Question

Suppose a 500 ml flask is filled with 0.10 mol of Bry, 1.6 mol of OCI, and 0.60 mol of BrCl. The following reaction becomes possible: Bry()+OCI (8) -BrOCI () + BrCl() The equilibrium constant K for this reaction is 0.770 at the temperature of the flask Calculate the equilibrium molarity of OCI. Round your answer to two decimal places x x 5 ?

Answer 1

volume = 500 mL = 0.5 L Molarity=mole/volume Molarity of Br2 = 0.10 / 0.50 = 0.20 M Molarity of oc2 = 1.60 / 0.50 = 3.20 M Molarity of BrCl = 0.60 / 0.50 = 1.20 M Construct ICA table to obtain their equilibrium concentrations: Reaction Br2 (g) + OC12 (g) + Broci (g) + Initial (M) 0.20 3.20 0.00 Change (M) -X +x Equilibrium (M) 0.20-X 3.20 -X 0.00 +x BrCl (g) 1.20 +x 1.20 +x Equilibrium expression: [Broci | Brci ] equilibrium constant Kc = [ Br2 I 012 ] 2 [ + 1.20+x ] 0.77 = [ 0.20 -> ] 3.20-x ] 0.230x2 + 3.818 x - 0.4928 = 0 By solving the above quadratic equation, we get x = equilibrium concentrations of | [ OC12 ] = 3.20 - 0.128 = 3.0719 M | [ oc2 ] = 3.07 M 0.128