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    Here it is


    Hydrogen chloride can be oxidised to chlorine by the Deacon process:
    4HCl(g) + O2(g) == 2Cl2(g) + 2H2O(g)
    0.800 mol of hydrogen chloride was mixed with 0.200 mol of oxygen in a vessel of volume 10 dm3. At equilibrium it was found that the mixture contained 0.200 mol of hydrogen chloride. Calculate Kc for the reaction



    apparently the answer is 1013 mol-1dm3


    Please show working! Thanks!
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    If 0.200 moles of HCl remain at the end, it means that 0.800-0.200 = 0.600 moles of HCl have reacted.

    If 0.600 moles of HCl have reacted, looking at the moles ratio from the balanced equation, then 0.600/4 = 0.150 moles of O2 have reacted.
    Also, it means that 0.600/2 = 0.300 moles of Cl2 have formed...and 0.600/2 = 0.300 moles of H20 have formed.

    0.200 - 0.150 = 0.050 moles of O2 remain at equilibrium.

    Now you need to find the concentrations of each reactant and product at equilibrium before putting their values into the equation for Kc, the equilibrium constant.
    Using concentration = moles/volume....
    conc of HCl = 0.200/10 = 0.0200 moles/dm^3
    conc of O2 = 0.050/10 = 0.0050 moles/dm^3
    conc of Cl2 = 0.300/10 = 0.0300 moles/dm^3
    conc of H2O = 0.300/10 = 0.0300 moles/dm^3

    And finally, Kc = [Cl2]^2 [H2O]^2 / [HCl]^4 [O2]

    Kc = 0.0300^2 x 0.0300^2 / 0.0200^4 x 0.0050
    = 1012.5 which rounds off to your answer
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    ok, so the Kc for the reaction is

    K_c=\dfrac{[Cl_2]^2[H_2O]^2}{[HCl]^4[O_2]}



    So take it from there... (rules of the forum are not to just provide people with answers, rather set them on their way)
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    guys, guys! don't post complete answers!
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    (Original post by Saunterer)
    If 0.200 moles of HCl remain at the end, it means that 0.800-0.200 = 0.600 moles of HCl have reacted.

    If 0.600 moles of HCl have reacted, looking at the moles ratio from the balanced equation, then 0.600/4 = 0.150 moles of O2 have reacted.
    Also, it means that 0.600/2 = 0.300 moles of Cl2 have formed...and 0.600/2 = 0.300 moles of H20 have formed.

    0.200 - 0.150 = 0.050 moles of O2 remain at equilibrium.

    Now you need to find the concentrations of each reactant and product at equilibrium before putting their values into the equation for Kc, the equilibrium constant.
    Using concentration = moles/volume....
    conc of HCl = 0.200/10 = 0.0200 moles/dm^3
    conc of O2 = 0.050/10 = 0.0050 moles/dm^3
    conc of Cl2 = 0.300/10 = 0.0300 moles/dm^3
    conc of H2O = 0.300/10 = 0.0300 moles/dm^3

    And finally, Kc = [Cl2]^2 [H2O]^2 / [HCl]^4 [O2]

    Kc = 0.0300^2 x 0.0300^2 / 0.0200^4 x 0.0050
    = 1012.5 which rounds off to your answer

    Thank you sooo much, you pointed out where i went wrong!

    Rep!

 
 
 
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