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} = c

_{0}(A) - a/d [D]

_{eq}[B ]

_{eq} = c

_{0}(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 / n

_{0} (A) = a/d * n(D)

_{eq} / n

_{0} (A)

or

yield, related to B = n(B) , transformed as desired / n

_{0} (B) = b/d * n(D)

_{eq} / n

_{0} (A)

would be my proposal

Separately, let's say we wanted to find the ratio of n_{0}(A) and n_{0}(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} = f

_{a(1)}* c

_{0}(A)

^{x} + f

_{a(2)}* c

_{0}(A)

^{(x-1)}..f

_{a(n)} c

_{0}(A) + k

_{a }+ f

_{b(1)}* c

_{0} (B)

^{y} + f

_{b(2)}* c

_{0}(B)

^{(y-1)}..f

_{b(m)} c

_{0}(B) + k

_{b }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}/c

_{0}(A)" and combine it with above expression.

for a desired yield y

_{des}, the result would look something like this:

y

_{des} = a/d* 1/c

_{0}(A) * (f

_{a(1)}* c

_{0}(A)

^{x} + f

_{a(2)}* c

_{0}(A)

^{(x-1)}..f

_{a(n)} c

_{0}(A) + k

_{a }+ f

_{b(1)}* c

_{0} (B)

^{y} + f

_{b(2)}* c

_{0}(B)

^{(y-1)}..f

_{b(m)} c

_{0}(B) + k

_{b }now, I would choose my initial concentration of A as per my wishes, and would result in something like

y

_{des} = a/d* 1/c

_{0}(A) * (const.

_{1}(c

_{0}(a)) + f

_{b(1)}* c

_{0} (B)

^{y} + f

_{b(2)}* c

_{0}(B)

^{(y-1)}..f

_{b(m)} c

_{0}(B) + k

_{b } this is a power series expression in c

_{o}(B) , hence can be calculated for

regards

Ingo