How Can Water Evaporate Despite Strong Hydrogen Bonds?

In summary, the process of water evaporation involves breaking strong intermolecular bonds between water molecules at the interface, facilitated by the concerted, ultrafast hydrogen-bond dynamics of interfacial water. The distribution of agitational energy for molecules within and between these clusters is not fully understood. This lack of knowledge highlights the complexity of water in the liquid phase and the need for further research in this area.
  • #1
Kazys
10
1
Some facts: Water does evaporate.
The average kinetic energy of a water molecule at 373 K is ~0.03 eV
The average strength of the H bond is ~0.24 eV
In the liquid phase the average H2O molecule is bound by 2 to 4 H bonds.
Therefore - it is bound in the liquid phase by bonds which are stronger then
the agitational (kinetic) energy of the molecules.
In order to evaporate these bonds must be broken. How is this possible?
An eV is an energy unit equal to 1.6E-19 J
 
Physics news on Phys.org
  • #2
Averages are meaningless here. What is important is that there is a distribution of kinetic energy, and some water molecules have sufficient energy to escape from the bulk of the liquid.

Note: this is not an A level question. I have lowered it to I.
 
  • Like
Likes russ_watters
  • #3
Kazys.. Have you wondered why evaporation cools liquids?
 
  • #4
It is an A level question because it has not been answered. Not just by you and me, by
anybody. Your answer just restates the question in other words. Yes, indeed, some
molecules do gain sufficient energy to escape, that is clear from the first stated
condition -water does evaporate. But how does this happen?? What is the distribution
of agitational energy of the water molecules? Strictly speaking not just the molecules
the "clusters" because the H-bond is sufficiently strong that water molecules do
act individually. A common answer is that the distribution is Maxwellian. That
is not just an unproved assumption, it is by definition false.
 
  • #5
Kazys said:
It is an A level question because it has not been answered.

An "A" level question requires graduate level knowledge of the subject matter. Your question does not indicate that level of knowledge.

Kazys said:
A common answer is that the distribution is Maxwellian.

The distribution does not have to be Maxwellian to explain why some water molecules have sufficient energy to escape, i.e., to explain how it is possible for water to evaporate. All that has to be true is that the distribution is such that there are some water molecules with kinetic energy larger than the energy binding them to adjacent water molecules. There are many, many possible distributions that have that property.

Also, even the requirement as I have stated it above is actually too strong. It is also possible for a water molecule to gain sufficient kinetic energy to escape as a result of a collision with an incoming molecule of N2 or O2 or some other gas contained in the air. So even if the distribution happened to be such that no water molecule, at some instant of time, had sufficient kinetic energy to escape due to the distribution, water molecules could still gain such energy as a result of collisions, and could therefore escape.

Kazys said:
That is not just an unproved assumption, it is by definition false.

I don't understand what you mean by this.
 
  • Like
Likes russ_watters
  • #6
Kazys said:
Strictly speaking not just the molecules
the "clusters" because the H-bond is sufficiently strong that water molecules do
act individually.
Was that supposed to be "do NOT act individually"?

In any case, do you have an accepted source or a calculation to back up the claim about the strength of the hydrogen bond between molecules?
 
  • #7
Kazys said:
it has not been answered. Not just by you and me, by anybody.

How hard did you look?

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.115.236102

Abstract
Evaporation is the process by which water changes from a liquid to a gas or vapor, and is a key step in Earth’s water cycle. At the molecular level, evaporation requires breaking at least one very strong intermolecular bond between two water molecules at the interface. Despite the importance of this process the molecular mechanism by which an evaporating water molecule gains sufficient energy to escape from the surface has remained elusive. Here, we show, using molecular dynamics simulations at the water-air interface with polarizable classical force field models, that the high kinetic energy of the evaporated water molecule is enabled by a well-timed making and breaking of hydrogen bonds involving at least three water molecules at the interface, the recoil of which allows one of the molecules to escape. The evaporation of water is thus enabled by concerted, ultrafast hydrogen-bond dynamics of interfacial water, and follows one specific molecular pathway.
 
