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## Main Question or Discussion Point

We started to learn about chemical equilibrium and equilibrium constants a few weeks ago and something has been bugging me. I don't understand why solids and liquids are not included in the equilibrium constants for some reactions.

Here's a hypothetical situation with the following reaction

[tex] A_{(s)} + B_{(g)} \leftrightarrow C_{(g)}[/tex]

Now the K expression would normally be written as [tex] K = \frac{[C]}{

Here's a hypothetical situation with the following reaction

[tex] A_{(s)} + B_{(g)} \leftrightarrow C_{(g)}[/tex]

Now the K expression would normally be written as [tex] K = \frac{[C]}{

**} [/tex]**

with the solid left out since it has constant concentration (density) at a constant temperature.

At equilibrium the forwards and backwards reaction rate must also be the same. Now from my understanding, heterogeneous reactions only occur at the surface of the solid or liquid, therefore, the rate of the forwards reaction would be affected by the surface area of the "A" solid. If you increase the surface area of "A" by adding more solid or by crushing it into a powdery form wouldn't it increase the rate of the forwards reaction while leaving the backwards reaction rate unaffected and thereby shift the equilibrium to the right? However, according to the K expression changing the surface area of the solid would have no effect on the equilibrium concentrations of either "B" or "C".

Also I have another related equilibrium question. Would an unsaturated solution be considered to be in dynamic equilibrium? My teacher says it isn't since you can't observe any solid solute in such a condition and all the species in a reaction must be present for a dynamic equilibrium to be established. However, don't microscopic amount of solid solute still exist even in an unsaturated solution?with the solid left out since it has constant concentration (density) at a constant temperature.

At equilibrium the forwards and backwards reaction rate must also be the same. Now from my understanding, heterogeneous reactions only occur at the surface of the solid or liquid, therefore, the rate of the forwards reaction would be affected by the surface area of the "A" solid. If you increase the surface area of "A" by adding more solid or by crushing it into a powdery form wouldn't it increase the rate of the forwards reaction while leaving the backwards reaction rate unaffected and thereby shift the equilibrium to the right? However, according to the K expression changing the surface area of the solid would have no effect on the equilibrium concentrations of either "B" or "C".

Also I have another related equilibrium question. Would an unsaturated solution be considered to be in dynamic equilibrium? My teacher says it isn't since you can't observe any solid solute in such a condition and all the species in a reaction must be present for a dynamic equilibrium to be established. However, don't microscopic amount of solid solute still exist even in an unsaturated solution?