Why is rate of dissolution constant?

In summary, the conversation is discussing the dissociation of A into ions and the rate law that predicts a constant rate. However, the rate law does not take into account the decreasing surface area of the solid, which would result in a lower rate. There are several possibilities for this discrepancy, such as using a more sophisticated expression for rate of dissolution or dealing with a solid that dissolves very slowly. The conversation also touches on the introduction of equilibrium constants and solubility product constants in relation to rate laws, and the realization that the rate law in question is not accurate. Ultimately, it is concluded that once a solution is saturated, the amount of solid or its surface area does not affect the rate of dissolution.
  • #1
sodium.dioxid
51
0
A(s) <---> B(aq)
For the dissociation of A into ions, the rate law predicts a constant rate:
rate = k [solid], where the concentration of the solid is constant.
But this doesn't make sense because as the solid is becoming smaller, surface area is decreasing. Less surface area means less molecules breaking off at any interval of time (lower rate). What gives?
 
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  • #2
Hmmm. Could be a few possibilities, right off the top of my head -

1.) There are more sophisticated expressions for the rate of dissolution that take surface area into account - where did you find this one?;

2.) one could be working with some sort of normalized expression for the rate of dissolution which already has taken surface area into account; or

3.) one is dealing with a solid that dissolves very slowly, so the change in surface area is negligible under the time scales involved of the measurements.
 
  • #3
Mike H said:
Hmmm. Could be a few possibilities, right off the top of my head -

1.) There are more sophisticated expressions for the rate of dissolution that take surface area into account - where did you find this one?;

2.) one could be working with some sort of normalized expression for the rate of dissolution which already has taken surface area into account; or

3.) one is dealing with a solid that dissolves very slowly, so the change in surface area is negligible under the time scales involved of the measurements.

The rate law in question is not something I got from anywhere. I just used my previous knowledge of formulating rate laws and applied it in this situation. From consulting many textbooks, they introduce equilibrium constant and it's relation to the rate law for gases, but solubility product constant is randomly presented without it's relation to solution-related rates. Since equilibrium constants and rate laws were introduces in context with gases, one would think that the same relationship further applies to solutes, especially when there is a lack of warning against this exact practice on the book's part. In short, I have realized that the rate law is not what I have written.
 
  • #4
Dissolution before saturation is not an equilibrium process, so the solubility products doesn't apply. Once the solution is saturated amount of solid (or rather its surface) doesn't matter, as both dissolution and precipitation occur at the same surface - so for both processes surface is identical.
 
  • #5
Borek said:
Dissolution before saturation is not an equilibrium process, so the solubility products doesn't apply. Once the solution is saturated amount of solid (or rather its surface) doesn't matter, as both dissolution and precipitation occur at the same surface - so for both processes surface is identical.

Thank you Borek. I just looked up a textbook by Zumdahl and it had the same explanation.
 

1. Why is understanding the rate of dissolution important in scientific research?

The rate of dissolution is important in scientific research because it allows us to understand how quickly a substance dissolves in different solvents. This information is crucial in fields such as pharmaceuticals, where the rate of dissolution can impact the effectiveness of a medication.

2. What factors influence the rate of dissolution?

The rate of dissolution is influenced by several factors, including surface area, temperature, agitation, and the properties of the solvent and solute. For example, increasing the surface area of a solid solute can speed up the rate of dissolution, while a higher temperature can also increase the rate.

3. Why is the rate of dissolution constant?

The rate of dissolution is constant because it is determined by the properties of the solvent and solute, and these do not change during the dissolution process. However, the rate can be affected by external factors such as temperature and agitation.

4. How does the rate of dissolution relate to the solubility of a substance?

The rate of dissolution is closely related to the solubility of a substance. Solubility is a measure of how much of a solute can dissolve in a given amount of solvent, while the rate of dissolution is a measure of how quickly this dissolution process occurs. Generally, substances with higher solubility will have a faster rate of dissolution.

5. How is the rate of dissolution measured in experiments?

The rate of dissolution is typically measured by monitoring the concentration of the solute in the solvent over time. This can be done using various techniques such as spectrophotometry or titration. The change in concentration over time allows us to calculate the rate of dissolution and determine any factors that may be affecting it.

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