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Offline Big-Daddy

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Equilibrium yield
« on: August 28, 2013, 11:21:52 PM »
Is a mathematical definition of chemical yield for an equilibrium reaction, that Δc(LR) = -yield * c0(LR) where LR is the limiting reagent reacting species? I have a question which asks, what is the initial mass ratio of reactants required to produce a 50% yield - I presume this means to react 50% of the limiting reagent and transfer that into products, rather than 50% of any other reactant?

There is also a follow-up question: am I right to think that, if we want to minimize the necessary mass ratio for a certain yield, we should choose the reactant with the highest value of Mr(Species) * v(Species), where Mr(Species) is the molecular weight of the reactant and v(Species) is the stoichiometric coefficient of that reactant in the reaction equation, as the LR and then calculate the mass ratio required to produce the given yield from that?

Offline Big-Daddy

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Re: Equilibrium yield
« Reply #1 on: September 07, 2013, 01:24:10 PM »
Hello, can someone clarify this?

Offline magician4

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Re: Equilibrium yield
« Reply #2 on: September 08, 2013, 09:30:20 AM »
Quote
Is a mathematical definition of chemical yield for an equilibrium reaction,(...)
at times, "yield" needs clarification what it should be related to.
without further explanations, it usually is related to the limiting factor, and 100% of it's initial value.
considerations of equilibrium situations to the best of my knowledge aren't taken into account here explicitly (as all reactions are equilibrium reactions: chemistry always can be reversed), hence for example "52% yield of limiting factor" well could be 100% of what you could get at all, given the situation to be considered.

furthermore, "Δc(LR) = -yield * c0(LR)" would require (a) that all of the educt would be transformed to the desired product and nothing else (i.e. "Δc(LR) >> -yield * c0(LR)" doesn't occur: no side reactions allowed), and (b) that you had for example no change of volume to be considered at all.
hence, I wouldn't take this to be a good definition of "yield"

in my opinion, still the only valuable definition of "yield" would be the ratio of " n(educt) to be considered, transformed as desired : n0(educt) to be considered "

Quote
There is also a follow-up question: am I right to think that, if we want to minimize the necessary mass ratio for a certain yield, we should choose the reactant with the highest value of Mr(Species) * v(Species), where Mr(Species) is the molecular weight of the reactant and v(Species) is the stoichiometric coefficient of that reactant in the reaction equation, as the LR and then calculate the mass ratio required to produce the given yield from that?
if I understood you correctly, this only would be true if Π(c)f of educts and products would be of same order in the law of mass action expression.
if, however, a low Mr substance is of huge stoechiometrical influence  (like, for example in Te + 3 F2  :rarrow: TeF6 :  K = [TeF6]1/a(Te) * [F2]3), influencing this system for optimum mass ratio well could mean that you'd have to increase fluorine, not tellurium

regards

Ingo
« Last Edit: September 08, 2013, 09:43:40 AM by magician4 »
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Offline Big-Daddy

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Re: Equilibrium yield
« Reply #3 on: September 08, 2013, 11:37:40 AM »
Quote
Is a mathematical definition of chemical yield for an equilibrium reaction,(...)
at times, "yield" needs clarification what it should be related to.
without further explanations, it usually is related to the limiting factor, and 100% of it's initial value.
considerations of equilibrium situations to the best of my knowledge aren't taken into account here explicitly (as all reactions are equilibrium reactions: chemistry always can be reversed), hence for example "52% yield of limiting factor" well could be 100% of what you could get at all, given the situation to be considered.

