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Yield question

  1. Aug 20, 2013 #1
    If I want to calculate the number of moles ratio of reactants which, in a reversible reaction of equilibrium constant K, will give me a certain yield 'f', what is the mathematical definition of the yield in this case?

    For instance, in the reaction A + B ⇔ C + D, let's say we can denote extent by ΔA such that K=(C0-ΔA)*(D0-ΔA)/((A0+ΔA)*(B0+ΔA)). Then is yield f=-ΔA/A0 (i.e. 90% yield would suggest that ΔA=-0.9*A0)? The problem becomes that then this is not 90% of everything, because B0 may not be the same as A0, in which case 90% of B0 would not have been converted, only 90% of A0. And if this "90% yield" was done with ΔB instead then our definition would involve 90% of B0 being converted but not 90% of A0, which provides two conflicting understandings of the "90% yield" statement.
  2. jcsd
  3. Aug 21, 2013 #2
    You have to take into account the limiting reagent. If you have 1 mol of A and 2 mol of B in a flask and the reaction is allowed to go to completion you will get 100% yield if you have (at the end) 0 mol of A, 1 mol of B, 1 mol of C and 1 mol of D. If you instead get something like 0.5 mol of C/D then your yield is 50% etc.

    Wiki is great.
  4. Aug 21, 2013 #3
    Ok so "90% yield" means "90% of the number of moles of the limiting reagent (i.e. whichever of the reactants has the smallest ratio of number of moles/coefficient in the reaction) were transferred out to products, leaving 10% of the number of moles of that reagent".

    So if a question says "work out the initial mass ratio of reactants that will produce a 90% yield" for a reaction aA + bB + cC -> products, there are actually 3 ratios that will work - one found by saying that C is the limiting reagent (so ΔC=-0.90*C0), one by saying A is the limiting reagent (ΔA=-0.90*A0) and one by saying B is the limiting reagent (ΔB=-0.90*B0). The higher the stoichiometric coefficient and Mr for the limiting reagent you choose, the lower the mass ratio (of each non-limiting reagent to the limiting reagent) will turn out to be (by comparison to the mass ratios found by considering the other reactants as limiting reagents).
    Last edited: Aug 21, 2013
  5. Aug 21, 2013 #4
    No. The limiting reagent is not the reactant with the smallest stoichiometric coefficient. Consider your above equation with a = 5 and b = 1, and you mix 2 mol of A and 10 mol of B. Surely the B is not limiting?

    To know the mass ratio you'd need knowledge of the molecular weight of each reactant.

    Also, the way I think about yield, you cannot MAKE the yield different by changing the limiting reagent. The yield of a reaction considers a theoretical yield which predicts how much you get if the reaction proceeded completely. Maybe you can take the equilibrium constant into account, if known, and calculate a theoretical yield from that but I am not certain that that is actually done (it seems redundant to me to do that). Regardless, if you change the limiting reagent you change the theoretical yield as well. So you cannot change the yield of a reaction by changing how much of each reagent you add.

    Changing yield usually involves playing with the conditions of the reaction. For example, in reactions with small equilibrium constants, you can change the amount of reactants/products, temperature, pressure etc. This is frequently done in an effort to drive certain reactions in certain directions.

    I don't exactly understand what you mean here. What is Mr?
  6. Aug 21, 2013 #5
    As I said, "smallest ratio of number of moles/coefficient in the reaction". Maybe I could have been clearer. Whichever of the reactants has a smaller value of n[Species]/v[Species] where v[Species] is the stoichiometric coefficient of that species in the reaction and n[Species] is the number of moles of that species present, is the limiting reagent. Ok?

    This is an equilibrium question. So when they say "work out the initial mass ratio of reactants that will produce a 90% yield", they are asking you to work out m[Reactant]/m[Limiting Reagent] which will produce a "90% yield". Originally my question was, what mathematically is this "90% yield". But now I think I have my answer: it is the fraction of number of moles of the limiting reagent which is transferred out. In the equilibrium calculation, ΔA=-yield*A0 where A0 is the initial number of moles of A.

    Mr is molecular weight. What I am saying here is making the prediction that choosing as your limiting reagent a higher Mr reactant with a higher stoichiometric coefficient in the reaction will end up giving you a lower mass ratio of m[Reactant]/m[Limiting Reagent] where m[Reactant] is any one of the other reactants besides the limiting reagent. Do you agree?
  7. Aug 27, 2013 #6
    Sorry, been really busy lately and haven't had much chance to play on PF.

    Very quickly, everything you have written makes sense but I'm not quite sure that yield is calculated in that way. I think yield assumes the reaction goes to completion, then when you have done the actual experiment you figure out how much product you actually made and do actual/theoretical ratio.

    Then again I only ever really see yield discussed in synthesis type work and I never paid much attention to how they actually calculate it, I've always assumed that yield means actual/theoretical where theoretical = to completion. Maybe someone with more experience can answer definitely on this issue.

    What you are talking about above (namely the ΔA) is called extent of reaction, you can google it to see how it can be used to get some useful information.
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