It turns out that /„. ;?> A'oho.JO], so the loss of O, is dominated by photolysis, while ^o+otIOj] ^ ^o+o,!^?], so the loss of O atoms is dominated by reaction with 02 to form Os.

The lifetime of O, is 110& times greater than the lifetime of O in the stratosphere. Other constituents have lifetimes of days, weeks, months, or longer—many orders of magnitude longer than the lifetime for O,. In addition, many phenomena of interest, such as the Antarctic ozone hole, mid-latitude trends, and perturbations from volcanoes, occur on time-scales of months to decades. As a result, the "ozone" problem involves time-scales ranging over 15 orders of magnitude. This leads to both conceptual and computational difficulties.

Let us define a new constituent, odd oxygen (O,.), to be the sum of O, and O (each of these has an odd number of oxygen atoms, hence the name). The number density of odd oxygen, [□,], is defined to be the sum of the number densities of |03] and [O]. Similarly, the volume mixing ratio (VMR) of O, is defined to be the sum of the VMRs of O, and O.

To understand why this is useful, let us write the continuity equation for O, and

arOii

-*o«o2[0][02] - ¿.„„¿OHO,] + ./o,[0,] + 2/o,[02]

Note that we have neglected transport in this example. Summing these equations, we obtain a continuity equation for O,:

dt dt dt

The right-hand side of Equation (3.9) has two terms. The first term, 2/0,[02], is equal to twice the rate at which O , is photolyzed to form O (reaction (3.1)). This is the rate at which Or is being produced. The second term, [0][0J, is twice the rate at which Q, and O react to form 02. This is the rate at which O, is being destroyed. The factor of 2 in front of these terms accounts for the fact that each of these reactions creates or destroys two members of O,. The net rate of change of 0, is the difference between these terms.

Notable by their absence in Equation (3.9) are the terms representing the reaction between O and O, to form O, (reaction (3.2)) and the photolysis of O, to form 0-, and O (reaction (3.4)). These reactions destroy one member of Ox, but create another. Reaction (3.2), for example, results in the loss of an O atom but the creation of an O, molecule. Because the abundance of O, is the sum of the abundances of O and 0-, these two reactions produce no net change in the abundance of Ov.

Let us now calculate the lifetime of 0, in the lower stratosphere. When calculating the lifetime of a chemical family, one cannot generally calculate it as t/L because the loss rate of the family is generally not proportional to the abundance of the family. For Or, the loss rate is 2A0+() (0,110). Instead, we calculate the lifetime as the abundance of O, divided by the total loss rate:

We have used the approximation [0,| = [0,J to go from the first expression for T(), to the second; we will show this approximation is accurate in the next section. It turns out that this estimate of the lifetime of O, is too large by a factor of about 10-20 because we have neglected several reactions that destroy O,—these will also be discussed in the next section. A more accurate calculation would reveal that the lifetime of (), in the lower stratosphere is several months, decreasing to days iit the upper stratosphere.

To review: O, is created by the photolysis of 02 (reaction (3.1)) and destroyed by the reaction of O, and O (reaction (3.5)) on time-scales of months in the lower stratosphere, decreasing to less than a day in the upper stratosphere. On time-scales that are short compared with this, the sum of the abundances of O, and O is constant. While their sum is constant, however, O, and O are inter-converting on a time-scale of tens of minutes or less. Figure 3.1 shows a schematic of this.

IOJ 1

photolysis photolysis

03 oxygen o

Was this article helpful?

## Post a comment