Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
 Pages:  1  2
Author: Subject: Solids in rate laws and equilibrium
unionised
International Hazard
*****




Posts: 4364
Registered: 1-11-2003
Location: UK
Member Is Offline

Mood: No Mood

[*] posted on 16-5-2006 at 11:20


It still does and it still increases the rate of the forward and backward reaction so the effects cancel.
View user's profile View All Posts By User
guy
International Hazard
*****




Posts: 982
Registered: 14-4-2004
Location: California, USA
Member Is Offline

Mood: Catalytic!

[*] posted on 16-5-2006 at 12:55


Why does it not cancel out for gases?



View user's profile View All Posts By User
unionised
International Hazard
*****




Posts: 4364
Registered: 1-11-2003
Location: UK
Member Is Offline

Mood: No Mood

[*] posted on 18-5-2006 at 11:16


Good question.
Let's start with a couple of relatively real examples so I can remember what I'm on about. H2+I2 <--> 2 HI
and CO2 <--> C +CO
Under the right conditions these 2 reactions are both reversible. Heated HI can decompose to the elements and heated CO can deompose to C and CO2. In one case, all the materials are gases (the H +I reaction only really works if it's pretty hot) so what needs to happen is for 2 HI molecules to bump into eachother (or (simultaneously) into the wall or into an H2 or an I2 or even another HI) with enough vigour to ensure that the activation barrier is overcome and they will react to give H2 +I2. This can happen throughout the volume of the gas.

For CO decomposition the reaction can go by one of two paths. It can take place entrirely in the gas phase leaving CO2 and a sinlge atom of C as the vapour.
On the other hand, it can take place at the surface of the solid carbon.
That gives 2 similar equations
2CO --> C(g) +CO2(g) or
2CO --> C(s) +CO2(g).
The difference in the energy here is the energy needed to vaporise an atom of carbon. Since carbon is a solid it is relatively difficult to vaporise, so the second path, giving the much more stable solid, is much easier.
Practically all the reaction thus takes place at the surface of the solid. For the gas phase reaction the whole of the mixture is available for the reaction to take place.
I think that's the basis of the difference.
View user's profile View All Posts By User
guy
International Hazard
*****




Posts: 982
Registered: 14-4-2004
Location: California, USA
Member Is Offline

Mood: Catalytic!

[*] posted on 18-5-2006 at 22:05


Yeah I think you are on to something. It is true that rate IS proportional to surface area and NOT the mass of a solid. So therefore, equilibrium is related to the surface area of a solid. So equilbrium is reached when there is enough surface area for rates to balance out. I am correct so far? Now I think I'm close to the conclusion but I can't completely figure it out so maybe can come up with one for me? :) I can't figure out the rest of this conclusion: "Adding more solid increases mass but...."

[Edited on 5/19/2006 by guy]




View user's profile View All Posts By User
daeron
Hazard to Self
**




Posts: 74
Registered: 26-3-2005
Member Is Offline

Mood: cancerogenic

[*] posted on 19-5-2006 at 05:04


the rxn rate isnt proportional but dependent on the surface area of the solid, the larger the surface are the faster the mass transfer. If the rxn is mass transfer limited that increasing the surface area will increase the rxn rate, however if its limited by the kinetics than one doesnt need to increase the surface area. another thing one can do is, if theres a case of a g-l-s system:
-increase the mixing power input(better g-l, l-s mass transfer)
-higher presssure and/or temperature, but P is of more importance here if you want to influence the mass transfer&equib.
-look into the gas holdup, solid holdup,etc

however a chem. equilibrium is rarely affected much by the increase of surface area, since that will only affect the rxn rate in one direction, and since it is a system at chem. equilibrium one gets as a result faster rxn rate in the opposite direction, hence silight or no change in equilibrium.

so what im tryin to say is that in order to shift the equlibrium one must first look into the system conditions, the as will just speed up all the rxn rates in the chem equilibrium, and a equilibrium will be reached in a shorter period(i.e. the equilib. conversion rate, concentrations and all)

...erm...

what was the question again?:P

[Edited on 19-5-2006 by daeron]




Humans with their brains are like a caveman with a supercomputer. Yes there is a chance that the caveman will use it to calculate the trajectories and the momentums of the celestial bodies, but the chances are that he will just crack open a coconut with it.
View user's profile View All Posts By User
 Pages:  1  2

  Go To Top