1.0875 x 10-2 atm
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval
From the parameters given, the rate of appearance of O2 is 1.1 * 10^-2.
The equation of the reaction is;
2O3(g) → 3O2(g)
We can see from the equation that; [tex]-\frac{1}{2} \frac{d[O3]}{dt} = \frac{1}{3} \frac{d[O2]}{dt}[/tex]
Hence it follows that;
[tex]-\frac{3}{2} \frac{d[O3]}{dt} = \frac{d[O2]}{dt}[/tex]
Since we already have the rate of disappearance of O3 from the problem as 7.25 × 10^-3, the rate of appearance of O2 is now given by;
[tex]\frac{d[O2]}{dt} = \frac{3}{2} * 7.25 * 10^-3[/tex]
[tex]\frac{d[O2]}{dt} = 1.1 * 10^-2[/tex]
Learn more: https://brainly.com/question/20339399
