Chemistry - Is the average kinetic energy of evaporating water molecules (at room temperature) equivalent to the average kinetic energy of boiling water?

Solution 1:

At gas/liquid phase equilibrium, average kinetic energy of evaporating molecules, i.e. those just passed to a gas phase, is equal to average kinetic energy of condensing molecules.

The latter is then approximately proportional to $T$.

If these average values were not equal, the system would not be in thermal equilibrium.

Feedback to comments:

@theorist In case the gas phase consists exclusively ( or almost ) from water molecules, then majority of gas molecules collisions with liquid water leads for them to becoming a part of the liquid.

But still, if the mean kinetic energy of condensing molecules differs from the mean kinetic energy of molecules that during the same time interval evaporated liquid, Houston has a problem.

Those collisions that are not part of mass exchange would have to cause nonzero thermal transfer, what is not possible when temperatures of phases is equal.

BTW, I do not think, imply nor say that liquid leaving probability is KE independent. I suppose there is a KE threshold to leave liquid, to overcame the mean bonding energy. OR, perhaps, rather a threshold of the normal component of the velocity vector ( or equivalent energy for 1 degree of freedom ), as the molecule motion has to be properly oriented.

I have created a quick and dirty model to estimate the percentage of vapour molecules that condense after hitting the liquid phase.

I assumed the Boltzman energy distribution for 1 degree of freedom - the normal direction wrt the liquid surface. Then I recalculated the molar water enthalpy of evaporation to 1 water molecule as the energy to overcome when evaporating and calculated the fraction of molecules with enough energy in this degree of freedom.

The water molar enthalpy of evaporation: $\Delta H_{w,mol,evap}=\text{40.7 kJ/mol}$

The Boltzmann statistic: $\exp{\left(-\frac{\Delta H_{w,mol,evap}}{N_\mathrm{A} \cdot kT}\right)} \approx \exp{\left(-\frac{4895}{T}\right)}$

Than I have calculated saturated vapour pressure and related vapour density from the Clausius-Clapeyron equation.

Then I estimated the condensing fraction as the Boltzmann fraction multiplied by water/vapour density ratio.

But still, it seems to me the system would be just much closer to equilibrium than the prior assuming of all condensing vapour molecules, that is not in equilibrium.

t$\pu{[^{\circ}C]}$ T$\pu{[K]}$ x(Boltzmann) p$\pu{[Pa]}$ Vap. Density$\pu{[kg/m^3]}$ Rel. density x condensing
0 273.15 1.65E-08 831 0.0066 6.60E-06 0.0025
10 283.15 3.11E-08 1566 0.0120 1.20E-05 0.0026
20 293.15 5.60E-08 2824 0.0209 2.09E-05 0.0027
30 303.15 9.71E-08 4899 0.0350 3.50E-05 0.0028
40 313.15 1.63E-07 8205 0.0568 5.68E-05 0.0029
50 323.15 2.64E-07 13309 0.0892 8.92E-05 0.0030
60 333.15 4.16E-07 20972 0.1364 1.36E-04 0.0030
70 343.15 6.38E-07 32182 0.2032 2.03E-04 0.0031
80 353.15 9.56E-07 48200 0.2957 2.96E-04 0.0032
90 363.15 1.40E-06 70603 0.4213 4.21E-04 0.0033
100 373.15 2.01E-06 101325 0.5884 5.88E-04 0.0034

Interesting would be the result, if

  • The Maxwell-Boltzmann statistics would be taken as the energy distribution model for liquid phase molecules
  • It would be taken its subset, fitting the Boltzmann 1-degree-of-freedom energy threshold criterium
  • It would be calculated the mean kinetic energy of this subset, then subtracted the heat of evaporation.
  • Finally, it would be calculated the equivalent temperature.

Solution 2:

[OP] the average kinetic energy of evaporating water molecules

You have to specify whether you are talking about the kinetic energy just before the water molecule breaks the hydrogen bonds to its neighbors or just afterwards. A millisecond before or after the event, of course, the average kinetic energy will be determined by the bulk temperature.

One way to picture this is two water molecules "colliding" at the surface of the liquid. In order for one water molecule to leave the liquid phase, the collision needs to have at least sufficient energy to break the hydrogen bonds (and perhaps some more to overcome an activation energy). When that happens, the total kinetic energy of the molecules in the collision will be lowered (because energy is conserved). What exactly we mean by "collision" in the liquid phase is not so important because the temperature-dependence of the kinetics of the process is still governed by the Arrhenius equation (and the Boltzmann distribution of collision energies) even when we go from simple mono-atomic gases to wicked-complicated liquids.

The energy required to break the water loose (~40 kJ/mol, according to Poutnik's answer) is much higher than the median collision energy at room temperature (~2 kJ/mol). Whether the water molecule has more or less kinetic energy than the average after it enters the gas phase is not so important. The important part is that it had an unusually high energy, and most of that excess energy went into breaking the hydrogen bonds.

[OP] Assuming we could measure the kinetic energy of multiple evaporating water molecules just as they left the surface of liquid water (at standard temp and pressure) over time, would the average kinetic energy of the sample of evaporating water molecules equal the average kinetic energy of boiling water (at 101.3kpa)?

The kinetic energy of the water molecules that just evaporated is not exceptionally high or low. As the molecule leaves the liquid phase, the available energy is somehow partitioned between that water molecules and the ones it leaves behind. Your question boils down to whether the evaporative cooling of the liquid is temperature-dependent. I'm not sure if there is an experiment that would address this, and whether it has been done already.

[OP] Diagram for discussion

The diagram has a lot of problems, so I would not use it for discussion. The curve does not change that much with temperature, the shaded area is too big by far (according to Poutnik's answer, a molecule on the surface of the liquid has about a one in a million chance per collision to get into the gas phase at 373 K). Here is a diagram showing how the energy available from collisions changes when the temperature is raised from room temperature to body temperature. You can see that the changes for the bulk of energies are subtle. If you zoom in to the regions of exceptionally high collision energies, however, you can see that the changes are substantial.

enter image description here