Leaving aside the point on the optimum mass ratio for now. I don't understand, how can 100% be suggested by "52% yield of limiting reagent"?

furthermore, "Δc(LR) = -yield * c0(LR)" would require (a) that all of the educt would be transformed to the desired product and nothing else (i.e. "Δc(LR) >> -yield * c0(LR)" doesn't occur: no side reactions allowed), and (b) that you had for example no change of volume to be considered at all.
hence, I wouldn't take this to be a good definition of "yield"

in my opinion, still the only valuable definition of "yield" would be the ratio of " n(educt) to be considered, transformed as desired : n0(educt) to be considered "[/quote]

I understand point (b), on the volume change, but not point (a). How about, say, the definition "yield = n0(LR) - neq(LR) / n0(LR) where LR is the limiting reagent and neq is the equilibrium number of moles of LR" so that we could then note that yield = -Δn(LR) / n0(LR) ... does this work?

Offline magician4

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Re: Equilibrium yield
« Reply #4 on: September 08, 2013, 12:29:43 PM »
Quote
Leaving aside the point on the optimum mass ratio for now. I don't understand, how can 100% be suggested by "52% yield of limiting reagent"?
let's consider a simple esterification, like "acetic acid + ethanol  ::equil:: acetic acid ethanoic ester + water"

a typical K value value for this would be 4 , and the LMA would look like this:

[tex] K = \frac {[AcOEt] \cdot [H_2O]}{[AcOH] \cdot [EtOH]} [/tex]

if, for simplification purposes, we say that ONLY this reaction should occur (which, in reality, it doesn't), and if we had equal concentrations of acetic acid, ethanol for starters (and if the total volume was considered not to change in a relevant way), it follows:

[tex]c_0(AcOH) = c_0(EtOH) ; \ c(AcOEt) = c(H_2O) = x ; \ c(AcOH) = c(EtOH) = c_0(AcOH) - x[/tex]

therefore:

[tex] K = 4 = \frac {x^2}{(c_0 - x)^2} ; \ \ x =  \frac {4 c_0 }{3 } - \sqrt { \left(  \frac {4 c_0 }{3 } \right) ^2 - \frac {4 c_0^2}{3} } [/tex]

if, for judging this result,   we'd for example choose c0= to be 1 mol/L , x would result in ~ 0.89 mol/L

hence,even with optimum reaction performance you couldn't get a yield better than 89% from this, which, however, was ALL that you could possibly get, i.e. 100% of what you possibly could get - you can't possibly become better than this (given the initial conditions, that is)

however, you could get worse, by goofing here and there, for example

Quote
How about, say, (...)
let's consider the photochlorination of propane at 25°C , with 2 - chloro propane being the desired product

real experimental results are, that you had next to 100% propane consumption in equilibrium, BUT only 57% of it becoming 2-chloropropane, the remaining 43% mainly being 1-chloropropane (asides from some minor mulitchlorination, elimination and that like)

so, the decrease in educt would NOT equal genesis of desired product, hence n0(LR) - neq(LR) does NOT equal n (product from LR , desired)


regards

Ingo
« Last Edit: September 08, 2013, 01:37:45 PM by magician4 »
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Offline Big-Daddy

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Re: Equilibrium yield
« Reply #5 on: September 08, 2013, 03:14:53 PM »
let's consider the photochlorination of propane at 25°C , with 2 - chloro propane being the desired product

real experimental results are, that you had next to 100% propane consumption in equilibrium, BUT only 57% of it becoming 2-chloropropane, the remaining 43% mainly being 1-chloropropane (asides from some minor mulitchlorination, elimination and that like)

so, the decrease in educt would NOT equal genesis of desired product, hence n0(LR) - neq(LR) does NOT equal n (product from LR , desired)

I see. So there 2 points are raised: the "equilibrium theoretical yield" may be different from the "equilibrium actual yield" due to poor (i.e. imperfect) technique; and, more relevant to this discussion (we are having on equilibrium theoretical yield), side reactions may mean that this definition of yield is insufficient.

