Increasing the contract surface area between hydrocarbons raises boiling point, so hexane should have a higher boiling point than propane. This doesn't really make sense to me. What is the difference between pulling apart a pair of propane twice, versus one pair of hexane? Why would the total amount of energy needed differ? It's seems to go against the concept of pressure.

  • $\begingroup$ It is also about entropy. If you pull two propane molecules out of liquid propane you get two molecules with more freedom to move as opposed to just one when it is hexane. Using your argument, you would get two different boiling points for hexane, depending whether to pull apart a single hexane (lower boiling point) or two at a time (higher boiling point). $\endgroup$ – Karsten Theis Apr 16 '19 at 21:22
  • $\begingroup$ Sorry, Karsten, I don't really agree with you. Entropy has little to do with this process, which can actually be explained only in terms of enthalpy. The boiling point of a molecule is unique ,that means that you can't have two boiling points for water, for instance, or for any other substance. If you pull apart hexane, then you are dissociating the molecule into two different fragments, which is different than separating two molecules using heat. $\endgroup$ – Pier Apr 16 '19 at 21:55
  • $\begingroup$ @Pier You are using entropy implicitly in your answer as you compare the enthalpy per molecule to disrupt dispersion forces. You could have compared it per mass, in which case all hydrocarbons would have similar boiling points. There is a nice article on Wikipedia, en.wikipedia.org/wiki/Entropy_of_vaporization. The (molar) entropy of vaporization happens to be very similar for hydrocarbons irrespective of size, so you get good results just comparing enthalpies of vaporization. $\endgroup$ – Karsten Theis Apr 17 '19 at 4:40

Some data on size of hydrocarbons and boiling point

I found complete data on the butane/octane pair but not the propane/hexane pair, so I am using those as illustration (I think they are taken at a temperature near the normal boiling point, respectively).

$$ \begin{array}{|c|c|c|c|c|} \hline \text{substance} & \Delta H_{vap} & \Delta S_{vap} & T_b & \text{ref}\\ \hline \text{butane} & \pu{22.4 kJ/mol} & \pu{82.30 J/mol K}& 273 K & [1]\\ \hline \text{octane} & \pu{39.1 kJ/mol} & \pu{98.3 J/mol K}& 398.7 K & [2]\\ \hline \end{array} $$

Ref. 1: https://webbook.nist.gov/cgi/cbook.cgi?ID=C106978&Mask=4

Ref. 2: https://file.scirp.org/pdf/OALibJ_2016091314302030.pdf

The enthalpy of vaporization is determined by measuring equilibrium vapor pressure over a range of temperatures (Clausius Clapeyron equation). Then, the entropy of vaporization is calculated from the boiling point and the enthalpy of vaporization ($\Delta G_{vap} = 0$ at the boiling point):

$$ \Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$$

What is the difference between pulling apart a pair of propane twice, versus one pair of hexane?

As far as enthalpy is concerned, there is not much difference. You have to disrupt about the same dispersion forces removing one molecule of hexane from liquid hexane or removing two molecules of propane.

However, if enthalpy were the only aspect you had to consider, liquids would never turn into gases because you lose all interactions in the gas phase. Experience shows that if you increase the temperature sufficiently, at one point you disrupt the interactions between molecules in a liquid and get a gas.

Why would the total amount of energy needed differ?

What determines whether a process happens or not is not the change in enthalpy alone, but the change in Gibbs energy. The Gibbs energy contains an enthalpy component and a highly temperature-dependent entropy component. The entropy change of a molecule from the liquid to the gas state is about the same for butane and octane (also see Trouton's rule), but the enthalpy change is different by a factor of two. That's why the two substances have different boiling points.

$$ \Delta G_{vap} = \Delta H_{vap} - T_b \Delta S_{vap} = 0$$

$$ T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}}$$

If you ask for the analysis for evaporating a kilogram each of butane and octane, the enthalpy change for the entire sample would be similar, but the entropy change for the entire sample would be off by a factor of about 2. This is because the molar entropy of vaporization is similar for butane and octane, but a kilogram of butane contains about twice the number of molecules compared to a kilogram of octane.

It's seems to go against the concept of pressure.

This is a bit of a puzzling statement. There are two things that require energy when a liquid turns into a gas at constant pressure. The first is disrupting the intermolecular forces in the liquid. The second is pushing against the pressure to increase the volume of the sample as liquid turns into gas. The latter is abbreviated as P-V-work. If you look at internal energy changes, you have to consider it. If you look at enthalpy changes, however, it is taken out of the equation because enthalpy is defined as $$ H = U + P V$$ As we are discussing Gibbs energy and enthalpy, we don't have to worry about P-V-work. ("It is not part of the bill").

  • 1
    $\begingroup$ I understand and agree with your explanation, but I feel that your last paragraph is a bit confusing as you use kg not mole as your comparison in the first 2 lines. You say ' ..is not the change in enthalpy' in line 3 but 'the enthalpy change is different by a factor of 2 ' in the last line. You write also in para 1 that dispersion forces are similar 'there is not much difference' but then use this difference to explain the change in boiling point in the last para. $\endgroup$ – porphyrin Apr 17 '19 at 6:58
  • $\begingroup$ @porphyrin - I added some data and broke the paragraph you mentioned into two parts. $\endgroup$ – Karsten Theis Apr 18 '19 at 12:30
  • 1
    $\begingroup$ very clear now :) $\endgroup$ – porphyrin Apr 18 '19 at 13:48

To answer your question, keep in mind that for hydrocarbons the main attractive force at play is dispersion interactions. They increase as the size of the molecule increases. In other words, since the molecule of hexane is "larger" than the molecule of propane, it means that it has a larger surface capable of interacting with the surface of another molecule of the same kind. The more surface you have, the more intense is this interaction, and the more energy you need to tear the two molecules apart.

Take a look at the table listed in this website: https://crippen.education.ufl.edu/calculators/activities/BP.html

You see that the larger the mass (or surface of the molecule), the higher the boiling point. Can you tell us a little bit more about your last question? Why should it go against the pressure? I hope that this helps you a little.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.