[k42490]

I DID AN EXPERIMENT TO FIND THE RATE LAW FOR THE REACTION:
S203 ^2- + 2H⁺ -> S + SO2 + H20

WE FOUND THAT THE ORDER OF THE REACTION IS 2 WITH RESPECT TO THIOSULFATE AND 1 WITH RESPECT TO H+. I HAVE NO IDEA IF THIS IS RIGHT?? I TRIED TO MAKE A MECHANISM THAT WOULD WORK FOR THE REACTION, DOES THIS WORK:

2S203 ^2- + H+ -> HSO4- + 2S + SO2 (SLOW)
S + HSO4- + H+ -> H20 + 2S203 ^2- (FAST)

i DON'T KNOW IF THIS WORKS, HSO4- WOULD BE AN INTERMEDIATE BUT WOULD THE S BE AN INTERMEDIATE TOO. ARE THERE ANY OTHER WAYS TO MAKE THIS RATE LAW WORK?? tHANKS

[Borek]

Please read forum rules. Don't abuse capital letters.

[k42490]

Sorry.... it's easier to write the chemical formulas if everything is in caps

[LQ43]

Not sure your rate law is correct

But its unlikely, you have a trimolecular elementary process in both steps -
doubtful that sulfur would be an intermediate and a product

HSO3- might be more likely as an intermediate with H+ involved in both steps one at a time.

[Rabn]

I've made this sequence under the assumption that your reaction is correct.  take a look at that.  When trying to figure out the kinetics of a reaction it is best to fully draw each structure.  There is one neg charge on one oxygen, which is delocalized over all three oxygens which provides about 1.75 bonds to each oxygen.  The negative charge on the sulfur really isn't a negative charge per say because the electronegativity of the 3 oxygens pretty much take most of the charge. 
Your slow step requires a lot of bond shifting, there is a lot going on there.  One thing to keep in mind at all times when considering chemical reactions is this: 2 species running into each other is statistically probable, 3 things running into each other at the same time is not likely but it happens, if more than 3 species need to run into each other and form a complex intermediate you more than likely far off.

What I propose is this:

The slow step is the production of water, your data shows this by your result of an order of 1, it is likely as shown, a consecutive addition of 2 H+ to one O, this requires 2 species colliding twice which is more likely than 3 at once.

The fast step is the negative charge moving to the central S and the expelling of the S adduct, in fact my guess is that it happens so quickly as to be almost instantaneous.

What I'd like you to do is look at this and think about it. Ask questions. I feel bad for giving you the answer but I feel like if I show you and it is discussed it will help your understanding more than if I were to try help you along.  Kinetics is tricky, having a guide to help you explore may be better than oracle.

[LQ43]

Rabn, just wondering, I couldn't see the right hand side of your drawing  but with the mechanism shown could not the formation of water be written as two steps colliding twice as you mentioned.

S2O32-  +  H+ --> S2O3H - 

S2O3H- + H+  --> S2O2  +  H2O

if the formation of water is the slow step, wouldnt this add up to first order in S2O32- and 2nd order in H+, even in a concerted fashion and not then consistent with the rate law found?

WE FOUND THAT THE ORDER OF THE REACTION IS 2 WITH RESPECT TO THIOSULFATE AND 1 WITH RESPECT TO H+. I HAVE NO IDEA IF THIS IS RIGHT??

[Rabn]

The right side of the picture is just an arrow, you aren't missing any information.  To answer your question. You can think of the addition of H+ as 2 separate steps. The help you see why it is first order in H+ you can imagine it like this: 

You have a set amount of the S₂O₃ and then add H+, the more H+ there is in the solution the more "pressure" the H+ has. This may seem weird, but imagine the amount of S₂O₃ as the number of bathrooms at a sports stadium.  Each bathroom can hold 2 people. Once two people have entered the bathroom, the door closes and the bathroom turns into a spaceship escape pod and both the people and bathroom fly off.  The more H+ you add, the bigger the line, the more pushing there is to get into the bathrooms, the faster people shuffle into them. THat is the more H+ you add the more energy the solution has, the faster the H+ adds to the oxygen.

So why is it that the S₂O₃ is second order. Using the same analogy, imagin there are a ton of people that need to use the bathroom. the more bathrooms you add the faster the people get in them. Now you can see that it doesn't matter how many people you add, if the number of bathrooms is fixed the line can only move so fast. But if you add bathrooms, the people can get to them faster. I'm not sure if that's clear enough...let me know. 

[k42490]

thanks this has been a great help for me to visualize.

[LQ43]

Sorry this is bothering me. I can see the pseudo first order in H+ but not second order in S2O32-. As the mechanism is written, it is still first order S2O32- . As elementary processes there needs to be a second S2O32- involved either as a fast dimerization step or in a subsequent step for second order. 

Which is why I wondered about the rate law results and how they were determined. Graphically a second order curve might be seen in a first order plot if the scatter is large or the scale is not optimized.