Why would 1kW power supplies fail around Nitric acid (HNO3) gas?

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](http://en.wikipedia.org/wiki/Nitric_acid). 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.




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.](http://dx.doi.org/10.1002/9783527610433.chb02210) 


**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](http://www.jmcproducts.com/wp-content/uploads/2012/02/comparative_analysis.pdf) 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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