Relations on the Kinetic velocities in a cyclic reaction

[Ssnow]

Hi to all, I whant to ask a question about theoretical chemistry. Let us consider a cyclic reaction $\alpha A\rightarrow \beta B\rightarrow \gamma C\rightarrow \alpha A$ where $\alpha,\beta;\gamma$ are the stochiometric coefficients and $A,B,C$ chemical molecules ... there are relations between the three Kinetic velocities $v_{1}=k_{1}[A]^{\alpha}$, $v_{2}=k_{2}\left[B\right]^{\beta}$, $v_{3}=k_{3}[C]^{\gamma}$ ?
Ssnow

 

[HAYAO]

Could this be a homework question? Just in case it is, I am only going to give you guidelines, and not the answer.

I am unsure what you are trying to ask because of the poor grammar, but are you asking what rate equations can be derived based on the chemical equation $\alpha A\rightarrow \beta B\rightarrow \gamma C\rightarrow \alpha A$ ?

Of course there are relations between the kinetic velocities because the end product is the initial reactant. Think about how [A] evolves. Obviously, you have the first equation that shows kinetic velocity of reaction from [A], which shows that [A] decreases in the rate $k_{1}$. But at the same time, it should increase due to the reaction from [C], right? So what modification do you need for the first equation $v_{1}=\frac{d[A]}{dt}=k_{1}[A]^{\alpha }$?

 

[Ssnow]

I am unsure what you are trying to ask because of the poor grammar

Sorry HAYAO, my mistake " I *want to ask ... "

In fact was only a curiosity, it is not a homework question, I thought (from an algebraic point of view) to this relation on velocities: $v_{1}+v_{2}+v_{3}=0$,
this statement has sense ?
Thanks,
Ssnow

PS. I am not an expert in the field...

 

[HAYAO]

Hmm.

Well, let's see, the equations are actually,
$$v_{1}=\frac{d[A]}{dt}=k_{1}[A]^{\alpha }-k_{3}[C]^{\gamma }$$
$$v_{2}=\frac{d[B ]}{dt}=k_{2}[B ]^{\beta }-k_{1}[A]^{\alpha }$$
$$v_{3}=\frac{d[C]}{dt}=k_{3}[C]^{\gamma }-k_{2}[B ]^{\beta }$$
So yes, you are right.
$$v_{1}+v_{2}+v_{3}=0$$

 

[mjc123]

v1+v2+v3=0

This will not be true generally. At equilibrium, if the system comes to equilibrium, v₁ = v₂ = v₃ = 0. This does not mean that no reactions are occurring; they are, but all at the same rate, so that there is no net change in any of the concentrations.

 

[mjc123]

You have to be a bit careful. In reaction 1, α As react to give β Bs, while in reaction 3, γ Cs react to give α As. So
d[A]/dt = - αk₁[A]^α + αk₃[C]^γ
Moreover, I take OP to mean that v₁ is the rate of reaction 1, etc. so v₁ ≠ d[A]/dt.
What you get is
1/α*d[A]/dt + 1/β*d[B ]/dt + 1/γ*d[C]/dt = 0
If v₁ is defined as k₁[A]^α, etc., then all the v's are positive (none of the reactions is shown as reversible) and their sum must be positive (except when it is zero at equilibrium).

 

[Ssnow]

This will not be true generally. At equilibrium, if the system comes to equilibrium, v₁ = v₂ = v₃ = 0. This does not mean that no reactions are occurring; they are, but all at the same rate, so that there is no net change in any of the concentrations.

1) Ok this was my dubt, in fact $v_{1},v_{2},v_{3}$ are as @HAYAO wrote before and not how I supposed in my question ...
2) My situation is not an equilibrium situation
3)Thank you @mjc123 for detail and clarifications ...

Ssnow

 

[HAYAO]

You have to be a bit careful. In reaction 1, α As react to give β Bs, while in reaction 3, γ Cs react to give α As. So
d[A]/dt = - αk₁[A]^α + αk₃[C]^γ

I totally forgot about the order of reaction. Sloppy mistake. Thanks.

