Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Big-Daddy on April 29, 2013, 04:43:08 PM
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I don't propose to go into the maths in my case, unless someone wants to point me towards it, but I'd like to ask a question that's on my mind: I know that it is possible to write a precise differential equation for any 1 equilibrium which (after integration) will yield a result of being able to calculate the concentration of any species involved in the equilibrium at any time, given the forward and backward rate constants, equilibrium constant, and initial concentrations of all species.
Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
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This question is very abstract. Can you ... er... deabstractify it?
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I'll try. http://en.wikipedia.org/wiki/Rate_equation#Equilibrium_reactions_or_opposed_reactions shows for the most simple possible equilibrium A ::equil:: B, how a differential equation can be obtained in terms of the forward rate constant k1 and backward rate constant k2, as well as the equilibrium constant K, which can be solved for the concentration at any time t of any species present in the reaction. This is covered through to the end by the Wikipedia link, for the incredibly basic case of A ::equil:: B.
Now let's say that instead of one equilibrium we have many. They do not necessarily "start" at the same time, e.g. I might place HCl in a solution and then at some time t later add K2HPO4, each of which will start more equilibria off. Is it in theory always possible to summarize all the equilibria in a differential equation which, when solved, calculates the concentration of every species that we now have in our system (e.g. for the above example, it would be H+, OH-, H2O, K+, and all the phosphate forms), given all the initial concentrations of these species, the time at which we put them in, and all needed constants?
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You are asking if rate for every mechanism you could think of is exactly solvable? I don't see why not. At least, I can't think of a case off the top of my head where it isn't.
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Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
Yes. :)
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Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
Yes. :)
At what stage would I begin to get into the maths involved? :P
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Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
Yes. :)
At what stage would I begin to get into the maths involved? :P
Didn't understand your question. You could start right away if you were so inclined.
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Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
Yes. :)
At what stage would I begin to get into the maths involved? :P
Didn't understand your question. You could start right away if you were so inclined.
I am so inclined! What should I look into?
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Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
Yes. :)
At what stage would I begin to get into the maths involved? :P
Didn't understand your question. You could start right away if you were so inclined.
I am so inclined! What should I look into?
http://www.youtube.com/watch?v=-vuJnQLLkf4&feature=youtu.be
It's really not that hard. Good bookkeeping followed by a reliable simultaneous ODE solver.
If you only want equilibrium concentrations its just a bunch of simultaneous non-linear equations.
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Is it possible, in theory, to write a differential equation for any large system of equilibria, which can then be solved to calculate the concentration of any species involved in the system at a certain time, given the rate constants for every equilibrium, every equilibrium constant, and the initial concentrations of all species involved?
Yes. :)
At what stage would I begin to get into the maths involved? :P
Didn't understand your question. You could start right away if you were so inclined.
I am so inclined! What should I look into?
http://www.youtube.com/watch?v=-vuJnQLLkf4&feature=youtu.be
It's really not that hard. Good bookkeeping followed by a reliable simultaneous ODE solver.
If you only want equilibrium concentrations its just a bunch of simultaneous non-linear equations.
Ok that video makes sense on the whole. But it seems to apply to reactions going to completion. Will the equilibrium balances be written in the same way?
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Ok that video makes sense on the whole. But it seems to apply to reactions going to completion.
Nope.
Will the equilibrium balances be written in the same way?
Yes.
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Ok that video makes sense on the whole. But it seems to apply to reactions going to completion.
Nope.
Will the equilibrium balances be written in the same way?
Yes.
If it's an equilibrium shouldn't we be concerned with a forward rate constant and a backward rate constant for each reaction? There only appears to be a forward rate constant. And no term for the equilibrium constants either ...
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Ok that video makes sense on the whole. But it seems to apply to reactions going to completion.
Nope.
Will the equilibrium balances be written in the same way?
Yes.
If it's an equilibrium shouldn't we be concerned with a forward rate constant and a backward rate constant for each reaction? There only appears to be a forward rate constant. And no term for the equilibrium constants either ...
A ::equil:: B with k1 & k2
is the same as
A :rarrow: B with k1
and
B :rarrow: A with k2
No equilibrium constant is needed. If k1 and k2 are correct then Keq=k1 / k2 automatically.
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If it's an equilibrium shouldn't we be concerned with a forward rate constant and a backward rate constant for each reaction? There only appears to be a forward rate constant. And no term for the equilibrium constants either ...
Out of pure curiosity, what is your current level? HS? College? etc.
Again, pure curiosity. Have you exposure to how to solve ODE's simultaneously numerically?
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Ok that video makes sense on the whole. But it seems to apply to reactions going to completion.
Nope.
Will the equilibrium balances be written in the same way?
Yes.
If it's an equilibrium shouldn't we be concerned with a forward rate constant and a backward rate constant for each reaction? There only appears to be a forward rate constant. And no term for the equilibrium constants either ...
A ::equil:: B with k1 & k2
is the same as
A :rarrow: B with k1
and
B :rarrow: A with k2
No equilibrium constant is needed. If k1 and k2 are correct then Keq=k1 / k2 automatically.
Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:
A+B ::equil:: C with k1
C ::equil:: 2E with k2
But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?
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Out of pure curiosity, what is your current level? HS? College? etc.
Again, pure curiosity. Have you exposure to how to solve ODE's simultaneously numerically?
Officially high school, but I read college books in my spare time ::)
No I have no knowledge of solving ODEs simultaneously.
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Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:
A+B ::equil:: C with k1
C ::equil:: 2E with k2
But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?
Typically approximations are used, such as the steady-state approximation. The validity of such approximations depends a lot on the relative values of the various rate constants. Without approximations, the math rapidly becomes unwieldy, even for apparently simple systems.
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Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:
A+B ::equil:: C with k1
C ::equil:: 2E with k2
But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?
Ok, I'll try again:
A ::equil:: B k1f & k1r
B ::equil:: C k2f & k2r
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot c_A + k_{1r} \cdot c_B \\
\frac{dc_B}{dt}=k_{1f} \cdot c_A - k_{1r} \cdot c_B -k_{2f} \cdot c_B + k_{2r} \cdot c_C \\
\frac{dc_C}{dt}=k_{2f} \cdot c_B - k_{2r} \cdot c_C \\
[/tex]
At t=0 c_A, c_B, c_C all are known.
Solve as an initial value problem.
These are your "final ODE's" and they contain no explicit equilibrium constants. No matter how many equations the same approach applies. You can write one ODE for each species.
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Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:
A+B ::equil:: C with k1
C ::equil:: 2E with k2
But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?
Ok, I'll try again:
A ::equil:: B k1f & k1r
B ::equil:: C k2f & k2r
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot c_A + k_{1r} \cdot c_B \\
\frac{dc_B}{dt}=k_{1f} \cdot c_A - k_{1r} \cdot c_B -k_{2f} \cdot c_B + k_{2r} \cdot c_C \\
\frac{dc_C}{dt}=k_{2f} \cdot c_B - k_{2r} \cdot c_C \\
[/tex]
At t=0 c_A, c_B, c_C all are known.
Solve as an initial value problem.
These are your "final ODE's" and they contain no explicit equilibrium constants. No matter how many equations the same approach applies. You can write one ODE for each species.
OK I think I've got it.
How would we write the term, in the ODE for A, for a reaction in which the participation is nA ::equil:: B (i.e. stoichiometric coefficient)?
[tex]
-n k_{1f} \cdot c_A
[/tex]
Or what? And what if the rate of this equilibrium with respect to A is not first order?
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nice thread. thanks for sharing that information. it was very informative.
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OK I think I've got it.
Good.
How would we write the term, in the ODE for A, for a reaction in which the participation is nA ::equil:: B (i.e. stoichiometric coefficient)?
[tex]
-n k_{1f} \cdot c_A
[/tex]
Yes.
And what if the rate of this equilibrium with respect to A is not first order?
If it isn't, it isn't.
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If it isn't, it isn't.
OK let's take a look at a more complicated example because I'm not sure how this will work.
In a general case 3A+5B ::equil:: C, forward rate is 2nd-order in A and 3rd-order in B (not normal I know but let's just go with it). How do we write the parts of the ODE for cA that would be contributed by this equilibrium? My concern is that this time rate depends on both cA and cB ...
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In a general case 3A+5B ::equil:: C, forward rate is 2nd-order in A and 3rd-order in B (not normal I know but let's just go with it).
Write me a rate expression. You know how to do it for single equaions, right?
My concern is that this time rate depends on both cA and cB ...
Doesn't matter.
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Write me a rate expression. You know how to do it for single equaions, right?
Well I've never seen it before so I'm probably wrong but I'll give it a shot. :( If it were just 3A ::equil:: C rate equation is:
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot (c_A)^x + \frac{1}{3} k_{1r} \cdot (c_C)^n \\
[/tex]
So in the case of aA + bB ::equil:: cC + dD, I might suggest:
[tex]
\frac{dc_A}{dt}=-\frac{b}{a} k_{1f} \cdot (c_A)^x \cdot (c_B)^y + \frac{c}{a} \frac{d}{a} k_{1r} \cdot (c_C)^n \cdot (c_D)^m \\
[/tex]
x is the order with respect to A, y with respect to B, n with respect to C, m with respect to D; a, b, c, d are the stoichiometric coefficients on A to D respectively.
