Question

At a certain temperature the equilibrium constant, Kc, 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?

Answer 1

Ans. Initial [I₂] = Moles / Volume of reaction flask

                                    = 0.45 mol/ 2.0 L

                                    = 0.225 M

Initial [Cl₂] = 0.45 mol/ 2.0 L = 0.225 M

# Since there is no ICl initially, the reaction would go to the left to from ICl so restore equilibrium.

That is, we have to proceed with I₂ + Cl₂ --------> 2ICl

When a reaction is reversed, the equilibrium constant is reciprocated.

So, new Keq’ for backward reaction = 1 / Keq for forward reaction

            Or, Keq’ = 1 / 0.11 = 9.091

# Create an ICE table with the initial concentrations as shown in picture-

Now,

            Equilibrium constant, keq’ = [ICl]² / ([I₂] [Cl₂])

            Or, 9.091 = (2X)² / [(0.225 - X) (0.225 - X)]

            Or, 9.091 x (0.050625 + X² - 0.45X) = 4X²

            Or, 0.460231875 + 9.091X² - 4.09095X - 4X² = 0

            Or, 5.091X² - 4.09095X + 0.460231875 = 0

Solving the quadratic equation we get following two roots-

            X1 = 0.669                             ; X2 = 0.135

Since X can’t be greater than 0.225, reject X1.

Therefore, X = 0.135

# [ICl] at equilibrium = 2X = 1 x 0.135 = 0.270

Hence, equilibrium [ICl] = 0.270 M