Sciencemadness Discussion Board - Solid Reactants and the Rate Equation

Solid Reactants and the Rate Equation

markEssex - 26-11-2014 at 03:03

I'm currently doing my school chemistry project on the kinetics of the reaction between Magnesium and Sulfuric acid however having some real trouble finding out weather solid reactant (magnesium) is include in the rate equation. Obviously being a solid it doesn't exactly have a concentration but is it still included on the basis of it's surface area?

Thanks in advance

blogfast25 - 26-11-2014 at 05:52

A complete empirical model of the reaction rate of the reaction between Mg and H2SO4 should make provision for the surface area of the Mg, unless it is kept rigorously constant (but then the model will provide no information on the influence of the surface area of the Mg, obviously).

So your model would take the form of:

rate = k S [H2SO4] with k the rate constant, S the surface area of the Mg and [H2SO4] the concentration of H2SO4.

[Edited on 26-11-2014 by blogfast25]

j_sum1 - 26-11-2014 at 06:51

Rate laws as they are normally expressed are functions of the concentration of the reactants. Of course, concentration is not the only factor that contributes to the reaction rate. However, all other conditions being equal, the rate law shows the relationship between the concentrations and the rate. Once derived it can also give some insight into the reaction mechanism.

In my experience, rate laws are often confused with equilibrium equations which might be what you are thinking of. Both use the letter k for a constant. (The two are somewhat related in that for a single stage equilibrium reaction the equilibrium equation can be derived by equating the rate laws for the forwards and backwards reactions.) The key difference is this. An equilibrium equation gives absolutely no information on how fast the reaction will proceed. If you work out the units for the K, there is no time component.
The convention for equilibrium equations is to omit concentrations of any species that removes itself from the reaction environment through a phase change. So for example when considering solubility, there is no attempt made to calculate the concentration of the solid when considering the Ksp value. Likewise gas bubbles and immiscible liquids that form are also omitted.

In your case, the situation is simple. You would expect the rate to be proportional to the surface area of the solid. (This is difficult to quantify and will change during the course of the reaction. However at low acid concentrations and with adequate stirring this will do as a first approximation.)
If the surface area and other factors are kept constant then the rate is related solely to the concentration. In your case, it is a first order proportionality but in more complex reactions the rate might be proportional to a concentration squared or cubed.

For the sake of simplicity, if you change the reaction conditions (surface area, shape of reaction vessel, amount of stirring, pressure, temperature or presence of catalysts) then you calculate a new rate law and a new rate constant. Yes these factors can be modelled (and are modelled.) A simple investigation would treat them as variables to be controlled.

Steam - 26-11-2014 at 06:55

I have had this problem before. see http://www.sciencemadness.org/talk/viewthread.php?tid=29856#...

Because the surface area of the Mg is constantly changing this problem is very hard to solve. The best advice that I got was to use a very large sphere of Mg. That way the change in surface area is negligible.

Just thinking about it, if you did know the rate at which the Mg was being dissolved, you could figure out the rate using a bit of calc. I posted an image of the way you might calculate this; however, I might very well be wrong about this. This is my best guess of what the rate would be with changing Surface Area.

I slightly simplified it by using the example of a cube, this can be exchanged with any shape, or even if the Mg was floating. In a perfect world, you would want to use a sphere of Mg, in an acid of low concentration. It is a shame we don't live in a perfect world!

[Edited on 26-11-2014 by Steam]

Steam - 26-11-2014 at 07:02

actually, I am totally off on the equation. Disregard it!

Steam - 26-11-2014 at 09:10

Sorry for yet another post, but I believe this is more correct. I am assuming for the sake of argument that the Mg is a sphere, but any shape will work.

blogfast25 - 27-11-2014 at 09:22

Quote: Originally posted by j_sum1

Rate laws as they are normally expressed are functions of the concentration of the reactants. Of course, concentration is not the only factor that contributes to the reaction rate. However, all other conditions being equal, the rate law shows the relationship between the concentrations and the rate. Once derived it can also give some insight into the reaction mechanism.

In my experience, rate laws are often confused with equilibrium equations which might be what you are thinking of. Both use the letter k for a constant.

