October 02, 2023, 02:52:28 PM
Forum Rules: Read This Before Posting

### Topic: Equilibrium Problem  (Read 16247 times)

0 Members and 1 Guest are viewing this topic.

#### Borek

• Mr. pH
• Deity Member
• Posts: 27424
• Mole Snacks: +1783/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: Equilibrium Problem
« Reply #15 on: March 13, 2013, 06:02:44 PM »
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.

Quote
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

• Sr. Member
• Posts: 1177
• Mole Snacks: +28/-94
##### Re: Equilibrium Problem
« Reply #16 on: March 13, 2013, 06:15:19 PM »
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.

Quote
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).

#### Borek

• Mr. pH
• Deity Member
• Posts: 27424
• Mole Snacks: +1783/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: Equilibrium Problem
« Reply #17 on: March 13, 2013, 06:39:01 PM »
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
##### Re: Equilibrium Problem
« Reply #18 on: March 14, 2013, 02:55:22 PM »
Sorry to interfere, but would the graph look like this?

• Sr. Member
• Posts: 1177
• Mole Snacks: +28/-94
##### Re: Equilibrium Problem
« Reply #19 on: March 14, 2013, 03:18:18 PM »
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%?

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Equilibrium Problem
« Reply #20 on: March 14, 2013, 03:30:11 PM »
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.

• Sr. Member
• Posts: 1177
• Mole Snacks: +28/-94
##### Re: Equilibrium Problem
« Reply #21 on: March 14, 2013, 04:05:17 PM »
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.

#### Borek

• Mr. pH
• Deity Member
• Posts: 27424
• Mole Snacks: +1783/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: Equilibrium Problem
« Reply #22 on: March 14, 2013, 04:36:05 PM »
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Equilibrium Problem
« Reply #23 on: March 14, 2013, 04:46:01 PM »
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.

• Sr. Member
• Posts: 1177
• Mole Snacks: +28/-94
##### Re: Equilibrium Problem
« Reply #24 on: March 14, 2013, 04:49:34 PM »
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?

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Equilibrium Problem
« Reply #25 on: March 14, 2013, 04:53:41 PM »
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.

#### Borek

• Mr. pH
• Deity Member
• Posts: 27424
• Mole Snacks: +1783/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: Equilibrium Problem
« Reply #26 on: March 14, 2013, 05:00:14 PM »
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
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Equilibrium Problem
« Reply #27 on: March 14, 2013, 05:07:25 PM »

Edit: basically that's what curiouscat stated. If that makes any sense

Does to me at least!

• Sr. Member
• Posts: 1177
• Mole Snacks: +28/-94
##### Re: Equilibrium Problem
« Reply #28 on: March 14, 2013, 05:56:52 PM »
Excuse my extremely bad drawing skills please ... is this the general idea?

#### Borek

• Mr. pH
• Deity Member
• Posts: 27424
• Mole Snacks: +1783/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: Equilibrium Problem
« Reply #29 on: March 14, 2013, 06:32:33 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info