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Topic: Calculation of reaction quotient with a change in pressure  (Read 2936 times)

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Offline plioe

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Calculation of reaction quotient with a change in pressure
« on: June 04, 2010, 06:37:10 AM »
Hi guys,

   I have some difficulty in answering this question. If anyone would enlighten me that would be great.
The question is this:

  The reaction of hydrogen production can be described by H2O(g) + CH4(g) <-----> CO(g) + 3H2(g) for which Kc = 0.26 at 1200C

The gaseous molecules H2O and CH4 are mixed in a reaction vessel of volume 0.32 dm^3. At equilibrium the vessel contains 0.26 mole of CO, 0.091 mole of H2 and 0.041 mole of CH4.

(a) What is the equilibrium concentration of H2O(g)?

     Kc = [CO][H2]^3 / [H2O][CH4]
     0.26 = [0.26][0.091^3] / [H2O][0.041]
Rearranging and solving, [H2O] = 0.018 mole

I'm not sure whether this is correct. Can anyone check it for me?

(b) If the volume of the flask at equilibrium were suddenly reduced to half its original volume(ie 16 dm^3) what would be the value of Qc?

   I have no idea how to solve this. The answer is 1.04 though.

(c) What is the value of Kp for the above equilibrium?

    I used the formula Kp = Kc(RT)^-n
    so Kp = 0.26*(8.314*1473)^2
            = 38.9 million
    The answer is 39000. What did i do wrong?
 

   

Offline Jorriss

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Re: Calculation of reaction quotient with a change in pressure
« Reply #1 on: June 13, 2010, 03:18:37 AM »
You're off by a factor of 1000, is it related to dm^3?

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