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Concentration of water vapor present at equilibrium

  1. Jul 1, 2004 #1


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    3 Fe (S) + 4 H2O (g) = Fe3O4 + 4 H2 (g)

    This equilibrium has a Kc of 4.6 @ 850 degree C. Determine the concentration of water vapor present at equilibrium if the reaction is initiated using 8.00 grams H2 and an excess of iron oxide, fe3O4, in a 16.0 liter container.

    This is what I have so far:

    Kc = [H2]^4 / [H20]^4

    4.6 = [0.25]^4 / X^4

    X = [0.170697] H2, Since this is a one:eek:ne reaction then the concentration of the water vapor would equal the concentration of H2, correct?
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  3. Jul 1, 2004 #2
    SO, from PV=nRT,
    I get 2.3 atm for the pressure of H2
    (using 1123.15 K, and R = 0.00821 L atm/mol K, V = 16 L)

    From that I get (2.3)^4 / 4.6 = p(H2O)^4

    using my calculator tells me that p(H20) is 3.37 atm.

    Because it is 1:1 in the equation means that IF the reaction went to completion, then there will be an amount of oxygen (in moles) equal to the former amount of hydrogen (in moles). But, the reaction did bot go to completion. That's why they gave you the equilibrium constant.

    edit: I don't know where I got O2 from :), I replaced it with H20, and apparently my calculator doesn't do quarter roots very well.
    Last edited: Jul 6, 2004
  4. Jul 1, 2004 #3


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    No !

    The moles of water vapor produced = moles of H2 consumed. If the initial partial pressure of H2 is P (this can be calculated from the Ideal Gas Equation), then the final pressures are P(H2) = P - X and P(H2O) = X.
  5. Jul 1, 2004 #4


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    The product of the reaction is water vapor and thus it should be the numerator in the equilibrium Kc equation. Also , I believe you need to include iron oxide in the equilibrium equation (unless in solid form). Due to these factors and also mathematical principles and situational variations it is not correct to say that the concentrations are equal.

    hope this helps

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