Proposing a reaction rate mechanism

In summary: So it is irrelevant.In summary, the rate law for the reaction is found from experiment to be-k_{1}C_{A}^2
  • #1
gfd43tg
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Homework Statement


The rate law for the reaction
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is found from experiment to be

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Suggest a mechanism consistent with the rate law.

Homework Equations

The Attempt at a Solution


Here is the webpage with the solution
http://www.umich.edu/~elements/07chap/frames.htm

Here is the solution
Rate Laws

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Relative Rates

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Net Rates: Rate of Formation of Product

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Need to find an expression for CA* because we cannot easily measure the concentration of A*, use PSSH to solve for CA*.

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Solving for CA*

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Substituting for CA* in Equation (4) the rate of formation of B is

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Relative rates overall

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My question is, why do they go about finding the rate law for B from the original reaction, instead of by steps. The way I thought to do this was to find ##r_{A}## from the elementary steps, exactly the same way they found ##r_{A^{*}}##.

##r_{1A} = -k_{1}C_{A}^2##
##r_{2A} = -k_{2}C_{A}C_{A^*}##
##r_{A} = r_{1A} + r_{2A}##

However, there is some complication with ##r_{A2}##. Since the reaction is ##A^* + A \rightarrow A + A##, How can I do the reaction rate of A assuming irreversible? I don't know if it should be ##r_{A2} = -k_{2}C_{A}C_{A^*}## or ##r_{A2} = -k_{2}C_{A}C_{A^*} + k_{2}C_{A}^2##
 
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  • #2
Sorry I cannot find the reaction, the proposed mechanism, the meaning of your symbols, and the experimental facts alluded to in your link. It would probably be best if you copied them to here. Though it looks like a mechanism of bimolecular collision followed by activation, which can be followed either by reversion/dissociation or onward retraction. But I'd need to be guessing what the problem is a bit less.
 
  • #3
The problem statement is written as is. The purpose of this problem is to illustrate some basic rules of thumb for using a rate law given to determine the mechanism. These are the rules of thumb given in the textbook.

Maybe this page will give you some of the additional information you want
http://www.umich.edu/~elements/07chap/frames.htm
 

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  • #4
Please state what is the hypothetical mechanism you are having trouble with. I could probably deduce what it probably is but this is unreasonable. You set out a series of equations, and after a bit I see you are using inconsistent terminology and your r1A is the same thing as rA1 etc, please correct. It now looks that there is a collisional activating step, but also a collisional deactivating step? (If so, at low concentrations the rate depends on collision frequency and the activated molecules react onwards because a second deactivating collison is rare, so overall rate is second-order. At high concentration most of the activated molecules are deactivated before they can react onwards, their concentration still increases with increasing A, but only linearly.)

Maybe your problem is only at the end you seem to have formulated two reactions as going the same way when they are opposed, and there is a - that should be a +, a sum that should be a difference?

Another guess - PSSH means something steady-state hypothesis?
 
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  • #5
Ok, so I fixed the part about the ordering of ##r_{1A}## vs. ##r_{A1}##, that was just a memory lapse. Apparently, the way to figure out the mechanism to a reaction is to memorize the table I have in my 2nd post. My problem is not memorizing the table. My issue is how they went on to find ##r_{A}## from their proposed mechanism. The proposed mechanism is

(1) ##A + A \xrightarrow {k_{1}} A^{*} + A## ##\hspace{0.5 in} r_{1A^{*}} = k_{1}C_{A}^2##
(2) ##A^{*} + A \xrightarrow {k_{2}} A + A## ##\hspace{0.5 in} r_{2A^{*}} = -k_{2}C_{A^{*}}C_{A}##
(3) ##A^{*} \xrightarrow {k_{3}} B + C ## ##\hspace{0.875 in} r_{3A^{*}} = -k_{3}C_{A^{*}}##

So then they sum ##r_{A^{*}} = r_{1A^{*}} + r_{2A^{*}} + r_{3A^{*}}##

Using the pseudo-steady state hypothesis, they say ##r_{A^{*}} = 0## and find an expression for ##C_{A^{*}}##. However, why is this step not repeated in finding ##r_{A}##? When they find ##r_{A}##, they simply say ##r_{B} = k_{3}C_{A^{*}}##, then substitute in ##C_{A^{*}}## found from ##r_{A^{*}}##, then using the original reaction ##A \rightarrow B + C##, they say ##r_{A} = -r_{B}##.