So then, what is a good mathematical way to interpret yield? Let's say I'm asked to calculate "the mole ratio of A:B required to produce a yield of 99% in a reaction with equilibrium constant K1, of form aA + bB  ::equil:: cC + dD, when A is also involved in another reaction, jA + eE  ::equil:: fF + gG with equilibrium constant K2 and fixed n0(E)". I would need a mathematical definition of the yield to proceed, how would I do it? (obviously you don't need to solve the equations, but if you could help me write them, that would be easily sufficient)

From the look of problems I've solved in the past, I'm aiming here to extend the definition of equilibrium theoretical yield from "yield = n0(LR) - neq(LR) / n0(LR)" that applies to single reactions without side-reactions, to a definition where we can treat the yield of one-reaction even as there are competing side reactions.

Offline magician4

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Re: Equilibrium yield
« Reply #6 on: September 08, 2013, 04:20:38 PM »
if you had two reactions a+b --> c and a+b --> d , with "d" being the desired product, and "c" being a side reaction product, and with two different K-values to denote them (let's for the purpose of simplification say that this was all there is, and let "a" by far be the limiting factor, i.e. db/dt ~ 0 , c0(b)>>c0(a)), you would set up a balance equation:

[tex] (i) \ \ c_0(a) = c_{eq}(a) + c(c) + c(d)  [/tex]

now , lets consider the two LMA expressions, including our side condition db/dt ~0:

[tex](ii) \ \ K_1 \ = \frac {[c]}{[a]_{eq} \cdot c_0(b)} \ \ \to \ \ [c] \ = \ K_1 \cdot [a]_{eq} \cdot c_0(b) [/tex]

[tex](iii) \ \ K_2 \ = \frac {[d]}{[a]_{eq} \cdot c_0(b)} \ \ \to \ \ [d] \ = \ K_2 \cdot [a]_{eq} \cdot c_0(b) [/tex]

introducing (ii) and (iii) to (i)

[tex] (iv) \ \ c_0 (a) = c_{ eq} (a) + \ K_1 \cdot [a]_{eq} \cdot c_0(b)  +   \ K_2 \cdot [a]_{eq} \cdot c_0(b) \ \ \to \ \ c_{ eq}(a) \ = \ \frac {c_o(a)}{c_0(b) \cdot \ ( K_1 + K_2)}[/tex]


if , other than  our initial condition here,  db/dt [itex]\neq \ \ \approx \ 0[/itex] , the expression would become slightly more complex - but in principle not that much (try to work it out for yourself, following the general systematics shown)

now then:
reintroducing (iv) to (iii)

[tex](v) \ \ [d] = \ K_2 \cdot \ \frac {c_o(a)}{c_0(b) \cdot \ ( K_1 + K_2) }    \cdot c_0(b)    =    \frac {K_2 \cdot c_0(a)}{ K_1 + K_2}[/tex]

if by now we define our yield to be "yield = n(d) / n0(a)" , and again for simplification purposes would set V = const. for the time being, it follows

[tex] \ \ yield \ = \ \frac {n(d)}{n_0(a) } \ = \ \frac {[d]}{ c_0(a)} = \ \frac {K_2}{ K_1 + K_2}[/tex]


that's it

regards

Ingo

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Offline magician4

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Re: Equilibrium yield
« Reply #7 on: September 08, 2013, 10:33:56 PM »
erratum:

I see I had a glitch in my derivation..

so (revised, from ~ the middle):



introducing (ii) and (iii) to (i)

[tex] (iv) \ \ c_0 (a) = c_{ eq} (a) + \ K_1 \cdot [a]_{eq} \cdot c_0(b)  +   \ K_2 \cdot [a]_{eq} \cdot c_0(b) \ \ \to \ \ c_{ eq}(a) \ = \ \frac {c_o(a)}{1+ [ c_0(b) \cdot \ ( K_1 + K_2)]}[/tex]


if , other than  our initial condition here,  db/dt [itex]\neq \ \ \approx \ 0[/itex] , the expression would become slightly more complex - but in principle not that much (try to work it out for yourself, following the general systematics shown)

now then:
reintroducing (iv) to (iii)