  • Like
Likes Orodruin, russ_watters, Bystander and 2 others
  • #8
To PeterDonis
It seems that I started this post improperly, for this I apologize. My intention was to illustrate that some common and very familiar
phenomena are still not completely understood. The evaporation of water is an example. Water in the liquid phase is a very complex
medium. The molecule is deceptively simple, but it has an exceptionally high dielectric constant, because it is so small and has
a high H to O ratio it forms strong cohesive H bonds. As a result, in the liquid phase it does not exist as separate molecules,
but as constantly reforming clusters of molecules. The distribution of agitational energy for molecules within these clusters
and/or between the clusters is not known. You reflect this lack of knowledge when you state: "there are many possible distributions",
Indeed so, but until the actual distribution is known the precise mechanism used by an escaping molecule cannot be determined.
Re. the Maxwellian distribution - it provides a good approximation for the distribution of kinetic energy of water molecules in the
vapor phase, by assumption it has often been applied also to the liquid phase, that is not correct. The "by definition" comment refers
to the conditions that Maxwell used to derive a distribution of kinetic energy for a statistically large number of interacting entities.
These are: interaction occurs between 2 entities at a time, these collisions are elastic, scattering is isotropic. None of
these conditions apply for liquid water.
 
  • #9
To: CWaters

Thank you very much for this reference. Knowledge exists at several levels of certitude. Knowledge obtained
by using simulations of virtual models, as is the case in the noted study, is valuable, however it
depends on the accuracy of the virtual model and the simulation process. It is reassuring that
the simulation result agrees with the conclusions reached in the following study, which is based on
experimental data of evaporation parameters at low pressures:

https://www.researchgate.net/publication/275027122_EvaporationCondensation_of_Water_Unresolved_Issues
 
Last edited by a moderator:
  • #10
To: Nugatory
My sincere apologies, a typo. It was meant to be "do NOT act etc",

Re. your question of sources for H-bond strength, I am reluctant to
quote a single one, the Wikipeda entry on "Hydorgen bond of water"
provides the data and a selection of sources.
 
  • #11
Kazys said:
My intention was to illustrate that some common and very familiar
phenomena are still not completely understood.

Saying that we do not know the exact way in which water molecules cluster, or the distribution of kinetic energy of the molecules, is not the same as saying we do not know how water can evaporate. Nothing that you are saying in any way invalidates the simple explanation: water evaporates because some water molecules have sufficient kinetic energy to escape.

In fact, that simple explanation is not precisely correct--but not for any reason that you gave. Consider: if some water molecules in the liquid have a high enough kinetic energy to escape, then some water molecules in the vapor should also have low enough kinetic energy to be captured. So what is actually going on is a dynamic process, with water molecules escaping from the liquid and being captured from the gas. In equilibrium, the rates of both processes will be the same, and macroscopically you have a stable state with a given number of molecules of liquid and a given number of molecules of vapor. Out of equilibrium, the rates are not the same: for example, when water boils, the rate of molecules going from liquid to gas is higher than the rate of molecules going from gas to liquid. When water vapor condenses, the opposite is true.

You could object, of course, that the processes involved in my more elaborate explanation are not completely understood either. But by the criterion you appear to be using, nothing is completely understood. All of our current theories are incomplete. All of them have elements that we have not derived from first principles but just assumed. So saying that a particular explanation is not complete is just repeating a commonplace fact about all scientific explanations.

Kazys said:
until the actual distribution is known the precise mechanism used by an escaping molecule cannot be determined.

This is false. We can know that a molecule of sufficient kinetic energy will escape, without knowing exactly which molecules have that amount of kinetic energy at a given instant. The mechanism of escape is obvious: the kinetic energy of the molecule is greater than the binding energy between it and neighboring molecules.

Kazys said:
Re. the Maxwellian distribution - it provides a good approximation for the distribution of kinetic energy of water molecules in the vapor phase, by assumption it has often been applied also to the liquid phase, that is not correct.

Can you give any examples of a mainstream reference assuming that the liquid phase has a Maxwellian distribution? And can you give mainstream references that show why that assumption is not correct?