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Write me a rate expression. You know how to do it for single equaions, right?
Well I've never seen it before so I'm probably wrong but I'll give it a shot. :( If it were just 3A ::equil:: C rate equation is:
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot (c_A)^x + \frac{1}{3} k_{1r} \cdot (c_C)^n \\
[/tex]
So in the case of aA + bB ::equil:: cC + dD, I might suggest:
[tex]
\frac{dc_A}{dt}=-\frac{b}{a} k_{1f} \cdot (c_A)^x \cdot (c_B)^y + \frac{c}{a} \frac{d}{a} k_{1r} \cdot (c_C)^n \cdot (c_D)^m \\
[/tex]
x is the order with respect to A, y with respect to B, n with respect to C, m with respect to D; a, b, c, d are the stoichiometric coefficients on A to D respectively.
Ok, fine. So we can still solve this set of ODE's as an initial value problem, once you write similar ones for changes of cB, cC and cD.
No problems. Essentially you've learnt now how to formulate unsteady state and equilibrium concentrations for a set of any number of reactions (so long as you know individual rate expressions).
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Ok, fine. So we can still solve this set of ODE's as an initial value problem, once you write similar ones for changes of cB, cC and cD.
No problems. Essentially you've learnt now how to formulate unsteady state and equilibrium concentrations for a set of any number of reactions (so long as you know individual rate expressions).
Thanks :) If were to solve for a very high value of t, or even take the limit t=∞, we would land up with the equilibrium concentrations? Meaning that, given the rate constants and initial concentrations, we can work out the equilibrium constant from how far the reaction goes?
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Thanks :) If were to solve for a very high value of t, or even take the limit t=∞, we would land up with the equilibrium concentrations? Meaning that, given the rate constants and initial concentrations, we can work out the equilibrium constant from how far the reaction goes?
Yes. Right.
Assuming an ideal world. Often measured rate constants are bad and kf/kr may not evaluate to predicted thermodynamic Keq.
Often the system is stiff and solving it is one gigantic headache. If you only want equilibrium concentrations sometimes n simultaneous non-linear equations is all you need. Tad less work than solving ODE's. (but not always)
Real life calculations are a mess.
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[
Thanks :) If were to solve for a very high value of t, or even take the limit t=∞, we would land up with the equilibrium concentrations? Meaning that, given the rate constants and initial concentrations, we can work out the equilibrium constant from how far the reaction goes?
Yes. Right.
Assuming an ideal world. Often measured rate constants are bad and kf/kr may not evaluate to predicted thermodynamic Keq.
Often the system is stiff and solving it is one gigantic headache. If you only want equilibrium concentrations sometimes n simultaneous non-linear equations is all you need. Tad less work than solving ODE's. (but not always)
Real life calculations are a mess.
Can any rate expression always be written in the format we discussed? Or are there some cases where even the differential rate equation can't be described like that?
Edit: And could the problem arise that the rate constant itself changes with time? Or the reaction orders? In which case Keq might not be that which we predict by taking the lim(t :rarrow: ∞). Whereas the mass balance/charge balance/equilibria system will always work for a system of equilibria ...
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Can any rate expression always be written in the format we discussed?
Yes. AFAIK.
Or are there some cases where even the differential rate equation can't be described like that?
No. AFAIK.
Edit: And could the problem arise that the rate constant itself changes with time?
Doesn't matter.
Or the reaction orders?
Again, doesn't matter. All we need is rate = fn(cA,cB,..cN,t). Normally no t.
In which case Keq might not be that which we predict by taking the lim(t :rarrow: ∞). Whereas the mass balance/charge balance/equilibria system will always work for a system of equilibria ...
I don't know what you mean by that.
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Can any rate expression always be written in the format we discussed?
Yes. AFAIK.
Or are there some cases where even the differential rate equation can't be described like that?
No. AFAIK.
What about the rate-laws under Section C (Nonelementary rate laws) Reaction (2), Section D and Section E here: http://www.umich.edu/~elements/course/lectures/three/ in the Section 1.2 - Power Model? Those are some very strange rate laws indeed, I can't even begin to understand them ...
Wait, so it is possible for reaction orders/the rate constant to change depending on the concentration? So how do we incorporate that into our model? We need to write the rate constant as a function of the concentrations, then the reaction order for each reactant/product as a function of the concentrations?
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What about the rate-laws under Section C (Nonelementary rate laws) Reaction (2), Section D and Section E here: http://www.umich.edu/~elements/course/lectures/three/ in the Section 1.2 - Power Model?
Even those. Doesn't matter.
Those are some very strange rate laws indeed, I can't even begin to understand them ...
When the reaction isn't elementary the law gets complicated.
Wait, so it is possible for reaction orders/the rate constant to change depending on the concentration?
You mentioned the possibility. I merely said it doesn't matter to our modelling strategy. ODE's will work. :)
But yes, I suppose for non-elementary reactions orders etc. might change. Why, even an entirely different form might apply to different concentration regimes.
So how do we incorporate that into our model? We need to write the rate constant as a function of the concentrations, then the reaction order for each reactant/product as a function of the concentrations?
Again, all I need is rate = function(time, concentrations). What function is the domain of determining reaction kinetics and rate laws. A different ball game. Could be empirical. Microkinetics. etc.
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What about the rate-laws under Section C (Nonelementary rate laws) Reaction (2), Section D and Section E here: http://www.umich.edu/~elements/course/lectures/three/ in the Section 1.2 - Power Model?
Even those. Doesn't matter.
Those are some very strange rate laws indeed, I can't even begin to understand them ...
When the reaction isn't elementary the law gets complicated.
Wait, so it is possible for reaction orders/the rate constant to change depending on the concentration?
You mentioned the possibility. I merely said it doesn't matter to our modelling strategy. ODE's will work. :)
But yes, I suppose for non-elementary reactions orders etc. might change. Why, even an entirely different form might apply to different concentration regimes.
So how do we incorporate that into our model? We need to write the rate constant as a function of the concentrations, then the reaction order for each reactant/product as a function of the concentrations?
Again, all I need is rate = function(time, concentrations). What function is the domain of determining reaction kinetics and rate laws. A different ball game. Could be empirical. Microkinetics. etc.
You said "even those" for the non-elementary, but then says it gets complicated. But those rate laws are clearly not expressed as r=kAxBy. OTOH I am not entirely convinced they are referring to a single reaction at all, it could be multiple reactions combined into a single law there? So if we express each reaction separately, will it always take the form r=kAxBy?
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OTOH I am not entirely convinced they are referring to a single reaction at all, it could be multiple reactions combined into a single law there?
What do you understand by non-elementary?
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OTOH I am not entirely convinced they are referring to a single reaction at all, it could be multiple reactions combined into a single law there?
What do you understand by non-elementary?
Multiple reactions combined into a single law :p But let's say we don't have the mechanism for a reaction, we only have the overall thing (but it still goes to completion). Can we express the law in the r=kAxBy form or does it need to be something more complicated?
My instinct when I see something non-elementary like A :rarrow: B :rarrow: C is that surely, just like the ODEs we just wrote, we can write 3 more now, one for each species. But if the rate law for each section of the ODE is not of form kAxBy then that means more work is needed...
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But let's say we don't have the mechanism for a reaction, we only have the overall thing (but it still goes to completion). Can we express the law in the r=kAxBy form or does it need to be something more complicated?
No general answer. Rate law for a non-elementary reaction could be practically anything.
Didn't you post a link with all kinds of complicated rate laws. Were they all in kAxBy form?
My instinct when I see something non-elementary like A :rarrow: B :rarrow: C is that surely, just like the ODEs we just wrote, we can write 3 more now, one for each species. But if the rate law for each section of the ODE is not of form kAxBy then that means more work is needed...
I don't understand what you want to say.
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But let's say we don't have the mechanism for a reaction, we only have the overall thing (but it still goes to completion). Can we express the law in the r=kAxBy form or does it need to be something more complicated?
No general answer. Rate law for a non-elementary reaction could be practically anything.
Didn't you post a link with all kinds of complicated rate laws. Were they all in kAxBy form?
My instinct when I see something non-elementary like A :rarrow: B :rarrow: C is that surely, just like the ODEs we just wrote, we can write 3 more now, one for each species. But if the rate law for each section of the ODE is not of form kAxBy then that means more work is needed...
I don't understand what you want to say.
OK, but even for a non-elementary law, the rate law will always be a function of the rate constants and concentrations (and maybe some other concentrations), so to put it in our ODE we just have to put that function in the space where I just now put the -k1f·(cA)x section etc. (the function will be put negative if it involves conversion of the substance for which I'm writing the ODE to something else, and will be put positive if it involves production of the substance for which I'm writing the ODE), correct?
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OK, but even for a non-elementary law, the rate law will always be a function of the rate constants and concentrations (and maybe some other concentrations), so to put it in our ODE we just have to put that function in the space where I just now put the -k1f·(cA)x section etc. (the function will be put negative if it involves conversion of the substance for which I'm writing the ODE to something else, and will be put positive if it involves production of the substance for which I'm writing the ODE), correct?
Yes. Correct.
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Does it occur, in your experience, that we have to use different rate equations in the middle of calculating concentration/time dependence for a single reaction? (i.e. either the reaction orders or the rate "constants" change at some point)
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Why should the kinetics of a reaction depend on the process of calculation?