[tex](v) \ \ [d] = \ K_2 \cdot \ \frac {c_o(a)}{ 1 + [ c_0(b)  \cdot \ ( K_1 + K_2) ]}    \cdot c_0(b)    =    \frac {K_2 \cdot c_0(a) \cdot c_0(b) }{  1 + [c_0(b) ) \cdot \ (K_1 + K_2)] }[/tex]

if by now we define our yield to be "yield = n(d) / n0(a)" , and again for simplification purposes would set V = const. for the time being, it follows

[tex] \ \ yield \ = \ \frac {n(d)}{n_0(a) } \ = \ \frac {[d]}{ c_0(a)} = \ \frac {K_2 c_0(b)}{ 1 + [ c_0(b) \cdot (K_1 + K_2)]}[/tex]


sorry for any confusion I might have caused

regards

Ingo
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Offline Big-Daddy

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Re: Equilibrium yield
« Reply #8 on: September 10, 2013, 02:52:27 PM »
Thanks, this does help.

Now let's suggest, if the reaction had some coefficients - so it was not simply A + B into D but aA + bB into dD - how should we redefine yield from the current definition of neq(D)/n0(A) to accommodate the stoichiometric coefficients?

Separately, let's say we wanted to find the ratio of n0(A) and n0(B) which produces a certain yield; that the coefficients for this are all 1, and that we can proceed with your previous yield definition. Then do we simply need to solve the expression

[tex]K = \frac{y^2 \cdot {n_0(LR)}^2}{n_0(LR) \cdot (1-y) \cdot (n_0(B) - y \cdot n_0(LR))}[/tex]

for the ratio n_0(B) / n_0(LR), and this is the solution? (where y is the yield desired, K is the equilibrium constant, and we assume volume to be constant or 1 dm3). Is that right?

Edit: I believe that this definition and the equation I wrote are entirely consistent, suggesting that your general definition of yield reduces to my original one in the case of a single reaction without side-reactions, which is what I expected and confirms that this one works. However I feel it needs to be further generalized; one issue, for instance, is the stoichiometric coefficients as I mentioned above.
« Last Edit: September 10, 2013, 05:30:40 PM by Big-Daddy »

Offline magician4

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Re: Equilibrium yield
« Reply #9 on: September 10, 2013, 05:58:16 PM »
Quote
Now let's suggest, if the reaction had some coefficients - so it was not simply A + B into D but aA + bB into dD - how should we redefine yield from the current definition of neq(D)/n0(A) to accommodate the stoichiometric coefficients?
I wouldn't want to redefine "yield" here , as I always would be interested in the ratio of what I could get relative to what I've invested  (no matter what the limiting factors were: in this case, the consequences of LMA for example)
all that would change was the way how to calculate for this maximum theoretical yield available, should only LMA rule and nothing else go wrong

to calculate your problem, I would begin with stating that for each d/a [D]eq present in equilibrium exactly 1 [A]0 had vanished, and by the same token that for each d/c [D]eq  exactly one [B ]0 had vanished.

therefore it follows:
[A]eq = c0(A) - a/d [D]eq

[B ]eq = c0(B) - b/d [D]eq

introducing this into the respective LMA:

[tex] K = \frac {[D]^d}{[A]^a \cdot [B ]^b } = \frac {[D]^d}{(c_0(A) - a/d [D] ) ^a  \cdot (c_0(B) - b/d [D] )^b } [/tex]

rearranging this equation should result in an power series expression with [D] of the highest order of a+b OR d , respectively (it depends on the real values of a,b and d)
this you'll have to solve, either analytically or numerically (depending on this very order) to gain an expression for [D] in equilibrium thereof

any by the end of the day (and with the volume considered to be a constant, for simplification purposes), again

yield, related to A  = n(A) , transformed as desired / n0 (A) =  a/d * n(D)eq / n0 (A)

or

yield, related to B  = n(B) , transformed as desired / n0 (B) =  b/d * n(D)eq / n0 (A)

would be my proposal

Quote
Separately, let's say we wanted to find the ratio of n0(A) and n0(B) which produces a certain yield;
speaking generally, there is no such "fixed" ratio  (even with a given K) for all concentrations in equilibrium reactions: only if enumerator and denominator were of same dimension in the respective LMA expression, i.e. in a very special subcase, this might occur.

