Exact Solutions for Kinetics of 2A + B → C + D Reaction with Rate Constant k

[Big-Daddy]

If I have a reaction where R(Gain of C)=k(Gain of C)*A²*B, where k is the rate constant and the reaction is 2A + B → C + D, how can I get two exact expressions, for the concentrations of A and B respectively, in terms of time, the initial values of A and B, and k?

This will be an ODE I imagine but beyond that I don't know how to get my answer.

 

[mfb]

You can express A and B in terms of C: $A=A_0-2C$, $B=B_0-C$.

This gives $\frac{dC}{dt}=k(A_0-2C)^2(B_0-C)$ if I understand your notation correctly.
This can be solved by separation of variables, but the expression could get messy.

 

[Big-Daddy]

You can express A and B in terms of C: $A=A_0-2C$, $B=B_0-C$.

This gives $\frac{dC}{dt}=k(A_0-2C)^2(B_0-C)$ if I understand your notation correctly.
This can be solved by separation of variables, but the expression could get messy.

Thanks. How should I go about getting the solution?

I have no objection to using a program or WolframAlpha to solve it, so long as my final expression is still in terms of the variables. Alternatively I don't mind learning how it's done.

 

[mfb]

Just type the equation into WolframAlpha (or similar programs) and hope ;).

Alternatively I don't mind learning how it's done.

http://en.wikipedia.org/wiki/Separation_of_variables
And I think you will need http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method as well.

 

[Big-Daddy]

You can express A and B in terms of C: $A=A_0-2C$, $B=B_0-C$.

This gives $\frac{dC}{dt}=k(A_0-2C)^2(B_0-C)$ if I understand your notation correctly.
This can be solved by separation of variables, but the expression could get messy.

But this equation will finally solve for the concentration of C with time. How do I go about solving for A or B with time?

 

[Big-Daddy]

You can express A and B in terms of C: $A=A_0-2C$, $B=B_0-C$.

This gives $\frac{dC}{dt}=k(A_0-2C)^2(B_0-C)$ if I understand your notation correctly.
This can be solved by separation of variables, but the expression could get messy.

I was wondering if we could use the following process:

We are given k(P,gain) - i.e. the rate constant for the gain of P (it makes no difference whether this is a reactant or product, only the value of k(P,gain) makes a difference). But we want to solve instead for the concentration of the product X in terms of k(P,gain), t, all initial concentrations. We have the orders of reaction with respect to each reactant ($n_R$) and the stoichiometric coefficients with respect to each substance (e.g. $v_X$ meaning the coefficient on X in the reaction equation).

Can we write this as:

$\frac{dP}{dt}=k*∏((R_0+((v_X/v_R)*(X_0-X)))^(n_R))$

(Assuming only reactants are present in the rate equation)
The ∏ shows my expression ([R]=[R]₀+((v_X/v_R)*([X]₀-[X])))^{order of reactant R} is being multiplied for every reactant (i.e. stick in this expression for every reactant in the rate equation and then multiply the expressions together), and then multiplied by k(P,gain), to reach the differential equation.

And then in theory solve this differential equation to get my expression in terms of the initial concentrations of each reactant, initial concentration of X, k(P,gain) and t (for set values of the stoichiometric coefficients and orders of reaction placed in)?

Thanks!

Edit: Would this imply that I'll get the same expression independent of which value of k I'm using (e.g. k(P,gain) or k(A,loss) or whatever)? Where does the species P come into this? Bearing in mind that I want to solve for the concentration of X, not P, with time, and all I have from P is the rate constant k(P,gain) and its initial concentration (with no desire to find out how that concentration changes, other than to work out how the concentration of X changes).

Edit 2: Or will I perhaps simply rewrite k(P,gain) as vX/vP*k(X,gain), if both are products, or -vX/vP*k(X,gain), if X is product and P is a reactant, and suddenly my equation is the same but in the place of what previously read dP/dt I now have dX/dt (and now I can get my solution in terms of the variables I mentioned)?

 

[mfb]

But this equation will finally solve for the concentration of C with time. How do I go about solving for A or B with time?

This is linked via your reaction: $A=A_0-2C$ and $B=B_0-C$.

I was wondering if we could use the following process:

I don't understand that approach, but the formula has some similarity to the one I posted.

Edit: Would this imply that I'll get the same expression independent of which value of k I'm using (e.g. k(P,gain) or k(A,loss) or whatever)?

You will need different equations for different k, as they correspond to different differentials and probably different factors on the right hand side as well.

 

[Big-Daddy]

This is linked via your reaction: $A=A_0-2C$ and $B=B_0-C$.

