Question

Suppose a 500. mL flask is filled with 0.90 mol of OC1,, 0.20 mol of BroCl and 1.4 mol of BrCl. The following reaction becomes possible: Br2(g) +OC12(g) =BroCl(g) +BrCl(g) The equilibrium constant K for this reaction is 0.798 at the temperature of the flask. Calculate the equilibrium molarity of Br2. Round your answer to two decimal places. IM xs ?

Answer 1

Initial concentration of OCl2 = mol of OCl2 / volume in L

= 0.90 mol / 0.500 L

= 1.8 M

Initial concentration of BrOCl = mol of BrOCl / volume in L

= 0.20 mol / 0.500 L

= 0.40 M

Initial concentration of BrCl = mol of BrCl / volume in L

= 1.4 mol / 0.500 L

= 2.8 M

ICE Table:

ICE Table: [0012] [Br2] [Brocul [Brcl] 1.8 0.4 2.8 initial change equilibrium +1x +1x -1x - 1x 1.8+1x +1x 0.4-1x 2.8-1x

Equilibrium constant expression is

Kc = [BrOCl]*[BrCl]/[OCl2]*[Br2]

0.798 = (0.4-1*x)(2.8-1*x)/((1.8 + 1*x)(1*x))

0.798 = (1.12-3.2*x + 1*x^2)/(1.8*x + 1*x^2)

1.436*x + 0.798*x^2 = 1.12-3.2*x + 1*x^2

-1.12 + 4.636*x-0.202*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -0.202

b = 4.636

c = -1.12

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 = 20.59

roots are :

x = 0.2442 and x = 22.71

x can't be 22.71 as this will make the concentration negative.so,

x = 0.2442

At equilibrium:

[Br2] = +1x = +1*0.2442 = 0.244 M

Answer: 0.24 M