in other words: there's a bunch of chemical reactions out there, where ratio of components  in state of equilibrium will shift, simply by diluting the whole situation with additional solvent, for example

however, calculating for a yield - dependant ratio with a given concentration of one of the two components can be done nevertheless:

as shown above, we needed an expression for [D]eq , being a solution for the respective LMA expression
this solution in general should look something like
[D]eq = fa(1)* c0(A)x + fa(2)* c0(A)(x-1)..fa(n) c0(A) + ka +  fb(1)* c0 (B)y + fb(2)* c0(B)(y-1)..fb(m) c0(B) + kb

now, if we're looking for a certain yield at a given volume (let's say with respect to A), I'd use my general, concentration related  yield expression "yield = a/d * [D]eq/c0(A)" and combine it with above expression.
for a desired yield ydes, the result would look something like this:
ydes = a/d* 1/c0(A) * (fa(1)* c0(A)x + fa(2)* c0(A)(x-1)..fa(n) c0(A) + ka +  fb(1)* c0 (B)y + fb(2)* c0(B)(y-1)..fb(m) c0(B) + kb

now, I would choose my initial concentration of A as per my wishes, and would result in something like
ydes = a/d* 1/c0(A) * (const.1(c0(a)) +  fb(1)* c0 (B)y + fb(2)* c0(B)(y-1)..fb(m) c0(B) + kb

 :rarrow: this is a power series expression in co(B) , hence can be calculated for

regards

Ingo
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Offline Big-Daddy

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Re: Equilibrium yield
« Reply #10 on: September 10, 2013, 07:14:12 PM »
Thanks, again this clears up that issue. Now, one final question that has occurred to me that complicates things once more. What if there are some moles of d present initially, or the product is contributed to from other reactions (with independently calculable n of the product being formed by the other reactions) - then would the concept of yield include the initial moles, or number of moles coming from external reactions, or not? And what does this generalized yield expression look like?

I see problems both ways - if you include the initial number of moles, then you are not considering the yield of this reaction alone; and if you don't, then you would end up with a different yield depending on which product you concern yourself with, if the products have different initial numbers of moles.

Offline magician4

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Re: Equilibrium yield
« Reply #11 on: September 10, 2013, 08:50:27 PM »
Quote
What if there are some moles of D present initially, (...)
if you had some D already present right from the start, this would be tantamount to an initial concentration c0 for D, and you would need to discriminate between c0(D) and additional D (or less: this depends on K) : I'd like to call this Dnew

the LMA expression thereof would look like this:

[tex] K = \frac {(c_0(D) + [D_{new}] )^d}{( c_0(A) - a/d [D_{new}])^a \cdot (c_0(B) - b/d [D_{new}])^b } [/tex]

Quote
or the product is contributed to from other reactions (..)
you think of something like A+B --> P ; C+D --> P in parallel, in the same solution, each reaction with its own K-value?
this of course would mean interacting equilibria.
those can be calculated for, too- but it might become somewhat more complex here, I agree
nevertheless, the basic principles of setting up equations, combining them, solving them... would remain the same as already shown


however, what you would need to discuss in those cases would be, what "yield" here should mean, and I guess that the more complicated the situation becomes, the more this definition depends on what you want it to mean, i, e. what you think would be meaningfull and suitable to the cause

therefore, I don't think that a generalized yield definition (esp. expressed in a mathematical way) would exist: you have to come up with one on your own, suited to the situation and what you wish it to be, and arrange your math accordingly

regards

Ingo
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Offline Big-Daddy

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Re: Equilibrium yield
« Reply #12 on: September 11, 2013, 03:56:58 PM »
I see. So being asked to calculate the initial mole ratios needed for a 90% yield could be interpreted in various reasonable ways if there are a certain number of moles of the product(s) present as well?

