Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Big-Daddy on March 11, 2013, 07:04:59 PM
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The gaseous substances A2 and B2 were mixed in a molar ratio 2:1 in a closed vessel at a temperature T1. When the equilibrium A2(g) + B2(g) ::equil:: 2AB(g) was established the number of heteronuclear molecules in a gas phase became equal to the total number of homonuclear molecules.
The question then asks the student to determine the equilibrium constant K1. I found it to be 7.2. I then answered quantitatively the remaining sections. The bit I need help on is this (I don't think any prior info is needed except perhaps the Kc1 of 7.2):
Consider the reaction yield n=neq(AB)/nmax(AB) as a function of the initial molar ratio A2 : B2 = x : 1 at any fixed temperature (nmax is the maximum amount calculated from the reaction equation). Answer the following questions qualitatively, without exact equilibrium calculations.
At what x the yield is extremal (minimal or maximal)?
What is the yield at: a) x tends to infinity; b) x tends to 0?
Draw the graph of n(x).
Now, consider the variable ratio A2 : B2 = x : 1 at a fixed total pressure.
At what x the equilibrium amount of AB is maximal?
I'm not really sure how to visualize this. For this last part, we should see the same minimal/maximal relationship as we see for the first part. The rest I need help on.
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Edit: Just realized that the last question is indeed not the same as the first. That makes me even more clueless than before. And since temperature and pressure are being mentioned, should I be considering Gibbs' free energy?
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Asked the same question, but no answer :(. I hope that someone answers this.
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What is the yield at: a) x tends to infinity; b) x tends to 0?
Perhaps try to start with these. In both cases you will have a huge excess of one of the reagents...
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What is the yield at: a) x tends to infinity; b) x tends to 0?
Perhaps try to start with these. In both cases you will have a huge excess of one of the reagents...
I am thinking that maybe, as x tends to infinity, neq(AB) will increase (simply Le Chatelier's Principle) but not as much as nmax(AB), since the equilibrium must obey a constant of conversion whereas nmax is from the reaction taken as irreversible. So as x tends to infinity n will tend to 0.
If x tends to 0 then nmax(AB) also will decrease (I hope I am defining nmax(AB) right, as in the infinite K case of complete conversion?), whereas neq(AB) will ... decrease less, maybe? I don't know, this feels like guessing. And when x=0 both will obviously be 0 (you cannot get AB in the absence of A!) so that doesn't help much.
What do you say so far?
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Yeah, there is the funny problem, if you started with 0 mole and got 0 mole what is the yield ;D?
"So as x tends to infinity n will tend to 0."
How do you mean that n is 0? B is the limiting reagent here and its amount determines nmax. You said also that neq(AB) will increase. How did you get 0 then?
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Yeah, there is the funny problem, if you started with 0 mole and got 0 mole what is the yield ;D?
"So as x tends to infinity n will tend to 0."
How do you mean that n is 0? B is the limiting reagent here and its amount determines nmax. You said also that neq(AB) will increase. How did you get 0 then?
Oh yes of course B is the limiting reagent so nmax does not vary directly with x but rather plateaus at x=1.
The equilibrium should still shift to the right if more A is added though (unless all B is consumed, which it cannot because then denominator of equilibrium constant becomes 0 and the constant would be infinite which it isn't), shouldn't it? So neq(AB) will still increase ...
Then I suppose, unless again I am not thinking clearly, that we could say that as x tends to infinity the numerator is increasing and the denominator constant, but how do we find the limit?
The list of advanced topics states formulae for Gibbs' energy etc., and this problem refers directly to conditions of temperature and pressure - can we really solve this without considering Gibbs' energy?
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Well, it says: Answer the following questions qualitatively, without exact equilibrium calculations.
Then I suppose, unless again I am not thinking clearly, that we could say that as x tends to infinity the numerator is increasing and the denominator constant, but how do we find the limit?
I am not totally sure, but I think that the yield is reaching 100%.
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The equilibrium should still shift to the right if more A is added though
Yes.
unless all B is consumed, which it cannot because then denominator of equilibrium constant becomes 0
Tends to zero and becomes zero are two different things.
It is a standard approach to use excess of one reagent to shift the equilibrium to the right to get the higher reaction yield, isn't it?
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The equilibrium should still shift to the right if more A is added though
Yes.
unless all B is consumed, which it cannot because then denominator of equilibrium constant becomes 0
Tends to zero and becomes zero are two different things.
