Question

Part A In a particular experiment at 300 ∘C, [NO2] drops from 0.0138 to 0.00886 M in 374 s . The rate of disappearance of NO2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0138 to 0.00886 in 374 . The rate of disappearance of for this period is ________ . −6.06×10−5 6.60×10−6 7.57×104 2.64×10−5 1.32×10−5

Answer 1

The rate of disappearance of
`NO_2` for this period is
`-1.32* 10^(-5) M/s`

Step-by-step explanation:

Initial concentration of
`NO_2` = x = 0.0138 M

Final concentration of
`NO_2` = y = 0.00886 M

Time elapsed during change in concentration = Δt = 374 s

Change in concentration ,
`\Delta [NO_2]`= y - x = 0.00886 - 0.0138 M = -0.00494 M

The rate of disappearance of
`NO_2` for this period is:

`(\Delta [NO_2])/(\Delta t)=(-0.00494 M)/(374 s)=-1.32* 10^(-5) M/s`