Suppose a 250.mL flask is filled with 1.9mol of Br2, 1.0mol of OCl2 and 1.6mol of BrOCl. The following reaction becomes possible: Br2g+OCl2g ----> BrOClg+BrClg The equilibrium constant K for this reaction is 0.115 at the temperature of the flask. Calculate the equilibrium molarity of OCl2. Round your answer to two decimal places. M

Answer 1

1.9mol of Br2, 1.0mol of OCl2 and 1.6mol of BrOCl in 250 mL flask give molarities of M =
(Br2) = 1.9/0.250 =7.60; (OCl2) = 1/0.250 = 4; (BrOCl) = 1.6/0.250 = 6.4
....................Br2(g) + OCl2(g) ----> BrOCl(g) + BrCl(g)
I..................7.60.........4....................6.4...............0
C....................-x..........-x....................+x..............+x
E.................7.60-x......4-x.................6.4+x............x
K = (BrOCl)(BrCl)/(Br2)(OCl2) = 0.115
I've filled in the E line for the reaction. You should solve for x in the K expression I've set up. then evaluate 4-x and that will be the equilibrium concentration of OCl2.
Post your work if you get stuck. I've done the chemistry part and I've left the math part for you to do.