Question

Suppose a 500 ml flask is filled with 0.40 mol of Bry, 1.3 mol of OCI, and 1.1 mol of BrOCI. The following reaction becomes possible: Bry()+OCI,() BrOCI()+BCI) The equilibrium constant for this reaction is 2.95 at the temperature of the flask. Calculate the equilibrium molarity of Br. Round your answer to two decimal places.

Answer 1

Initial concentration of Br2 = mol of Br2 / volume in L

= 0.40 mol / 0.500 L

= 0.80 M

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

= 1.3 mol / 0.500 L

= 2.6 M

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

= 1.1 mol / 0.500 L

= 2.2 M

ICE Table:

Brocl [Br2] [OCL2] [BrCl] initial 0.8 2.6 2.2 change -1x -1x +1x +1x equilibrium 0.8-1x 2.6-1x 2.2+1x +1x

Equilibrium constant expression is

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

2.95 = (2.2 + 1*x)(1*x)/((0.8-1*x)(2.6-1*x))

2.95 = (2.2*x + 1*x^2)/(2.08-3.4*x + 1*x^2)

6.136-10.03*x + 2.95*x^2 = 2.2*x + 1*x^2

6.136-12.23*x + 1.95*x^2 = 0

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

a = 1.95

b = -12.23

c = 6.136

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 = 1.017*10^2

roots are :

x = 5.722 and x = 0.5499

x can't be 5.722 as this will make the concentration negative.so,

x = 0.5499

At equilibrium:

[Br2] = 0.8-1x = 0.8-1*0.5499 = 0.250 M

Answer: 0.250 M