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Why should the kinetics of a reaction depend on the process of calculation?
Why does this matter? I'm asking whether we observe a single rate law to work for all reactions until the end, or whether in your experience you actually have observed that reactions do not follow the same exact rate law until the end (the end being when all the concentrations are equilibrium concentrations - might be 0 for a reactant in an equilibrium where backward rate constant is 0, i.e. one which goes to completion).
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What I meant was: the rate law is an artificial construct based upon a specified mechanism. If experimental reaction rate data deviate from the prediction of the rate law at any point, it's probably not because of any bizarre behavior, but because the specified mechanism isn't correct.
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What I meant was: the rate law is an artificial construct based upon a specified mechanism. If experimental reaction rate data deviate from the prediction of the rate law at any point, it's probably not because of any bizarre behavior, but because the specified mechanism isn't correct.
OK, so in your experience the deviation would never occur because the rate constants or reaction orders are themselves functions of concentration, but rather because the rate law was inaccurately determined. The question comes down to this:
If we can break up the mechanism into a (correct) set of single steps, each of which will therefore have an elementary rate law, can this rate law always be expressed in the form r=k[A]x[ B]y where k, x and y are constants, with complete accuracy as far as your experience suggests?
Directed to anyone who can answer.
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If we can break up the mechanism into a (correct) set of single steps, each of which will therefore have an elementary rate law, can this rate law always be expressed in the form r=k[A]x[ B]y where k, x and y are constants, with complete accuracy as far as your experience suggests?
With complete accuracy? Never.
Rate laws in this commonly used form (r=k[A]x[ B]y) for elementary steps are yet an approximation. A model.
This isn't like Newton's law that nature is forced to obey it. More like Ohm's Law. Mostly true and a good approximation.
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@BD
What do you envision the reaction orders would vary as a function of?
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With complete accuracy? Never.
Rate laws in this commonly used form (r=k[A]x[ B]y) for elementary steps are yet an approximation. A model.
This isn't like Newton's law that nature is forced to obey it. More like Ohm's Law. Mostly true and a good approximation.
Huh ... ok, let me ask a different question then: do we ever observe the rate law to dramatically change at a certain point in the experiment (i.e. discrete change)? Or is it always just a continuous change which we can model in terms of species concentration?
@BD
What do you envision the reaction orders would vary as a function of?
Concentration of any and all species. OR something discrete, if you have ever observed that this occurs (e.g. at some point in the concentration of the reactants or products, the rate law suddenly changes from r=k[A]2[ B]3 to r=k[A]3[ B]2).
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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?
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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
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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.
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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)?
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(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.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fb%2Fb0%2FMichaelis-%3Cbr+%2F%3EMenten_saturation_curve_of_an_enzyme_reaction_LARGE.svg&hash=fd76d092bbf697fde91dfdee77379297052c4e05)
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I'm afraid you are drifting into the realm of speculation / fantasy.
As usual.
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(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.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fb%2Fb0%2FMichaelis-%3Cbr+%2F%3EMenten_saturation_curve_of_an_enzyme_reaction_LARGE.svg&hash=fd76d092bbf697fde91dfdee77379297052c4e05)
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)?
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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 ...
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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.
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Ouch! :-[ I only meant to give an example.
Choose better examples?
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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.
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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.
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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.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fmath%2F5%2Fa%2Ff%2F5af85b040c6390c90bbe99cb509887a9.png&hash=b54d913acd8c2f9267fa8198030da3bb64f13469)
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.
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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.
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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.
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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.
Could be. I don't know. All I can tell you is you've frustrated at least two of us here.
Maybe I am wrong about my advice. I might have been too harsh. Sorry.
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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.
Could be. I don't know. All I can tell you is you've frustrated at least two of us here.
Maybe I am wrong about my advice. I might have been too harsh. Sorry.
Yeah, don't worry, I have got the main picture. I think the advice is right. I will come back later.
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Meh, explaining difficult concepts - and asking pointed, clear questions - is not always easy over an internet forum. I have no issues with Big Daddy's lines of conversation. However BD I do think you would be better off focusing on perfecting your understanding of the basics rather than trying to pinpoint all the places and scenarios where widely used models fail. Don't get me wrong - there is value to this at times. However if you push too hard all you'll end up doing is confusing yourself. It does become counterproductive.
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To be honest, I feel I have learnt quite a lot in my time here, from the three of you (and others). I think I have gained a lot from the discussions which seem to you like running round in circles, and though I may take a long while to understand the concept, when I do I usually won't ask again.
Nevertheless clearly you are frustrated, and I can see what you mean, that asking good questions is an art. As a (reasonably immature :P) teenager that is my biggest problem. I will try to improve it.
If you can explain what problem exactly there is in my questions that frustrates you (as Corribus has done), it would help me a lot. I know you feel that discussion is running around in circles but it has been a long time since I felt that on this forum.
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in the case of aA + bB ::equil:: cC + dD, I might suggest:
[tex]
\frac{dc_A}{dt}=-\frac{b}{a} k_{1f} \cdot (c_A)^x \cdot (c_B)^y + \frac{c}{a} \frac{d}{a} k_{1r} \cdot (c_C)^n \cdot (c_D)^m \\
[/tex]
x is the order with respect to A, y with respect to B, n with respect to C, m with respect to D; a, b, c, d are the stoichiometric coefficients on A to D respectively.
I'm thinking now this doesn't work. The stoichiometric coefficients should be in the brackets:
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot (c_A)^x \cdot (\frac{b}{a} \cdot c_B)^y + k_{1r} \cdot (\frac{c}{a} \cdot c_C)^n \cdot (\frac{d}{a} \cdot c_D)^m \\
[/tex]
No? I just get the feeling that if an order goes to 0 the coefficient shouldn't be having an effect on the rate either.
I am not sure yet what I did wrong to frustrate you, but still your help would be much appreciated.
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I'm thinking now this doesn't work. The stoichiometric coefficients should be in the brackets:
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot (c_A)^x \cdot (\frac{b}{a} \cdot c_B)^y + k_{1r} \cdot (\frac{c}{a} \cdot c_C)^n \cdot (\frac{d}{a} \cdot c_D)^m \\
[/tex]
No? I just get the feeling that if an order goes to 0 the coefficient shouldn't be having an effect on the rate either.
I am not sure yet what I did wrong to frustrate you, but still your help would be much appreciated.
I'll try to help but I really don't understand what you are saying here.
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I'll try to help but I really don't understand what you are saying here.
If you look at the two ODEs I've written:
Version 1:
[tex]
\frac{dc_A}{dt}=-\frac{b}{a} k_{1f} \cdot (c_A)^x \cdot (c_B)^y + \frac{c}{a} \frac{d}{a} k_{1r} \cdot (c_C)^n \cdot (c_D)^m \\
[/tex]
Version 2:
[tex]
\frac{dc_A}{dt}=-k_{1f} \cdot (c_A)^x \cdot (\frac{b}{a} \cdot c_B)^y + k_{1r} \cdot (\frac{c}{a} \cdot c_C)^n \cdot (\frac{d}{a} \cdot c_D)^m \\
[/tex]
Which is correct, for the single equilibrium aA+bB ::equil:: cC+dD (where the order with respect to A is x, to B is y, to C is n, to D is m)? Although the first one originally seemed OK I now am leaning towards the second because I would guess that the stoichiometric coefficients should also end up cancelling down to 1 (as they will if they are inside the brackets to the power of order, but not if they are outside) if the species is zero-order. Not sure though.
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Which is correct, for the single equilibrium aA+bB ::equil:: cC+dD (where the order with respect to A is x, to B is y, to C is n, to D is m)? Although the first one originally seemed OK I now am leaning towards the second because I would guess that the stoichiometric coefficients should also end up cancelling down to 1 (as they will if they are inside the brackets to the power of order, but not if they are outside) if the species is zero-order. Not sure though.
Don't think it matters. They only differ in their constants. Use either.
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Don't think it matters. They only differ in their constants.
That would mean I would be writing different equations for the rate of change of [A] with time. Imagine for instance that the rate is not dependent on [B] so that y=0 (from my equations). In the first version of the equation, the stoichiometric ratio (b/a) continues to affect the rate. In the second version of the equation, the stoichiometric ratio with respect to B also cancels out. So d[A]/dt is just dependent on [A], no more on [B].
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That would mean I would be writing different equations for the rate of change of [A] with time. Imagine for instance that the rate is not dependent on [B] so that y=0 (from my equations). In the first version of the equation, the stoichiometric ratio (b/a) continues to affect the rate. In the second version of the equation, the stoichiometric ratio with respect to B also cancels out. So d[A]/dt is just dependent on [A], no more on [B].
Write down both equations for y=0. k1f in first form is not equal to k1f in the second form. They are related by a constant.
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k1f in first form is not equal to k1f in the second form. They are related by a constant.
Ah, got it now. That constant that relates them is the product of the stoichiometric coefficients that are not cancelled.
But if I look up values of the rate constant, which of the equations do they normally refer to? My second equation, I'm guessing/hoping ... (My available values of rate constants don't tell me whether they are corrected for stoichiometry or not.)
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Ah, got it now. That constant that relates them is the product of the stoichiometric coefficients that are not cancelled.
Right
But if I look up values of the rate constant, which of the equations do they normally refer to?
