At 400 k, the equilibrium constant for the reaction is kc = 7.0. br2 (g) + cl2 (g) 2brcl (g) a closed vessel at 400 k is charged with 1.00 m of br2 (g), 1.00 m of cl2 (g), and 3.00 m of brcl (g). a. write the equilibrium expression for this reaction. b. find q and determine if the reaction shifts to the left or right. c. draw an ice chart for the reaction.

a) when Kc = concentration of products / concentration of reactants
  So according  to the reaction equation:
Br2(g) + Cl2(g) → 2BrCl(g)

∴ Kc =[BrCl] ^2 / [Br2][Cl2]

b) when q = [BrCl]^2 / [Br2][Cl2]
and we have [BrCl] = 3 m 
[Br2] = 1 m 
[Cl2] = 1 m
So by substitution:
q= 3^2 / 1*1 = 9 

- and we can see that q > Kc 
the reaction is not at equilibrium that means there are more products and the reaction shifts to the left to increase the reactants and decrease the products to achieve equilibrium.

C) by using ICE table:

              Br2(g) + Cl2(g) → 2BrCl (g)
initial       1               1               3
change  -X              -X            +X
Equ       (1-X)          (1-X)         (3+X)

when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X)  by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m 
    [BrCl] = 3+0.215 = 3.215 m