[Corribus]

What do you envision could bring about such a discrete, sudden change?  I mean, let's pretend this discrete sudden change happens when one reactant is at 1.0 M.  What happens between 0.99999 M and 1.0 M to bring about this discrete sudden change?

[Big-Daddy]

What do you envision could bring about such a discrete, sudden change?  I mean, let's pretend this discrete sudden change happens when one reactant is at 1.0 M.  What happens between 0.99999 M and 1.0 M to bring about this discrete sudden change?

Not that sudden :p I meant more like, the rate law can be modelled very accurately as r=k[A][B ]³ for the majority of the process, and then perhaps when one reactant has a very low concentration its rate law begins to shoot up, either continuously, or quickly reaches maybe r=k[A]²[B ] and then stays there.

mod edit fixing the bold tags

[curiouscat]

What do you envision could bring about such a discrete, sudden change?  I mean, let's pretend this discrete sudden change happens when one reactant is at 1.0 M.  What happens between 0.99999 M and 1.0 M to bring about this discrete sudden change?

Not that sudden :p I meant more like, the rate law can be modelled very accurately as r=k[A]³ for the majority of the process, and then perhaps when one reactant has a very low concentration its rate law begins to shoot up, either continuously, or quickly reaches maybe r=k[A]²[ B] and then stays there.

Ok, first, can you come up with an elementary reaction where the rate law  is  r=k[A]³[ B]² or even k[A]²[ B]³

Forget transitions and whether they are sudden or not. Give me an elementary step of this form first.

I'm afraid you are drifting into the realm of speculation / fantasy.

[Big-Daddy]

Ok, first, can you come up with an elementary reaction where the rate law  is  r=k[A]³[ B]² or even k[A]²[ B]³

Forget transitions and whether they are sudden or not. Give me an elementary step of this form first.

I'm afraid you are drifting into the realm of speculation / fantasy.

That example was intended as hyperbole so we might make do with r=k[A][B] into just r=k[A] instead, but actually I don't mind the argument if it's to do with non-elementary rate laws instead. And of course we can hypothesize a non-elementary rate law of r=k[A]³[ B]².

So do we ever observe these discrete changes (obviously not instant; constant rate law to within very good approximation for most of the process, then rate law changes, over a small gap in concentration, to a new rate law), perhaps when the concentration of one reactant gets very low or when a product's concentration gets very high (or when we near equilibrium, for the case of equilibria)?

[curiouscat]

(obviously not instant; constant rate law to within very good approximation for most of the process, then rate law changes, over a small gap in concentration, to a new rate law

Low conc. : r = k1  c  Order =1

High conc. r = k2      Order = 0

Non elementary.

[Borek]

I'm afraid you are drifting into the realm of speculation / fantasy.

As usual.

[Big-Daddy]

(obviously not instant; constant rate law to within very good approximation for most of the process, then rate law changes, over a small gap in concentration, to a new rate law

Low conc. : r = k1  c  Order =1

High conc. r = k2      Order = 0

Non elementary.

So it does happen. If we wanted to model this in a differential equation, would you find a continuous function of concentration for the rate constant and reaction order (i.e. write k as f([C]) and n as f([C])), with functions which change only a little up till that point then jump sharply, or would you have to solve the two sections separately (i.e. solve the differential equations with the first rate law until a high enough concentration is reached, then choose a discrete point in concentration at which to use the other rate law to model from then on, then solve the next differential equation using the second rate law to model from then on)?

[Big-Daddy]

I'm afraid you are drifting into the realm of speculation / fantasy.

As usual.

Ouch!  I only meant to give an example. Also, I realize those laws are non-elementary but that doesn't make them fantasy ...

[curiouscat]

So it does happen. If we wanted to model this in a differential equation, would you find a continuous function of concentration for the rate constant and reaction order (i.e. write k as f([C]) and n as f([C])), with functions which change only a little up till that point then jump sharply, or would you have to solve the two sections separately (i.e. solve the differential equations with the first rate law until a high enough concentration is reached, then choose a discrete point in concentration at which to use the other rate law to model from then on, then solve the next differential equation using the second rate law to model from then on)?

Why don't you give it a shot? Try solving it. Let's see what you get.

[curiouscat]

Ouch!  I only meant to give an example.

Choose better examples?