And where will you be looking it up? Show me a literature example that uses your contrived variable order rate equation.
e.g. If I meet a unicorn will it be right handed or left handed? :)
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And where will you be looking it up? Show me a literature example that uses your contrived variable order rate equation.
Don't know what rate equation the constant below uses. All I want is to be able to formulate a rate equation in terms of the constants available, which I imagine relate to the reaction itself with no stoichiometry.
Sucrose (aq) :rarrow: glucose (aq) + fructose (aq) [acidic conditions, 298.15 K]
k=6.0·10-5 s-1
Better example, to look at the stoichiometry, might be:
2 N2O5 (g) :rarrow: 4 NO2 (g) + O2 (g) [298.15 K]
k=3.38·10-5 s-1
Both values come from Atkins' "Physical Chemistry", 8th edition, page 1020 (Table 22.1: Kinetic data for first-order reactions).
Are you saying that the definition of the rate constant (i.e. whether corrected for stoichiometry or not) differs from source to source? Doesn't that lead to massive confusion, when the rate constant is not accompanied by something clarifying whether it is corrected or not?
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These are simple 1st order reactions. Where's your confusion?
Write down (equations, not words) which different ways you think these equations might be misinterpreted?
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Are you saying that the definition of the rate constant (i.e. whether corrected for stoichiometry or not) differs from source to source? Doesn't that lead to massive confusion, when the rate constant is not accompanied by something clarifying whether it is corrected or not?
There's no "massive" confusion, but yes things can get tricky at times. Which is why it is considered good practice to write down rate expressions explicitly when potential for confusion exists.
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Write down (equations, not words) which different ways you think these equations might be misinterpreted?
OK, take example 2:
2 N2O5 (g) :rarrow: 4 NO2 (g) + O2 (g) [298.15 K]
k=3.38·10-5 s-1
[tex]
\frac{dc_(N2O5)}{dt}=-k_{1} \cdot (c_(N2O5)) \\
[/tex]
Or:
[tex]
\frac{dc_(N2O5)}{dt}=-k_{1} \cdot 2 \cdot (c_(N2O5)) \\
[/tex]
In fact, in neither of my equations did I account for the second possibility...
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Write down (equations, not words) which different ways you think these equations might be misinterpreted?
OK, take example 2:
2 N2O5 (g) :rarrow: 4 NO2 (g) + O2 (g) [298.15 K]
k=3.38·10-5 s-1
[tex]
\frac{dc_(N2O5)}{dt}=-k_{1} \cdot (c_(N2O5)) \\
[/tex]
Or:
[tex]
\frac{dc_(N2O5)}{dt}=-k_{1} \cdot 2 \cdot (c_(N2O5)) \\
[/tex]
In fact, in neither of my equations did I account for the second possibility...
Typically the second (though I have seen people use it the other way too). That's the reason explicit equations work better.
This is the only major way of confusing things that one must be careful about. Your earlier examples etc. were not very relevant.
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This is the only major way of confusing things that one must be careful about. Your earlier examples etc. were not very relevant.
I think I need one more case, one which deals with at least reactant having higher reaction order, to understand.
Let's say we have the reaction 2A + B + 3C :rarrow: D, second-order in A, first-order in B, zero-order in C. (This might not exist but I cannot think of another way of finding out the principle.) With the rate constant k as normally available, are these the ODEs we want to write:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_B}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_C}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
And there is nothing in there for C, because the rate is not dependent on C (zero-order). Is this right? If not, can you please write the 3 ODEs we want to solve simultaneously?
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The best way for me to check this is to provide you with a concrete example:
Let's say we have the reaction 2A + B + 3C :rarrow: D, second-order in A, first-order in B, zero-order in C. (This might not exist but I cannot think of another way of finding out the principle.) With the rate constant k as normally available, is this is the ODE we want to write for [A]:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
And there is nothing in there for C, because the rate is not dependent on C (zero-order). Is this right?
I'd use this:
[tex]
\frac{dc_A}{dt}=-2 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
But I cannot claim with certainty that this is indeed how a majority of readers would interpret it.
Again, it's only a matter of constants.
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I think I need one more case, one which deals with at least reactant having higher reaction order, to understand.
Let's say we have the reaction 2A + B + 3C :rarrow: D, second-order in A, first-order in B, zero-order in C. (This might not exist but I cannot think of another way of finding out the principle.) With the rate constant k as normally available, are these the ODEs we want to write:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_B}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_C}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
[/tex]
Is this right?
No. Cannot be. That'd mean rates of change of A, B, C are all same. Obviously, they are not. (stoichiometry). Your three ODE's are mutually inconsistent. A mass balance would never work out.
The forgivable confusion is merely in writing one rate (the first one). The other species have to follow from it. No freedom there.
If not, can you please write the 3 ODEs we want to solve simultaneously?
Try again.
And there is nothing in there for C, because the rate is not dependent on C (zero-order).
Yes. That's right.
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[tex]
\frac{dc_A}{dt}=-2 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
Here's the confusion: you've got 2, the stoichiometric coefficient of cA, but it's not together with cA itself (inside the brackets). By the same logic we could write a rate equation:
[tex]
\frac{dc_A}{dt}=-2 k_{1} \cdot ( c_A)^2 \cdot (c_B)^1 \cdot 3 \cdot (c_C)^0 \\
[/tex]
Now explicitly specifying all coefficients and rate orders. But even as cC disappears from the rate expression our overall ODE is still different:
[tex]
\frac{dc_A}{dt}=-6 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
I'm very confused now ... should I just put it down to the need to know what rate law the constant is coming from, and leave it there?
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Try again.
There has to be some logic behind this. From what I can understand, either they should be written:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (c_A)^2 \cdot (\frac{1}{2} \cdot c_B) \\
\frac{dc_B}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
\frac{dc_C}{dt}=-k_{1} \cdot (\frac{2}{3} \cdot c_A)^2 \cdot (\frac{1}{3} \cdot c_B) \\
[/tex]
Or:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (c_A)^2 \cdot (2 \cdot c_B) \\
\frac{dc_B}{dt}=-k_{1} \cdot (\frac{1}{2} \cdot c_A)^2 \cdot (c_B) \\
\frac{dc_C}{dt}=-k_{1} \cdot (\frac{3}{2} \cdot c_A)^2 \cdot (3 \cdot c_B) \\
[/tex]
Please tell me it is one of the two above (and which one it is)...
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Try again.
There has to be some logic behind this. From what I can understand, either they should be written:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (c_A)^2 \cdot (\frac{1}{2} \cdot c_B) \\
\frac{dc_B}{dt}=-k_{1} \cdot (2 \cdot c_A)^2 \cdot (c_B) \\
\frac{dc_C}{dt}=-k_{1} \cdot (\frac{2}{3} \cdot c_A)^2 \cdot (\frac{1}{3} \cdot c_B) \\
[/tex]
Or:
[tex]
\frac{dc_A}{dt}=-k_{1} \cdot (c_A)^2 \cdot (2 \cdot c_B) \\
\frac{dc_B}{dt}=-k_{1} \cdot (\frac{1}{2} \cdot c_A)^2 \cdot (c_B) \\
\frac{dc_C}{dt}=-k_{1} \cdot (\frac{3}{2} \cdot c_A)^2 \cdot (3 \cdot c_B) \\
[/tex]
Please tell me it is one of the two above (and which one it is)...
Neither.
Ok, answer this: When 1 mole of A reacts how many moles of B and C react?
Ergo, what's the mathematical relationship between
[tex] \frac{dc_A}{dt} [/tex] and [tex] \frac{dc_B}{dt} [/tex] .
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what's the mathematical relationship between
[tex] \frac{dc_A}{dt} [/tex] and [tex] \frac{dc_B}{dt} [/tex] .
[tex] \frac{dc_A}{dt} = \frac{2}{3} \cdot \frac{dc_C}{dt} = 2 \cdot \frac{dc_B}{dt}[/tex]
would be my attempt.
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what's the mathematical relationship between
[tex] \frac{dc_A}{dt} [/tex] and [tex] \frac{dc_B}{dt} [/tex] .
[tex] \frac{dc_A}{dt} = \frac{2}{3} \cdot \frac{dc_C}{dt} = 2 \cdot \frac{dc_B}{dt}[/tex]
would be my attempt.
Good.
Now do either of your latest attempts at ODE's satisfy these ratios. Calculate your ratios. Should be easy. Arithmetic.
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Now do either of your latest attempts at ODE's satisfy these ratios. Calculate your ratios. Should be easy. Arithmetic.
No, they don't.
I was thinking:
[tex]
\frac{dc_A}{dt}=-6 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
Where 6 is the product of the stoichiometric coefficients of all the reactants. But then we would have to write the same equation for each species, which obviously would be wrong.
Can you show the answer? (For all 3 ODEs) I will seek the internal logic from the answer. (Then, if I get it, pose myself an abstract problem and solve it :P )
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My answer is:
[tex]
\frac{dc_A}{dt}= - 2 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_B}{dt}= - k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_C}{dt}= - 3 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
I could be wrong.
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My answer is:
[tex]
\frac{dc_A}{dt}= - 2 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_B}{dt}= - k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
[tex]
\frac{dc_C}{dt}= - 3 k_{1} \cdot ( c_A)^2 \cdot (c_B) \\
[/tex]
I could be wrong.