Whereas - I need to check my understanding here - if we can say that nx(P) moles of product P are contributed by a reaction other than the one for which we are given a certain desired yield, then the yield of this reaction can be expressed as

[tex]yield = \frac{(n_{eq}(P) - n_x(P)) \cdot v(LR)}{n_0(LR) \cdot v(P)}[/tex]

v(LR) is the stoichiometric coefficient on LR and v(P) is the stoichiometric coefficient on the product. And if we had multiple side-reactions each contributing some number of moles to P, we simply need to subtract all of these so that we have in the numerator of our yield expression, the number of moles of P, at equilibrium, produced by the reaction for which this yield is given. This value, our yield as defined above, will be the same for all products in a single reaction, so if we had both P and Q being products, we could also write

[tex]yield = \frac{(n_{eq}(P) - n_x(P)) \cdot v(LR)}{n_0(LR) \cdot v(P)} = \frac{(n_{eq}(Q) - n_x(Q)) \cdot v(LR)}{n_0(LR) \cdot v(Q)}[/tex]

where nx(Q) is really the sum of number of moles of Q produced by all reactions other than this one. Am I right?

Offline magician4

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Re: Equilibrium yield
« Reply #13 on: September 12, 2013, 07:06:23 AM »
Quote
I see. So being asked to calculate the initial mole ratios needed for a 90% yield could be interpreted in various reasonable ways if there are a certain number of moles of the product(s) present as well?
in my opinion, yes
Quote
(..)then the yield of this reaction can be expressed as (...)
I don't see a reason to introduce those stoichiometric factors here, as those only would come into the picture with LMA expressions.

as I said, if you had two reactions resulting the desired product, and if they were interconnected via the concentration of the product, you'd have to connect the two LMA expressions, and solve everything for the desired value (i.e. equilibrium concentration of P)
those expressions could become quite complex (depending on the very LMA's to consider), and I don't think that your proposal would fit the bill, esp. not in a generalized way

for example:

let's consider two reactions leading to the same product, for some odd reason carried out in the same system (those things in reality do happen for example in biochemistry, where the amount of P might be regulated by two different processes in a cell)

process (I) : A + B  :rarrow: P  (let's say an esterification)
process (II) : D  :rarrow: P (let's say an isomerization)

so we could write the respective LMA expressions:

[tex] K_{(I)}  =  \frac {[P_{total}]}{[A]  \cdot  [B ]} =  \frac {[P_{(I)}] + [P_{(II)}]} {(c_0(A) - [P_{(I)}])  \cdot  (c_0(B) - [P_{(I)}]) } \to [P_{(II)}] =  K_{(I)}  \cdot  (c_0(A) - [P_{(I)}]) \cdot (c_0(B) - [P_{(I)}]) - [P_{(I)}] [/tex]

[tex] K_{(II)} =  \frac {[P_{total}]}{[D]}  =  \frac {[P_{(I)}] + [P_{(II)}]}{c_0(D) - [P_{(II)}]}  \to  [P_{(II)}] =  \frac { K_{(II)} \cdot  c_0(D) - [P_{(I)}]} { 1 + K_ {(II)}} [/tex]

and combine them
[tex] K_{(I)}  \cdot  (c_0(A) - [P_{(I)}]) \cdot (c_0(B) - [P_{(I)}]) - [P_{(I)}] =  \frac { K_{(II)} \cdot  c_0(D) - [P_{(I)}]} { 1 + K_ {(II)}} [/tex]

this equation can be rearranged to result an expression in [P(I)]2 , and solved thereafter
reintroducing the result into above original expression for K(II) will give you [P(II)] , and [P(I)]+[P(II)] finally the [Ptotal]

now you have "one half" of a yield expression, and the only thing remaining would be to decide for a meaningfull educt-c0 / combination of educt-c0s it should be related to


regards

Ingo
« Last Edit: September 12, 2013, 07:34:50 AM by magician4 »
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Offline Big-Daddy

