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### Topic: Equilibrium kinetics  (Read 118193 times)

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#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #135 on: May 26, 2013, 01:15:22 PM »
Quote
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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##### Re: Equilibrium kinetics
« Reply #136 on: May 26, 2013, 01:35:57 PM »
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

$$c_{0,He}$$

as before, why not?

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #137 on: May 26, 2013, 01:45:49 PM »

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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##### Re: Equilibrium kinetics
« Reply #138 on: May 26, 2013, 02:22:41 PM »
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?

#### Borek

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##### Re: Equilibrium kinetics
« Reply #139 on: May 26, 2013, 02:27:17 PM »
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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#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #140 on: May 26, 2013, 02:33:33 PM »
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?

Mathematics is the language of science, and for a good reason too.

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #141 on: May 26, 2013, 02:35:11 PM »
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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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##### Re: Equilibrium kinetics
« Reply #142 on: May 26, 2013, 02:47:46 PM »
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.

#### Borek

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##### Re: Equilibrium kinetics
« Reply #143 on: May 26, 2013, 02:57:35 PM »
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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##### Re: Equilibrium kinetics
« Reply #144 on: May 26, 2013, 05:56:25 PM »
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 ...
« Last Edit: May 26, 2013, 07:11:41 PM by Big-Daddy »

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #145 on: May 26, 2013, 11:55:48 PM »
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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##### Re: Equilibrium kinetics
« Reply #146 on: May 27, 2013, 10:05:38 AM »
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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##### Re: Equilibrium kinetics
« Reply #147 on: May 31, 2013, 12:59:08 PM »
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  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 B, B+C 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?

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #148 on: May 31, 2013, 02:58:19 PM »
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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##### Re: Equilibrium kinetics
« Reply #149 on: May 31, 2013, 03:13:14 PM »
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?