For the reaction 2NO2→N2O2+O2 rate expression is as follows −d[NO2]dt =K [NO2]n where K=3×10−2mol−1 Lsec−1. If the rate of formation of oxygen is 1.5×10−4mol Litre−1Sec−1 then the molar concentration of NO2 in mole Litre−1 is

2NO2→N2O2+O212[−d[NO2]dt]=[+d[N2O2]dt]=+d[O2]dt−d[NO2]dt=2×d[O2]dtFrom the units of K it is second order reaction ∴ n=2.−d[NO2]dt=K[NO2]22×d[O2]dt=K[NO2]22×1.5×10−4=3×10−2[NO2]2, [NO2]2=10−410−2=10−2;[NO2]=10−1=0.1M

For the reaction 2NO2→N2O2+O2 rate expression is as follows −d[NO2]dt =K [NO2]n where K=3×10−2mol−1 Lsec−1. If the rate of formation of oxygen is 1.5×10−4mol Litre−1Sec−1 then the molar concentration of NO2 in mole Litre−1 is

2NO2→N2O2+O212[−d[NO2]dt]=[+d[N2O2]dt]=+d[O2]dt−d[NO2]dt=2×d[O2]dtFrom the units of K it is second order reaction ∴ n=2.−d[NO2]dt=K[NO2]22×d[O2]dt=K[NO2]22×1.5×10−4=3×10−2[NO2]2, [NO2]2=10−410−2=10−2;[NO2]=10−1=0.1M

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For the reaction $2 N O_{2} \to N_{2} O_{2} + O_{2}$ rate expression is as follows $\frac{- d [N O_{2}]}{d t}$ =K ${[N O_{2}]}^{n}$ where K= $3 \times 10^{- 2} m o l^{- 1}$ ${Lsec}^{- 1} .$
If the rate of formation of oxygen is $1.5 \times 10^{- 4} m o l L i t r e^{- 1} S e c^{- 1}$ then the molar concentration of $N O_{2}$ in mole $L i t r e^{- 1}$ is

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Detailed Solution

$2 N O_{2} \to N_{2} O_{2} + O_{2}$
$\frac{1}{2} [\frac{- d [N O_{2}]}{d t}] = [\frac{+ d [N_{2} O_{2}]}{d t}] = + \frac{d [O_{2}]}{d t}$
$\frac{- d [N O_{2}]}{d t} = 2 \times \frac{d [O_{2}]}{d t}$
From the units of K it is second order reaction $∴$ n=2.

$\frac{- d [N O_{2}]}{d t} = K {[N O_{2}]}^{2}$
$2 \times \frac{d [O_{2}]}{d t} = K {[N O_{2}]}^{2}$
$2 \times 1.5 \times 10^{- 4} = 3 \times 10^{- 2} {[N O_{2}]}^{2}, {[N O_{2}]}^{2} = \frac{10^{- 4}}{10^{- 2}} = 10^{- 2}; [N O_{2}] = 10^{- 1} = 0.1 M$