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### Topic: Equilibrium Problem  (Read 18957 times)

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##### Re: Equilibrium Problem
« Reply #30 on: March 14, 2013, 08:56:30 PM »
That WAS my idea.

But I had nothing better to do so I did some calculations. Turns out I was wrong. There is a sharp corner and the function is not continuous. Most likely that's because limiting reagent changes.

Not that exact shape matters much. What is important is that the yield goes up to 100% at zero and infinity, and there is a minimum. We got to these conclusions without a need for calculations.

Can you provide the function you used for n in terms of x?

Thanks for clarifying on the shape of the graph by the way. Turns out Raderford was right after all!

#### curiouscat

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##### Re: Equilibrium Problem
« Reply #31 on: March 15, 2013, 01:15:18 AM »
That WAS my idea.

But I had nothing better to do so I did some calculations. Turns out I was wrong. There is a sharp corner and the function is not continuous. Most likely that's because limiting reagent changes.

Not that exact shape matters much. What is important is that the yield goes up to 100% at zero and infinity, and there is a minimum. We got to these conclusions without a need for calculations.

Interesting. Just goes to show how much our intuition can be off.

I never expected that sharp corner.

#### curiouscat

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##### Re: Equilibrium Problem
« Reply #32 on: March 15, 2013, 01:16:53 AM »

There is a sharp corner and the function is not continuous. Most likely that's because limiting reagent changes.

May I  nitpick? It is continuous but not differentiable?

I'll add another point that took me by surprise: I would have expected a zero derivative at x=0. Apparently isn't.

Wonder what that value is and if it has a significance.

#### Borek

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##### Re: Equilibrium Problem
« Reply #33 on: March 15, 2013, 04:57:58 AM »
Can you provide the function you used for n in terms of x?

There was no simple function - I put the data into spreadsheet, columns contained respectively: A, B (always 1), stoichiometric yield, equilibrium yield, ratio of the two.

May I  nitpick? It is continuous but not differentiable?

No idea. I tried to cover the corner with higher accuracy, hoping for a nice curve, but it was always sharp.

Quote
I'll add another point that took me by surprise: I would have expected a zero derivative at x=0. Apparently isn't.

Actually it can be - remember scale of the plot is such that the range 0..0.01 is covered by just a pixel or two, I have not tried to switch to some logarithmic scale to see how it really behaves. There is plenty of place for a nice asymptote.
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##### Re: Equilibrium Problem
« Reply #34 on: March 15, 2013, 04:04:49 PM »

There is a sharp corner and the function is not continuous. Most likely that's because limiting reagent changes.

May I  nitpick? It is continuous but not differentiable?

I'll add another point that took me by surprise: I would have expected a zero derivative at x=0. Apparently isn't.

Wonder what that value is and if it has a significance.

I think it must have a zero derivative at x=0 and the only reason we don't see it is because it's impossible to get a very high level of precision on how things are changing that close to 0 (on a normal-sized graph at least). We can try plotting x=0 to x=1 and seeing what happens then with smaller and smaller intervals.

To Borek: what function of [A]0 and [B]0 did you use to get equilibrium yield?

I always try to express [AB] in terms of these two and Δ[A2] (extent), but if [AB]=-2*Δ[A2] then when we square this expression due to the numerator of Keq we'll end up with 4*(Δ[A2])2 (which could come from a positive) and this gives me finally wrong results.

Edit: Nevermind, I've got an expression that works to do the calculation (e.g. Δ[A2]=-0.75 when [A]0 and [B]0 are 2 and 1 respectively and Kc is 7.2, which I think is the correct solution to the first part of this prep problem). But the graph found is very different from yours - can we compare functions?
« Last Edit: March 15, 2013, 04:32:34 PM by Big-Daddy »

#### Borek

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##### Re: Equilibrium Problem
« Reply #35 on: March 15, 2013, 04:54:49 PM »
Sorry, I have not saved the spreadsheet and I have not solved the problem on paper, so I don't have notes. I started with

$$7=\frac {x^2}{(A-x)(B-x)}$$

and solved for x. Differences between 7 and 7.2 are negligible.

Solving becomes pretty simple if you assume k=1, and the plot shape is preserved, it is just much more shallow.
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##### Re: Equilibrium Problem
« Reply #36 on: March 16, 2013, 08:10:01 AM »
Sorry, I have not saved the spreadsheet and I have not solved the problem on paper, so I don't have notes. I started with

$$7=\frac {x^2}{(A-x)(B-x)}$$

and solved for x. Differences between 7 and 7.2 are negligible.

Solving becomes pretty simple if you assume k=1, and the plot shape is preserved, it is just much more shallow.

A and B are initial concentrations of A2 and B2 I take it. Then shouldn't we have (2x)2 in the numerator, given that the equation is A2 + B2  2AB (we get 2 moles of AB produced for every mole of A2 converted out, and 1 mole of B2 converted for every mole of A2 converted, so your expression is right except we should have 4x2 where you have x2)?

This is similar to what I did except I originally took x as positive (Δ[A2]). This came out with a value of -0.75 if the constant is Kc and A=2, B=1, which is also what I got from the first problem of this prep question. Your current equation gets 0.89 (with Kc as 7.2 not 7) which is not what I got as my first answer.

#### Borek

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##### Re: Equilibrium Problem
« Reply #37 on: March 16, 2013, 09:40:28 AM »
Yes, you are right about (2x)2. It doesn't change much, general shape of the plot is still the same.
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##### Re: Equilibrium Problem
« Reply #38 on: March 16, 2013, 10:38:55 AM »
Yes, you are right about (2x)2. It doesn't change much, general shape of the plot is still the same.

I have something finally similar.

$$K_c=\frac {(-2\Delta A)^2}{(A_0+\Delta A)(B_0+\Delta A)}$$

When this is solved for ΔA we get a function for ΔA in terms of A0, B0 and Kc which we can then multiply by -2 to reach the equilibrium concentration of AB. This is neq(AB). nmax(AB)=2*A0 for A0<B0 (A0 moles of reactants converted to products, so 2*A0 moles of products produced, as you get 2 moles of AB for each mole of A2 used) and 2*B0 for B0<A0 (B0 moles of reactants converted, so 2*B0 moles of products formed, as you get 2 moles of AB for each mole of B2 used). The final result looks similar to your graph.
« Last Edit: March 16, 2013, 12:03:46 PM by Borek »