0
$\begingroup$

If $\ce{NO}$ is unstable, why does it have an atmospheric lifetime of 4 days? I know that it has a lone pair and one unpaired electron making it highly reactive. Therefore, shouldn't it react immediately in the atmosphere?

$\endgroup$
  • 3
    $\begingroup$ Before you ask "why", ask "if". $\endgroup$ – Ivan Neretin Jun 6 '17 at 19:27
  • $\begingroup$ Four days seems long, and there are other variables involved, mainly ozone concentration as NO + O3 is a major sink process for NO. Could you give a source for your stated value? Thx. $\endgroup$ – airhuff Jun 6 '17 at 21:13
7
$\begingroup$

It could be possible that you're thinking of the overall lifetime of $\ce{NO_x}$ ($\ce{NO2 + NO}$), which indeed does have a longer atmospheric lifetime than $\ce{NO}$ individually and which even, at around 5 km altitude, has a lifetime of ~4 days. The two interconvert during the day due to photo-chemistry, and the ratio of the two is dependent on the photolysis rate of $\ce{NO2}$ ($j_{\ce{NO2}}$) and on the rate coefficient of the reaction between $\ce{NO}$ and $\ce{O3}$.

As $j_{\ce{NO2}}$ is constant with altitude, the ratio between $\ce{NO}$ and $\ce{NO2}$ is dependent upon the reaction rate coefficient between $\ce{NO}$ and $\ce{O3}$. This, in turn, is governed by the temperature, and under lower temperatures, the reaction rate is slower. Combined with a decrease in the number concentration of ozone (due to the decrease in the number concentration of air with increasing altitude, if we assume a constant mixing ratio of ozone throughout the troposphere, of course), the ratio favours $\ce{NO}$ over $\ce{NO2}$.

At higher altitudes, then, you do see a situation where there is more $\ce{NO}$ than $\ce{NO2}$, and where the overall lifetime of $\ce{NO_x}$ can be higher than the 1 to 2 days found at the surface (reaching, sometimes, almost 12 days in the upper troposphere).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.