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Topic: Reaction rate Definition - Please help me understand.  (Read 4112 times)

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Offline user11

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Reaction rate Definition - Please help me understand.
« on: July 09, 2014, 10:19:40 AM »
Hi, I have a question regarding Reaction rate, and I’d really appreciate some help.

Question concerning reaction rate definition:

The rate of the reaction for an elementary rection aA+bB --> pP+qQ is defined as
-(1/a)*d[A]/dt = -(1/b)*d[B.]/dt = (1/p)*d[P]/dt= -(1/q)*d[Q]/dt

Or by the equation
r=K[A]^a[B.]^b

Does this mean that the rate of the reaction is The same as The rate at which ONE of the 2 reactants (in this example) converts into product? That’s what the first equation says. So the reaction rate is not the sum of the rate of the two reactants but the rate of ONE reactant? Is this because they have the same order (their reaction rates are the same)

Question concerning complex reactions and reaction rate

When we have complex reactions: then I would image that reaction rate can’t just be the rate of one of the reactants (as it seems in simple reactions), so what is meant by “The reaction rate” in complex reactions? Is there a definition for this?


The last thing I don’t understand with respect to complex reactions: In a reaction A+B --> C, How can 2 reactants  (A and B) have different reaction rates, if the two reactants must combine in order to from C? I would assume they always have the same rate of collision in order to form one product?

Offline Borek

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Re: Reaction rate Definition - Please help me understand.
« Reply #1 on: July 09, 2014, 01:39:49 PM »
It is defined so that it is identical for each substance - you don't want to have different rates of the same reaction when observing different involved substances.
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Offline user11

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Re: Reaction rate Definition - Please help me understand.
« Reply #2 on: July 09, 2014, 04:27:57 PM »
It is defined so that it is identical for each substance - you don't want to have different rates of the same reaction when observing different involved substances.
I'm not sure i entirely understand that - can you please elaborate on that?

Offline Borek

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Re: Reaction rate Definition - Please help me understand.
« Reply #3 on: July 10, 2014, 02:49:19 AM »
I am not sure what more to add.

Reaction rate is a property of the reaction. Thus, it should not depend on whether you observe changes of concentration of A, or B, or P, or Q - it should be identical in each case. That is guaranteed by the use of 1/a, 1/b, 1/p and 1/q coefficients.
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Offline Irlanur

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Re: Reaction rate Definition - Please help me understand.
« Reply #4 on: July 10, 2014, 09:00:58 AM »
The thing is: you cant form a molecule P without loosing a Molecule A. Thus the rate of Formation and the rate of Loss are not independent and one variable (the reaction rate) is sufficient.

Quote
r=K[A]^a[B.]^b

Thats not a definition, but it might be observed in some cases. It is not always true.

Offline user11

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Re: Reaction rate Definition - Please help me understand.
« Reply #5 on: July 10, 2014, 09:41:55 AM »
The thing is: you cant form a molecule P without loosing a Molecule A. Thus the rate of Formation and the rate of Loss are not independent and one variable (the reaction rate) is sufficient.

Quote
r=K[A]^a[B.]^b

Thats not a definition, but it might be observed in some cases. It is not always true.

So what you are saying is that the rate at which A decreases must equal the rate at which P forms (this makes good sense to me) but Can the rate at which B decreases be less or more than the rate at which A decreases? Or would  the lowest rate reactant then then be a "limiting rate factor" such that The reaction rate of both Reactants would still be equal same?

So is it true that this relationship -(1/a)*d[A]/dt = -(1/b)*d[B.]/dt = (1/p)*d[P]/dt= -(1/q)*d[Q]/dt  Holds true for all reactions?

So this equation r=K[A]^a[B.]^b is just an experimentally derived equation that just holds true in some cases?

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Re: Reaction rate Definition - Please help me understand.
« Reply #6 on: July 10, 2014, 03:07:37 PM »
Can the rate at which B decreases be less or more than the rate at which A decreases?

No. (That is, after using the stoichiometry coefficients, that's why they are put in the definition).
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