Question

In a particular experiment at 300 ∘C, [NO2] drops from 0.0100 to 0.00800 M in 100 s. The rate of appearance of O2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0100 to 0.00800 in 100 . The rate of appearance of for this period is ________ . 4.0×10−3 2.0×10−3 2.0×10−5 4.0×10−5 1.0×10−5

Answer 1

Answer:

$1\times{10}^{-5}\frac{M}{s}$

Explanation:

The stoichiometry for this reaction is  

$2NO_2\rightarrow2NO+O_2$

The rate for this reaction can be written as  

$-r_{NO_2}=-\frac{d\left[NO_2\right]}{dt}=\frac{(0.01-0.008)M}{100s}=2\times{10}^{-5}\frac{M}{s}$

This rate of disappearence of $NO_2$ can be realated to the rate of appearence of $O_2$ as follows  (the coefficients of each compound are defined by the stoichiometry of the reaction)

$-r_{O_2}=-r_{NO_2}\times\frac{coefficient\ O_2\ }{coefficient\ NO_2}=2\times{10}^{-5}\frac{M}{s}\times\frac{1\ mole\ O_2\ }{2\ mole\ NO_2}=1\times{10}^{-5}\frac{M}{s}$