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

The rate of appearance of O2 for this period is -1.0x10^-5 M/s.

Step-by-step explanation:

The rate of appearance of O2 for a given period can be determined using the change in concentration of NO2 over that period. In this case, the concentration of NO2 drops from 0.0100 to 0.00800 M in 100 s. First, calculate the change in concentration of NO2:

Δ[NO2] = [NO2]final - [NO2]initial = 0.00800 M - 0.0100 M = -0.00200 M

Since the reaction is 2NO2 → 2NO + O2, the stoichiometric ratio between NO2 and O2 is 2:1. Therefore, the change in concentration of O2 is half of the change in concentration of NO2:

Δ[O2] = -0.00200 M ÷ 2 = -0.00100 M

Finally, divide the change in concentration of O2 by the time period to find the rate of appearance of O2:

Rate = Δ[O2] ÷ time = -0.00100 M ÷ 100 s = -1.0x10-5 M/s

Answer 2

`-r_(O_2)=1*{10}^(-5)(M)/(s)`

Step-by-step explanation:

The decomposition of
`NO_2` follows the equation

`2NO_2\rightarrow2NO+O_2`

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

`-r_(NO_2)=(d\left[NO_2\right])/(dt)=(0.010-0.008\ M)/(100\ s)=2*{10}^(-5)(M)/(s)`

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

`-r_(O_2)=-r_(NO_2)*\frac{1\ mole\ \ O_2}{2\ mole\ \ {\rm NO}_2}=1*{10}^(-5)(M)/(s)`