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]
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.
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
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.
To find the equilibrium-constant expression for a reaction that has been halved and then reversed, we need to use a combination of the stoichiometry and the concept of equilibrium. Here are the steps to find the new equilibrium-constant expression:
Step 1: Halve the equation:
2NO + 2H2 --> N2 + 2H2O
The halved equation becomes:
NO + H2 ---> N + H2O
Step 2: Reverse the halved equation:
N + H2O --> NO + H
Step 3: Write the equilibrium-constant expression for the reversed equation:
[NO][H]
------
[N][H2O]
So, the equilibrium-constant expression for the reaction that has been halved and then reversed is:
[NO][H]
------
[N][H2O]