# Time required for a thin layer of ethanol to evaporate?

If someone were to pour about 5 ml of 94% pure ethanol on a flat surface and at room temperature, approximately how much time would it take for it to evaporate?

EDIT: In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

• Is there any background for the experiment (for example, you're trying to make an ideal eyeglass cleaner or something)? It might help someone to give you a better answer. Otherwise, you could certainly give this a try at home with a lower purity ethanol and see if you can extrapolate. – jonsca Nov 19 '12 at 2:17
• What is the surface? Is the surface wettable by ethanol? Evaporation time of a thin layer is strongly dependent on the layer thickness/specific surface area. Suffice to say, ethanol evaporates quite rapidly with respect to water due to its relatively low specific heat capacity and high vapor pressure. – Richard Terrett Nov 19 '12 at 12:15
• Basically, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes? – samseva Nov 19 '12 at 16:36
• Google for "evaporation number" and its definition – Georg Apr 16 '13 at 11:09
• bayblab.blogspot.jp/2009/01/… – user2617804 Jul 7 '14 at 7:14

In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

It be simpler and faster to do the experiment than to try to predict the time needed.

The kinetic molecular theory explains why liquids evaporate at temperatures lower than their boiling points. At any temperature, the molecules in a liquid have a range of kinetic energies described by a Boltzmann distribution. Some percentage of the molecules have enough kinetic energy to escape into the gaseous phase. These molecules contribute to the vapor pressure of the liquid. As the temperature increases, more molecules are above the gas threshold and the vapor pressure increases. As the vapor pressure increases.

In a closed system, equilibrium would be established with an unchanging ratio of liquid to vapor. Molecules would be exchanged between the liquid and the vapor, but their relative amounts would remain constant. The surface you describe is not closed - the vapor molecules can wander away by diffusion or convection. Equilibrium is disrupted, and Le Châtelier's Principle tells us that the equilibrium shifts to compensate. As more ethanol molecules escape, more vaporize to replace them until there are no molecules remaining in the liquid. In the following equation, $K$ is the equilibrium constant, $p$ is the partial pressure of ethanol vapor (the vapor pressure) and [$\ce{C2H6O}$] is the concentration of ethanol in the liquid.

$$\ce{C2H6O (l) <=> C2H6O (g)}$$ $$K=\frac{p_{\ce{C6H6O}}}{[\ce{C2H6O}]}$$

After establishment of equilibrium (fast), the rate-determining step of the evaporation is probably the diffusion of the gaseous ethanol molecules away. The average kinetic energy of a gas particle can be expressed as a function of mass ($m$, in kg) and root mean square velocity ($v^2_\text{rms}$) and separately as a function of temperature ($T$, in kelvin) times the Boltzmann constant ($k_\text{B} = 1.38 \times 10^{-23} \frac{\text{J}}{\text{K}}$). We can derive a formula for rms velocity.

$$\overline{E_\text{k}}=\frac{1}{2}mv^2_\text{rms}$$ $$\overline{E_\text{k}}=\frac{3}{2}k_\text{B} T$$ $$v^2_\text{rms}=\frac{3k_\text{B} T}{m}$$ $$v_\text{rms} = \sqrt{\frac{3k_\text{B} T}{m}}$$

You could calculate the velocity of ethanol particles escaping. If you set an arbitrary distance (one meter is probably fine), then you can calculate a time it takes one particle to travel that distance (on average). If we know the equilibrium constant, we can determine how much vapor is above the liquid and then calculate the amount of time it takes to move.

But how do we know the equilibrium constant? It varies with temperature! The value $\Delta G^\circ_\text{vap}$ in the equation below is the free energy change of vaporization of ethanol in the standard thermodynamic state. $R$ is the ideal gas constant.

$$K=\mathrm{e}^{-\frac{\Delta G^\circ_\text{vap}}{RT}}$$

The model above ignores complicating factors like mean free path, the fact that evaporation is endothermic (meaning the liquid cools as it evaporates and the vapor pressure decreases with time), the temperature and heat capacity of the surface govern how much kinetic energy is available to the liquid to begin with, and that any amount of air current in the vicinity will move the vapor away significantly faster than diffusion.

A complete model includes consideration of the following:

Constants

• the Boltzmann constant
• The ideal gas constant
• the free energy change of vaporization of ethanol (not actually constant, but it varies only slightly over the temperature range)
• the vapor pressure of ethanol as a function of temperature
• the mass of an ethanol molecule
• the heat capacity of the surface

variables

• temperature of the air
• volume of ethanol
• temperature of the surface
• atmospheric pressure (needed for mean free path corrections)
• velocity of air currents

So, in principle, you could do it. Even so, the best answer cannot be had without some serious calculus. In practice, it would be faster to do the experiment (if often is). We often forget that science is empirical.

• -1 for mixing equilibrium thermodynamics into a question for a kinetic process. Horrible – Georg Apr 16 '13 at 11:11
• The point was to demonstrate that the theoretical prediction was hard while the experiment was trivial. – Ben Norris Apr 16 '13 at 13:36

Essentially random.

The air movement anywhere is essentially random and the greater the air currents above the sample the quicker it will evaporate. Do the experiment: spill two samples of $5~\mathrm{ml}$ simultaneously, blow onto one of the two and check how much quicker it evaporates.

For a much more detailed answer, upvote Ben’s!