Question

Nitrogen dioxide $(NO_{2})$ dissociates into nitric oxide $(NO)$ and oxygen $(O_{2})$ as follows:
$2NO_{2}\to 2NO + O_{2}$
If the rate of decrease of concentration of $NO_{2}$ is $6.0 \times 10^{-12} mol \,L^{-1}\, s^{-1}$ What will be the rate of increase of concentration of $O_{2}$ ?

• A. $3 \times 10^{-12} mol\, L^{-1} s^{-1}$
• B. $6 \times 10^{-12} mol\, L^{-1} s^{-1}$
• C. $1 \times 10^{-12} mol\, L^{-1} s^{-1}$
• D. $1.5 \times 10^{-12} mol\, L^{-1} s^{-1}$

Answer 1

Answer: (A)

For the reaction, $2NO_{2} \rightarrow 2NO +O_{2}$
$-\frac{1}{2}\frac{d\left[NO_{2}\right]}{dt}=\frac{1}{2}\frac{d\left[NO\right]}{dt}=\frac{d\left[O_{2}\right]}{dt}$
$-\frac{d \left[NO_{2}\right]}{dt}=6\times10^{-12} mol \, L^{-1}\, s^{-1}$
$\frac{d\left[O_{2}\right]}{dt}=3\times10^{-12} mol\, L^{-1}\, s^{-1}$