Kazys said:
The "by definition" comment refers
to the conditions that Maxwell used to derive a distribution of kinetic energy for a statistically large number of interacting entities.
These are: interaction occurs between 2 entities at a time, these collisions are elastic, scattering is isotropic. None of
these conditions apply for liquid water.

This is true; but it does not show that the distribution of kinetic energies in a liquid is not Maxwellian. It only shows that Maxwell's derivation of that result cannot be applied to a liquid, at least not in the way it applies to a gas.
 
  • Like
Likes russ_watters
  • #12
I think that this statement:
Kazys said:
My intention was to illustrate that some common and very familiar phenomena are still not completely understood. The evaporation of water is an example.
Is not compatible with this one:
Kazys said:
Knowledge obtained by using simulations of virtual models, as is the case in the noted study, is valuable, however it depends on the accuracy of the virtual model and the simulation process.
If our model is so good that even a virtual simulation is accurate, then we do understand it. Maybe you can use "completely" to make it impossible to achieve, but then it is just a tautology and not a real limitation on our understanding.
 
  • Like
Likes russ_watters
  • #13
To PeterDonis

Thank you for your comments. Re, the question -

Can you give any examples of a mainstream reference assuming that the liquid phase has a Maxwellian distribution? And can you give mainstream references that show why that assumption is not correct?

A number can be quoted, a good one is the following:

Marek, R, J. Straub “Analysis of the evaporation coefficient and the condensation coefficient of water”, Int. J. Heat Mass Transfer 44 (2001) pp 39-53

The study analyses the correction factors which are used to force agreement between measurements and computed results which use the
assumption that the agitational energy distributions of both the vapor and the liquid phases are Maxwellian. The authors review a very large
number of publications (close to hundred, as I recall) and show that the correction factors can differ by orders of magnitude. Empirical results
do usually differ, measurement errors are unavoidable, but this is clearly out of line. It shows that the theory is in error. The distribution of
the agitational energy of water liquid molecules is NOT Maxwellian.
 
  • #14
Kazys said:
Empirical results
do usually differ, measurement errors are unavoidable

But do the actual empirical results--the measurements--differ substantially? It doesn't seem so; it seems, from what you are describing, that the issue is not the measurements themselves, but trying to match up the theoretical prediction with the measurements.

Kazys said:
It shows that the theory is in error. The distribution of
the agitational energy of water liquid molecules is NOT Maxwellian.

I agree that this indicates that the theory being used is incomplete. However, I don't think that means we don't understand how water can evaporate. I think it means we don't understand the specific distribution of energies of water molecules in the liquid phase. But we don't need to understand that to understand how water can evaporate. We would if we wanted to predict the exact rate at which water would evaporate given specific conditions (temperature, pressure, humidity of the air, etc.).
 

Related to How Can Water Evaporate Despite Strong Hydrogen Bonds?

1. What is evaporation?

Evaporation is the process by which a liquid, in this case water, changes into a gas. It occurs when the molecules of water gain enough energy to break free from the liquid and turn into water vapor.

2. What factors affect the rate of water evaporation?

The rate of water evaporation is affected by several factors, including temperature, humidity, surface area, and air movement. Warmer temperatures, lower humidity levels, larger surface area, and increased air movement can all lead to faster evaporation.

3. How does water evaporate from the ocean?

Water evaporates from the ocean through the same process as any other body of water. The sun's heat warms the surface of the ocean, causing the water molecules to gain energy and turn into water vapor. This water vapor then rises and cools, eventually forming clouds and potentially leading to precipitation.

4. Can water evaporate at any temperature?

Yes, water can evaporate at any temperature, as long as there is enough energy for the water molecules to break free from the liquid and turn into vapor. However, the rate of evaporation is higher at higher temperatures due to the increased energy.

5. What is the purpose of water evaporation in the water cycle?

Water evaporation is an essential part of the water cycle, which is the continuous movement of water on, above, and below the Earth's surface. It helps to transfer water from the Earth's surface to the atmosphere, where it can then be redistributed as precipitation, completing the water cycle.

Similar threads

Replies
32
Views
2K
Replies
1
Views
918
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
4K
Back
Top