So I would solve for C first and then work out A and B. Thanks :)

 

[Big-Daddy]

How would we treat the system if it were an equilibrium? Let's consider the reaction

N₂ + 3H₂ ⇔ 2NH₃

What steps would we take to make a differential equation out of this with which we can calculate the concentration of one of the species at any time? (Given all starting data, like the forward and backward rate constants, and all initial concentrations.) I'm not too worried about solving the differential equation by hand :P

 

[epenguin]

I think you have less complication of formulae if you express all in terms of A and the constants A₀ and B₀.

Then the equation becomes

- dA/dt = k A²[A/2 + (B₀ - A₀/2)]

(work through to check though).

Then carefully integrate using partial fractions.


Re your final question I think you can still reduce all to a single differential equation in which there will appear several constants, initial concentrations and two rate constants or a rate constant and an equilibrium constant.

You do not give any information of where this is all coming from. Trying to read between the lines I wondered if you are under the misapprehension that a reaction with that stoichiometry has to have that rate law, which is far from true.

 

[Big-Daddy]

I think you have less complication of formulae if you express all in terms of A and the constants A₀ and B₀.

Then the equation becomes

- dA/dt = k A²[A/2 + (B₀ - A₀/2)]

(work through to check though).

Then carefully integrate using partial fractions.


Re your final question I think you can still reduce all to a single differential equation in which there will appear several constants, initial concentrations and two rate constants or a rate constant and an equilibrium constant.

You do not give any information of where this is all coming from. Trying to read between the lines I wondered if you are under the misapprehension that a reaction with that stoichiometry has to have that rate law, which is far from true.

Certainly not (regarding the last paragraph), as you will see in the equation I posted I have separate terms for each stoichiometric coefficient and for each reaction order (i.e. order with respect to each reactant). ;)

How would I go about doing it for equilibria? I understand that I can write k₁ for the forward rate constant and k₂ for the backward rate constant but other than that I don't know where to start (I'm assuming because it's an equilibrium that the process is actually different to that for a standard reaction).

 

[epenguin]

Certainly not (regarding the last paragraph), as you will see in the equation I posted I have separate terms for each stoichiometric coefficient and for each reaction order (i.e. order with respect to each reactant). ;)

How would I go about doing it for equilibria? I understand that I can write k₁ for the forward rate constant and k₂ for the backward rate constant but other than that I don't know where to start (I'm assuming because it's an equilibrium that the process is actually different to that for a standard reaction).

I don't know if we are talking about the same thing.
I am saying that other rate eqations such as -dA/dt = k[A] or k[A]² or k[A] and others are equally possible for reactions with your stoichiometry.

Could you solve the differential equations for those cases?

The equilibrium one can be done if you can do the irreversible one. But I think you would do better to do simpler cases before the more complicated one.

 

[Big-Daddy]

I don't know if we are talking about the same thing.
I am saying that other rate eqations such as -dA/dt = k[A] or k[A]² or k[A] and others are equally possible for reactions with your stoichiometry.

Could you solve the differential equations for those cases?

I can certainly write differential equations for those cases easily. Solving them is a problem I'm willing to leave until I learn separation of variables independently.

The equilibrium one can be done if you can do the irreversible one. But I think you would do better to do simpler cases before the more complicated one.

Agreed, but I can write the differential equations for irreversible reactions. My general formula, which works (I can provide the justification if needs be), should be evidence of this. That's why I'd like to move on to equilibrium reactions. As I said, solving is a computational issue really and I'll wait till I learn separation of variables in some detail before attempting it.

 

[epenguin]

I can certainly write differential equations for those cases easily. Solving them is a problem I'm willing to leave until I learn separation of variables independently.
Agreed, but I can write the differential equations for irreversible reactions. My general formula, which works (I can provide the justification if needs be), should be evidence of this. That's why I'd like to move on to equilibrium reactions. As I said, solving is a computational issue really and I'll wait till I learn separation of variables in some detail before attempting it.

Sorry, this is not 'separation of variables' - I have written, but you should check, the equation in one variable which is all that is necessary.

Then that is strictly a differential equation, but excercises of the form dA/dt = f(A) i.e. dt = dA/f(A) usually come in the books in the chapter 'Integration', earlier than differential equations. And I have told you the method of solution requires Partial Fractions - have you ever done them? (If not they are pretty easy - I only add that denominators with square or other repeated factors in the denominator are just slightly more tricky than others. ) )

I am just trying to point you to the fastest way to get there, however the site rules require the student to show his steps that way.

If you can do it for the irreversible reaction you should then be able to do it for the reversible one, but if you can't do the first you won't be able to do the second and will get lost. In fact it is looking like you need to do first some simpler irreversible ones.

But if you want the broad answer for reversible, the principle is the best though maybe not the only way is to formulate it in terms of a 'distance from equilibrium' variable ([A]-[A_eq]). Again you have just a single variable, but the expressions involve the intitial concentrations of all reactants and two rate constants (or equivalently a rate constant and an equilibrium constant).

