(CO) = 1.30/28 mols/6.00 L = 0.00773
(Cl2) = 2.30/71 moles/6.00 L = 0.0054
...................CO(g) + Cl2(g) ↔ COCl2(g)
I...............0.00773....0.0054.........0
C...................-x............-x..............+x
E...........0.0773-x....0.0054-x...........x
Substitute the E line into the Kc expression and solve for x, then evaluate CO from 0.00773-x.
Kc expression = (COCl2)/(CO)(Cl2)
Post your work if you get stuck.
Consider 1.30 mol of carbon monoxide and 2.30 mol of chlorine sealed in a 6.00 L container at 476 oC. The equilibrium constant, Kc, is 2.50 (in M-1) for
CO(g) + Cl2(g) ↔ COCl2(g)
Calculate the equilibrium molar concentration of CO
2 answers
That's my response above with no name.
2 answers