As promised, I will attempt an abstraction. aA+bB+cC :rarrow: dD+eE+fF (I may not write all 6 ODEs), order is x wrt A, y wrt B, z wrt C.
[tex]
\frac{dc_A}{dt}= - a k_{1} \cdot (c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
[tex]
\frac{dc_B}{dt}= - b k_{1} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
[tex]
\frac{dc_C}{dt}= - c k_{1} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
[tex]
\frac{dc_D}{dt}= d k_{1} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
[tex]
\frac{dc_E}{dt}= e k_{1} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
The important thing to note is:
[tex]
\frac{dc_{Species}}{dt}= v_{Species} \cdot k_{1} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
Where vSpecies is the stoichiometric coefficient on that species in the reaction, multiplied by -1 if the species is a reactant and +1 if the species is a product.
The equations apply regardless of the values of x,y,z, including when they equal 0?
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This makes a lot of sense actually.
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The important thing to note is:
[tex]
\frac{dc_{Species}}{dt}= v_{Species} \cdot k_{1} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z \\
[/tex]
Where vSpecies is the stoichiometric coefficient on that species in the reaction, multiplied by -1 if the species is a reactant and +1 if the species is a product.
The equations apply regardless of the values of x,y,z, including when they equal 0?
Yep. Now you are making sense. :)
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Yep.
And if it were an equilibrium then:
[tex]
\frac{dc_{Species}}{dt}= v_{Species} \cdot k_{1f} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z - v_{Species} \cdot k_{1r} \cdot (c_D)^n \cdot (c_E)^m \cdot (c_F)^p \\
[/tex]
vSpecies continuing to be the stoichiometric coefficient on the species you're writing this ODE for, and n,m,p being orders of the reverse reaction wrt D, E, F respectively. Same equation whatever x,y,z,n,m,p are, same equation whatever the stoichiometric coefficients of other species are. Correct?
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And if the species is involved in multiple rate equations, you would just add on more sections like those above to the ODE for that species, with vSpecies for each section being the stoichiometric coefficient of the species in that reaction.
So if in addition to the equilibrium above I had 3A :rarrow: G :rarrow: H happening at the same time, which is now first-order in A, I'd have as my total ODE for A:
[tex]
\frac{dc_{Species}}{dt}= v_{Species} \cdot k_{1f} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z - v_{Species} \cdot k_{1r} \cdot (c_D)^n \cdot (c_E)^m \cdot (c_F)^p - 3 \cdot k_{2} \cdot ( c_A) \\
[/tex]
(vSpecies here is the stoichiometric coefficient on A in the equilibrium reaction discussed above.)
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Yes. Mostly right.
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Yes. Mostly right.
Good. I will take "mostly right" as "all right so far". ;D
To check: the bit we add to the ODE for cA, for the reaction 3A :rarrow: G :rarrow: H, still follows an elementary rate law (given that A's conversion to G is elementary, so we can really break this up into 3A :rarrow: G, G :rarrow: H), and if it's first order we can write this contribution
[tex]-3 \cdot k_{2} \cdot (c_A)[/tex]
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Yes. Mostly right.
Good. I will take "mostly right" as "all right so far". ;D
So long as you add a k2f...
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So long as you add a k2f...
Ahh! Sorry, silly mistake. Edited in :P
So even if 3A :rarrow: G :rarrow: H overall must be non-elementary, the process 3A :rarrow: G is elementary and thus we can always write its contribution (to the ODE for A) in the form:
[tex]-k_{2} \cdot 3 \cdot (c_A)^t[/tex]
Where t is the order of this new reaction 3A :rarrow: G wrt to A.
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Yes.
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Yes.
Thanks very much for the help.
96 posts! You must be exhausted :P Sorry if I have frustrated you too much over the course of the last few dozen posts.
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96 posts! You must be exhausted :P Sorry if I have frustrated you too much over the course of the last few dozen posts.
I wish there was a prize for the longest attempt at the Socratic Method. ;D
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I wish there was a prize for the longest attempt at the Socratic Method. ;D
Sorry! :D I learn best by example.
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Another query:
If the volume of the system is changing, how do we incorporate that? Something to do with molar volumes, and their dependence on pressure/temperature, for each species?
Edit: I'm not sure if this is what Borek would call a secondary or irrelevant field, but I think I have learnt the main matter so unless this is vastly too complicated for me I would like to give it a try.
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Another query:
If the volume of the system is changing, how do we incorporate that? Something to do with molar volumes, and their dependence on pressure/temperature, for each species?
For variable volume instead of using
[tex]
\frac{dc_A}{dt}= R_A \\
[/tex]
Use the more general form:
[tex]
\frac{d \left ( c_A \cdot V \right ) }{dt}= R_A \cdot V \\
[/tex]
V is reactor volume, R is rate of production for that particular species.
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but I think I have learnt the main matter
Brave. Very brave. :)
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[tex]
\frac{d \left ( c_A \cdot V \right ) }{dt}= R_A \cdot V \\
[/tex]
V is reactor volume, R is rate of production for that particular species.
Looks like we're now working in number of moles. That seems reasonable.
Since the volume is changing, does the volume have to be expressed as a function of the different concentrations or number of moles of the different species? Or is it simply its own term, not expressed in terms of the species? After all, how would you predict how much the volume changes for a certain change in number of moles, without having expressed the volume as a function of the number of moles ...
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Since the volume is changing, does the volume have to be expressed as a function of the different concentrations or number of moles of the different species? Or is it simply its own term, not expressed in terms of the species? After all, how would you predict how much the volume changes for a certain change in number of moles, without having expressed the volume as a function of the number of moles ...
You tell me. Why is your volume changing? Each case is different. Express it as a function of whatever is making it change.
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Why is your volume changing?
Because the numbers of moles of each gas are changing (for example), and a certain number of moles of one gas takes up more volume than the same number of moles of another gas. How should I express the volume then? Perhaps as the sum of the number of moles of each species multiplied by the molar volume of that species, this seems completely general to me (and then express the molar volumes in terms of pressure and temperature, if they are changing too).
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Why is your volume changing?
Because the numbers of moles of each gas are changing (for example), and a certain number of moles of one gas takes up more volume than the same number of moles of another gas. How should I express the volume then? Perhaps as the sum of the number of moles of each species multiplied by the molar volume of that species, this seems completely general to me (and then express the molar volumes in terms of pressure and temperature, if they are changing too).
Sounds good. Do it that way. Use your gas laws.
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Do it that way. Use your gas laws.
OK, so V can always be defined as the number of moles of each species, multiplied by Vm, summed. So we can write it as such, assuming the change in volume is referring to the effects of the reaction itself (no. of moles changing) rather than, say, the container volume being externally changed.
Then the question is what is Vm: if it's constant, then I just write it in as such; otherwise, using the gas laws, it will be a function of pressure, temperature and nothing else (and then gas constants, depending on which gas law is used). If pressure is non-constant (as a result of changing number of moles of gas), anything I can do to model it?
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OK, so V can always be defined as the number of moles of each species, multiplied by Vm, summed.
For gases, yes.
So we can write it as such, assuming the change in volume is referring to the effects of the reaction itself (no. of moles changing) rather than, say, the container volume being externally changed.
Then the question is what is Vm: if it's constant, then I just write it in as such; otherwise, using the gas laws, it will be a function of pressure, temperature and nothing else (and then gas constants, depending on which gas law is used). If pressure is non-constant (as a result of changing number of moles of gas), anything I can do to model it?
Please define your system carefully first. Is it isothermal? Isobaric? Isochoric? etc.
Think physically. Is it a fixed volume container? If not, what?
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Please define your system carefully first. Is it isothermal? Isobaric? Isochoric? etc.
For the time being, we could model it as isothermal, with both pressure and volume non-constant.
Think physically. Is it a fixed volume container? If not, what?
Well I imagine for example adding acid to a solution of a soluble carbonate (e.g. sodium). CO2 gas gets released, increasing both the volume of the system (because CO2 gas takes up more volume per mole than the carbonate anion in solution) and the pressure. The temperature would also change but I am happy for now to consider the system as isothermal.
If it is little additional difficulty to consider it "really", i.e. temperature, pressure and volume can all change, then I would want to learn to do that. Otherwise - isothermal.
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Please define your system carefully first. Is it isothermal? Isobaric? Isochoric? etc.
For the time being, we could model it as isothermal, with both pressure and volume non-constant.
So how'd you construct such a system? Practically. Please describe it.
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So how'd you construct such a system? Practically. Please describe it.
I don't know. In the case of adding acid to carbonate solution (i.e. pressure, volume and temperature are all non-constant) there doesn't seem to be anything to describe, it's clear why all 3 will change. I don't know how you might keep it isothermal though.
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So how'd you construct such a system? Practically. Please describe it.
I don't know. In the case of adding acid to carbonate solution (i.e. pressure, volume and temperature are all non-constant) there doesn't seem to be anything to describe, it's clear why all 3 will change. I don't know how you might keep it isothermal though.
If you take a strong closed container why would volume change? Doesn't seem "clear" to me.
You assume too much.
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If you take a strong closed container why would volume change?
Well, because not all the volume is being taken up by reacting species at the beginning (it's a solution, to which acid is added - a gas is then released, which thereby adds to the volume).
