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Negative reaction orders

by MathewsMD
Tags: negative, orders, reaction
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MathewsMD
#1
Mar24-14, 09:14 PM
P: 295
We were just going over examples of reaction orders in class and I was told 2 things:

1.) Negative reaction orders are possible
2.) An order of 0 says the reaction rate is independent of the reactant concentration(s)

I had a few questions, though.

1.) What exactly constitutes a negative order reaction if reaction orders are determined by the RDS in elementary steps? Are reactants being produced when the forward reaction occurs? What exactly is going on for this to occur?

2.) If the order of the reaction is 0, but one reactant has an order of -1/2 and the other 1/2, doesn't this still mean that the reaction rate is dependent on reactant concentration despite having an overall order of 0?

Any clarification please? Thanks!
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Yanick
#2
Mar25-14, 04:51 PM
P: 382
Maybe you need to clarify if these statements refer to overall rate laws or order with respect to some participant.

Overall rate laws are determined by experiment and can get very freaky and complicated when all is said and done.

See the Wikipedia article: http://en.m.wikipedia.org/wiki/Order_of_reaction

Or refer to a decent physical chemistry text for more info.
MathewsMD
#3
Mar25-14, 06:43 PM
P: 295
Quote Quote by Yanick View Post
Maybe you need to clarify if these statements refer to overall rate laws or order with respect to some participant.

Overall rate laws are determined by experiment and can get very freaky and complicated when all is said and done.

See the Wikipedia article: http://en.m.wikipedia.org/wiki/Order_of_reaction

Or refer to a decent physical chemistry text for more info.
Both, for elementary steps and the overall reaction order. I was told in class, with no specific reference even after speaking with the professor, that negative reaction orders are possible (i.e. -1/2). This is only possible if the elementary step also has a negative order, correct? I'm just unsure on how a reaction proceeding in the forward direction could have a negative order. Any clarification please since for elementary steps, the order is related to the reactant coefficients, and the reactant must be produced in order for the forward reaction to have a negative reaction order, correct?

Yanick
#4
Mar25-14, 07:38 PM
P: 382
Negative reaction orders

I certainly don't know every single possibility that can occur in kinetics but my gut tells me that for an elementary step you shouldn't have negative reaction orders since the coefficients are the exponents. Maybe someone with more knowledge can help for the elementary step case. For an overall rate law I can see non integer as well as negative orders being possible because overall rate laws, as I mentioned, can be freaky looking.

Think about what [itex]\frac{dy}{dt}=k x^{-1} [/itex] means. It doesn't mean that anything necessarily goes backwards, just that the instantaneous change of y as a function of time gets smaller as x gets larger.
epenguin
#5
Mar28-14, 10:59 AM
HW Helper
epenguin's Avatar
P: 1,985
A full rate equation may include concentration not only of 'reactants' but of products (this may be, but not necessarily, full back-reaction) and they will usually be in the denominator of the equation for rate of the forward reaction, so could be said to correspond to a negative reaction order. And in this sense must be more rule than exception.

But don't worry about such things beyond maybe first, second and zero orders, worry about mechanisms and the equations they generate, or about observed kinetics and the inferences therefrom of mechanism. If you come to examples you come to them, but otherwise it would be a slightly wordish preoccupation IMHO.


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