Nitrogen dioxide NO2 dissociates into nitric oxide NO and oxygen O2 as follows- 2NO2- 2NO - O2If the rate of decrease of concentration of NO2 is 6-0- 10-12 mol L-1 s-1- What will be the rate of increase of concentration of O2-

The correct option is B 3× 10^ 12 mol L^ 1s^ 1 For the reaction, 2NO_2→ 2NO + O_2 12d[NO_2]dt = 12d[NO]dt = d[O_2]dt d[NO_2]dt = 6× 10^ 1 ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

Nitrogen diox...

Question

Nitrogen dioxide $(N O_{2})$ dissociates into nitric oxide $(N O)$ and oxygen $(O_{2})$ as follows:
$2 N O_{2} \to 2 N O + O_{2}$
If the rate of decrease of concentration of $N O_{2}$ is $6.0 \times 10^{- 12} m o l L^{- 1} s^{- 1}$ . What will be the rate of increase of concentration of $O_{2}$ ?

3 \times 10^{- 12} m o l L^{- 1} s^{- 1}

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6 \times 10^{- 12} m o l L^{- 1} s^{- 1}

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1 \times 10^{- 12} m o l L^{- 1} s^{- 1}

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1.5 \times 10^{- 12} m o l L^{- 1} s^{- 1}

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Solution

The correct option is B $3 \times 10^{- 12} m o l L^{- 1} s^{- 1}$
For the reaction, $2 N O_{2} \to 2 N O + O_{2}$
$- \frac{1}{2} \frac{d [N O_{2}]}{d t} = \frac{1}{2} \frac{d [N O]}{d t} = \frac{d [O_{2}]}{d t}$
$- \frac{d [N O_{2}]}{d t} = 6 \times 10^{- 12} m o l L^{- 1} s^{- 1}$
$\frac{d [O_{2}]}{d t} = 3 \times 10^{- 12} m o l L^{- 1} s^{- 1}$ .

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