In any case that probably isn't very relevant to think about. OK, how do we deal with a constant volume ("isochoric"?) system with changing pressure and temperature?
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If you take a strong closed container why would volume change?
Well, because not all the volume is being taken up by reacting species at the beginning (it's a solution, to which acid is added - a gas is then released, which thereby adds to the volume).
In any case that probably isn't very relevant to think about. OK, how do we deal with a constant volume ("isochoric"?) system with changing pressure and temperature?
Same as before. If V doesn't change, where's your problem?
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Same as before. If V doesn't change, where's your problem?
Pressure and temperature are both changing. Beyond affecting the rate constants, this shouldn't make a difference to the ODEs for concentration, right? But let's say that I want to be able to calculate the total pressure in terms of time, or the temperature at a certain point in time - how do I do that? (I imagine pressure will rely on temperature and number of moles, but I'm unclear how to model it for inclusion into our ODE system; temperature will probably rely on Gibbs' energy changes and number of moles, but I'm still unclear how to include this into the ODE system)
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Same as before. If V doesn't change, where's your problem?
Pressure and temperature are both changing. Beyond affecting the rate constants, this shouldn't make a difference to the ODEs for concentration, right? But let's say that I want to be able to calculate the total pressure in terms of time, or the temperature at a certain point in time - how do I do that? (I imagine pressure will rely on temperature and number of moles, but I'm unclear how to model it for inclusion into our ODE system; temperature will probably rely on Gibbs' energy changes and number of moles, but I'm still unclear how to include this into the ODE system)
Do an energy balance. Equate heat of reaction to specific heat capacity. Assuming adiabatic conditions.
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Do an energy balance. Equate heat of reaction to specific heat capacity. Assuming adiabatic conditions.
OK, this definitely seems beyond my current thermodynamics knowledge. I'll have to read up a bit more on thermo before we can tackle this.
Let's say the temperature is constant as well as volume, or maybe the temperature is varied at a controlled rate externally. How do we deal with just calculating the (total) pressure (which should be a function of temperature as well, mind) at any point in time?
What I'm trying to do here, across the last few posts, is understand how to include volume into the calculations, as well as pressure (which would probably affect the volume, and vice versa) as well as temperature, by understanding how to write the volume function in terms of number of moles, pressure and temperature (already been explained), then how to write the pressure function in terms of number of moles and temperature. Then I would consider how to think about temperature being varied.
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Let's say the temperature is constant as well as volume, or maybe the temperature is varied at a controlled rate externally. How do we deal with just calculating the (total) pressure (which should be a function of temperature as well, mind) at any point in time?
So long as V is constant previous ODE applies.
[tex]
\frac{dc_A}{dt}= R_A \\
n = c \cdot V \\
P=\frac{nRT}{V} \\
[/tex]
If T varies as an input you already know T=fn(t).
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[tex]
\frac{dc_A}{dt}= R_A \\
n = c \cdot V \\
P=\frac{nRT}{V} \\
[/tex]
Ah of course ... if V is constant, then we just find the number of moles of each gas, sum them and this is n, which we can then use in a gas law. As you wrote below we can write T as a function of time if necessary.
I really want to find out what to do if pressure and volume are both changing as a result of number of moles of gas changing. Temperature is either constant, or can be written as a function of time. In principle we need to write Vm as a function of pressure (and temperature), but then pressure cannot be a function of V (because Vm is itself being used to find V) ...
Every question comes down to this: is there any way to express P without V (when P is changing due to changing number of moles of gas)?
If T varies as an input you already know T=fn(t).
Got that one, thanks. Makes sense.
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I really want to find out what to do if pressure and volume are both changing as a result of number of moles of gas changing. Temperature is either constant, or can be written as a function of time.
That problem seems under-determined. AFAIK it cannot be solved; rather it does not make sense as a problem itself.
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Think of it like this:
Say you have a circular stack of bricks of known density. If you keep adding bricks I can tell you how tall the stack will get if you specify the diameter of the stack. Alternatively if you told me the height I could tell you how wide the stack is when you add a certain amount of bricks to it.
But if you say, 'I added 2000 bricks now tell me both the height and diameter of my brick stack!", then I think that's an unanswerable question.
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That problem seems under-determined. AFAIK it cannot be solved; rather it does not make sense as a problem itself.
OK. So now:
I've learnt how to formulate the ODEs when volume is constant. Total pressure at any time is found from the gas laws so long as volume is constant. I suppose that if we want partial pressures at any time, we just need to substitute in the partial pressure expression from the gas law (e.g. if we're using ideal gas law, then substitute CA=PA/RT if we want to be able to get partial pressure for species A at time t). I've learnt how to calculate V if V is changing, and include it into the ODEs (as a function of P and T, but obviously then neither P nor T can be a function of V).
Is there any way of defining or calculating total pressure without V? Perhaps as the sum of the partial pressures of all species?
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Is there any way of defining or calculating total pressure without V? Perhaps as the sum of the partial pressures of all species?
And how would you calculate partial pressures?
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And how would you calculate partial pressures?
By replacing all concentration terms with Pspecies/(RT) as I suggested (assuming you know the initial partial pressures)?
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And how would you calculate partial pressures?
By replacing all concentration terms with Pspecies/(RT) as I suggested (assuming you know the initial partial pressures)?
Don't know.
My advice is try a specific numerical example.
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My advice is try a specific numerical example.
OK, I will try. Where can I find a simultaneous ODE solver? And should I assume that, if I get a numerical result, it's correct? (I'm not in a position to experimentally verify these things right now, nor would I know how to experimentally find the concentration at a certain time in the middle of the reaction. So we need a known answer.)
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OK, I will try. Where can I find a simultaneous ODE solver?
Matlab. Octave. Mathematica.
Many many others.
And should I assume that, if I get a numerical result, it's correct?
Sure.
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Matlab. Octave. Mathematica.
Many many others.
OK, I have Mathematica available. But to reach a level where I can write in simultaneous ODEs will probably take a while. Let me get back to you once I've learnt the language to a high enough level to try this problem, and then seen if we get numerical answers.
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Matlab. Octave. Mathematica.
Many many others.
OK, I have Mathematica available. But to reach a level where I can write in simultaneous ODEs will probably take a while. Let me get back to you once I've learnt the language to a high enough level to try this problem, and then seen if we get numerical answers.
Fine by me.
Discussing a problem is always easier than actually attempting to solve it.
PS. You don't even necessarily need to use Mathematica. Try solving them analytically. If you are brave enough to attempt a solution to Schrodinger's Eq. you are definitely ready for this stuff.
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PS. You don't even necessarily need to use Mathematica. Try solving them analytically.
You mean a closed-form algebraic solution? Is that even possible? Quote from my physical chemistry textbook: "Many of the differential equations that describe physical phenomena are so complicated that their solutions cannot be cast as functions." And here we're not looking at one differential equation but a system. I couldn't even solve it numerically much less algebraically!
I will be happy to simply see if writing P=sum of Pgas/(RT) for all gases (replacing all concentration terms for gases with Pgas/(RT)) gets a numerical answer out of Mathematica. May take a while, need to learn the language (I'm told it's very useful anyway in undergrad years so now is a good opportunity).
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You mean a closed-form algebraic solution? Is that even possible?
Yes. Sometimes.
Quote from my physical chemistry textbook: "Many of the differential equations that describe physical phenomena are so complicated that their solutions cannot be cast as functions."
Important word is "many". Not "all".
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OK. And what if there is an inert gas in the container, which does not react - is there another way to write an ODE for it (given that its number of moles is constant, but partial pressure changing as the system has other gases whose number of moles are changing)?
Or do we simply exclude this gas, and say that the V refers to the volume taken up by the species which are reacting (so we have equations for them and can calculate the partial pressures of them at any time), and P to the total pressure exerted by these gases rather than abjectly "total pressure" in container? (Keeping in mind that this relies on molar volume for a given species being a function of partial pressure of that species rather than total pressure, which I'm not so sure about!) After which we could still get the right values for partial pressure?
Or just concede that there is no way of computing with a system like this, so we need to remove the inert gas and try again for the system to work? (i.e. the reacting species need to be isolated for these calculations to hold)
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OK. And what if there is an inert gas in the container, which does not react - is there another way to write an ODE for it (given that its number of moles is constant, but partial pressure changing as the system has other gases whose number of moles are changing)?
Write me a specific problem. This is too vague to discuss in generalities.
Or just concede that there is no way of computing with a system like this, so we need to remove the inert gas and try again for the system to work? (i.e. the reacting species need to be isolated for these calculations to hold)
Isolated how?
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Write me a specific problem. This is too vague to discuss in generalities.
Here goes.
The reaction 2 N2O5 (g) ::equil:: 4 NO2 (g) + O2 (g) is initiated in a reactor which also contains some unreactive He gas (the initial concentration/partial pressure of this gas is known, and by definition of being unreactive we know that the number of moles of He do not change). Can we calculate the concentrations or partial pressures of the 4 constituents of the system (N2O5, NO2, O2 and He) at any time t? If yes, what ODEs do we need to write to do so? (We will need one more ODE than the 3 based on the 3 reacting species, an ODE for He)
At best I would like to consider the isothermal case directly, but otherwise I'd be happy to look at the constant volume, constant temperature case and see if we can come up with any ideas from there.
