Question

The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) + Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.306 moles of CO and 0.306 moles of Cl2 are introduced into a 1.00 L vessel at 600 K. [CO] = M [Cl2] = M [COCl2] = M

Answer 1

The initial concentration of CO = [CO]_i = 0.306 mol/1 L = 0.306 M

The initial concentration of Cl₂ = {Cl₂]_i = 0.306 mol/1 L = 0.306 M

Let's say the concentration of COCl₂ at equilibrium = [COCl₂]_e = x M

i.e. [CO]_e = (0.306-x) M and [Cl₂]_e = (0.306-x) M

According to the definition of Kc, Kc = [COCl₂]_e/[CO]_e *[Cl₂]_e

i.e. 77.5 = x/(0.306-x)*(0.306-x)

i.e. (0.306-x)²*77.5 = x

i.e. 0.093636 + x² - 0.612x = x

i.e. x²-1.612x+0.093636 = 0

This is a quadratic equation. By solving, you will get, x = 0.06

Therefore, [COCl₂]_e = 0.06 M,

[CO]_e = (0.306-0.06) M = 0.246 M and [Cl₂]_e = 0.246 M