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Re: Equilibrium yield
« Reply #14 on: September 12, 2013, 02:21:55 PM »
I don't see a reason to introduce those stoichiometric factors here, as those only would come into the picture with LMA expressions.

If we didn't include the stoichiometric factors in the definition of the yield, we could easily end up with a 200% yield. Also, since "the yield" for a reaction is fixed in definition whichever product we refer to, if we have products with different stoichiometric coefficients, we will need to make up for the difference, as different numbers of moles of each will be present at equilibrium despite the same yield of reaction applying to both. Hence I think there must be a place in the definition of yield (in terms of equilibrium number of moles and initial number of moles) for the stoichiometric coefficients.

as I said, if you had two reactions resulting the desired product, and if they were interconnected via the concentration of the product, you'd have to connect the two LMA expressions, and solve everything for the desired value (i.e. equilibrium concentration of P)
those expressions could become quite complex (depending on the very LMA's to consider), and I don't think that your proposal would fit the bill, esp. not in a generalized way

for example:

let's consider two reactions leading to the same product, for some odd reason carried out in the same system (those things in reality do happen for example in biochemistry, where the amount of P might be regulated by two different processes in a cell)

process (I) : A + B  :rarrow: P  (let's say an esterification)
process (II) : D  :rarrow: P (let's say an isomerization)

so we could write the respective LMA expressions:

[tex] K_{(I)}  =  \frac {[P_{total}]}{[A]  \cdot  [B ]} =  \frac {[P_{(I)}] + [P_{(II)}]} {(c_0(A) - [P_{(I)}])  \cdot  (c_0(B) - [P_{(I)}]) } \to [P_{(II)}] =  K_{(I)}  \cdot  (c_0(A) - [P_{(I)}]) \cdot (c_0(B) - [P_{(I)}]) - [P_{(I)}] [/tex]

[tex] K_{(II)} =  \frac {[P_{total}]}{[D]}  =  \frac {[P_{(I)}] + [P_{(II)}]}{c_0(D) - [P_{(II)}]}  \to  [P_{(II)}] =  \frac { K_{(II)} \cdot  c_0(D) - [P_{(I)}]} { 1 + K_ {(II)}} [/tex]

and combine them
[tex] K_{(I)}  \cdot  (c_0(A) - [P_{(I)}]) \cdot (c_0(B) - [P_{(I)}]) - [P_{(I)}] =  \frac { K_{(II)} \cdot  c_0(D) - [P_{(I)}]} { 1 + K_ {(II)}} [/tex]

this equation can be rearranged to result an expression in [P(I)]2 , and solved thereafter
reintroducing the result into above original expression for K(II) will give you [P(II)] , and [P(I)]+[P(II)] finally the [Ptotal]

now you have "one half" of a yield expression, and the only thing remaining would be to decide for a meaningfull educt-c0 / combination of educt-c0s it should be related to


regards

Ingo

That's fine, I agree we would need to connect the yield expressions. This would be the only way of getting reasonable values of nx(P) through the simultaneous equations. But after doing so, I don't see the problem of regarding "yield of reaction A" as equilibrium number of moles of a product P of reaction A, minus the number of moles of P which were produced by reactions other than A, divided by the initial number of moles of the limiting reagent for A and corrected for stoichiometry? As far as I can see, this boils down to the same as your initial proposal of " number of moles of limiting reagent transferred out, divided by number of moles of limiting reagent initially present", post #3.

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