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Question: Nitrogen dioxide \((NO_{2})\) dissociates into nitric oxide (NO) and oxygen \((O_{2})\) as follows: ...

Nitrogen dioxide (NO2)(NO_{2}) dissociates into nitric oxide (NO) and oxygen (O2)(O_{2}) as follows:

2NO2→2NO+O22NO_{2} \rightarrow 2NO + O_{2}

If the rate of decrease of concentration of NO2NO_{2} is 6.0×10−12molL−1s−16.0 \times 10^{- 12}molL^{- 1}s^{- 1}. What will be the rate of increase of concentration of O2?O_{2}?

A

3×10−12molL−1s−13 \times 10^{- 12}molL^{- 1}s^{- 1}

B

6×10−12molL−1s−16 \times 10^{- 12}molL^{- 1}s^{- 1}

C

1×10−12molL−1s−11 \times 10^{- 12}molL^{- 1}s^{- 1}

D

1.5×10−12molL−1s−11.5 \times 10^{- 12}molL^{- 1}s^{- 1}

Answer

3×10−12molL−1s−13 \times 10^{- 12}molL^{- 1}s^{- 1}

Explanation

Solution

For the reaction, 2NO2→2NO+O22NO_{2} \rightarrow 2NO + O_{2}

−12d[NO2]dt=12d[NO]dt=d[O2]dt- \frac{1}{2}\frac{d\lbrack NO_{2}\rbrack}{dt} = \frac{1}{2}\frac{d\lbrack NO\rbrack}{dt} = \frac{d\lbrack O_{2}\rbrack}{dt}

−d[NO2]dt=6×10−12molL−1s−1- \frac{d\lbrack NO_{2}\rbrack}{dt} = 6 \times 10^{- 12}molL^{- 1}s^{- 1}

d[O2]dt=3×10−12molL−1s−1\frac{d\lbrack O_{2}\rbrack}{dt} = 3 \times 10^{- 12}molL^{- 1}s^{- 1}