Upper atmospheric (stratospheric) ozone protects us from harmful ultraviolet radiation coming from t — Analytical Chemistry Chemistry Question
Theoretical Problem 26
Upper atmospheric (stratospheric) ozone protects us from harmful ultraviolet radiation coming from the Sun. On the other hand, ozone in the lower atmosphere (troposphere) damages the human respiratory system as well as crops and other plants, since ozone is a strong oxidizing agent.
Ozone formation in urban areas can be attributed mainly to the two-step reaction mechanism:
NO2 + hv → NO + O k1 (1)
O + O2 → O3 k2 (2)
The first step is the photolysis of NO2, which is a common air pollutant in cities. (N.B.: k1 includes the intensity of the solar radiation on an average sunny day.) The second step is the reaction of the O atom formed with atmospheric oxygen. Under atmospheric conditions reaction (2) is orders of magnitudes faster than reaction (1).
Let us suppose that a small amount of NO2 (e.g. 10^-7 mole fraction of air) is introduced into the atmosphere and only reactions (1) and (2) take place.
Which species is going to be in a quasi steady-state? Write down the algebraic equation to calculate its concentration after an initial (so-called induction) period.
Model Answer
It is the O atom, because its formation is much slower than its removal, which fulfills the condition of being in a quasi steady-state.
d[O]/dt = k1 [NO2] - k2 [O] [O2] = 0
which gives
[O] = (k1 [NO2]) / (k2 [O2])
Write down the differential and integral rate equations describing ozone formation.
Model Answer
Differential:
d[O3]/dt = k2 [O] [O2]
Since [O] is in quasi steady-state, the problem simplifies to a zeroth order differential equation (the molecular oxygen concentration will stay approximately constant, only a small fraction of it will react). Initially there is no ozone, i.e. [O3]0 = 0.
The integral form (i.e. the solution) is: [O3] = k2 [O] [O2] t
The rate coefficient of reaction (1) is 0.0070 s^-1. The initial NO2 concentration is 2.5 x 10^12 molecule cm^-3.
What is the ozone concentration after 1 minute?
Model Answer
Substituting into the integral form for ozone concentration the equation for O atom concentration we get:
[O3] = k2 [O] [O2] t = k1 [NO2] t
and NO2 decays in a first-order process:
[NO2]t = [NO2]0 exp(-k1 t) = 1.64 x 10^12 molecule cm^-3
The oxygen atoms formed in the first reaction practically all form ozone:
[O3] = [NO2]0 - [NO2]t = 8.6 x 10^11 molecule cm^-3
What is the half-life of NO2?
Model Answer
t1/2 = ln 2 / k1 = 99 s
What effect does temperature have on the rate of ozone formation? Why?
Model Answer
We have shown that the ozone concentration is a function of NO2 concentration and k1. Since photolysis rates (such as k1) are usually less sensitive to temperatures than thermal processes (such as the bimolecular reaction (2)), only small changes in the ozone concentration are expected.
Ozone is not only formed but also removed from the troposphere, mainly by its reaction with NO:
NO + O3 → NO2 + O2 k3 (3)
Assuming reactions (1 - 3), O3, NO and NO2 will be in equilibrium.
k3 = 1.8 x 10^-14 cm3 molecule^-1 s^-1
What is the [NO]/[NO2] ratio, if the equilibrium ozone concentration is 9 x 10^11 molecule cm^-3?
Model Answer
Equilibrium means no net production or loss for any of the species:
d[NO2]/dt = -k1 [NO2] + k3 [NO] [O3] = 0
k1 [NO2] = k3 [NO] [O3]
[NO]/[NO2] = k1 / (k3 [O3]) = 0.432
Assuming the same equilibrium ozone concentration, how does the above ratio change if we raise the temperature from 10 °C to 25 °C? The activation energy of reaction (3) is 10.8 kJ mol^-1.
Model Answer
The ratio will decrease, since according to the Arrhenius expression, k3 increases with temperature.
k3(T2)/k3(T1) = exp[-Ea/R * (1/T2 - 1/T1)] = 1.26
The concentration ratio has to be multiplied by 1/1.26 = 0.79, so the new value is 0.342.