# $NO_{2}$ required for a reaction is produced by the decomposition of $N_{2}O_{5}$ in $CCl_{4}$ as per the equation, $2N_{2}O_{5}(g)\rightarrow 4NO_{2}(g)+O_{2}(g)$.The initial concentration of $N_{2}O_{5}$ is $3.00 mol L^{-1}$ and it is $2.75 mol L^{-1}$ after 30 minutes. The rate of formation of $NO_{2}$ is: Option 1) $4.167\times 10^{-3}molL^{-1}min^{-1}$       Option 2) $1.667\times 10^{-2}molL^{-1}min^{-1}$ Option 3) $8.333\times 10^{-3}molL^{-1}min^{-1}$ Option 4) $2.083\times 10^{-3}molL^{-1}min^{-1}$

($1.667\times 10^{-2}molL^{-1}min^{-1}$)

$2N_{2}O_{5}(g)\rightarrow 4NO_{2}(g)+O_{2}(g)$

$t=0$        $3.0 M$

$t=30$        $2.75 M$

so, $\frac{-\Delta [N_{2}O_{5}]}{\Delta t}=\frac{-[2.75-3.0]}{30-0}=\frac{0.25}{30}$

from the reaction

$-\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}$

$\frac{\Delta [NO_{2}]}{\Delta t}=\frac{0.25}{30}\times 2=1.66\times 10^{-2}M/min$

$=1.66\times 10^{-2}molL^{-1}min^{-1}$

Option 1)

$4.167\times 10^{-3}molL^{-1}min^{-1}$

Option 2)

$1.667\times 10^{-2}molL^{-1}min^{-1}$

Option 3)

$8.333\times 10^{-3}molL^{-1}min^{-1}$

Option 4)

$2.083\times 10^{-3}molL^{-1}min^{-1}$

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