Why is octane more volatile than water while having a higher boiling point?

Question

Octane has a boiling point of 120 °C. Water has a boiling point of 100 °C. The definition of boiling point is, "the temperature which the liquid substance's saturated vapor pressure equals the atmospheric pressure". Volatile substances have higher saturated vapor pressure at a given temperature, than the lesser volatile substances. When temperature increased (heated), the saturated vapor pressure increases rapidly and non-linearly. So, the volatile substance's saturated vapor pressure reaches atmospheric pressure quicker than the lesser volatile substance, hence, reaching boiling pint at a lower temperature.

However, water which is a less volatile substance than octane (if you put two separate liquid spots from water and octane, octane evaporates faster) has a lower boiling point than octane at atmospheric pressure. Is there something I am missing? If my reasoning is wrong please correct me.

Answer 1

One thing you are missing is that air contains water, but usually does not contain octane. So for water, the process is:

$$\ce{H2O(l) <=> H2O(g)}$$

and for octane, it is

$$\ce{C8H18(l) <=> C8H18(g)}$$

For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.

If you look up the vapor pressure at room temperature, you will find that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).

Answer 2

The are three major factors, affecting observed relative volatility, ordered by assumed importance:

• The molar mass. The liquid with higher molar mass evaporates faster than a liquid with the same vapor pressure but lower molar mass. This has two reasons:

  • That is because for the given vapor pressure, the mass concentration of vapor is proportional (as ideal gas approximation) to its molar mass: $\rho = \frac{pM}{RT}$
  • The higher molar mass leads to lower specific enthalpy of evaporation, so it need lower heat transfer for the same evaporation rate.: $ΔH_\text{evap,spec}=\frac{ΔH_\text{evap,molar}}{m}$

• Another factor Karsten mentioned (and I as a former enlisted military meteorologist should remember to write) is water vapor presence, which slows down evaporation. Air already contains typically $\pu{20 to 100\%}$ of possible water vapor amount, what slows down water evaporation. That is not the case for other liquids.

• The third factor is the molar evaporation enthalpy. The higher is the molar evaporation enthalpy of the liquid, the faster is the vapor pressure decrease with temperature.

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The typical other case is toluene with the boiling point $\pu{111 ^\circ C}$, but significantly more volatile than water.

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Note the van't Hoff equation about the dependence of liquid vapor pressure on absolute temperature in the integral form:

$$\ln{\frac{p_2}{p_1}=\frac{ΔH_\mathrm{evap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})}$$

Substance	Molar evaporation enthalpy ($\pu{kJ/mol}$)	Molar mass ($\ce{g/mol}$)
Water	40.65	18.02
Toluene	36.08	92.14
n-Octane	39.1	114.23

Answer 3

It is a case of careful observation and awareness of the construction of matter. Water and octane are molecular so a given volume with similar densities [1 vs 0.8] have very different # of molecules. The molar heats of vaporization are similar ~10Kcal/mole, [I leave it to you to look up the exact data]. Approximately similar masses with the same available energy and similar vapor pressures the fewer molecules will evaporate faster. As the MW difference increases the differences will change. Compare kerosene or cetane.

Also drop shape and surface area are important when comparing rates. Heat transfer is important.

Answer 4

Firstly, there are 18 different isomers of octane, each having a unique boiling point.

The gasoline standard is that "100 octane" corresponds to pure iso-octane (2,2,4-trimethylpentane).

Iso-octane has a boiling point of 99°C.

Secondly, the mass-rate of evaporation, according to the Hertz-Knudsen equation, given everything else is equal, is proportional to the square root of molecular weight. Therefore, if the vapor pressure of water and iso-octane are approximately the same, the mass-rate of evaporation of octane should be greater.

See also problem 1b here.

Thirdly, since the density of octane is about 0.7 relative to water, if equal-volume drops are being evaporated, the rate of evaporation will be even greater than the mass-rate.

Answer 5

The boiling point is when it starts making bubbles inside.

"Liquids may change to a vapor at temperatures below their boiling points through the process of evaporation. Evaporation is a surface phenomenon in which molecules located near the liquid's edge, not contained by enough liquid pressure on that side, escape into the surroundings as vapor. On the other hand, boiling is a process in which molecules anywhere in the liquid escape, resulting in the formation of vapor bubbles within the liquid."

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"a liquid will evaporate until the surrounding air is saturated"

Evaporation is a type of vaporization that occurs on the surface of a liquid as it changes into the gas phase. High concentration of the evaporating substance in the surrounding gas significantly slows down evaporation, such as when humidity affects rate of evaporation of water.

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All things are not equal.

The boiling point is an important property because [all things being equal, e.g., in a vacuum] it determines the speed of evaporation. Small amounts of low-boiling-point solvents like diethyl ether, dichloromethane, or acetone will evaporate in seconds at room temperature, while high-boiling-point solvents like water or dimethyl sulfoxide need higher temperatures, an air flow, or the application of vacuum for fast evaporation.