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### Topic: Rate Law for Reversible Reaction Without Reaction Mechanism  (Read 702 times)

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#### RH111

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• Mole Snacks: +0/-0 ##### Rate Law for Reversible Reaction Without Reaction Mechanism
« on: April 09, 2021, 03:41:54 AM »
Hi all,

I have been stuck on a chemical kinetics question for quite some time (the question is #26 from the 2021 USNCO local exam).

We are given a reversible reaction A + B C and are told the forward reaction is kf[A]. We are asked to find the rate law for the reverse reaction.

I think there is not information to identify the reverse rate law because (from what I understand) we know nothing about the reaction mechanism, which is necessary for us to use the steady-state approximation.

My questions are on the following right now:
1. Is the provided information enough for us to identify the reaction mechanism ourselves?
2. Do we even need to know the reaction mechanism to identify the reverse rate? If not, how do I approach this? Is there some theory involved?

Any help is appreciated, thanks!
~RH111

#### Orcio_Dojek ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #1 on: April 10, 2021, 10:44:00 AM »
In thermodynamical equilibrium rate of forward reaction is equal to rate of the reverse reaction.

Since forward reaction is dependent on concentration of A, this excludes zero-order reaction for degradation of C.

In theory it is possible that reverse reaction is second-order reaction (v ~ [C]2), but there are no points directing that.

In my opinion A is the most accurate answer.

#### RH111

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• Mole Snacks: +0/-0 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #2 on: April 11, 2021, 04:25:47 PM »
Thank you for your response! If it helps to work backward, the correct answer according to USNCO is B: Rate = kr*[C]/[.B].

After some more research, I came across an equation (in Raymond Chang's AP Chemistry) that says the kf/kr = KC. Using this yields the correct answer. Is this equation applicable to any reversible equation, and does anyone know why it holds true? Also, how can we determine this without knowing the reaction mechanism?

Thanks!
RH111

#### Orcio_Dojek ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #3 on: April 11, 2021, 05:26:19 PM »
Sorry for misdirection. I just assumed earlier that since forward reaction doesn't depend on B, B must be constant. I think that answers A and B is just a different way how the rate constant is written, but of course maybe B is more accurate.

Quote
Is this equation applicable to any reversible equation, and does anyone know why it holds true? Also, how can we determine this without knowing the reaction mechanism?

It is applicable, but only for one intermediate stage.

For overall reaction it is K = (kf1 x kf2 x kf3 x ...) / (kr1 x kr2 x kr3 x ...), where kf1 is 1st step forward reaction constant, kr1 is 1st reverse reaction consant and K - equilibrium (thermodynamical) constant for reaction as overall.
« Last Edit: April 11, 2021, 08:26:52 PM by Orcio_Dojek »

#### Orcio_Dojek ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #4 on: April 12, 2021, 04:17:40 AM »
Sorry for misdirection. Reverse reaction is dependent on B, since B doesn't occur in forward reaction rate (this would give different rate constants and Kc = kf / kr won't be true).

#### RH111

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• Mole Snacks: +0/-0 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #5 on: April 12, 2021, 04:30:39 AM »
No worries at all! Thank you for the explanation.

It is applicable, but only for one intermediate stage.
What exactly do you mean by intermediate stage, and how can we tell that applies to the given reversible reaction?

Thanks again!

#### Orcio_Dojek ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #6 on: April 12, 2021, 04:45:54 AM »
I mean:

K = kf / kr

for every one step, for example:

1. A + B -> C + D

K1 = (C x D) / (A x B)

2. C + D -> E

K2 = E / (C x D)

So:

K = K1 x K2 = E / (A x B)

Here (in your test) K is something like K1 x K2, without investigating exact mechanism.

#### RH111

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• Mole Snacks: +0/-0 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #7 on: April 12, 2021, 04:47:11 AM »
Ohh okay, that makes sense. Thank you.

#### mjc123

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• Mole Snacks: +275/-12 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #8 on: April 12, 2021, 09:54:46 AM »
Quote
Thank you for your response! If it helps to work backward, the correct answer according to USNCO is B: Rate = kr*[C]/[.B].

