Video: https://www.youtube.com/watch?v=_qFN1hDZwp8

If you choose not to view the video, in the first demonstration a paper towel is soaked in isopropanol. As expected, the paper towel burns. However, in the second demonstration the paper towel is soaked in isopropanol and water. Surprisingly, the towel burns without being damaged.

My chemistry professor tells me that it has something to do with the energy cost of a phase change $\left[\ce{H2O}\ (l) \longrightarrow \ce{H2O}\ (g)\right]$; however, I am still very confused.

Could someone please elaborate? Thank you.


I think your professor is incorrect, and it has to do with the thermodynamics and kinetics of the combined vaporization/combustion process.

The heat of vaporization of water $\left(40.7\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite is considerably less than the heat of combustion of isopropanol $\left(1830\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite. These numbers indicate that approximately $45$ water molecules must evaporate for each isopropanol molecule combusted in order to hold the temperature constant. This corresponds to a $\sim2\%$ mol/mol solution, which (a) is not consistent with the $\sim50\%$ v/v solution he actually used, and (b) would probably be too dilute to burn at room temperature the way it does in the video. (Someone more motivated than I could probably confirm this using flash point data.)

I think, instead, the behavior depends on the following:

  1. Vapor-liquid equilibrium stipulates that a liquid phase can never be at a temperature greater than its boiling point$^\dagger$.
  2. It is the vapor of combustible substances that actually burns, not the liquid (see, e.g., this OSHA training document)
  3. Up to a concentration of about $60\%$ alcohol, the vapor phase in equilibrium with a water-isopropanol liquid mixture has a considerably higher concentration of isopropanol as compared to the liquid (source):

Phase diagram

For the present purpose, to read the diagram choose a temperature on the $y$-axis and track across the $x$-axis, noting where you cross the red and blue lines. These correspond to the equilibrium liquid and gas phase compositions, respectively, at that temperature, given a mixture that's not too lean or too rich in isopropanol. For example, take $82\ ^\circ\mathrm C$: you hit the red line (liquid) at about $20\%$ isopropanol, and the blue line (vapor) at about $53\%$ isopropanol.

Since the boiling point of isopropanol is $82.6\ ^\circ\mathrm C$, #1 means that as long as there's liquid left soaked into the paper (even $100\%$ isopropanol!), the paper can't burn. You can actually see this in the video: he ignites the paper at around 0:57, but it doesn't actually start to char until about 1:08, ten seconds later.

The reason why the paper does then start to char in spots is that those spots have locally run out of liquid isopropanol to keep them cool. There remains a cloud of burning gaseous isopropanol all around the paper (clearly!), which is more than enough to burn the now-dry paper.

When there's water mixed with the isopropanol, though, the vapor-liquid equilibrium comes into play. Per point #3, based on the phase diagram, the isopropanol likes to be in the gas phase a lot more than the water, so once the vapor is ignited and burning (point #2), the heat produced tends to evaporate more of the alcohol than the water. This feeds the flame for a while, but eventually it sputters out, with plenty of water left on the paper to keep it cooled well below its char point.

$^\dagger$ Mixtures of liquids technically don't have boiling points, as their behavior is more complex, but the concept is useful for making the point.


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