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I experienced a situation where some electrical power supplies (rated over 1kW) were exposed to nitric acid ($\mathrm{HNO_3}$) fumes at roughly 9ppm for half a day; these power supplies intermittently failed (perhaps 2 to 3 times per hour) in the presence of the fumes. Replacing the power supplies restored proper functionality to the equipment they were powering.

At first, I thought that perhaps there was arcing inside the power supply; however, nitric acid doesn't seem to have a flash point. I'm trying to come up with a reasonable explanation for why these particular components would fail. Maybe the fumes increased atmospheric conductivity, which shorted across components?

If there isn't a compelling reason from a chemistry perspective, that's also a fair response.

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  • $\begingroup$ Looking at the title, I can't help but think the more reasonable question would be "Why would they not?" $\endgroup$
    – Ilmari Karonen
    Commented Jan 22, 2015 at 14:46
  • $\begingroup$ You're missing the reality that other equipment next to it did not fail. The difference was actually air flow rates through the equipment in question. The exact reason for failure is important to nail down, so we can prioritize what things to replace first $\endgroup$
    – Mike Pennington
    Commented Jan 22, 2015 at 15:18

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Nitric acid corrodes copper. Drop a penny into some nitric acid (under a hood!) and you'll see that your power supplies really don't stand much of a chance in that environment. This happens with dilute nitric acid:

$$\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3^{−}}$$

When you open up one of the failed power supplies, do you see bluish crusty white stuff anywhere? That's nitric acid corrosion.

Addendum: Here are some quantitative corrosion rates:
Karl Hauffe and Roman Bender: Copper. In Corrosion Handbook. Wiley-VCH, 2008.

Addendum 2: Even though the concentration of nitric acid in the air is low, your power supplies are sucking a lot of air through them in 12 hours.

Let's suppose you have a power supply that dissipates 300 W of heat and you can tolerate a 10 degree Celsius temperature rise. Then according to this reference you need an airflow rate of about 53 cubic feet per minute.

That's 1.5 cubic meters of air per minute.

The density of air in a warmish (300 K) room is about 1.177 kilograms per cubic meter. So you're pulling

$$\frac{1.5~\mathrm{m^3}\ \text{air}}{1~\mathrm{min}} \frac{1.177~\mathrm{kg}\ \text{air}}{1~\mathrm{m^3}\ \text{air}} \frac{9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{kg}\ \text{air}} = 15.9~\frac{\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}}$$

which over 12 hours is

$$\frac{15.9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}} \frac{60~\mathrm{min}}{1~\mathrm{hour}} (12~\mathrm{hours}) = 11.4~\mathrm{g}\ \ce{HNO3}$$

If all of that reacted with copper in the power supply, this is how much copper would be converted into corrosion:

$$11.4~\mathrm{g}\ \ce{HNO3}\frac{1~\mathrm{mol}\ \ce{HNO3}}{63.01~\mathrm{g}\ \ce{HNO3}}\frac{3~\mathrm{mol}\ \ce{Cu}}{8~\mathrm{mol}\ \ce{HNO3}}\frac{63.546~\mathrm{g}\ \ce{Cu}}{1~\mathrm{mol}\ \ce{Cu}} = \fbox{4 g Cu }$$

...that's a lot of copper to lose (a penny is 2.5 g!). Even if only 10% of all the $\ce{HNO3}$ that's being sucked through the power supply reacts, it's still 0.4 grams of copper, certainly enough to significantly corrode electrical contact surfaces.

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    $\begingroup$ There is a nontrivial difference between 9ppm and the concentrated solution / penny in your answer. I'm not suggesting that this is wrong, but is 9ppm enough to reasonably explain failure within an hour of first exposure? $\endgroup$
    – Mike Pennington
    Commented Jan 21, 2015 at 21:44
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    $\begingroup$ All it has to do is spoil a contact surface somewhere. And it'll work on other metals, too. $\endgroup$
    – Fred Senese
    Commented Jan 21, 2015 at 21:46
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    $\begingroup$ Wow, now you're going to make me drop half my chemistry.se rep on a bonus... Good work, thank you :-) $\endgroup$
    – Mike Pennington
    Commented Jan 21, 2015 at 23:31
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    $\begingroup$ This is a very amazingly thorough answer! I have updated your math formulas, as I noticed you are still using the deprecated \rm (see meta for more details). We also use mhchem for chemistry markup, which is easy to use. Another helpful tutorial can be found here. $\endgroup$
    – Martin - マーチン ♦
    Commented Jan 22, 2015 at 5:03
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    $\begingroup$ Indeed a remarkable answer!+1 $\endgroup$
    – M.A.R.
    Commented Jan 22, 2015 at 11:35

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