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Topic: Equilibrium kinetics  (Read 113117 times)

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Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #75 on: May 16, 2013, 04:54:27 AM »
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...

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #76 on: May 16, 2013, 04:57:05 AM »
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.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #77 on: May 16, 2013, 08:44:00 AM »
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?
« Last Edit: May 16, 2013, 09:32:11 AM by Big-Daddy »

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #78 on: May 16, 2013, 09:31:50 AM »

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.

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #79 on: May 16, 2013, 09:36:24 AM »

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.

Quote
If not, can you please write the 3 ODEs we want to solve simultaneously?

Try again.

Quote
And there is nothing in there for C, because the rate is not dependent on C (zero-order).

Yes. That's right.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #80 on: May 16, 2013, 09:46:16 AM »
[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?

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #81 on: May 16, 2013, 09:52:39 AM »
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)...

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #82 on: May 16, 2013, 09:55:50 AM »
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] . 

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #83 on: May 16, 2013, 10:20:11 AM »
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.

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #84 on: May 16, 2013, 10:26:20 AM »
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.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #85 on: May 16, 2013, 10:30:17 AM »
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 )

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #86 on: May 16, 2013, 10:40:27 AM »
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.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #87 on: May 16, 2013, 11:09:47 AM »
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?

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #88 on: May 16, 2013, 11:16:08 AM »
This makes a lot of sense actually.

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #89 on: May 16, 2013, 11:17:20 AM »

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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