My proposal is to do exactly what they did for ##r_{A^{*}}##. I have an issue with reaction (1). How would you write ##r_{1A}## for that?

##r_{1A} =- k_{1} C_{A}^2##, or maybe it would be ##r_{1A} = -k_{1}C_{A}^2 + k_{1}C_{A}C_{A^{*}}##, or even perhaps something else? How do you write an elementary rate law with respect to a given species when the species is both a product and reactant, and the reaction is not reversible?

This issue is not relevant with respect to ##A^{*}## because that intermediate species is always either solely a reactant or solely a product, never both simultaneously in any of the 3 reaction steps.
 
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  • #6
OK, re my comment about + and -, sum and difference, I see they have put signs into your equations 1, 2, 3 and so your sum is an algebraic sum, which I think is slightly unusual presentation but OK now I know. In words the rate of change of [A*] is its rate of formation minus its rate of disappearance via two processes.

For your question check back to what they say about the PSSH. Assume that [A*] is small compared with [A]. Then
([A] + [B ]) is constant is just conservation of mass for that reaction, and rA = - rB is just saying the same thing (a molecule of A is converted into one of B).

B is created in a unimolecular elementary irreversible reaction from A*, step (3) and Eq. 3 is the physical consequence of this first-order process.

Why not do for rA what they did for rA* and equate it to 0? You notice if we do that we have no kinetics and no problem, that it would be saying A is not undergoing any reaction? A and B are at the ends of the scheme, A* is in the middle. Think of it like say a despatcher who sends a large amount from A on to B but in his own warehouse has only a small amount at any time. Because it is small the change of absolute amounts are small, it is A and B that are changing a lot.

Re notation I reccomend you where possible to write double subscripts for rates and rate constants telling what into what, and square molarity brackets [A] etc. which are hardly more cumbersome than subscripts like CA etc.* That way when you see a formula kA*B[A*] you and a reader instantly know what it refers to , instead of having to say, ah k3, now let's see, what was that? Or often species are labeled with numbers 1, 2, 3,... and in your mechanism you'd have k12, k21 and k23. It'll save you some errors and confusions,. Some journals were starting to reccomend or require this for publications when I was last following them much.

* a bit tricky to do in this forum as B inside [ ] gets read as instruction for bold text!
 
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  • #7
Alright thank you, I'm not sure why you are thinking that these reactions are unusual. The + and - dictate whether the concentration of the species is increasing or decreasing. The notation common in chemical engineering is ##C_{A}##, rather than ##[A]##, so I just follow the convention in my discipline.
 

Related to Proposing a reaction rate mechanism

1. What is a reaction rate mechanism?

A reaction rate mechanism is a detailed step-by-step process that explains how a chemical reaction occurs. It involves identifying the intermediates and transition states involved in the reaction and their respective rate constants.

2. Why is it important to propose a reaction rate mechanism?

Proposing a reaction rate mechanism is important because it allows us to understand and predict the rate of a chemical reaction. This information is crucial in various fields such as pharmaceuticals, materials science, and environmental studies.

3. How do scientists propose a reaction rate mechanism?

Scientists propose a reaction rate mechanism by conducting experiments and analyzing the data to determine the rate law of the reaction. They also use theoretical models and computer simulations to propose a mechanism that is consistent with the experimental data.

4. What are the challenges in proposing a reaction rate mechanism?

There are several challenges in proposing a reaction rate mechanism, including the complexity of the reaction and the difficulty in identifying all the intermediates and transition states involved. Additionally, the lack of accurate experimental data can also make it challenging to propose a mechanism.

5. How can a proposed reaction rate mechanism be validated?

A proposed reaction rate mechanism can be validated by conducting additional experiments to confirm the predicted rate law and by comparing the predicted reaction rate with the actual experimental data. Computer simulations and theoretical models can also be used to validate the proposed mechanism.

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