It is a standard approach to use excess of one reagent to shift the equilibrium to the right to get the higher reaction yield, isn't it?
Ah ok I understand now: yield is tending to 100% as x tends to infinity (it will never reach 100% because not all B can be converted, but we are tending to all B being converted nonetheless).
As x tends to 0, A is limiting so of course nmax(AB) drops off to 0 as well. As x decreases, the equilibrium should shift to the left so neq(AB) decreases as well ... so are we tending to 0% here? Or will there be something like the equilibrium decreases less than direct conversion so the yield still tends to 100% somehow?
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As x tends to 0, A is limiting so of course nmax(AB) drops off to 0 as well. As x decreases, the equilibrium should shift to the left so neq(AB) decreases as well ... so are we tending to 0% here? Or will there be something like the equilibrium decreases less than direct conversion so the yield still tends to 100% somehow?
Think it over. Imagine x is just 0.0001. What is B2/A2 ratio?
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As x tends to 0, A is limiting so of course nmax(AB) drops off to 0 as well. As x decreases, the equilibrium should shift to the left so neq(AB) decreases as well ... so are we tending to 0% here? Or will there be something like the equilibrium decreases less than direct conversion so the yield still tends to 100% somehow?
Think it over. Imagine x is just 0.0001. What is B2/A2 ratio?
My picture is that we will have very little equilibrium if there is too little A2. It's clear that as x tends to 0, neq(AB) and nmax(AB) both tend to 0 as well. But I cannot conceptualize whether the equilibrium moves slowly or not (or how slowly it moves/needs to move) compared to the direct conversion.
At x=0.0001, the ratio of B2/A2 is 10,000:1 and my calculation suggests that 2.2*10-7 moles (taking initial B2 as 1 mole) will be produced of AB in equilibrium. Meanwhile it's clear that a direct reaction will produce much more comparatively - 0.0002 moles of AB - so then, the yield tends to 0 (as the numerator, neq(AB), falls much faster than the denominator, nmax(AB)? But I arrived at that with a calculation - how should I deduce qualitatively that as A2 drops, this would happen?
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At x=0.0001, the ratio of B2/A2 is 10,000:1
Isn't it exactly the same situation? Huge excess of one reagent?
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One very different way to think about this is symmetry considerations. Your original problem is symmetric in A and B.
Based on that alone I'd wager that whatever efficiency you get for the case "x tends to infinity" has to be the same as for the case "x tends to 0".
What else can we say?
(1) Efficiency is constrained to always be between 0 and 1.
(2) Symmetry says that the curve has to have a line of symmetry at x=1; so x=1 is a probable extremum too (minima or maxima)
Not a 100% sure about my reasoning but think about it.
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Replying to both:
If x=1 we will probably have a minimum as much less is converted out in equilibrium than in direct conversion.
By your logic then it's clear that the yield should be going to infinity as x goes to infinity. Can we thus draw from this that the greater in excess one of the reagents is, the greater percentage of the reactants combined overall will be transformed to make products; the more in excess one of the reagents is, the greater the percentages of both which are transferred to the products will be ...?
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By your logic then it's clear that the yield should be going to infinity as x goes to infinity.
No, yield can't get greater than 1.
Can we thus draw from this that the greater in excess one of the reagents is, the greater percentage of the reactants combined overall will be transformed to make products; the more in excess one of the reagents is, the greater the percentages of both which are transferred to the products will be ...?
No, you are limited by the stoichiometry - and one of the reagents is limiting. You can use it to the last molecule, but the one in excess will be still present.
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By your logic then it's clear that the yield should be going to infinity as x goes to infinity.
No, yield can't get greater than 1.
Can we thus draw from this that the greater in excess one of the reagents is, the greater percentage of the reactants combined overall will be transformed to make products; the more in excess one of the reagents is, the greater the percentages of both which are transferred to the products will be ...?
No, you are limited by the stoichiometry - and one of the reagents is limiting. You can use it to the last molecule, but the one in excess will be still present.
Oops, I meant 1.
I see what you mean about the stoichiometry limiting us. What sort of conclusion can we draw from this then? After all, this is a preparatory problem from the IChO - an answer must mean something about equilibria in general (which will also be inherent in the types of problems that are going to come up in the real thing).
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I don't think it is meant to draw any generalized conclusions - more like it is trying to make you flexible about analyzing such problems and seeing how the basic principles can be applied in different ways. Or perhaps how the seemingly different cases are in fact identical and can be solved with the same principles.
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Sorry to interfere, but would the graph look like this?
