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    4HCl + O2 = 2Cl2 + 2H2O

    0.4 mol of HCl and 0.1 mol of O2 were placed in a 4 dm3 vessel
    At eqm only 0.04 mol of HCl was present. Find Kc

    I said it was 164025 mol-1 dm3

    But the book said it was 1.6 mol-1 dm3

    I am fairly sure they are wrong
    could someone clarify please
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    (Original post by jsmith6131)
    4HCl + O2 = 2Cl2 + 2H2O

    0.4 mol of HCl and 0.1 mol of O2 were placed in a 4 dm3 vessel
    At eqm only 0.04 mol of HCl was present. Find Kc

    I said it was 164025 mol-1 dm3

    But the book said it was 1.6 mol-1 dm3

    I am fairly sure they are wrong
    could someone clarify please
    Strange.. I got the answer to be 41006.25 mol^-1dm^3
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    are you sure you used the correct mol ratio?
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    (Original post by jsmith6131)
    are you sure you used the correct mol ratio?
    Yeah.. what values did you use?
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    (Original post by jsmith6131)
    4HCl + O2 = 2Cl2 + 2H2O

    0.4 mol of HCl and 0.1 mol of O2 were placed in a 4 dm3 vessel
    At eqm only 0.04 mol of HCl was present. Find Kc

    I said it was 164025 mol-1 dm3

    But the book said it was 1.6 mol-1 dm3

    I am fairly sure they are wrong
    could someone clarify please
    initially

    4HCl + O2 = 2Cl2 + 2H2O
    0.4 ..... 0.1 ..... 0.0 .....0.0

    If we assume it's water vapour and it takes part...

    at equilibrium

    4HCl + O2 = 2Cl2 + 2H2O
    0.04 ..... 0.01 ..... 0.18 .....0.18

    Concentrations in 4 litre vessel

    4HCl + O2 = 2Cl2 + 2H2O
    0.01 ..... 0.0025 ..... 0.045 .....0.045

    equilibrium law

    kc = [Cl2]2[H2O]2/[HCl]4[O2]

    kc = (0.002025 * 0.002025)/1x10-8 * 0.0025

    kc = 164025

    I agree with you
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    ok thanks
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    (Original post by NutterFrutter)
    Yeah.. what values did you use?

    (Original post by charco)
    initially

    4HCl + O2 = 2Cl2 + 2H2O
    0.4 ..... 0.1 ..... 0.0 .....0.0

    If we assume it's water vapour and it takes part...

    at equilibrium

    4HCl + O2 = 2Cl2 + 2H2O
    0.04 ..... 0.01 ..... 0.18 .....0.18

    Concentrations in 4 litre vessel

    4HCl + O2 = 2Cl2 + 2H2O
    0.01 ..... 0.0025 ..... 0.045 .....0.045

    equilibrium law

    kc = [Cl2]2[H2O]2/[HCl]4[O2]

    kc = (0.002025 * 0.002025)/1x10-8 * 0.0025

    kc = 164025

    I agree with you

    I agree but I got values closer to nutter when I used 0.225 instead of 0.18 because I tried to make the number of moles equate which seemed wrong.
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    (Original post by charco)
    initially

    4HCl + O2 = 2Cl2 + 2H2O
    0.4 ..... 0.1 ..... 0.0 .....0.0

    If we assume it's water vapour and it takes part...

    at equilibrium

    4HCl + O2 = 2Cl2 + 2H2O
    0.04 ..... 0.01 ..... 0.18 .....0.18

    Concentrations in 4 litre vessel

    4HCl + O2 = 2Cl2 + 2H2O
    0.01 ..... 0.0025 ..... 0.045 .....0.045

    equilibrium law

    kc = [Cl2]2[H2O]2/[HCl]4[O2]

    kc = (0.002025 * 0.002025)/1x10-8 * 0.0025

    kc = 164025

    I agree with you
    I see my mistake.
 
 
 
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