For the reaction N2O5g→2NO2g+1/2O2g the value of rate of disappearance of N2O5 is given as 6.25×10−3mol L−1s−1. The rate of formation of NO2 and O2 is given respectively as :

−dN2O5dt=+12dNO2dt=2dO2dtGiven −dN2O5dt=6.25×10−3∴dNO2dt=26.25×10−3=1.25×10−2dO2dt=6.25×10−32=3.125×10−3molLL−1S−1

For the reaction N2O5g→2NO2g+1/2O2g the value of rate of disappearance of N2O5 is given as 6.25×10−3mol L−1s−1. The rate of formation of NO2 and O2 is given respectively as :

−dN2O5dt=+12dNO2dt=2dO2dtGiven −dN2O5dt=6.25×10−3∴dNO2dt=26.25×10−3=1.25×10−2dO2dt=6.25×10−32=3.125×10−3molLL−1S−1

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For the reaction $N_{2} O_{5 (g)} \to 2 N O_{2 (g)} + 1 / 2 O_{2 (g)}$ the value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$ . The rate of formation of $N O_{2}$ and $O_{2}$ is given respectively as :

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$6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$ and $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$

$1.25 \times 10^{- 2} m o l L^{- 1} s^{- 1}$ and $3.125 \times 10^{- 3} m o l L^{- 1} s^{- 1}$

$6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$ and $3.125 \times 10^{- 3} m o l L^{- 1} s^{- 1}$

$1.25 \times 10^{- 2} m o l L^{- 1} s^{- 1}$ and $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$

answer is B.

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Detailed Solution

$\frac{- d [N_{2} O_{5}]}{d t} = + \frac{1}{2} \frac{d [N O_{2}]}{d t} = \frac{2 d [O_{2}]}{d t}$
Given $\frac{- d [N_{2} O_{5}]}{d t} = 6.25 \times 10^{- 3}$
$∴ \frac{d [N O_{2}]}{d t} = 2 (6.25 \times 10^{- 3}) = 1.25 \times 10^{- 2}$
$\begin{aligned} \frac{d [O_{2}]}{d t} = \frac{6.25 \times 10^{- 3}}{2} \\ = 3.125 \times 10^{- 3} molL L^{- 1} S^{- 1} \end{aligned}$