[Big-Daddy]

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)  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 K_c1 of 7.2):

Consider the reaction yield n=n_eq(AB)/n_max(AB) as a function of the initial molar ratio A₂ : B₂ = x : 1 at any fixed temperature (n_max 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.

[Big-Daddy]

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?

[Rutherford]

Asked the same question, but no answer  . I hope that someone answers this.

[Borek]

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...

[Big-Daddy]

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, n_eq(AB) will increase (simply Le Chatelier's Principle) but not as much as n_max(AB), since the equilibrium must obey a constant of conversion whereas n_max is from the reaction taken as irreversible. So as x tends to infinity n will tend to 0.

If x tends to 0 then n_max(AB) also will decrease (I hope I am defining n_max(AB) right, as in the infinite K case of complete conversion?), whereas n_eq(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?

[Rutherford]

Yeah, there is the funny problem, if you started with 0 mole and got 0 mole what is the yield  ?

"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 n_max. You said also that n_eq(AB) will increase. How did you get 0 then?

[Big-Daddy]

Yeah, there is the funny problem, if you started with 0 mole and got 0 mole what is the yield  ?

"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 n_max. You said also that n_eq(AB) will increase. How did you get 0 then?

Oh yes of course B is the limiting reagent so n_max 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 n_eq(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?

[Rutherford]

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%.

[Borek]

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?

[Big-Daddy]

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 n_max(AB) drops off to 0 as well. As x decreases, the equilibrium should shift to the left so n_eq(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?

[Borek]

As x tends to 0, A is limiting so of course n_max(AB) drops off to 0 as well. As x decreases, the equilibrium should shift to the left so n_eq(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?

[Big-Daddy]

As x tends to 0, A is limiting so of course n_max(AB) drops off to 0 as well. As x decreases, the equilibrium should shift to the left so n_eq(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, n_eq(AB) and n_max(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, n_eq(AB), falls much faster than the denominator, n_max(AB)? But I arrived at that with a calculation - how should I deduce qualitatively that as A2 drops, this would happen?

[Borek]

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?

[curiouscat]

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.

[Big-Daddy]

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 ...?