 

[Big-Daddy]

Sorry, this is not 'separation of variables' - I have written, but you should check, the equation in one variable which is all that is necessary.

Then that is strictly a differential equation, but excercises of the form dA/dt = f(A) i.e. dt = dA/f(A) usually come in the books in the chapter 'Integration', earlier than differential equations. And I have told you the method of solution requires Partial Fractions - have you ever done them? (If not they are pretty easy - I only add that denominators with square or other repeated factors in the denominator are just slightly more tricky than others. ) )

I am just trying to point you to the fastest way to get there, however the site rules require the student to show his steps that way.

If you can do it for the irreversible reaction you should then be able to do it for the reversible one, but if you can't do the first you won't be able to do the second and will get lost. In fact it is looking like you need to do first some simpler irreversible ones.

Again, I am not looking for solutions to the differential equations now. I do not need to solve them. That can come later; all I want to do for just now is understand how to reach them.

The equation in the format I wrote definitely works for irreversible reactions. Thus I am ready to move on to writing (not solving) differential equations for reversible reactions.

As for separation of variables, mfb also wrote his solution in terms of only one real variable - C - but apparently it still needs to be solved by separation of variables? Read his first post ...

But if you want the broad answer for reversible, the principle is the best though maybe not the only way is to formulate it in terms of a 'distance from equilibrium' variable ([A]-[A_eq]). Again you have just a single variable, but the expressions involve the intitial concentrations of all reactants and two rate constants (or equibablently a rate constant and an equilibrium constant).

I think I may see where you're coming from, but to reach a differential equation I may require a bit more prodding.

Wikipedia (here: http://en.wikipedia.org/wiki/Rate_equation#Equilibrium_reactions_or_opposed_reactions) shows a few basic pointers for the general equilibrium sA+tB ⇔ uX+vY, but then it seems to make a mistake: it suggests that the coefficient on A, i.e. s, is the same as the order of the reaction with respect to A (and the same for the other substances) by writing the rate law as r=k₁*A^s*B^t-k₂*X^u*Y^v when s, t, u and v have just been specified as stoichiometric coefficients, not orders of reaction. Is this some kind of approximation? Is it possible to proceed without taking it?

Perhaps it would be better to start from the simple case listed below (A ⇔ B) as even there I don't know how the differential equation was arrived at, but first I'd like to understand the few basic pointers given about the general equilibrium sA+tB ⇔ uX+vY, and why it is making the statement that stoichiometric coefficients = orders of reaction.

 

[epenguin]

Again, I am not looking for solutions to the differential equations now. I do not need to solve them. That can come later; all I want to do for just now is understand how to reach them.

The equation in the format I wrote definitely works for irreversible reactions. Thus I am ready to move on to writing (not solving) differential equations for reversible reactions.

As for separation of variables, mfb also wrote his solution in terms of only one real variable - C - but apparently it still needs to be solved by separation of variables? Read his first post ...



I think I may see where you're coming from, but to reach a differential equation I may require a bit more prodding.

Wikipedia (here: http://en.wikipedia.org/wiki/Rate_equation#Equilibrium_reactions_or_opposed_reactions) shows a few basic pointers for the general equilibrium sA+tB ⇔ uX+vY, but then it seems to make a mistake: it suggests that the coefficient on A, i.e. s, is the same as the order of the reaction with respect to A (and the same for the other substances) by writing the rate law as r=k₁*A^s*B^t-k₂*X^u*Y^v when s, t, u and v have just been specified as stoichiometric coefficients, not orders of reaction. Is this some kind of approximation? Is it possible to proceed without taking it?

Perhaps it would be better to start from the simple case listed below (A ⇔ B) as even there I don't know how the differential equation was arrived at, but first I'd like to understand the few basic pointers given about the general equilibrium sA+tB ⇔ uX+vY, and why it is making the statement that stoichiometric coefficients = orders of reaction.

You say you don't know how the simplest of the equations was arrived at, but you proposed a more complicated one which you said you have arrived at.

Nothing could be simpler than how the equations of kinetics are arrived at. Either a molecule by itself changes. E.g. it isomerises or splits up independent of any other reactants. Each molecule has the same chance of doing this at any time so the total rate is simply proportional to the number of molecules present or concentration. E.g. - d[A]/dt = k[A]. That is a unimolecular elementary process and a first-order rate equation.
Or else the reaction depends on the encounter of two molecules, which happens at a rate proportional to the product of their concentrations (though only a fraction of such encounters is productive usually), so that would give you -d[A]/dt = -d/dt = k[A], or if it is two molecules of the same kind -d[A]/dt = k[A]². Bimolecular reactions with second order rate equations.

