Chemical Kinetics: 2A to C Reaction Rate Law

[Amok]

So, let's say we have an elementary reaction:

2A --> C

The rate of this reaction is (according to IUPAC: http://goldbook.iupac.org/R05156.html)

- \frac{1}{2} \frac{d[A]}{dt} = k [A]^2

If we integrate this we get a certain integrated rate law (second order)

However, if we multiply all of the stoichiometric coefficients by 2:

4A --> 2C

And do the same procedure, we'll get a different rate law, right?

Does this mean that the stoichiometric coefficients should be taken as the smallest possible integers in a rate equation? I can't find this statement anywhere. Help me understand this if you can.

 

[Borek]

Same problem with equilibrium (reaction quotient) - depending on the stoichiometric coefficients you will get different values for equilibrium constant. Nothing wrong with that - that's an obvious consequence of different representation of the process. However, if everyone uses different representation of the process, it is difficult to compare results. To avoid that we use - as an accepted convention - smallest possible integer coefficients in reaction equations. Always.

Now, it doesn't mean you can't use fractions during balancing, or even when doing stoichiometry problems, when only ratio counts, but when presenting your results it is always better to follow the convention. It makes communication easier.

 

[Amok]

Same problem with equilibrium (reaction quotient) - depending on the stoichiometric coefficients you will get different values for equilibrium constant. Nothing wrong with that - that's an obvious consequence of different representation of the process. However, if everyone uses different representation of the process, it is difficult to compare results. To avoid that we use - as an accepted convention - smallest possible integer coefficients in reaction equations. Always.

Now, it doesn't mean you can't use fractions during balancing, or even when doing stoichiometry problems, when only ratio counts, but when presenting your results it is always better to follow the convention. It makes communication easier.

Yes, that's exactly what I though. Thank you so much. I was having this argument with some people who are supposed to know this stuff, but I didn't find a place where this stuff was actually written.

 

[Ygggdrasil]

Since we're dealing with elementary reactions, the correct representation is the one that represents the reaction mechanism. If the reaction proceeds through two molecules of A coming together to form B, then 2A --> B is the correct way to represent the elementary reaction. If the reaction instead requires 4 molecules of A to come together, the 4A --> 2B representation would be correct.

 

[Amok]

Since we're dealing with elementary reactions, the correct representation is the one that represents the reaction mechanism. If the reaction proceeds through two molecules of A coming together to form B, then 2A --> B is the correct way to represent the elementary reaction. If the reaction instead requires 4 molecules of A to come together, the 4A --> 2B representation would be correct.

Huh, that's an even better answer. In the case where this is not a an elementary answer then I guess you'd just have to know the partial oders of the reaction experimentally.

You guys are cool. Thank you.