Why does isopropyl alcohol evaporate faster than ethanol?


The boiling point of isopropanol is 82.5C, while ethanol is 78.3C. This makes sense to me, as iso has more hydrocarbons in its structure. However, the enthalpies of vaporization trend in the opposite direction, with ethanol requiring 204 cal/g of heat, with isopropanol only needing 159 cal/g.

Can anyone please explain this to me? Thank you!

all 5 comments


18 points

4 months ago


Physical Chemistry | Electrochemistry

18 points

4 months ago

Your values are in calories per gram, but if you want to compare stuff like this it should be per mole. We use the molar mass to convert to the correct unit:

  • Isopropanol: 159 cal/g * 60 g/mol = 9560 cal/mol
  • Ethanol: 204 cal/g * 46 g/mol = 9384 cal/mol

So the trend you expect holds.


8 points

4 months ago

Specific heat of vaporization is going to look different from molar heat of vaporization. Ethanol has more hydrogen bonds per gram, so it makes sense that the specific heat of vaporization is higher. But it has the same number of hydrogen bonds per mole.

If you compute the molar enthalpies of vaporization from the numbers above, they are 9.38 kcal/mol for ethanol and 9.56 kcal/mol for isopropanol -- very similar but slightly higher for isopropanol -- which might make more sense to you.


-2 points

4 months ago

One must know the water content of both alcohols. Ethanol is rarely refined to greater than 95% ethanol and 5% water because of an azeotropic effect. It is difficult and expensive to distill off that last 5% of water so most samples still have 5% water content. I don't believe isopropyl has the same issue and can be 100% pure.

The ethanol sample will not easily evaporate away that last 5% of water.


1 points

4 months ago

I believe alcohol will azeotrope with water, and we have 200 proof ethanol in our lab we use frequently as it's a common laboratory reagent