Chemical Equilibrium and Probability

  • Thread starter flybynight
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  • #1
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I am looking at this equilibrium:
H2O + D2O <-> 2DHO
(D = Deuterium)
I have 2.0 mol of H2O, 1.0 mol of D2O and 2.0 mol of DHO.
What are the equilibrium amounts of each of the molecules?

What I tried:
6 mol of H, 4 mol of D
H2O: P(X=H) & P(Y=H) = 9/25
D2O: P(X=D) & P(Y=D) = 4/25
HDO: P(X=D) & P(Y=H) + P(X=H) & P(Y=D) = 12/25

Moles of H2O: 9/25 * 10 = 3.6 mol
Moles of D2O: 4/25 * 10 = 1.6 mol
Moles of DHO: 12/25 * 10 = 4.8 mol

However, that makes 12 mol H and 8 mol D. I know I probably should have divided somewhere, but I don't know where.

Thanks,
Peter
 

Answers and Replies

  • #2
Ygggdrasil
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Moles of H2O: 9/25 * 10 = 3.6 mol
Moles of D2O: 4/25 * 10 = 1.6 mol
Moles of DHO: 12/25 * 10 = 4.8 mol

It should be prob * 5 moles, since you started with 2 mol H2O + 1 mol D2O + 2 mol DHO = 5 moles.
 

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