Molar Concentration at Equilibrium

Question

At high temperatures, iodine molecules can dissociate into iodine atoms. For the reaction I₂(g) ⇌ 2 I･(g) Kc = 1.27 × 10⁻⁴ at 900 °C. A 65.5 g sample of I₂ is placed in a 1.00 L vessel and heated to 900 °C. What will be the molar concentration of iodine atoms, [I･], in the vessel when the system comes to equilibrium?

Answer 1

Convert mass of I₂ to moles and then to M

65.5 g I2 x 1 mol I2 / 253.8 g = 0.258 mol

0.258 mol / 1 L = 0.258 M

Set up an ICE table

I₂(g) ⇌ 2 I･(g) ... Kc = 1.27 × 10⁻⁴

0.258....0......Initial

-x........+2x....Change

0.258-x...2x..Equilibrium

Write the Kc expression

Kc = [I-]² / [I₂]

Substitute values and solve for [I-]

1.27x10^-4 = (2x)² / 0.258-x

4x² = 3.28x10^-5 - 1.27x10^-4x

4x² - 3.28x10^-5 + 1.27x10^-4x

x = 4.10x10^-6 M = [I-] @ equilibrium