Jon P. answered • 02/24/15
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The equilibrium constant would be equal to [HCONH2] / ([CO][NH3]).
(No powers are required because the coefficients of all reactants are 1.
[CO][NH3] = 1 * 2 = 2.
So [HCONH2] / 2 = 0.84.
Therefore [HCONH2] = 1.68 M
Jon P.
tutor
Sorry, you're right -- I misunderstood how to do the problem. Please ignore my answer, I have to think about it some more.
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02/24/15
Jon P.
tutor
OK...
Suppose x moles of CO react with x moles of NH3 to lead to the equilibrium. This will result in the creation of x moles of HCONH2. At that point, there will be 1-x moles of CO and 2-x moles of NH3. So the equilibrium constant will be equal to:
[HCONH2] / ([CO][NH3]) = x / ((1-x)(2-x)) which will be equal to 0.84.
Let's solve this...
x / ((1-x)(2-x)) = 0.84
x = 0.84 (1-x)(2-x)
x = 0.84(2 -3x +x2)
x = 1.68 - 2.52x + 0.84x2
0 = 1.68 -3.52x + 0.84x2
x = 3.52 ± √((-3.52)2 -4(0.84)(1.68))
-------------------------------------
2*0.84
= 3.52 ± √2(12.3904 -5.6448)
---------------------------------
1.68
= (3.52 ± 2.60) / 1.68 = 3.64 or 0.55
3.64 is impossible because that is greater than the maximum possible concentration of HCONH2. So 0.55 is the correct value of x, and that's the concentration of HCONH2.
Is that better, or still wrong somewhere?
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02/24/15
Leigh A.
Yes, that one was right! Thank you so much for the help! (:
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02/24/15
Leigh A.
02/24/15