# Can rate laws have negative exponents?

Discussion in 'MCAT Study Question Q&A' started by itsover26, Jul 15, 2008.

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1. ### itsover26Banned

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Jul 10, 2008
I guess they could, but when would you know to make the exponent negative? I don't really understand why they would have - exponents...i guess negative means that substrate simply decreases the rate? how would that appear in a problem?

apparently, the exponents can also be fractions...that makes a lil more sense since there's still positive effects on the rate.

so, - exponents would simply imply negative effects on the rate?

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3. ### engineeredoutLightning Ballseeker 7+ Year Member

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May 11, 2008
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No negative order reactions that you'd have to be concerned about on the mcat. A reaction similar to what you describe would be an autocatalytic reaction. In an autocatalytic reaction the products of the reaction act as catalysts, so the rate of the reaction gets faster and faster as the reaction proceeds. Technically the reactants themselves don't slow the reaction, but the more reactants in an autocatalytic reaction the slower it proceeds.

Doubt you'd need to know autocatalytic reactions for the mcat, but theres some background info for you.

If you're really interested about the rare negative order cases, I can look up some stuff for you from my kinetics textbook when I get home. Its not something my professor even bothered focusing on in my kinetics class.

You'd be correct though, that a negative order would mean that increasing one of the reactants would decrease the rate of the reaction. In an industrial application it means that there is a limit as too the mass transfer rate of the reactants

Actually come to think about it, negative order reactions and autocatalytic reactions are opposites of each other.

You know what, maybe its not a bad thing to know for the mcat at all.

Last edited: Jul 15, 2008
4. ### engineeredoutLightning Ballseeker 7+ Year Member

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May 11, 2008
Pennsylvania
Yeah this guy got banned, but I figured I'd post up the material I PMed to him in case anyone else was interested.

Anyone for negative order kinetics, I have a nice little graph I took a picture of from my camera but I don't have access to webspace to upload it. If you give me your e-mail address I can send it to you. It shows a comparison of -1, 0 and 1/2 order rates.

Basically what ends up happening in rate equations with orders less than 1 is that the normal equations really get screwed with, because extrapolating from them leads to a negative reactant concentration, which is impossible. What you'll see in the graph is instead of the concentrations going down like shown, they eventually end up even-ing out because there are limits put on the equations, specifically that t <= CAo/k, but again that gets into math way too complicated for mcatty.

To make it even more confusing, negative order rate laws have the partial pressure of the reactants as a variable, and remember that since the concentration of the reactant changes, all of this has to be integrated over time. What happens in the reactions that my book shows is that it will end up with one partial pressure displaying first order kinetics, and one equation displaying negative order kinetics.

This occurs in packed bed catalytic reactors, which are reactors full of catalyst pellets in them. Lets say that we have two species. We have Reactant A and product B. At the beginning of the reaction, the rate is dependant on the amount of surface area of the catalyst that reactant A covers. As product B is produced, it begins to absorb the catalyst and block reactant A. As this happens, the reactant A keeps losing catalyst area, so the reaction slows down. Therefor A operates with positive order kinetics, and B operates with negative order kinetics. This ends up giving you a rate law something like r = KaPa/KbPb.

There are also "poisons" to a reaction. In this case, it would mean throwing another species C into the mix, whose only purpose is to take up catalyst space.

Crazy isn't it? Probably way too much detail then you needed but there it is.