March 01, 2024, 08:50:35 AM
Forum Rules: Read This Before Posting


Topic: Equilibrium kinetics  (Read 113115 times)

0 Members and 1 Guest are viewing this topic.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #120 on: May 19, 2013, 09:12:01 AM »


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.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #121 on: May 19, 2013, 09:26:20 AM »
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.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #122 on: May 19, 2013, 10:02:20 AM »
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?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #123 on: May 19, 2013, 10:04:56 AM »
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?

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #124 on: May 19, 2013, 10:17:46 AM »
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)?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #125 on: May 19, 2013, 02:55:21 PM »
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.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #126 on: May 20, 2013, 12:43:42 PM »
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.)

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #127 on: May 20, 2013, 02:16:50 PM »


OK, I will try. Where can I find a simultaneous ODE solver?

Matlab. Octave. Mathematica.

Many many others.

Quote
And should I assume that, if I get a numerical result, it's correct?

Sure.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #128 on: May 20, 2013, 03:05:23 PM »
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.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #129 on: May 21, 2013, 10:42:18 AM »
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.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #130 on: May 21, 2013, 02:30:01 PM »
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).

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #131 on: May 21, 2013, 02:55:47 PM »
You mean a closed-form algebraic solution? Is that even possible?

Yes. Sometimes.


Quote
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".


Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #132 on: May 23, 2013, 12:40:57 PM »
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)
« Last Edit: May 23, 2013, 12:57:37 PM by Big-Daddy »

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #133 on: May 23, 2013, 12:58:43 PM »
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.

Quote
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?

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #134 on: May 26, 2013, 07:56:35 AM »
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?

Sponsored Links