The equilibrium constant Kc for the reaction N2(g) + 3H2(g) -> <- 2NH3 at 450 degrees celcius is 0.159. Calculate the equilibrium compostion when 1.00 mol N2 is mixed with 3.00 mol H2 in a 5.00-L vessel.

Answer 1

.............N2 + 3H2 ==> 2NH3
begin(mols)..1.0..3.00.....0
change........-x...-3x.....+2x
final.......1.0-x..3.00-3x..+2x

Kc = (NH3)^3/(N2)(H2)^3
Substitute the ICE chart values into the Kc expression and solve for x and the other values. (x is i moles).
Then moles/5.00 L = M

Note the correct spelling of celsius.