Which Reaction Mechanism Aligns with the Rate Law?

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In summary: Concentration of B will increase.I believe epenguin tries to ask you whether there is any dependency between amount of O2 and O present (at least initially).##k_{eq}=\frac {[O][O_2]}{[O_3]}##Is one such relation quantitatively. If ##[O]## is increased then ##[O_2]## will decrease (Le chatelier principle) and vice versa. Are you giving me a hint about something else?Yes, about something else.
  • #1
mooncrater
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Homework Statement


This is the question:
Consider the reaction: $$2NO_2(g)+O_3 (g)\longrightarrow N_2O_5 (g)+O_2 (g) $$
The reaction of nitrogen dioxide and ozone is first order in ##NO_2(g)## and in ##O_3 (g)##. Which of these possible reaction mechanisms is consistent with the rate law?
MECHANISM I :$$ NO_2(g)+O_3 \longrightarrow NO_3 (g)+O_2 (g)$$------(slow)
$$NO_3 (g)+NO_2 (g) \longrightarrow N_2O_5$(g) $$------(fast)
MECHANISM II:$$O_3 (g)\longrightarrow O_2(g)+[O] $$(this one is in equilibrium)(fast)
$$NO_2 (g)+[O] \longrightarrow NO_3 (g) $$-------(slow)
$$NO_3 (g)+NO_2 (g) \longrightarrow N_2O_5 (g) $$-------(fast)
And the options are :
(A) I only
(B) II only
(C) Both I and II
(D) Neither I not II

Homework Equations

The Attempt at a Solution


The rate equation from the original equation:$$rate=k [NO_2][O_3] $$
And from mechanism 1:$$ r_1=k_1 [O_3][NO_2] $$
And from the mechanism 2 :
$$r_2=k_2 [NO_2][O] $$
$$k_{eq}=\frac {[O][O_2]}{[O_3]} $$
So keq[O3]/[O2]=[O]
putting this in ##r_2##
$$r_2=\frac {k _2k_{eq}[O_3][NO_2]}{[O_2]} $$
Which doesn't correspond with the gas law. So only 1 should be correct and the answer should be (A) but the given answer is (C)... but ##r_2## doesn't correspond to ##r##. So where am I wrong?
 
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  • #2
mooncrater said:

Homework Statement


This is the question:
Consider the reaction: $$2NO_2(g)+O_3 (g)\longrightarrow N_2O_5 (g)+O_2 (g) $$
The reaction of nitrogen dioxide and ozone is first order in ##NO_2(g)## and in ##O_3 (g)##. Which of these possible reaction mechanisms is consistent with the rate law?
MECHANISM I :$$ NO_2(g)+O_3 \longrightarrow NO_3 (g)+O_2 (g)$$------(slow)
$$NO_3 (g)+NO_2 (g) \longrightarrow N_2O_5$(g) $$------(fast)
MECHANISM II:$$O_3 (g)\longrightarrow O_2(g)+[O] $$(this one is in equilibrium)(fast)
$$NO_2 (g)+[O] \longrightarrow NO_3 (g) $$-------(slow)
$$NO_3 (g)+NO_2 (g) \longrightarrow N_2O_5 (g) $$-------(fast)
And the options are :
(A) I only
(B) II only
(C) Both I and II
(D) Neither I not II

Homework Equations

The Attempt at a Solution


The rate equation from the original equation:$$rate=k [NO_2][O_3] $$
And from mechanism 1:$$ r_1=k_1 [O_3][NO_2] $$
And from the mechanism 2 :
$$r_2=k_2 [NO_2][O] $$
$$k_{eq}=\frac {[O][O_2]}{[O_3]} $$
So keq[O3]/[O2]=[O]
putting this in ##r_2##
$$r_2=\frac {k _2k_{eq}[O_3][NO_2]}{[O_2]} $$
Which doesn't correspond with the gas law. So only 1 should be correct and the answer should be (A) but the given answer is (C)... but ##r_2## doesn't correspond to ##r##. So where am I wrong?

I think it is OK as far as you got, but you have not got to the end. You need to get rid of the [O2]] in your last equation and express in terms of initial reactants only.

That said it seems to me still incompatible with mechanism 2.
 
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  • #3
epenguin said:
I think it is OK as far as you got, but you have not got to the end. You need to get rid of the [O2]] in your last equation and express in terms of initial reactants only.

That said it seems to me still incompatible with mechanism 2.
But we don't have any other equation in which ##O_2## is present. (In ##2^{nd}## mechanism. ) So I think that this ##O_2## won't go.
 
  • #4
mooncrater said:
But we don't have any other equation in which ##O_2## is present. (In ##2^{nd}## mechanism. ) So I think that this ##O_2## won't go.

