What is a homogeneous catalyst? What are the advantages/disadvantages of it over heterogeneous catalysts?
A homogeneous catalyst is a catalyst that is capable of dissolving in solution, because it by definition is in the same phase as the rest of the reactants in the solution.
Here are the principles of homogeneous catalysts that I see in my textbook (Inorganic Chemistry, Shriver, Atkins, Ch. 25):
- Homogeneous catalysts are effective at being highly selective towards producing the desired product.
- If a reaction is exothermic, it will release a lot of heat. It is easier to release heat from a solution (as one would do for a homogeneous catalyst) than if one were to use a heterogeneous catalyst, which tends to be an insoluble solid in solution that adsorbs reaction participants onto it.
- Reactants can easily access the homogeneous catalyst, because it is in solution already! This promotes high catalytic activity.
- Species in solution are easier to characterize (e.g. with various spectroscopy techniques) than species adsorbed onto a solid surface.
- A catalyst capable of dissolving in solution is going to have to be separated later if it is to be recycled for re-usage (as per the principles of green chemistry).
- The homogeneous catalyst may have issues at high temperatures, since we don't want our solution (which contains this catalyst) and any volatile reactants to evaporate, even though high temperatures usually promotes faster reactions.
Maybe this will help as well? Here's an example of how a homogeneous catalyst might be used in context.
HOMOGENEOUS CATALYST EXAMPLE: MONSANTO PROCESS, THE CARBONYLATION OF METHANOL
The catalytic cycle for this is shown below (Inorganic Chemistry, Shriver, Atkins, Ch. 25). I recreated the image, and annotated it.
I circled the overall organic reaction as:
#\mathbf(color(blue)("CH"_3"OH" + "CO" -> "CH"_3"COOH"))#
Take the time to digest this catalytic cycle, if you wish to get the context. Here is what is happening:
The homogeneous catalyst is the square-planar
#\mathbf(color(blue)("CH"_3"OH"))#, we add #"HI"#to release water (acid-catalyzed #"S"_N2#from organic chemistry) to yield #"CH"_3"I"#.
When we add
#"CH"_3"I"#in this rate-determining step, it is an oxidative addition (as is typical of square planar complexes) to give an octahedral complex where the rhodium oxidized from #stackrel(+1)("Rh")#to #stackrel(+3)("Rh")#.
You can determine this by knowing that
#"CH"_3#as a free ion is #"CH"_3:^(-)#and #"I"#as a free ion is #"I"^(-)#, adding #-2#to the overall charge, but retaining the overall charge, thus giving rhodium a #+3#oxidation state.
This is an interesting reaction called a "1,1-insertion", which by empirical findings is a methyl migration.
The methyl group cis to the carbonyl (
#"CO"#) migrates over and creates an acyl group ( #("RC"="O")"CH"_3#). This leaves the oxidation state of rhodium alone, since #"CO"#contributes no charge.
#\mathbf(color(blue)("CO"))#, our second organic reactant, we perform a ligand association to regain an octahedral complex without changing the oxidation state of rhodium because #"CO"#contributes no charge.
Reductive elimination of the acyl group and the iodide generates
#"CH"_3("C"="O")-"I"#, called acyl iodide.
This reduces the oxidation state of rhodium from
#+3#to #+1#, regenerating the starting homogeneous catalyst! That's because the acyl group as a free ion has a charge of #-1#, and similarly, iodide as a free ion is #"I"^(-)#, but the overall charge of the complex remained the same.
The acyl iodide is hydrolyzed via a water intermediate to make the organic product,
This also regenerates
#"HI"#as a result of the acid iodide hydrolysis, which we had used earlier while reacting with #"CH"_3"OH"#in step 1.
This catalytic cycle can repeat, now that we regenerated the homogeneous catalyst (