Solids and Liquids in Hetrogenous Equilibrium

In summary, at equilibrium the forwards and backwards reaction rates must be the same, but increasing the surface area of a solid has no effect on the equilibrium concentrations of either "B" or "C".
  • #1
Dragynfyre
20
0
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]}{} [/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?
 
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  • #2
Dragynfyre;3630290Also 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 said:
No. If microscopic amounts of solid existed then it would be saturated. The very definition of "unsaturated" implies that there is insufficient amounts of the compound dissolved to make it saturated.
 
  • #3
Dragynfyre said:
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".

Increasing the surface area of a solid would increase both the forward and backward rate equally. This happens because both the forward and backward reactions must take place on the surface.
 
  • #4
Ygggdrasil said:
Increasing the surface area of a solid would increase both the forward and backward rate equally. This happens because both the forward and backward reactions must take place on the surface.

But in the hypothetical reaction I posted there is no solid in the backwards reaction.
 
  • #5
Let's say the reaction is crystallization of a salt from solution. Addition of ions from solution to the crystal at the crystal's surface is much faster than nucleating new crystals at a different location in the solution (this is why solutions can be supersaturated). Therefore, the rate of deposition will depend on the surface area of the solid.
 
  • #6
Ygggdrasil said:
Let's say the reaction is crystallization of a salt from solution. Addition of ions from solution to the crystal at the crystal's surface is much faster than nucleating new crystals at a different location in the solution (this is why solutions can be supersaturated). Therefore, the rate of deposition will depend on the surface area of the solid.

In that situation I would agree that the backwards reaction would depend on the surface area of the solid. However, the solid crystal is essentially placed on both sides of the reaction equation in that situation because in order for the deposition to occur the ion has to come in contact with the crystal's surface. Therefore, I don't believe the logic in that situation will apply to all situations of heterogeneous equilibrium.

Now let me adjust the hypothetical reaction to be a bit easier to visualize:

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

Notice that I added a second gaseous substance on the right side. Now in this case I don't see how the backwards reaction could possibly change depending on the surface area of the "A" solid since the backwards reaction takes place between two gases.
 

1. What is a heterogeneous equilibrium?

A heterogeneous equilibrium refers to a situation where more than one phase or state of matter is present at the same time, and the reaction occurs between these phases. For example, a solid and liquid coexisting in a closed system is a common example of a heterogeneous equilibrium.

2. How is equilibrium reached in a heterogeneous system?

In a heterogeneous equilibrium, the forward and reverse reactions proceed at different rates due to the different phases involved. Eventually, the rates of the two reactions will become equal, and equilibrium is reached. This means that the amount of each substance present will remain constant over time, although the reactions are still occurring.

3. How does temperature affect a heterogeneous equilibrium?

Temperature has a direct impact on the equilibrium of a heterogeneous system. An increase in temperature will generally shift the equilibrium towards the endothermic side of the reaction (usually the side with more gas molecules), while a decrease in temperature will shift it towards the exothermic side (usually the side with fewer gas molecules).

4. Can the concentration of a substance affect a heterogeneous equilibrium?

Yes, the concentration of a substance can also affect a heterogeneous equilibrium. If the concentration of one of the reactants or products is changed, the equilibrium will shift in order to maintain a balance. This is known as Le Chatelier's Principle.

5. How do catalysts influence a heterogeneous equilibrium?

Catalysts do not directly affect the position of equilibrium in a heterogeneous system, but they can increase the rate at which equilibrium is reached. This is because they lower the activation energy required for the reaction to occur, allowing it to reach equilibrium faster.

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