Moreover, I take OP to mean that v₁ is the rate of reaction 1, etc. so v₁ ≠ d[A]/dt.

Hmm. That's a good point. I was not entirely sure how OP defined the rate of reaction.

2) My situation is not an equilibrium situation

Whether it is an equilibrium or not depends on the initial condition and how much time has elapsed. If your situation is not at equilibrium then it might be worth mentioning the initial condition and at what time t you are talking about. Otherwise, $v_{1}+v_{2}+v_{3}>0$.

For example, let's say you have [A]=1 mol/l, and [B ]=[C]=0 initially. When you let the reaction start, and when t→∞, you are going to have an equilibrium. Or, if your initial condition is already the concentration at the equilibrium, then you already have $v_{1}+v_{2}+v_{3}=0$.

 

[mjc123]

It was late and I was in a hurry; in fact by my definition v1+v2+v3 ≠ 0 at equilibrium as all the reactions are going forward, though there is no net change in concentrations. I was thinking about the situation with a reversible reaction, if you consider the reaction rate the difference of the rates of forward and back reactions, but that isn't the situation here.

By the alternative definition, v1 = d[A]/dt etc., the relation you get is
v1/α + v2/β + v3/γ = 0

 

[Ssnow]

v1/α + v2/β + v3/γ = 0

yes @mjc123, in fact I saw in some books that the velocity is defined with the ratio of the stoichiometric coefficient as $v=\frac{1}{\alpha}\frac{d [A]}{dt}$ instead to use only the derivative...
Thanks to all,
Ssnow

 

[epenguin]

Things like that do get considered in theoretical chemistry.
For further discussion I would suggest you forget the complication for the moment of the α, β γ, just let us suppose they are all 1, which is complicated enough to be going on with.

The thing to realize is that your scheme cannot be the complete representation of anything chemical.
The scheme shows A transforming into B and then back into A going through C. But if that were all, in normal chemistry B can transform back directly into A.So a normal chemical scheme would include the reverse arrows at each step, one from B back to A and from A directly to C. etc. For a normal chemical reaction there would be a constraint on these rate constants

k_abk_bck_ca = k_ack_cbk_ba

(obvious notation). This is called the "Principle of microscopic reversibility" - there are some quibbles about that and some prefer "principle of detailed balance".

The result of these principles is, you might say, boringness. Kinetics that could happen if it were not true do not happen normally.

So why think about it at all? As I said, it cannot be your schemes cannot be the complete representation of anything chemical. But it can represent something of which you have left out a part. for example , suppose one of the arrows represent a photochemical reaction (with constant illumination). Then the rate constants are not bound by the above equation. You have an input of energy which is not shown in the scheme. In a long time the system does not reach a true equilibrium, but a steady state.Unlike the approach to equilibrium, the approach to the steady state may not be monotonic. In many chemical mechanisms steady states are of interest and may be represented in figures looking like your scheme. Example, in enzyme kinetics you get schemes like illustrated below (sorry for quality). Here a substance, x, combines with enzyme E which exists in two interconvertible forms E and E', to form the interconvertible enzyme-substrate complexes EX, E'X which then convert into the reaction product P regenerating E and E'. Here E is produced from EX by two different processes – dissociation and catalytic breakdown. The effective rate constant for EX → E Is the sum of the two rate constants. We can observe reactions over a time where the total concentration of x is essentially constant, so we have a steady state. The microscopic reversibility relations hold for the disassociation rate constants only, not necessarily for the overall ones.
The equations for the resulting steady-state kinetics are complicated!

I can find some references for anyone interested. I would also be interested to hear from anyone else, firstly because I have the impression that this subject is dealt with within different specialities who do not know about each other, secondly because I would like to be brought up to date!