True you are not given it, ha ha.
I will let you osider just that first equilibrium by itself and see if you can come with it after all. :oldwink: :oldsmile:
 
  • #5
epenguin said:
True you are not given it, ha ha.
I will let you osider just that first equilibrium by itself and see if you can come with it after all. :oldwink: :oldsmile:
Sorry but I didn't understand what you mean to say.
 
  • #6
In general say younstart with substance A and it dissociates A ⇔ B + C

Can you say anything about [ B] at equilibrium?

OK, assume the equilibrium is to the left, I.e. [ B] is small compared to [A].
 
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  • #7
epenguin said:
In general say younstart with substance A and it dissociates A ⇔ B + C

Can you say anything about [ B] at equilibrium?

OK, assume the equilibrium is to the left, I.e. [ B] is small compared to [A].
Concentration of B will increase.
 
  • #8
I believe epenguin tries to ask you whether there is any dependency between amount of O2 and O present (at least initially).
 
  • #9
##k_{eq}=\frac {[O][O_2]}{[O_3]}##
Is one such relation quantitatively.
If ##[O]## is increased then ##[O_2]## will decrease (Le chatelier principle) and vice versa. Are you giving me a hint about something else?
 
  • #10
Yes, about something else.

Try to build an ICE table for the process, as if you were attempting to solve a simple equilibrium problem - what is the equilibrium concentration of O given K and initial [O3].
 
  • #11
Yes, you see you are trying to do something relatively clever and specialized while missing something very elementary and universal. Let me split my question and ask only for the first step.
epenguin said:
In general say younstart with substance A and it dissociates A ⇔ B + C
.
what is the ratio of [ B] and [C] at equilibrium? Or better, if they are not consumed in further reactions (which is an approximation you were assuming) what is their ratio at any time?
 
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  • #12
epenguin said:
Yes, you see you are trying to do something relatively clever and specialized while missing something very elementary and universal. Let me split my question and ask only for the first step.
what is the ratio of [ B] and [C] at equilibrium? Or better, if they are not consumed in further reactions (which is an approximation you were assuming) what is their ratio at any time?
1:1.
 
  • #13
So ##[O_2]=[O]##.
How Is it useful on removing that ##extra## ##[O_2] ##from the denominator? (##extra## as the numerator corresponds the the correct rate law... because of which we can't change that)
 
  • #14
mooncrater said:
So ##[O_2]=[O]##.

Can you use this information - together with the initial concentration of O3 and Keq - to calculate the concentration of O?

Don't bother about other things for now, we will see how they fit the problem later.
 
  • #15
Borek said:
Can you use this information - together with the initial concentration of O3 and Keq - to calculate the concentration of O?

Don't bother about other things for now, we will see how they fit the problem later.
Yup...
##[O]=\sqrt {K_{eq}[O_3]}##
 
  • #16
Can't you plug that into the rate equation?

And yes, that would mean mechanism 2 is not consistent with the observed rate (or I am missing something as well).

This is a little bit tricky, as - if I am reading these equation correctly - the observed rate equation will be different at the beginning (when there is no substantial amount of oxygen present) and later (when you can't ignore O2 presence).
 
  • #17
Borek said:
Can't you plug that into the rate equation?

And yes, that would mean mechanism 2 is not consistent with the observed rate (or I am missing something as well).

This is a little bit tricky, as - if I am reading these equation correctly - the observed rate equation will be different at the beginning (when there is no substantial amount of oxygen present) and later (when you can't ignore O2 presence).
Yes I can do that... But the real problem (as you are saying) that the mechanism 2 doesn't correspond to the rate law. But the answer is different from that. So can we reach the result that the given answer is wrong?
 
  • #18
We could do if that is what the equations tell us. But we haven't seen you tell use what the equation in terms of concentrations of starting reactants is yet. (Assume the overall reaction is irreversible.)
 

FAQ: Which Reaction Mechanism Aligns with the Rate Law?

What is consistency with the rate law?

Consistency with the rate law refers to the ability of an experimental data set to match the predicted rate law equation. In other words, it is the agreement between the observed and expected reaction rate based on the proposed rate law.

Why is consistency with the rate law important?

Consistency with the rate law is important because it allows us to validate the proposed rate law and understand the underlying reaction mechanism. It also helps in predicting the rate of a reaction under different conditions.

How do we determine consistency with the rate law?

To determine consistency with the rate law, we compare the experimental data with the predicted rate law equation. If the experimental and predicted values match closely, then the data is considered consistent with the rate law.

What factors can affect consistency with the rate law?

There are several factors that can affect consistency with the rate law, such as experimental errors, reaction conditions, and the complexity of the reaction mechanism. Any of these factors can lead to deviations from the predicted rate law.

What can we do if there is inconsistency with the rate law?

If there is inconsistency with the rate law, we can analyze the experimental data and try to identify the source of the inconsistency. This may involve adjusting experimental conditions, revising the proposed rate law, or conducting further experiments to obtain more accurate data.

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