Question

The equilibrium constant, Kc, for the following reaction is 9.49×10-4 at 533 K.COCl2(g) CO(g) + Cl2(g)When a sufficiently large sample of COCl2(g) is introduced into an evacuated vessel at 533 K, the equilibrium concentration of Cl2(g) is found to be 0.190 M.Calculate the concentration of COCl2 in the equilibrium mixture.M

Answer 1

38.0 M

Step-by-step explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO(g) + Cl₂(g)

To find the concentrations at equilibrium we will use an ICE chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with concentrations or changes in concentrations. Let's call unknown concentrations x and y.

COCl₂(g) ⇄ CO(g) + Cl₂(g)

I y 0 0

C -x +x +x

E y-x x x

The equilibrium concentration of Cl₂(g) is 0.190 M, that is, x = 0.190 M

The equilibrium constant (Kc) is:

`Kc=9.49 * 10^(-4) =([CO][Cl_(2)])/([COCl_(2)]) =(0.190^(2) )/(y-0.190) \\y = 38.2M`

The concentration of COCl₂ in the equilibrium mixture is

y - x = 38.2 M - 0.190 M = 38.0 M