Isolated how?
As in, if there are gases not involved in reactions which are still present in the same container, we cannot account for them or calculate anything for the system (as they would affect partial pressure/concentration calculations). Unless we can write an ODE for them too? Or there may be some other way?
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We will need one more ODE than the 3 based on the 3 reacting species, an ODE for He
What's the initial conc. of He? What's the final conc.?
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What's the initial conc. of He?
Why not keep it algebraic, like all the ODEs we've worked with so far?
What's the final conc.?
Do we need to know this to do the calculation? It isn't possible even to calculate the concentrations of the 3 reacting species, unless we know He's initial and final concentration? That's why, as I said, I was hoping we could form an ODE for He to solve simultaneously along with our 3 from the reaction. But I'm not sure if it's possible, or if it is, how to form it.
Let the initial concentration be
[tex]c_{0,He}[/tex]
as before, why not?
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Why not keep it algebraic, like all the ODEs we've worked with so far?
Because you are not seeing the obvious. He conc. is constant (assuming const. V).
So you don't need a non-trivial ODE for it.
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Because you are not seeing the obvious. He conc. is constant (assuming const. V).
So you don't need a non-trivial ODE for it.
Hmm - and Concentration=Partial Pressure/(RT), if the gases are ideal. So does this mean that the partial pressure of He also does not change, if the volume and temperature are constant, even as moles of the other gases change?
When put that way it seems simpler. The He present just does not affect our calculations. All other species' concentrations or partial pressures, we can calculate as before. If we want the total pressure, we can just calculate the partial pressure for all the reacting species, add them up and then add on He's partial pressure (which is the same as it was initially).
Even if volume is changing, should be ok: if we want to write volume as the sum of moles * molar volume of each species, we'd need to include He's number of moles, and then just find its molar volume (maybe a known function of the pressure and the temperature, maybe just a known value). Since He's number of moles is a constant, this will add as a constant to the volume (Vm[He] will never be a function of n(He), only of overall pressure and temperature).
Correct?
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So does this mean that the partial pressure of He also does not change, if the volume and temperature are constant, even as moles of the other gases change?
This is the very definition of the partial pressure.
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Because you are not seeing the obvious. He conc. is constant (assuming const. V).
So you don't need a non-trivial ODE for it.
Hmm - and Concentration=Partial Pressure/(RT), if the gases are ideal. So does this mean that the partial pressure of He also does not change, if the volume and temperature are constant, even as moles of the other gases change?
When put that way it seems simpler. The He present just does not affect our calculations. All other species' concentrations or partial pressures, we can calculate as before. If we want the total pressure, we can just calculate the partial pressure for all the reacting species, add them up and then add on He's partial pressure (which is the same as it was initially).
Even if volume is changing, should be ok: if we want to write volume as the sum of moles * molar volume of each species, we'd need to include He's number of moles, and then just find its molar volume (maybe a known function of the pressure and the temperature, maybe just a known value). Since He's number of moles is a constant, this will add as a constant to the volume (Vm[He] will never be a function of n(He), only of overall pressure and temperature).
Correct?
Probably right. Thought it's a tad hard wading through all your verbosity and still being sure.
Mathematics is the language of science, and for a good reason too.
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He present just does not affect our calculations.
That seems fishy. Don't think that's true.
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So, if the volume is changing (constant temperature), how do we calculate the partial pressure of He?
That seems fishy. Don't think that's true.
That's also what I thought, hence the question asking what effect it has. But, except for affecting total pressure (just add on the partial pressure), and total volume (if volume is changing), according to what you've said it shouldn't have any effect. Otherwise there should be an ODE for it.
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So, if the volume is changing (constant temperature), how do we calculate the partial pressure of He?
As if there was nothing else, trivial ideal gas type problem.
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As if there was nothing else, trivial ideal gas type problem.
But curiouscat was saying earlier that it can't be that the He present does not affect our calculations. It also makes sense to me that He, exerting pressure, must have some kind of effect on the reacting species.
In order for the He to affect our calculations, we need to write an ODE for it. That would deal with its effect on the reacting species, as well as their effect on He (as they change the volume, etc., for an isobaric system).
So what is that ODE? If we don't need to write one, then He must not be affecting the reacting species ...
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As if there was nothing else, trivial ideal gas type problem.
But curiouscat was saying earlier that it can't be that the He present does not affect our calculations.
And, I was wrong. Note, I just said it sounded fishy. I was mentally using Le Chatelier as a heuristic, where for reactions with a mole change inerts do change equilibrium.
On actual calculations it doesn't work at constant volume.
Only shows to emphasise why you should do the calculations and not verbal hand-waving arguments.
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And, I was wrong. Note, I just said it sounded fishy. I was mentally using Le Chatelier as a heuristic, where for reactions with a mole change inerts do change equilibrium.
On actual calculations it doesn't work at constant volume.
Only shows to emphasise why you should do the calculations and not verbal hand-waving arguments.
OK, thanks everyone. The issue of inert gases is clear now.
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By the way, something obvious that I should have checked before has just occurred to me. If we have a system where one equilibrium (let's say, A ::equil:: B) takes place and we measure the rate constants and find the rate law, and then we want to model concentration with time for a system with multiple equilibria (e.g. A ::equil:: B, B+C ::equil:: D), can we use the same rate law with the same rate constants and reaction orders that we established from our one-equilibrium system, to calculate concentrations in our multiple-equilibria system? Or does the presence of other reactions mean we must change the rate law or constant for each equilibrium involved?
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Or does the presence of other reactions mean we must change the rate law or constant for each equilibrium involved?
Generally does not change. Equilibrium const. definitely does not change. Rate expressions can.
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Generally does not change. Rate expressions can.
Hmm - can the elementary rate laws for each step change?
So, ideally, we will have rate laws that we know to pertain directly to our system and all the equilibria it contains. Failing this, it is usually a good guess (with no other information) that a rate law will stay the same, including its constant, when the equilibrium occurs in a system of multiple equilibria. Right?
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Hmm - can the elementary rate laws for each step change?
Yes.
So, ideally, we will have rate laws that we know to pertain directly to our system and all the equilibria it contains. Failing this, it is usually a good guess (with no other information) that a rate law will stay the same, including its constant, when the equilibrium occurs in a system of multiple equilibria. Right?
Yes. AFAIK. I could be wrong.
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Yes.
Well, not much we can do then - just use whichever rate laws we have available, be they system-specific or not. :P So we may as well assume they will work, after having tried to get system-specific ones.
There is no way to predict the rate law without experiment, is there?
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There is no way to predict the rate law without experiment, is there?
Sure there is. Do a literature search.
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Sure there is. Do a literature search.
I've done a search. All I could find is the fact that a mechanism can be used to predict the rate law, which I already knew.
Maybe I should respecify my question: can we predict theoretically what the new rate law for equilibrium A+B ::equil:: C+D will be in the system of equilibria (e.g. A+B ::equil:: C+D, A+D ::equil::C+E, B ::equil:: F) , given we know what the rate law for the equilibrium A+B ::equil:: C+D is, through experimental measurement or deduction from the mechanism, when this equilibrium is on its own.
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Maybe I should respecify my question: can we predict theoretically what the new rate law for equilibrium A+B ::equil:: C+D will be in the system of equilibria (e.g. A+B ::equil:: C+D, A+D ::equil::C+E, B ::equil:: F) , given we know what the rate law for the equilibrium A+B ::equil:: C+D is,
through experimental measurement
Usually, Yes.
or deduction from the mechanism
Typically, No.
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Usually, Yes.
This is by assuming the same rate law applies (or experimentally determining the new rate law for the multiple-equilibria system), right? Or is there an actual way to "correct" the experimentally determined law for the presence of various other equilibria, without actually doing the experiment with various other equilibria?
Typically, No.
What? So if we've worked out using a mechanism what the rate law should be for the A+B ::equil:: C+D case on its own, is it a poor assumption that the same rate law will hold for the system where there are several equilibria (e.g. A+B ::equil:: C+D, A+D ::equil::C+E, B ::equil:: F)?
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This is by assuming the same rate law applies (or experimentally determining the new rate law for the multiple-equilibria system), right? Or is there an actual way to "correct" the experimentally determined law for the presence of various other equilibria, without actually doing the experiment with various other equilibria?
Dunno. All I had in mind was do a fresh experiment. Then you have the right constants.
What? So if we've worked out using a mechanism what the rate law should be for the A+B ::equil:: C+D case on its own, is it a poor assumption that the same rate law will hold for the system where there are several equilibria (e.g. A+B ::equil:: C+D, A+D ::equil::C+E, B ::equil:: F)?
Not a poor assumption. Mostly it will work. For equilibrium constants it has to work.
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Not a poor assumption. Mostly it will work.
OK, good to know. On the other hand there is no way of finding the rate constants except from experimental measurements?
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On the other hand there is no way of finding the rate constants except from experimental measurements?
There. We turned a full circle.
Answer: Sure there is. Do a literature search.
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e.g.
Quantum mechanical transition state theory and a new semiclassical model for reaction rate constants
Ab Initio Evaluation of the Barrier Height. Theoretical Rate Constant of the NH3+ H :rarrow: NH2+ H2 Reaction
Ab Initio and DFT Direct Dynamics Studies on the Reaction Path and Rate Constant of the Hydrogen Abstraction Reaction: SiH3F+ H→ SiH2F+ H2
etc.
