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Posted by on Wednesday, April 14, 2010 at 10:02pm.

the equilibrium-constant expression for a reaction is
kc= [N2][H2O]2
----------
[NO]2[H2]
what is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

2NO + 2H2 --> N2 + 2H2O
NO + H2 ---> N + H2O
N + H2O --> NO + H
[NO][H]
------
[N][H2O]

  • chemistry check my work - , Wednesday, April 14, 2010 at 10:13pm

    Of course you realize you can't control the spacing on the board; therefore, you must time everything as a line format.

    When taking 1/2 the equation, Kc becomes the sqrt Kc. When reversing, Kc becomes 1/Kc.

  • chemistry check my work - , Wednesday, April 14, 2010 at 10:22pm

    sorry new to typing in math or chem problems so i don't really understand the first part of your answer. but is [NO][H]over[N][H2O] right or did i mess up somewhere

  • chemistry check my work - , Wednesday, April 14, 2010 at 10:36pm

    If this is the equation,
    2NO + 2H2 --> N2 + 2H2O

    and H2O is a gas, then the Kc expression is
    (N2)(H2O)^2/(NO)^2(H2)^2. I think you had (H2) and not (H2)^2

    If we take half the equation, we don't split the way we write molecules. It would look this way.
    1/2 NO + H2 ==>1/2 N2 + H2O

    Then Kc = (N2)^1/2*(H2O)/(NO)^1/2*(H2) and the new Kc is sqrt the old Kc.

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