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

Answer:

The rate of disappearance of $NO_2$ for this period is$-1.32\times 10^{-5} M/s$

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:

$\frac{\Delta [NO_2]}{\Delta t}=\frac{-0.00494 M}{374 s}=-1.32\times 10^{-5} M/s$