[lololoraine]

Given the ammonia equilibrium shown below.  Two gases are mixed in a flask at concentrations of 0.65 M N2 and 1.95 M H2  at about 500 C.  The following equilibrium occurs.  Determine the concentration of all chemical species at equilibrium (12 points).
Reaction Kc = 1.8 x 10-5 @ 500 0C

                2 NH3 (g)            N2 (g)    +   3 H2 (g)

[Astrokel]

Hey lololoraine

Set up an ICE table first and it should be simple.

12 points?   

Did you draw the equilibrium reaction incorrectly? should be the other way round?

[lololoraine]

Oh! I am sorry for that 12 poins thing.

It was given to us that the chemical equation is like that.

How do you solve it? I am confused with how I'll solve equilibrium problems.

Thank you for your *delete me*

[Astrokel]

Given the ammonia equilibrium shown below.  Two gases are mixed in a flask at concentrations of 0.65 M N2 and 1.95 M H2  at about 500 C.  The following equilibrium occurs.  Determine the concentration of all chemical species at equilibrium (12 points).
Reaction Kc = 1.8 x 10-5 @ 500 0C

                2 NH3 (g)            N2 (g)    +   3 H2 (g)

I suppose the reaction should be N2(g) + 3H2(g)  <---> 2NH3(g), well this is because you start off with N2 and H2.

Anyway its important to set up ICE Table. I = Initial, C = Change, E = Equilibrium Concentrations

    N2(g) + 3H2(g)  <---> 2NH3(g)
I   0.65      1.95             0
C    -x        -3x              +2x    (due to mole ratio)
E   (0.65-x)  (1.95-3x)     2x


I don't know the Kc refers to the equation you stated or refer to mine. I guess it refers to the equation i wrote above. Sub in values of [Equilibrium] to the Kc and you can solve for x.

Remember, always attempt ICE table! 

kelvin