$ in a closed vessel,the concentration of $NO_2$ increases by $2.0 \times 10^{-2} \ mol \ L^{-1}$ in $5 \ s$. Calculate the rate of change of concentration of $N_2O_5$.

  • A
    $4 \times 10^{-3}$
  • B
    $10^{-3}$
  • C
    $2 \times 10^{-3}$
  • D
    $10^{-2}$

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Similar Questions

The decomposition of $N_{2}O_{5}$ in $CCl_{4}$ at $318 \ K$ has been studied by monitoring the concentration of $N_{2}O_{5}$ in the solution. Initially the concentration of $N_{2}O_{5}$ is $2.33 \ mol \ L^{-1}$ and after $184 \ minutes$,it is reduced to $2.08 \ mol \ L^{-1}$. The reaction takes place according to the equation:
$2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$
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For the reaction $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate of disappearance of $O_2$,i.e.,$-\frac{d[O_2]}{dt}$.

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