For the equilibrium below at 400 K, Kc = 7.0.

Question

Br2(g) + Cl2(g) reverse reaction arrow 2 BrCl(g)
If 0.90 mol of Br2 and 0.90 mol Cl2 are introduced into a 1.0 L container at 400. K, what will be the equilibrium concentration of BrCl?

Answer 1

Br2 + Cl2 ==> 2BrCl

Kc = (BrCl)^2/(Br2)(Cl2)
Do an ICE chart.
initial:
Br2 = 0.9 mol/1 L - 0.9M
Cl2 = 0.9 M
BrCl = 0

change:
BrCl = +x
Cl2 = -x
Br2 = -x

equilibrium:
Br2 = 0.9-x
Cl2 = 0.9-x
BrCl = +x

Plug the equilibrium concns into the Kc expression and solve for x.

Answer 2

Br2 + Cl2 ==> 2BrCl

Kc = (BrCl)^2/(Br2)(Cl2)
Do an ICE chart.
initial:
Br2 = 0.9 mol/1 L = 0.9M
Cl2 = 0.9 M
BrCl = 0

change:
Br2 = -x
Cl2 = -x
BrCl = +2x

equilibrium:
Br2 = 0.9-x
Cl2 = 0.9-x
BrCl = +2x