I once reverese-powered a 74HC14 by a 1A 5V supply for about half an hour and - nothting happened! The thingy became extremely hot but worked properly afterwards.

How can this be??

In reverse direction, means from GND to Vcc the whole die looks like a large pn-junction. So, when being driven in reverse direction a large current will flow along this "pn-junction" and will cause a voltage drop of about 0.7V, depending on actual current, of course. So, if the current is coming from a 7805 and this current is limited to about 1.4A, then about 0.7V x 1.4A = 1W will heat up the die. That is much, but not necessarily too much to immediately destroy the chip.

Why?
The DIP package shows a heat resistance of 80K/W between junction and ambient, which means, that the die will be brought on a temperature of about 100°C, when the ambient is at about 20°C. But 100°C die temperature is well below the maximum die temperature of 150°C for a plastic package.

Also, many applications are powered by 5V supplies, which can only deliver currents way below the short circuit current of 1.4V. Not only because they eventually contain a weaker regulator like the 78M05 (500mA type) or even 78L05, but also because the 7805 can only deliver this high current, when being suffiently heat-sinked and when being fed by a strong enough transformer.

So, I guess, it's not unusual to see such a chip surviving this procedure, at least for a brief period.

But make no mistake: Having survived this torture does not mean that the chip is as intact as a new one. In any case, this chip must be replaced by a new one, because you never know, whether the chip has get any pre-damage!!!

Kai