After some more research, I came across an equation (in Raymond Chang's AP Chemistry) that says the kf/kr = KC. Using this yields the correct answer. Is this equation applicable to any reversible equation, and does anyone know why it holds true? Also, how can we determine this without knowing the reaction mechanism?

I am not happy with this answer. In fact, I am not happy with any attempt to give a kinetic explanation of thermodynamic phenomena. The laws of thermodynamics dictate that [C]/[A][B ] = constant, without any reference to the mechanism, which we may not know, and which may not be simple. Of course, the forward and reverse rates must be the same at equilibrium, and I suppose (without going into it in detail) the principle of microscopic reversibility must mean that the forward and reverse mechanisms are so related that equal rates result in the thermodynamic equilibrium condition. But I think it's quite dodgy to try to draw conclusions in the way the question asks you to.

The question says that "under certain conditions" the forward rate is kf[A]. It doesn't specify what the conditions are, but it would seem most likely that they involve using a large excess of B, so that [B ] is effectively constant throughout the reaction. In this case, [B ] is effectively hidden in kf - if the true rate law is kf'[A][B ], then kf = kf'[B ]. If you did another set of experiments with a different excess concentration of B, you would get the same rate law kf[A], but a different value of kf. In this scenario, the reverse rate law would not be kr[C]/[B ].

In short, it is dangerous to apply Keq = kf/kr without a clear understanding of the circumstances to which your data apply. (For example, in the above scenario with excess B, we could write K' = [C]/[A], where K' is constant for a given value of [B ], but varies with [B ]. Now what is kf/kr equal to - K or K'?)

In other words, it's a badly posed question to which the only reasonable answer is "there isn't enough information".

#### Babcock_Hall

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• Mole Snacks: +282/-21 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #9 on: April 12, 2021, 11:36:36 AM »
mjc123,

What is wrong with answer C?  That would have been what I chose.

#### RH111

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• Mole Snacks: +0/-0 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #10 on: April 12, 2021, 05:22:13 PM »
Thanks for your response.

it's a badly posed question to which the only reasonable answer is "there isn't enough information".

What throws me off, is that Option C is exactly that: not enough information. That's the option I went with when taking the test. It seems odd that they'd include that if, in fact, we cannot answer the question without knowing the mechanism. It seems that there must be something we're missing that allows them to come to the conclusion of Option B. After all, the American Chemical Society is a pretty reputable organization.

Also, would you be able to elaborate on the theory behind the kf/kr = KC equation? I'm not sure I understand how/why it works (albeit in only some situations).

Thanks!
RH111

#### Borek ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #11 on: April 12, 2021, 06:58:59 PM »
Assuming reaction

A + B C + D

(and doing several other assumptions) we have forward reaction rate of $k_f[A][ B]$ and reverse reaction rate of $k_r[C][D]$. At equilibrium forward and reverse reaction rates must be identical:

$$k_f[A][ B]=k_r[C][D]$$

hence

$$\frac{k_f}{k_r}=\frac{[C][D]}{[A][ B]}=K$$

This is known as a kinetic approach to equilibrium. As mjc already explained it is quite handwavy.
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#### RH111

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• Mole Snacks: +0/-0 ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #12 on: April 13, 2021, 04:42:44 PM »
Assuming reaction

A + B C + D

(and doing several other assumptions) we have forward reaction rate of $k_f[A][ B]$ and reverse reaction rate of $k_r[C][D]$.

Would anyone be able to elaborate on what these assumptions are? Are we assuming that the forward reaction and reverse reaction are first order in their respective reactants? If not, how can we say (without knowing the reaction mechanism) that Rate forward = $k_f[A][ B]$ and Rate reverse = $k_r[C][D]$?

Thanks!

#### Borek ##### Re: Rate Law for Reversible Reaction Without Reaction Mechanism
« Reply #13 on: April 13, 2021, 08:01:55 PM »
how can we say (without knowing the reaction mechanism) that Rate forward = $k_f[A][ B]$ and Rate reverse = $k_r[C][D]$?

We can't.

Thats more or less what these assumptions are, that's why this approach is handwavy.
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