[Big-Daddy]

So it does happen. If we wanted to model this in a differential equation, would you find a continuous function of concentration for the rate constant and reaction order (i.e. write k as f([C]) and n as f([C])), with functions which change only a little up till that point then jump sharply, or would you have to solve the two sections separately (i.e. solve the differential equations with the first rate law until a high enough concentration is reached, then choose a discrete point in concentration at which to use the other rate law to model from then on, then solve the next differential equation using the second rate law to model from then on)?

Why don't you give it a shot? Try solving it. Let's see what you get.

I have no idea what f([C]) contains, so I can't even express the ODE much less solve it. I'm asking whether we would go for something f([C]) which is continuous which varies sharply at the desired values of concentration (e.g. if the reaction order is a continuous f([C]), then a function whose value is extremely close to 1 at low concentrations and rapidly jumps to 0, and then stays close to 0, at high concentrations, is what we would want) or whether we would split the cases where one rate law applies and where the other one applies up, choosing some mid-point where the law changes discretely from one to the other, and then have two differential equations to deal with.

[Borek]

Ouch!  I only meant to give an example. Also, I realize those laws are non-elementary but that doesn't make them fantasy...

Problem is, every second thread you start ends drifting into some meaningless discussion of secondary details that are way beyond things you know and understand. Sorry to say that, but you are wearing us out for no gain. At the moment I am often ignoring your questions as I know answering will start an endless and fruitless discussion for which I have no time. You worked hard to put me in this position. I am far from being happy about that, but that's the only way for me to be still able to help others on the forum, otherwise I would waste all my time discussing for umpth time equilibrium calculations - which I explained to you already in October last year.

[curiouscat]

I have no idea what f([C]) contains, so I can't even express the ODE much less solve it.

Here's your rate law. Try solving now.

I'm asking whether we would go for something f([C]) which is continuous which varies sharply at the desired values of concentration (e.g. if the reaction order is a continuous f([C]), then a function whose value is extremely close to 1 at low concentrations and rapidly jumps to 0, and then stays close to 0, at high concentrations, is what we would want) or whether we would split the cases where one rate law applies and where the other one applies up, choosing some mid-point where the law changes discretely from one to the other, and then have two differential equations to deal with.

I see what you are asking but I want to see you work it out yourself.

[curiouscat]

Problem is, every second thread you start ends drifting into some meaningless discussion of secondary details that are way beyond things you know and understand. Sorry to say that, but you are wearing us out for no gain. At the moment I am often ignoring your questions as I know answering will start an endless and fruitless discussion for which I have no time. You worked hard to put me in this position.

I can sympathise with what Borek's saying here. Here's my grouse. I rarely see you do any of the grunt work: substituting values, solving equations, crunching numbers. Looking up references.  Most science is not a spectator sport. You cannot always have your answers on a platter.

 Let me offer you some sugesstions.  Asking good questions is an art. Try to learn it early. Take the effort to use Latex. We don't want to squint for ever to decode your equations. Quote relevant bits and not whole  blocks of previous replies. Precision is a virtue. Define your terms and symbols. Be quantitative. Eschew ambiguity.  et cetra.

When I answer a question I like to know that the person asking the question has put in a lot more effort in asking his question than I will in answering the same question.

[Big-Daddy]

Problem is, every second thread you start ends drifting into some meaningless discussion of secondary details that are way beyond things you know and understand.

Are they meaningless? Are they secondary? Are they beyond what I can understand? If I knew these things were secondary and irrelevant I would not have asked. I just want to reach a clear understanding on things.

Sorry to say that, but you are wearing us out for no gain. At the moment I am often ignoring your questions as I know answering will start an endless and fruitless discussion for which I have no time. You worked hard to put me in this position. I am far from being happy about that, but that's the only way for me to be still able to help others on the forum, otherwise I would waste all my time discussing for umpth time equilibrium calculations - which I explained to you already in October last year.

Ok, I see. I don't think I have asked for help on the same topic twice. But I see that my questions/responses are a cause for frustration to you.

I can sympathise with what Borek's saying here. Here's my grouse. I rarely see you do any of the grunt work: substituting values, solving equations, crunching numbers. Looking up references.  Most science is not a spectator sport. You cannot always have your answers on a platter.

In general, whenever you have asked I have always responded with the hard work. When you asked me to give attempts for the ODEs earlier this thread, I did, and used LaTeX as you suggest. Perhaps your last line - which I embolden - is the main idea though.