 

[Ssnow]

Ok, thank you to all for the answer but there is another thing not clear. @mjc123 said that in the equilibrium case all velocities are zero, why ? I think the sum will be zero but the concentrations are free to change in different ways in the direct or inverse direction, or it is wrong? Ssnow

 

[HAYAO]

The definition of equilibrium is when all competing processes are fully balanced and there is no net changes in the concentration of any of the species. It does not mean that the reaction is not occurring but since there is no net change, so the net reaction velocities are zero.

 

[mjc123]

@mjc123 said that in the equilibrium case all velocities are zero, why ?

As I said in post #9, I made that statement hurriedly and now think it is wrong.

 

[Ssnow]

It does not mean that the reaction is not occurring but since there is no net change, so the net reaction velocities are zero.

I understand that there is no net change (globally) and that the reactions take place, but for me it is not clear why locally we cannot consider a velocity different from zero (probably very small), the equilibrium $\alpha A \Leftrightarrow \beta B \Leftrightarrow \gamma C\Leftrightarrow \alpha A$ says that:

$v_{1d}=v_{1i} \wedge v_{2d}=v_{2i} \wedge v_{3d}=v_{3i}$

where $v_{id}$ is the direct velocity of the $i$-reaction and $v_{ii}$ is the inverse velocity of the $i$-reaction ... not that all are zeros...
Ssnow

 

[Ygggdrasil]

A few notes:
1. The cyclic reaction will not cycle as written (else it would be a perpetual motion machine). Consider the free energy of each of the states. Say α molecules of A has a free energy G₁, β molecules of B has a free energy of G₂ and γ molecules of C has a free energy of G₃. For αA --> βB to occur the free energy change associated with the reaction must be negative (ΔG ≤ 0 which would mean G₂ ≤ G₁). Applying this to the other reactions in the cycle, we get the following system of relations:

G₂ ≤ G₁
G₃ ≤ G₂
G₁ ≤ G₃

which has the solution G₁ = G₂ = G₃. Thus ΔG = 0 for all three reaction, and they are all at equilibrium. As discussed in the thread above, under these conditions, the net reaction velocity for all three reactions under these conditions is zero.

2. It would be possible to create a cycling reaction by coupling the reactions to a thermodynamically favorable process, say the conversion of D to E. For example, one could have a cycling system such as

A+D --> B
B --> C + E
C --> A

These types of cycles are common in various biological processes (such as the GTPase cycles that drive nucleocytoplasmic transport in cells)

3. There are many cases where there is a net creation and destruction of a molecule yet its concentration remains constant (e.g. we encounter these types of situations when studying metabolism) . While the concentration of a particular species may remain constant over time, this system is not in equilibrium because there is a net flux of molecules through the pathway. Instead, we refer to this situation as a steady state.

 

[DrStupid]

but for me it is not clear why locally we cannot consider a velocity different from zero (probably very small), the equilibrium $\alpha A \Leftrightarrow \beta B \Leftrightarrow \gamma C\Leftrightarrow \alpha A$ says that:

$v_{1d}=v_{1i} \wedge v_{2d}=v_{2i} \wedge v_{3d}=v_{3i}$

where $v_{id}$ is the direct velocity of the $i$-reaction and $v_{ii}$ is the inverse velocity of the $i$-reaction ... not that all are zeros...

You can do that. But for $v_{ii} = 0$ (that was the original assumption) this implies $v_{id} = 0$.

 

[Ssnow]

You can do that. But for $v_{ii} = 0$ (that was the original assumption) this implies $v_{id} = 0$.

Ok if $v_{ii}=0$ for every $i=1,2,3$ sure all velocities (direct and inverse) are zero because nothing change in concentration (in the equilibrium case). I think this a trivial situation.
Ssnow

 

[mjc123]

On a pedantic point, in English we talk about "forward" and "back" (or "reverse") reactions, not "direct" and "inverse". You may possibly cause confusion if you use these terms where the context is not clear (I encountered this recently in a post on chemicalforums).