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Sorry to interfere, but would the graph look like this?
x
Ok thanks, that would make sense.
I was thinking the same question as Raderford.
How should we go about drawing the graph of n(AB)? We'll draw (x,n(AB)) but the first question is, is it a sketch or a plot? I assume since there are no quantitative calculations it is a sketch.
I'm thinking this graph will be S-shaped where x is between 1 and infinity, and then a direct reflection of the S-shape between 1 and infinity is there between 0 and 1 (squashed down a lot more, obviously). We know there will be reflection at x=1 due to the symmetry of the reagent excess. The S will reach its peak at yield=100% when x tends to infinity - and its minimum when x=1 (calculation suggests about 50% of each reactant will get converted to products at equilibrium, so we will get around 50% yield as well) is 50%?
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Sorry to interfere, but would the graph look like this?
No sharp derivative discontinuity at x=1 I say.
Also, on the right it rises less steep since it goes to 1 only at ∞.
Just my guesses.
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Sorry to interfere, but would the graph look like this?
No sharp derivative discontinuity at x=1 I say.
Also, on the right it rises less steep since it goes to 1 only at ∞.
Just my guesses.
Can you have a look at my post? I already made these predictions and more.
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I believe plot is more or less OK (in terms of general shape). I agree with curiouscat it should not have a sharp corner (more like inverted Gaussian curve) and I don't think it will be symmetrical.
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Sorry to interfere, but would the graph look like this?
No sharp derivative discontinuity at x=1 I say.
Also, on the right it rises less steep since it goes to 1 only at ∞.
Just my guesses.
Can you have a look at my post? I already made these predictions and more.
Looked good to me.
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I believe plot is more or less OK (in terms of general shape). I agree with curiouscat it should not have a sharp corner (more like inverted Gaussian curve) and I don't think it will be symmetrical.
So it should be an inverted S shape, reaching minimum at yield = 50%, x=1, and then rising to an asymptote of yield = 100% at x = infinity in a forward S shape? I can't see another way to flatten out the sharp corner Raderford drew.
I thought it was discussed that it should be symmetrical? Where will be the differences, then? I'd rather not calculate the plot since it says "purely qualitatively" in the question.
When I say symmetrical, what I mean is the same shape from x=0 to x=1 as from x=1 to x = infinity, except reflected in the line x=1. A forward S shape for x=1 to x = infinity, the reflected S shape for x = 0 to x = 1. Is this right?
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I believe plot is more or less OK (in terms of general shape). I agree with curiouscat it should not have a sharp corner (more like inverted Gaussian curve) and I don't think it will be symmetrical.
So it should be an inverted S shape, reaching minimum at yield = 50%, x=1, and then rising to an asymptote of yield = 100% at x = infinity in a forward S shape? I can't see another way to flatten out the sharp corner Raderford drew.
I thought it was discussed that it should be symmetrical? Where will be the differences, then? I'd rather not calculate the plot since it says "purely qualitatively" in the question.
When I say symmetrical, what I mean is the same shape from x=0 to x=1 as from x=1 to x = infinity, except reflected in the line x=1. A forward S shape for x=1 to x = infinity, the reflected S shape for x = 0 to x = 1. Is this right?
I cannot parse your forward S and reverse S notation. Try drawing and posting and we might be able to comment better?
I was sloppy before when I called it symmetrical. It is symetrical under a one-to-one mapping of the 0-1 domain onto the larger 1-∞ domain. If that makes any sense.
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It can't be symmetrical if the change between minimum and 100% occurs on the distance of 1 on the left (0..1) and on the infinite distance on the right (1..∞).
Google for Gaussian curve - while it is symmetrical, it is kind of a shape I would expect, just stretched on the right side, after minimum.
Now that I reread your definition of symmetrical you are in a way right, it is just that your definition of symmetry is rather... unorthodox.
Edit: basically that's what curiouscat stated. If that makes any sense ;)
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Edit: basically that's what curiouscat stated. If that makes any sense ;)
Does to me at least! :)
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Excuse my extremely bad drawing skills please ... is this the general idea?
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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.
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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!
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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.
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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.
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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.
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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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?
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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
[tex]7=\frac {x^2}{(A-x)(B-x)}[/tex]
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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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
[tex]7=\frac {x^2}{(A-x)(B-x)}[/tex]
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 ::equil:: 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.
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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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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.
[tex]K_c=\frac {(-2\Delta A)^2}{(A_0+\Delta A)(B_0+\Delta A)}[/tex]
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.