Question

In a particular experiment at 300°C, [NO2] drops from 0.0100 to 0.00650 M in 100 s. The rate of appearance of O2 for this period is __________ M/s.

Answer 1

Rate of appearance of O2 = 1.75*10^-5 M/s

Step-by-step explanation:

NO2 decomposes to O2 as:

2NO2 → 2NO + O2

`Rate of reaction = -(1)/(2) (\Delta [NO_2])/(\Delta t) =(1)/(2) (\Delta [NO])/(\Delta t)=(1)/(1) (\Delta [O_2])/(\Delta t)`

minus sign signifies that NO2 disappear in the reaction.

Where,

`(\Delta [NO_2])/(\Delta t) = Rate of disappearance of NO_2`

`(\Delta [NO])/(\Delta t) = Rate of appearance of NO`

`(\Delta [O_2])/(\Delta t) = Rate of appearance of O_2`

Initial concentration of NO2 = 0.0100 M

After 100 s concentration of NO2 = 0.00650 M

Change in concentration of NO2 = (0.0100 - 0.00650)M

= 0.0035 M

`(1)/(2) (\Delta [NO_2])/(\Delta t) =(1)/(1) (\Delta [O_2])/(\Delta t)`

`(\Delta [O_2])/(\Delta t) = (1)/(2) (0.0035)/(100)`

`(\Delta [O_2])/(\Delta t) = 0.0000175 or 1.75 * 10^(-5) M/s`