Nothing points to him thinking of equilibrium equations. The letter k is not mandatory: the writer can specify any symbol to mean the 'rate constant', as long as he defines that clearly.

Ad it's very easy to engineer an experimental situation wherein the surface area is very close to a constant. No problem at all (and no 'spheres' needed either, although that could certainly work)

blogfast25 - 27-11-2014 at 09:54

Quote: Originally posted by Steam

Sorry for yet another post, but I believe this is more correct. I am assuming for the sake of argument that the Mg is a sphere, but any shape will work.

No, it's not correct either. I assume that [A] is [H2SO4] in the dissolution of Mg metal by dilute H2SO4.

For starters, "-[A]/dt" is not a rate. -d[A]/dt IS.

In an infinitesimal time interval dt, [A] is reduced by d[A]. Assuming the total volume of solution of H2SO4 was W, then

W d[A] = dn, the number of moles of Mg reacted away in the interval dt.

This corresponds to a volume dV = (M dn) / ρ, with M the atomic mass of Mg and ρ the density of magnesium.

As you pointed out, the surface area of a sphere is 4 Π R2. In the interval dt, the sphere loses a bit of volume dV and radius dR, i.e.:

dV = 4 Π R2 dR.

With W d[A] = dn and dV = ( M dn) / ρ and dV = 4 Π R2 dR, reworked:

d[A] = (4 Π ρ R2 dR)/(MW)

And - d[A]/dt = k [A] (4 Π ρ R2 dR)/(MW)

Also - ρ dR / (MW) = k [A] dt, so if [A] was to remain constant, the radius of the Mg sphere would decrease linearly with time.

Edit:

The equation - d[A]/dt = k [A] (4 Π ρ R2 dR)/(MW) has no solution when [A] is not constant. That's because the initial formulation of the problems lacks information.

The expression - d[A]/dt = k [A] S suggests that [A] and S are independent but they are not, in the case of non-constant [A].

When, say n moles of acid, and Mg have reacted away there is a clear relationship between the new concentration [A] and the new volume (and thus radius and surface area) of the remaining sphere. S has to be expressed as a function of ([A]0 - [A]) (with [A]0 the initial concentration as boundary condition).

For a sphere that quickly becomes complicated, unfortunately...

[Edited on 27-11-2014 by blogfast25]

j_sum1 - 27-11-2014 at 17:12

Quote: Originally posted by blogfast25

Quote: Originally posted by j_sum1

In my experience, rate laws are often confused with equilibrium equations which might be what you are thinking of.

Nothing points to him thinking of equilibrium equations. The letter k is not mandatory

Nothing says he wasn't. And conflating the two is common. The fact that he was enquiring about dropping one of the reactants from his calculations made me think it was worth mentioning.

I guess we need to wait until he returns before we know what the situation is and what depth of mathematical modelling is required. For a high school project, I suspect that the calculus is un-needed and a simple rate∝area relationship will suffice.

blogfast25 - 28-11-2014 at 05:19

Quote: Originally posted by j_sum1

Quote: Originally posted by blogfast25

Quote: Originally posted by j_sum1

In my experience, rate laws are often confused with equilibrium equations which might be what you are thinking of.

Nothing points to him thinking of equilibrium equations. The letter k is not mandatory

Nothing says he wasn't. And conflating the two is common.

So nothing says that he wasn't means to you that he was?? Conflating the two is common? A matter of your opinion. I don't agree. It's clear he was talking about reaction rate, not equilibria.

Calculus would not be needed at his level but I was responding to Steam's attempt, not the OP.

In the case where [A] is not constant (A reacts away significantly) even the simpler case of a cube with sides x (m), so S = 6 x2 and V = x3, becomes quite complicated.

For [A] = constant it is easy to work out a general model for the diminishing size of the Mg sphere (or cube).

j_sum1 - 28-11-2014 at 16:45

@blogfast
Maybe my context is relevant here.
I just marked a pile of high school chemistry exams where half the students assumed an equilibrium reaction for a particular question when none was stated. I think I can stand by my comment that conflating equilibrium and rate is common for high school students.
Other than that, I agree with everything you have stated.