Question

Suppose a 250. mL flask is filled with 2.0 mol of Br2, 1.1 mol of OCI, and 0.80 mol of BroCl. The following reaction becomes possible: Br2(g) + Oci (g) Broci (g) + Brci(g) The equilibrium constant K for this reaction is 0.624 at the temperature of the flask. Calculate the equilibrium molarity of BroCl. Round your answer to two decimal places. Пм xs ?

Answer 1

Initial concentration of Br2 = mol of Br2 / volume in L
= 2.0 mol / 0.250 L
= 8.0 M

Initial concentration of OCl2 = mol of OCl2 / volume in L
= 1.1 mol / 0.250 L
= 4.4 M

Initial concentration of BrOCl = mol of BrOCl / volume in L
= 0.80 mol / 0.250 L
= 3.2 M

ICE Table:
            

[Br2] [0012] [Brocul (Brci] initial 8.0 4.4 3.2 13:1 change -1x -1x +1x +1x equilibrium 8.0-1x 4.4-1x 3.2+1x +1x

Equilibrium constant expression is
Kc = [BrOCl]*[BrCl]/[Br2]*[OCl2]
0.624 = (3.2 + 1*x)(1*x)/((8-1*x)(4.4-1*x))
0.624 = (3.2*x + 1*x^2)/(35.2-12.4*x + 1*x^2)
21.96-7.738*x + 0.624*x^2 = 3.2*x + 1*x^2
21.96-10.94*x-0.376*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -0.376
b = -10.94
c = 21.96

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.527*10^2

roots are :
x = -30.98 and x = 1.886

since x can't be negative, the possible value of x is
x = 1.886

At equilibrium:
[BrOCl] = 3.2+1x = 3.2+1*1.886 = 5.09 M

Answer: 5.09 M