For the reaction $\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 2…

For the reaction
$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})$ the value
of rate of disappearance of $\mathrm{N}_{2} \mathrm{O}_{5}$ is given as $6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of formation ofv$\mathrm{NO}_{2}$ and $\mathrm{O}_{2}$ is given respectively as :
  1. $6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and
    $6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  2. $1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and
    $3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  3. $6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and
    $3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  4. $1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and
    $6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

Solution

$\begin{array}{l}
\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \\
-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{N}_{2} \mathrm{O}_{5}ight]=+\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{NO}_{2}ight]=2 \frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{O}_{2}ight] \\
\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{NO}_{2}ight]=1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \text { and } \\
\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{O}_{2}ight]=3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{array}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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