For the reaction $N_{2} O_{5} (g) \to {2NO}_{2} (g) {+ 1/2 O}_{2} (g)$ the value of rate of disappearance of . $N_{2} O_{5}$ is given as ${6.25 × 10}^{-3} {mol L}^{-1} s^{-1},$ The rate of formation of ${NO}_{2} {and O}_{2}$ is given respectively as :

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${1.25 × 10}^{-2} {mol L}^{-1} s^{-1} {and 3.125 × 10}^{-3} {mol L}^{-1} s^{-1}$

${1.25 × 10}^{-2} {mol L}^{-1} s^{-1} {and 6.25 × 10}^{-3} {mol L}^{-1} s^{-1}$

${6.25 × 10}^{-3} {mol L}^{-1} s^{-1} {and 3.125× 10}^{-3} {mol L}^{-1} s^{-1}$

${6.25 × 10}^{-3} {mol L}^{-1} s^{-1} {and 6.25 × 10}^{-3} {mol L}^{-1} s^{-1}$

answer is B.

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Detailed Solution

$N_{2} O_{5} (g) \to {2NO}_{2} (g) {+ 1/2 O}_{2} (g)$

$- \frac{d}{dt} [N_{2} O_{5}] = + \frac{1}{2} \frac{d}{dt} [{NO}_{2}] = 2 \frac{d}{dt} [O_{2}]$

$\frac{d}{dt} [{NO}_{2}] {= 1.25 × 10}^{-2} {mol L}^{-1} s^{-1} and$

$\frac{d}{dt} [O_{2}] {= 3.125 × 10}^{-3} {mol L}^{-1} s^{-1}$

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