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Chem II - Equilibrium

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For the reaction at 430 degree C, H2 + I2 <=> 2HI Kc = 54.3. Initially 0.714 moles of H2 and 0.984 miles of I2 and 0.886 moles of HI are placed in a 2.40 L vessel. The equilibrium concentration of H2 and I2 are _____.

Ok, I tried this by changing the moles to M by dividing the moles by 2.40 liters, and subing into the Kc formula, but I did not get any of the choices listed for the multiple choice. What am I doing wrong?

  • Chem II - Equilibrium - ,

    You were right to change the mols to molarity by mol/2.4 = M.
    H2 + I2 ==> 2HI
    Set up an ICE chart.
    initial:
    (H2) = 0.714/2.4 =
    (I2) = 0.984/2.4 =
    (HI) = 0.886/2.4 =

    Now do the reaction quotient =
    (HI)^2/(H2)(I2) = (0.369)^2/(0.2975)(0.410) = about 1.1 or so which means at equilibrium the reaction will have gone to the right.

    change:
    (HI)= +2y
    (H2) = -y
    (I2)= -y

    equilibrium:
    (HI) = 0.369+2y
    (H2) = 0.2975-y
    (I2) = 0.41-y

    Substitute equilibrium values into Keq and solve for y and add that back into the equilibrium set up to arrive at the final concns.

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