Answer:
The rate of appearence of O2 is 1.75 * 10^-5 M/s
Explanation:
Step 1: Data given
at 300 °C, [NO2] drops from 0.0100 to 0.00650 M in 100s.
Step 2: The balanced equation:
2NO2 ⇒ 2NO+ O2
Step 3: Calculate the rate of appearence of O2
Rate = Δconcentrtion )/Δtime
Rate of disappearence of NO2 = (0.01M - 0.0065M)/100s= 0.000035 M/s
The mol ratio NO2:O2 = 2:1
-1/2 *0.000035 = 1/1 * rate O2
Rate of appearence of O2 = 0.000035/2 =0.0000175 = 1.75 * 10^-5 M/s
The rate of appearence of O2 is 1.75 * 10^-5 M/s