[sodium.dioxid]

I think so. If I have a fish tank and put a solute the size of a rubik's cube into it, the solute dissolves at a rate r. If I start over with a new fish tank and put in two pieces of solid, one at each end of the tank, the solids surely must be dissolving at rate 2r. Thus, doubling the size doubles the rate.

[sodium.dioxid]

I think so. If I have a fish tank and put a solute the size of a rubik's cube into it, the solute dissolves at a rate r. If I start over with a new fish tank and put in two pieces of solid, one at each end of the tank, the solids together surely must be dissolving at rate 2r (each one is dissolving at r, but at the same time). Thus, doubling the size doubles the rate.

[juanrga]

I think so. If I have a fish tank and put a solute the size of a rubik's cube into it, the solute dissolves at a rate r. If I start over with a new fish tank and put in two pieces of solid, one at each end of the tank, the solids together surely must be dissolving at rate 2r (each one is dissolving at r, but at the same time). Thus, doubling the size doubles the rate.

The process is A(sol) ##\leftrigtharrow## A(aq)

The forward rate will be proportional to the amount of solid.

[Jorriss]

If you are taking about something like dissolution I am 'guessing' the rate must be proportional to surface area?

[sodium.dioxid]

Ok, good. We all realize that amount is important. But apparently, the rate law disagrees with our conclusion. For A(s) B(aq), the forward rate law is rate = k [A]. As you can see, it doesn't say anything about amount. Furthermore, the concentration of the solid is constant. This means that as long as there is solid present, it will run out at a constant rate without any regard to the amount present. The book says that as long as there is some solid (big or small), the rate is constant. But this clearly doesn't make sense in context of our fish tank example. As Jorriss pointed out, a larger solid has more surface area, so more molecules are breaking off into ions during any particular interval of time. What gives?

By the way, this discussion is about solutes. I thought I should mention that.

[sodium.dioxid]

*Don't ignore me, I am not impatient*

Basically, here is the problem:
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). Is the rate law wrong? Why?

[Jorriss]

*Don't ignore me, I am not impatient*

Basically, here is the problem:
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). Is the rate law wrong? Why?

That is not the correct reaction or rate law. Rate laws for solid-liquid reactions begin with adsorption of the solvent onto the solid, so it begins with A + B -> AB where AB is B adsorbed onto A. What is often measured is AB -> Products where AB is a saturated surface so all that one sees is reactants going straight to products by first order kinetics but the actual reaction is a bit more complicated.

Reading about langmuirs theory of gas-solid reactions may be insightful even if it is not this exact process. Or even enzyme catalyzed reactions - they are surprisingly similar in form.

[sodium.dioxid]

*Don't ignore me, I am not impatient*

Basically, here is the problem:
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). Is the rate law wrong? Why?

That is not the correct reaction or rate law. Rate laws for solid-liquid reactions begin with adsorption of the solvent onto the solid, so it begins with A + B -> AB where AB is B adsorbed onto A. What is often measured is AB -> Products where AB is a saturated surface so all that one sees is reactants going straight to products by first order kinetics but the actual reaction is a bit more complicated.

Reading about langmuirs theory of gas-solid reactions may be insightful even if it is not this exact process. Or even enzyme catalyzed reactions - they are surprisingly similar in form.

Great, thank you.

[juanrga]

Ok, good. We all realize that amount is important. But apparently, the rate law disagrees with our conclusion. For A(s) B(aq), the forward rate law is rate = k [A]. As you can see, it doesn't say anything about amount. Furthermore, the concentration of the solid is constant. This means that as long as there is solid present, it will run out at a constant rate without any regard to the amount present. The book says that as long as there is some solid (big or small), the rate is constant. But this clearly doesn't make sense in context of our fish tank example. As Jorriss pointed out, a larger solid has more surface area, so more molecules are breaking off into ions during any particular interval of time. What gives?

By the way, this discussion is about solutes. I thought I should mention that.

The rate expression "k [A]" is derived assuming that volume is constant and that all molecules of A are statistically equivalent. This is right for most gas and liquid phase reactions, but the volume of the solid dissolving is not constant and not all the molecules in the solid can be treated as equivalent (only molecules at the surface of the solid can dissolve).

If we compute the rate of the dissolution process

A(s) ##\leftrightarrow## A(aq)

we obtain something as the Noyes-Whitney Equation, which depends on the surface of the solid and thus on the size of the solid.