# Chemistry

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I'm working with a reaction of the form:

A + B -> C

Problem: IF the equilibrium constant is 4.2 and an equimolar mixture of A + B are reacted, verify that there will be an equilibrium yield of 67%.

Here is my attempt thus far.

4.2 = [C] / ([A]*[B])
x = starting molar quantity of A/B
y = final molar quantity of C
v = volume of reaction mixture

4.2 = y/v / ((x-y)^2/v^2)
4.2 = yv / (x^2-2xy+y^2)

And I'm stuck. Any help is greatly appreciated.

• Chemistry - ,

I have struggled with this problem and I believe there is an inherent fallacy in the problem. For example, if the reaction is
A + B ==> C, then
K = (C)/(A)(B) = 4.2

BUT, by Le Chatelier's Principle, if we increase A and/or B, the reaction will be shifted to the right and more C will be formed.
Try it with 1 M A and 1 M B and you end up with 0.617 C and 1-0.617 = 0.383 A and B for a yield of 61.7%(100*0.617/1).

With 2 M A and 2 M B, we end up with 1.42 C and 2-1.42 = 0.58 A and B for a yield of 71% (100*1.42/2).

With 3 M A and 3 M B, we end up with 2.265 C and 3-2.265 = 0.734 for a yield of 73.4% (100*2.265/3).

So the percentage changes because the ratio changes even though I started with equimolar amounts of A and B.

Working backwards.
A + B ==> C
Start with 1 M A and 1 M B, you can't get 4.2 for K if 67% is used for the yield (if I understand what equilibrium yield is) but 4.2 is obtained if we use 61.7%. The same is true for the other amounts of A and B initially.

If you find that equilibrium yield is something different than (C)/(A)at equilibrium please let me know.

• Chemistry - ,

Thanks so much. I thought I was missing something, but you are right. Either I'm misreading the problem or it can't be solved from the information provided. I'll turn that in. Thanks!

• Chemistry - ,

Since the answer I got is a bit different from the one given, here is my complete take on this:

Assume 1 mole of A and 1 mole of B reacting (This is legitimate since the yield should be independent of the actual number of moles used as long as they are equal). Let's represent the moles of A and B reacting by "x". Then we reach the equilibrium amounts as follows:

A B C
1 1 0 Initial
-x -x +x Change
(1-x) (1-x) x Final (Equil.)

x / [(1-x)^2] = 4.2
x = 4.2(1-x)^2
x / (1 -2x + x^2) = 4.2
x = (4.2)(1 -2x + x^2)
x = 4.2 - 8.4x + 4.2x^2
0 = 4.2 - 9.4x + 4.2x^2
x = 0.62 and 1.62
Since x must be less than 1 mole, x = 0.62. We started with 1 mole A, 1 mole B. The 0.62 moles of either that reacted represents 62% of the initial amount of of 1 mole.

• Chemistry - ,

The 62% you obtained is the same as my 61.7% for 1 M (except I'm not allowed that many significant figures and I obtained that value the same way) but if we use 2 or 3 initially, it isn't independent of concn. The 2 works out to be 72% and 3 works out to be 75%. Higher values gives still higher percentages. But isn't that correct? Larger values of reactants should give larger values of percent (and a shift to the right).

• Chemistry - ,

I think the problem was constructed with the false assumption that the yield is independent of concentration and I accepted it uncritically. Le Chatelier's Principle still rules. Glad you were so alert. I checked [A]=[B]=3M with my setup and got a 75% yield, farther to the right than 1M. Close enough to yours. Cheers.

• Chemistry - ,

The problem I have with this is, that if I change the initial 1 concentration to something else, like 2 for example, then x/2 solves to a different yield % (71%). This leads me to believe that there is no constant yield percent for the given problem.

• Chemistry - ,

There isn't a constant yield and there shouldn't be. Notice I used 2 M in one of my calculations and obtained 71%. 3M gives 75% and higher values initially (4,5,6 etc) gives still higher percentages. That's why I said in my response that the problem was based on a fallacy. As GK has pointed out, Le Chatelier's Principle still rules and an increase in A or an increase in B forces the equilibrium to the right, producing more C and a higher percentage yield (with the caveat that I may not be using the same definition of equilibrium yield as the person who constructed the problem).