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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

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mairarmol

Answer:

[tex]-r_{O_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

Explanation:

The decomposition of [tex]NO_2[/tex] follows the equation  

[tex]2NO_2\rightarrow2NO+O_2[/tex]

By definition, the rate of a chemical reaction can be expressed by  

[tex]-r_{NO_2}=\frac{d\left[NO_2\right]}{dt}=\frac{0.010-0.008\ M}{100\ s}=2\times{10}^{-5}\frac{M}{s}[/tex]

The rate of appearance of [tex]O_2[/tex] is related to the rate of disappearance of [tex]NO_2[/tex] by the stoichiometry. This means that, for each mole of [tex]O_2[/tex] that appears 2 moles of [tex]NO_2[/tex]  are consumed. So  

[tex]-r_{O_2}=-r_{NO_2}\times\frac{1\ mole\ \ O_2}{2\ mole\ \ {\rm NO}_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

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