10
$\begingroup$

In Vapor-Liquid Equilibrium, it is my understanding that molecules exchange between the gas and liquid continuously; even when no difference in temperature or partial pressure exists.

enter image description here

Regarding Solid-Liquid Equilibrium; is there a similar continuous exchange of molecules? For example on a single crystal of ice submerged in pure water at perfect and constant 0°C, will molecules on the surface of this crystal exchange with the liquid? In the past I have assumed this to be the case, but now realize I have not seen a reference that explicitly details this like is common for vapor-liquid equilibrium. Are there any references that calculate the rate at which this occurs? Side question, is there also a continuous exchange for solid-gas equilibrium?

$\endgroup$
4
  • 4
    $\begingroup$ Yes, and yes. All solids have some vapor pressure, and (often negligible) solubility. $\endgroup$
    – Mithoron
    Commented Jun 5, 2022 at 20:48
  • 2
    $\begingroup$ Generally, molecules/atoms/ions have no means for awareness the system is in equilibrium, being "instructed to do nothing". Additionally, they do not have temperature. A single kinetic energy value belongs to a wide range of possible system temperatures, so they are changing a phase at will. $\endgroup$
    – Poutnik
    Commented Jun 6, 2022 at 8:20
  • 1
    $\begingroup$ Your question was already answered chemistry.stackexchange.com/questions/73344/… $\endgroup$ Commented Jun 8, 2022 at 3:55
  • 3
    $\begingroup$ Does this answer your question? Proof of Dynamic Nature of Equilibrium $\endgroup$ Commented Jun 8, 2022 at 3:55

1 Answer 1

15
$\begingroup$

Even two solids in contact may exchange atoms, such as gold in contact with lead. Accidental cold-welding is a serious problem in space technology, in particular.

You can prove to yourself that ice/water, in equilibrium, exchanges molecules: immerse an ice-cube on a string in water at 0°C, and the cube will slowly change shape. "No chemist ever observes the same ice cube twice..." as Heraclitus or Heisenberg stated (I'm not certain).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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