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

Answer:

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

Explanation:

NO2 decomposes to O2 as:

2NO2  → 2NO + O2

$Rate of reaction = -\frac{1}{2} \frac{\Delta [NO_2]}{\Delta t} =\frac{1}{2} \frac{\Delta [NO]}{\Delta t}=\frac{1}{1} \frac{\Delta [O_2]}{\Delta t}$

minus sign signifies that NO2 disappear in the reaction.

Where,

$\frac{\Delta [NO_2]}{\Delta t} = Rate of disappearance of NO_2$

$\frac{\Delta [NO]}{\Delta t} = Rate of appearance of NO$

$\frac{\Delta [O_2]}{\Delta t} = Rate of appearance of O_2$

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

$\frac{1}{2} \frac{\Delta [NO_2]}{\Delta t} =\frac{1}{1} \frac{\Delta [O_2]}{\Delta t}$

$\frac{\Delta [O_2]}{\Delta t} = \frac{1}{2} \frac{0.0035}{100}$

$\frac{\Delta [O_2]}{\Delta t} = 0.0000175 or 1.75 \times 10^{-5} M/s$