That is all there is! Except there are more complicated equations that arise from the fact that not all reactions are elementary, but many consist of more than one elementary step, and it is one of the jobs of kinetics to sort out the probable steps using hypothetical mechanisms (= sequences of elementary steps) from which you deduce equations putting together the above rate laws for each step, and comparing the predictions of the resulting equations with experiment.

Re your critique of a 'mistake': it is not a mistake because they say "assuming each one is elementary". If anything is misleading it is in their unnecessary generality. They intend the arrow to represent a mechanistic elementary chemical step. If so it is misleading to lead you to think that their s, t, u, v are ever anything but 1!

(The article is OK if you understand it, but a bit pedantic IMO. I don't know who wrote it but I can think of someone who could have written it!)

Note how, in their treatment of a simple reversible reaction, they write

\ \frac{dx}{dt} = \frac{k_b[A]_0}{x_e} (x_e - x)

which is an example of what I mentioned - using a 'distance from equilibrium' variable, more easily recognised as what I said in last post if you write it

- \ \frac{d(x - x_e)}{dt} = \frac{k_b[A]_0}{x_e} (x - x_e)

 

[Big-Daddy]

You say you don't know how the simplest of the equations was arrived at, but you proposed a more complicated one which you said you have arrived at.

Nothing could be simpler than how the equations of kinetics are arrived at. Either a molecule by itself changes. E.g. it isomerises or splits up independent of any other reactants. Each molecule has the same chance of doing this at any time so the total rate is simply proportional to the number of molecules present or concentration. E.g. - d[A]/dt = k[A]. That is a unimolecular elementary process and a first-order rate equation.
Or else the reaction depends on the encounter of two molecules, which happens at a rate proportional to the product of their concentrations (though only a fraction of such encounters is productive usually), so that would give you -d[A]/dt = -d/dt = k[A], or if it is two molecules of the same kind -d[A]/dt = k[A]². Bimolecular reactions with second order rate equations.

That is all there is! Except there are more complicated equations that arise from the fact that not all reactions are elementary, but many consist of more than one elementary step, and it is one of the jobs of kinetics to sort out the probable steps using hypothetical mechanisms (= sequences of elementary steps) from which you deduce equations putting together the above rate laws for each step, and comparing the predictions of the resulting equations with experiment.

Re your critique of a 'mistake': it is not a mistake because they say "assuming each one is elementary". If anything is misleading it is in their unnecessary generality. They intend the arrow to represent a mechanistic elementary chemical step. If so it is misleading to lead you to think that their s, t, u, v are ever anything but 1!

(The article is OK if you understand it, but a bit pedantic IMO. I don't know who wrote it but I can think of someone who could have written it!)

Note how, in their treatment of a simple reversible reaction, they write

\ \frac{dx}{dt} = \frac{k_b[A]_0}{x_e} (x_e - x)

which is an example of what I mentioned - using a 'distance from equilibrium' variable, more easily recognised as what I said in last post if you write it

- \ \frac{d(x - x_e)}{dt} = \frac{k_b[A]_0}{x_e} (x - x_e)

What is x there?

The problem is I'm not really sure how to start in terms of combining the rate constants with the equilibrium constant in general; I've heard K=k₁/k₂, but I can't see that this would be true if the reaction were not "elementary" (i.e. if stoichiometric coefficients do not necessarily equal orders of reaction). So on the whole, I don't know how to start! (with reaching my differential equation)

The equation I wrote is of the same form as mfb's, merely including a further constant (where he put -C, I put C₀-C to denote the possibility of C₀ not being 0).

 

[epenguin]

What is x there?

The problem is I'm not really sure how to start in terms of combining the rate constants with the equilibrium constant in general; I've heard K=k₁/k₂, but I can't see that this would be true if the reaction were not "elementary" (i.e. if stoichiometric coefficients do not necessarily equal orders of reaction). So on the whole, I don't know how to start! (with reaching my differential equation)

The equation I wrote is of the same form as mfb's, merely including a further constant (where he put -C, I put C₀-C to denote the possibility of C₀ not being 0).

The formulae in which x appeared are just taken from the Wikipedia article that you quoted. Since I was saying that this was just the same 'distance from equilibrium' variable viz. ([A]-[A_eq]) from my post #14 where I was using your terminology I thought you would recognise.

K=k₁/k₂ is true of the elementary mechanism you have. That mechanism has only two rate constants. If the mechanism is not elementary it will have more rate constants, practically by defnition. The equlibrium constant is then the product of all the forward constants, divided by the product of all the backwards constants, as you can easily deduce by the same sort of reasoning as for the elementary reactions in the Wiki article.

Now don't ask me how you do the calculations for even more complicated mechanisms, because the answer is to do it for simpler ones first and if you can do it for them you can do it for more complicated ones, the only new thing is longer formulae, but no novelty of principle.