Question

The equilibrium constant Kc for the reaction: 2HCl(g) ⇄ H2(g) + Cl2(g) is 0.0213 at 400 oC. If 20.0 moles of HCl(g) are heated at 400 oC, what amounts of HCl(g), H2(g) and Cl2(g) would be present in the equilibrium mixture?

Answer 1

HCl H2 Cl2
initial mole 20 0 0
final mole 20-2x x x
normally, if the question only give you the mole of one reagent instead of the concentration, you need to consider the molar volume of each continents in the equation:
(x/N)(x/N)/((20-2X)/N)^2, however, the N is eliminate in the equation.

So the equation is (x*x)/(20-2x)^2=0.0213
final mol of HCl H2 Cl2
15.48 2.259 2.259

Answer 2

wow. thanks a bunch :)