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Thread: Please help with Kc calculation problem... 10 easy points...?

  1. #1 Please help with Kc calculation problem... 10 easy points...? 
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    There is this really simple question regarding calculating the Kc value at equilibrium.

    I will show you how I worked it out, and you can tell if what I did is correct, because I'm told by my friend that I should get a different answer.

    Chemical Equation: I2 + H2 = 2HI ; H<0 (i.e. forward reaction is exothermic)

    Initial:
    2 mol of Iodine, 0 mol of Hydrogen and 8 mol of HI.

    At equiblibrium:
    It is found that there is 4 mol HI left.

    Volum = 0.5dm^3
    WORKING OUT:

    HI:

    Inital = 8
    Change = -4
    Mol at equilibrium = 4
    Concentration at equilibrium: 8M


    I2:

    Inital = 2
    Change = +2
    Mol at equilibrium = 4
    Concentration at equilibrium: 8M


    H2:

    Inital = 0
    Change = +4
    Mol at equilibrium = 2
    Concentration at equilibrium: 4M


    Therefore, Kc = [products]/[reactants].

    Therefore, Kc = (8^2)/(8*4)

    Kc = 2.


    That is what I got as my answer.


    However, someone tells me that the answer should be 0.5 (i.e. 1 / my answer), because apparently the reverse reaction is favoured.

    However, I've never heard of this regarding the calculation of Kc.

    Why would it be [I2][H2] / [HI]^2

    That doesn't make sense, or am I missing something?




    Please help.

    Thanks


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  3. #2  
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    H2 Change is +2, not +4.

    That was just a typo, never mind that.


    Basically, the problem is that I get Kc = 2, and my friend gets Lc = 0.5 (i.e. 1 / my answer, because he says the reverse reaction is favoured).

    Who is correct?


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