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I have a question about the value obtained by Rate Law. For example, in this case:

image.png.1578c4b3389d63b0099360453fc7310f.png

the value of rate obtained by this equation is the global rate of reaction, right, and not the rate of a specific specie, I know that using this global rate I can find the species rate, multiplying for his stoichiometric numbers.

Is my assumption correct? Because if the answer is YES, this question no make sense for me.

I was trying to solve this question:

image.png.817ad57fa9fff94a79864f3a9331f729.png

I created this graph (Title is in Portuguese):

image.png.2397b133f205a9dee6d62389bc09ca77.png

I used "1/[HI] x t," which indicates a second-order reaction with a rate law equal V = k[HI]^2, so, the slope of the graph is numeric equal for k, AND, if my previous assumption is correct, the value obtained for V is the global rate reaction, as a consequence, the constant (k) 0.0078 (L.mol-1s-1) represent the unique rate law and not of specific reactant (HI).

However, the gab, says that 0.0078 represents the rate constant (K) for the HI

image.png.f8c818e4d7814172e54bc9993a7db6c8.png

For me, it makes no sense, because, if it is correct, the value of V obtained by rate laws is the rate of HI and not for the global reaction since k refers to him and not for the global reaction.

Posted (edited)
On 7/26/2023 at 8:19 PM, Scienc said:

I have a question about the value obtained by Rate Law. For example, in this case:

image.png.1578c4b3389d63b0099360453fc7310f.png

the value of rate obtained by this equation is the global rate of reaction, right, and not the rate of a specific specie, I know that using this global rate I can find the species rate, multiplying for his stoichiometric numbers.

Is my assumption correct? Because if the answer is YES, this question no make sense for me.

I was trying to solve this question:

image.png.817ad57fa9fff94a79864f3a9331f729.png

I created this graph (Title is in Portuguese):

image.png.2397b133f205a9dee6d62389bc09ca77.png

I used "1/[HI] x t," which indicates a second-order reaction with a rate law equal V = k[HI]^2, so, the slope of the graph is numeric equal for k, AND, if my previous assumption is correct, the value obtained for V is the global rate reaction, as a consequence, the constant (k) 0.0078 (L.mol-1s-1) represent the unique rate law and not of specific reactant (HI).

However, the gab, says that 0.0078 represents the rate constant (K) for the HI

image.png.f8c818e4d7814172e54bc9993a7db6c8.png

For me, it makes no sense, because, if it is correct, the value of V obtained by rate laws is the rate of HI and not for the global reaction since k refers to him and not for the global reaction.

I may be out of date when it comes to modern terminology, but I don't understand what you mean by "global" rate of reaction, nor do I understand what is meant by a "unique" rate law.

A reaction rate can be expressed in terms of the rate of consumption of any of the reactants, or the rate of generation of any of the products, surely? No one of these is more fundamental or "global" than another, so far as I can see.  

Where do you get this rate expression for ozone decomposition from? It looks a bit strange, especially the inverse dependence on O2 concentration.  

 

Edited by exchemist

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