Question

At a certain temperature the equilibrium constant, K_c, equals 0.11 for the reaction:

2 ICl(g) ↔ I₂(g) + Cl₂(g).

What is the equilibrium concentration of ICl if 0.45 mol of I₂ and 0.45 mol of Cl₂ are initially mixed in a 2.0-L flask?

At a certain temperature the equilibrium constant, K_c, equals 0.11 for the reaction:

2 ICl(g) ↔ I₂(g) + Cl₂(g).

What is the equilibrium concentration of ICl if 0.45 mol of I₂ and 0.45 mol of Cl₂ are initially mixed in a 2.0-L flask?

0.14 M

0.27 M

0.34 M

0.17 M

Answer 1

Initial [I₂] = Moles /Volume = 0.45 mol/ 2 L = 0.225 M

Initial [Cl₂] = 0.45 mol/2.L = 0.225 M

Keq = 1/0.11 = 9.091

Constructing the ICE table.

                                           I₂            +      Cl₂       -------------->      2ICl

Initial                                0.225                0.225                                0

Change                              -x                       -x                                   +2x

Equilibrium                    (0.225-x)          (0.225-x) +2x

Keq = [ICl]²/[I₂][Cl₂]

9.091 = (2x)² /(0.225 - x)2

5.091 x² - 4.09 x + 0.4602 = 0

On solving

x = 0.135

Equilibrium [ICl] = 2x = 2 x 0.135 = 0.270 M