Question

Suppose a 250.mL flask is filled with 0.90mol of Br2 , 0.60mol of BrOCl and 1.9mol of BrCl . The following reaction becomes possible: Br2(g)+OCl2(g) +BrOCl(g)+BrCl(g) The equilibrium constant K for this reaction is 0.571 at the temperature of the flask. Calculate the equilibrium molarity of OCl2 . Round your answer to two decimal places.

Answer 1

molarity of Br2 = no of moles/volume in L

                        = 0.9/0.25   = 3.6M

molarityBrOCl     =   no of moles/volume in L

                   = 0.6/0.25   = 2.4M

molarity of BrCl = no of moles/volume in L

             = 1.9/0.25   = 7.6M

                             Br2(g)+OCl2(g)-------------->BrOCl(g)+BrCl(g)

I                             3.6         0                              2.4    7.6

C +x    +x    -x    -x

E 3.6+x +x    2.4-x 7.6-x

   K = [BrOCl][BrCl]/[B2][OCL2]

0.571 = (2.4-x)(7.6-x)/(3.6+x)*x

0.571*(3.6+x)*x = (2.4-x)(7.6-x)

    x = 1.60

   [OCL2]   = x   = 1.60M >>>>answer