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Hmm not really. There was this:
There is no way to predict the rate law without experiment, is there?
To which the only answer I could find was by deducing the mechanism. But this finds the rate law in terms of k. k itself must still come from an experiment as far as this method teaches me.
And now there is rate constant k:
there is no way of finding the rate constants except from experimental measurements?
Thanks for the recommended articles. I will try to get a hold of some of them. How is this article: http://schwartzgroup1.arizona.edu/steve/files/book_chapter_2005.pdf?
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Looks good. If you can understand it.
I skimmed. It seems pretty advanced.
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Looks good. If you can understand it.
I skimmed. It seems pretty advanced.
I agree. It seems much too advanced for me right now. But it's just encouraging to know there are methods to do it out there.
One more thing to check. In our ODE system, we use concentrations, because rate laws depend on concentrations. But to be more precise, should we be using activities? Do rate laws depend finally on activities, or concentrations?
If it is concentrations, then even if we need activities as our final answer it's easy to just calculate all the concentrations at the given time and then use them to find the activities.
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One more thing to check. In our ODE system, we use concentrations, because rate laws depend on concentrations. But to be more precise, should we be using activities? Do rate laws depend finally on activities, or concentrations?
If it is concentrations, then even if we need activities as our final answer it's easy to just calculate all the concentrations at the given time and then use them to find the activities.
We'll cross that river once we get to it.
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We'll cross that river once we get to it.
When will we get to it? Anything else I need to do first...
An Internet search doesn't give anything suggesting we should actually be using activities. It would also require a modification from the RA*V ODE if we use activities instead of concentrations (multiplying activity by volume does not reach number of moles). On the other hand it seems logical that rate laws should be based around activities instead of concentrations, but I wouldn't know.
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It would also require a modification from the RA*V ODE if we use activities instead of concentrations (multiplying activity by volume does not reach number of moles).
Why would you multiply activity by volume.
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Why would you multiply activity by volume.
You can't, that's my point. Though our ODE might still be fine at constant volume if we replace all concentration terms directly with activity terms, it's not so simple to replace the concentration terms with activity terms in the non-constant volume form of the ODE:
[tex]
\frac{d \left ( c_A \cdot V \right ) }{dt}= R_A \cdot V \\
[/tex]
RA was the function in terms of rate that we spent so long figuring out.
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No clue.
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OK, let's leave the activities thing for the time being unless someone does know the answer.
Often the system is stiff and solving it is one gigantic headache.
You said this a while back. Is that when you're trying to find analytical solutions? The equations as we wrote them now are exact, I think, so there must be some approximations you make most of the time to reach analytical solutions.
Or even when looking for numerical solutions, the ODEs still throw up some problems?
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OK, let's leave the activities thing for the time being unless someone does know the answer.
Often the system is stiff and solving it is one gigantic headache.
You said this a while back. Is that when you're trying to find analytical solutions?
What do you understand by a "stiff" system?
The equations as we wrote them now are exact, I think, so there must be some approximations you make most of the time to reach analytical solutions.
Quasi Steady State Assumption often. Equilibriated steps sometimes. Depends.
Or even when looking for numerical solutions, the ODEs still throw up some problems?
Quite often.
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What do you understand by a "stiff" system?
Something that resists solution by commonly known or straightforward (/obvious) methods.
Quite often.
Ah. Can you give an example of one which provides problems for numerical solution?
Equilibriated steps sometimes.
Where can I look this particular one up? (what assumption is taken, when and why we can take it, how it makes the solution easier) The steady state approximation is already known to me.
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What do you understand by a "stiff" system?
Something that resists solution by commonly known or straightforward (/obvious) methods.
Why don't you read it up. That answer is practically useless.
Quite often.
Ah. Can you give an example of one which provides problems for numerical solution?
If I gave you one what will you do with it.
Equilibriated steps sometimes.
Where can I look this particular one up?
Google's a good start.
The steady state approximation is already known to me.
Good.
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Why don't you read it up. That answer is practically useless.
What I read up contained some quite detailed maths which I mostly haven't come across. But from several internet searches it seems like actually quite a complicated field. Overall, my impression on a basic level is that we call the ODE system stiff if, of the different mathematically possible real solutions, a very wide range and/or difference exists (e.g. one solution is 109, another 10-9) meaning that in predicting estimates for iteration we will find it very cumbersome to yield both answers which could be correct.
If I gave you one what will you do with it.
Mainly just wanted to see if the ODE itself looked any different from the ones we've already studied. Or if there is a visibly obvious reason from the ODE why it should be stiff. I guess I could just have asked. Sorry, if that would have been simpler for you.
Google's a good start.
I Googled "chemical kinetics equilibriated steps" and found nothing which seems to be of relevance. There was some stuff I found very interesting on a Wikipedia page that came up as a result (http://en.wikipedia.org/wiki/Detailed_balance), which is vaguely to do with kinetics (it postulates that each reaction should actually be written as an equilibrium, i.e. reversible, and then throws up some details, interesting for me at least, about what is or is not possible if we treat reactions are irreversible), but I don't see what ostensibly is the approximation you mean by 'equilibriated steps'. On the Wikipedia page, the "approximation" is to make a series (cycle) of irreversible reactions into equilibria so that they are actually possible, but this is of course the opposite of an approximation, and in any case you've already shown me how to write the ODEs for equilibria as opposed to irreversible, the page doesn't go into approximations.
My question was a little open-ended before so I'll govern it now. Are there any more approximations I should be aware of, e.g. in the first few years of undergraduate study of kinetics? (Other than steady state) I bring this up because some of the stuff on the Wikipedia page I linked to is beyond me. If that is what you mean by equilibriated steps maybe it's better to ignore it for now.
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Why don't you read it up. That answer is practically useless.
What I read up contained some quite detailed maths which I mostly haven't come across. But from several internet searches it seems like actually quite a complicated field. Overall, my impression on a basic level is that we call the ODE system stiff if, of the different mathematically possible real solutions, a very wide range and/or difference exists (e.g. one solution is 109, another 10-9) meaning that in predicting estimates for iteration we will find it very cumbersome to yield both answers which could be correct.
Well, somewhat OK I guess but in any case that should answer your own question "You said this a while back. Is that when you're trying to find analytical solutions? "
If I gave you one what will you do with it.
Mainly just wanted to see if the ODE itself looked any different from the ones we've already studied. Or if there is a visibly obvious reason from the ODE why it should be stiff. I guess I could just have asked. Sorry, if that would have been simpler for you.
No it won't look any different. A wide variation in magnitudes of constants, yes. More pertinently, I've been trying (quite unsuccessfully I admit) to coax you away from the idea that you can do much quantitative science by just "looking". The only good ways to appreciate kinetics, ODE's, stiffness etc. is by actually getting your hands dirty. Of all the questions you've asked I don't recall many where you've actually been attempting any numerical solution?
My question was a little open-ended before so I'll govern it now. Are there any more approximations I should be aware of, e.g. in the first few years of undergraduate study of kinetics? (Other than steady state) I bring this up because some of the stuff on the Wikipedia page I linked to is beyond me. If that is what you mean by equilibriated steps maybe it's better to ignore it for now.
Sometimes fast steps are assumed to be at equilibrium. Can't think of any others.
If solving numerically, you don't need approximations mostly.
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No it won't look any different. A wide variation in magnitudes of constants, yes. More pertinently, I've been trying (quite unsuccessfully I admit) to coax you away from the idea that you can do much quantitative science by just "looking". The only good ways to appreciate kinetics, ODE's, stiffness etc. is by actually getting your hands dirty. Of all the questions you've asked I don't recall many where you've actually been attempting any numerical solution?
Solving ODEs simultaneously is going to be quite difficult for me. Until I can do it on a computer I'm not going to do it by hand. Although I can handle very basic cases analytically, these are practically trivial compared with what we've been talking about, and I don't see much point in going through a rote solution using a computer which just requires me to type out what we've been talking about here and then plugging in some initial values? I'll get there eventually but not sure I see the hurry in comparison to understanding the formulation. I wouldn't even know whether the answer that comes out is actually correct.
Sometimes fast steps are assumed to be at equilibrium. Can't think of any others.
If solving numerically, you don't need approximations mostly.
OK, thanks.
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No it won't look any different. A wide variation in magnitudes of constants, yes. More pertinently, I've been trying (quite unsuccessfully I admit) to coax you away from the idea that you can do much quantitative science by just "looking". The only good ways to appreciate kinetics, ODE's, stiffness etc. is by actually getting your hands dirty. Of all the questions you've asked I don't recall many where you've actually been attempting any numerical solution?
Solving ODEs simultaneously is going to be quite difficult for me. Until I can do it on a computer I'm not going to do it by hand. Although I can handle very basic cases analytically, these are practically trivial compared with what we've been talking about, and I don't see much point in going through a rote solution using a computer which just requires me to type out what we've been talking about here and then plugging in some initial values? I'll get there eventually but not sure I see the hurry in comparison to understanding the formulation.
To me that's absolutely the wrong, totally inefficient approach. But you are entitled to having your opinion.
And if you think a computer solution is "a rote solution", I urge you to correct that misconception. There's nothing rote about solving a good kinetics problem, with or without a computer.