Question

Consider the reaction:
CO + Cl2 ↔ COCl2 KC = 1.2 × 103

Suppose we start with 0.3500 mol CO, 0.5500 mol COCl2, and no Cl2 in a 3.00 L flask. How many moles of Cl2 will be present at equilibrium?

Answer 1

concentration of CO = 0.35 / 3 = 0.117 M

concentration of CoCl2 = 0.55 / 3 = 0.1833 M

CO    +     Cl2    -------------> COCl2       

0.1166       0                        0.1833

0.1166-x    x                         0.1833+x

Kc = [COCl2] / [CO][Cl2]

1.2 x 10^3 = (0.1833 + x ) / x (0.1166 - x)

x = 0.1144

moles of Cl2 present = 0.3432 mol