[The Chairman]

Hi all, I was wondering if any one could explain to me in some detail how a catalyst weakens the bonds of a molecule by forming a reactive intermediate? I understand the basics of heterogeneous and homogenous catalysts (I'm currently taking AP Chemistry at my high school at the moment so my understanding might be a bit more basic) but I can't stop wondering about why bonds are weakened when the intermediate is formed. Bond polarities possibly? In any event, all help towards answering this question is greatly appreciated.

[John Cuthber]

There are lots of sorts of catalysis so there are lots of ways they reduce energy barriers.
One of them is illustrated here
http://en.wikipedia....neous_catalysis
Another here
http://en.wikipedia..../Acid_catalysis
and so on.
Did you have any particular catalyst in mind?

[The Chairman]

I guess heterogeneous catalysts are the ones I'm wondering about at the moment. I've done research but all anyone one site I've found so far says is that molecules bind to the surface of the catalyst and in doing so, the bonds the atoms are weakened thus lowering the required amount of kinetic energy required. I am specifically interested in why the bonds weaken after bonding to the catalyst.

[mississippichem]

The Chairman, on 5 February 2012 - 05:11 PM, said:

I guess heterogeneous catalysts are the ones I'm wondering about at the moment. I've done research but all anyone one site I've found so far says is that molecules bind to the surface of the catalyst and in doing so, the bonds the atoms are weakened thus lowering the required amount of kinetic energy required. I am specifically interested in why the bonds weaken after bonding to the catalyst.

In some cases some of the bonding electron density of the adsorbed molecule gets donated to the metallic surface's conduction band. The bond gets weaker as a result of the bonding molecular orbitals being less occupied.

[The Chairman]

mississippichem, on 5 February 2012 - 05:14 PM, said:

In some cases some of the bonding electron density of the adsorbed molecule gets donated to the metallic surface's conduction band. The bond gets weaker as a result of the bonding molecular orbitals being less occupied.

Seems like a reasonable explanation but do you mind elaborating and perhaps explaining in a more basic way? My chemistry understanding is again somewhat limited and so the terminology is a bit confusing.

[Suxamethonium]

Um... I'll take a stab at simplifying for you- but it's not really my area (only think i've seen in detail recently was Pd chemistry... and I forget most of that lol).

Ok, so- imagine you have a Pd surface- it is a metal, so there are electrons delocalised over the entire surface. Next you have an ethylene molecule- its C=C bond is electron rich, and has easy to access pi- electrons. The pi orbitals interact with the Pd surface and the electrons in the two systems can interact. Diagramatically (I hope the spaces come out):

Diagram 1

So the new bonds between the carbon and palladium are weaker than the pi bond. At the same time, Pd has an affinity for H in much the same way:

Diagram 2

The Pd surface now has C₂H₄ and 2H clusters all positioned relatively close to each other- and the final reaction can proceed forming ethane:

Diagram 3

So, the energy of the total reaction is the same, but the initial energy required is less because the energy to form the transition state is less. As to why this is less, I will let someone else pick up on it- as I'm not all that sure myself (though I have a few ideas).

Edit: Spaces didn't work so included pretty picture!

Attached thumbnail(s)

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This post has been edited by Suxamethonium: 8 February 2012 - 02:15 AM

[mississippichem]

Suxamethonium, on 8 February 2012 - 02:00 AM, said:

Um... I'll take a stab at simplifying for you- but it's not really my area (only think i've seen in detail recently was Pd chemistry... and I forget most of that lol).

Ok, so- imagine you have a Pd surface- it is a metal, so there are electrons delocalised over the entire surface. Next you have an ethylene molecule- its C=C bond is electron rich, and has easy to access pi- electrons. The pi orbitals interact with the Pd surface and the electrons in the two systems can interact. Diagramatically (I hope the spaces come out):

Diagram 1

So the new bonds between the carbon and palladium are weaker than the pi bond. At the same time, Pd has an affinity for H in much the same way:

Diagram 2

The Pd surface now has C₂H₄ and 2H clusters all positioned relatively close to each other- and the final reaction can proceed forming ethane:

Diagram 3

So, the energy of the total reaction is the same, but the initial energy required is less because the energy to form the transition state is less. As to why this is less, I will let someone else pick up on it- as I'm not all that sure myself (though I have a few ideas).

Edit: Spaces didn't work so included pretty picture!

That pretty well captures the majority of what I was trying to convey. I tend to play fast with jargon as I'm usually surrounded by my kind. Your explanation is a good one though and better communicated than mine.