H2O (g) + Cl2O (g) ↔ 2 HOCl (g) (a) Initially, 0.0555 mol H2O and 0.0230 mol Cl2O are mixed in a 1.00 L flask. At equilibrium, there is found to be 0.0200 mol of HOCl (g). Calculate the concentrations of H2O (g) and Cl2O (g) at equilibrium. (b) Using your results from part (a), calculate the equilibrium constant, Kc. (c) 1.0 mol pure HOCl is placed in a 2.0 L flask. Use your Kc from part (b) to calculate the equilibrium concentrations of H2O (g) and Cl2O (g).

a)

[H₂O] = 0.0455M

[Cl₂O] = 0.0130M

[HOCl] = 0.0200M

b) Kc = 0.676

c) [H₂O] = [Cl₂O] = 0.177M

Explanation:

Based on the reaction:

H₂O (g) + Cl₂O (g) ⇄ 2HOCl (g)

Kc is defined as:

Kc = [HOCl]² / [H₂O][Cl₂O] For molar concentrations in equilibrium

As volume of the flask is 1.00L, the initial molar concentrations of H₂O and Cl₂O is 0.0555M and 0.0230M, respectively.

In equilibrium, the concentrations are:

[H₂O] = 0.0555M - X

[Cl₂O] = 0.0230M - X

[HOCl] = 2x = 0.0200M → X = 0.0100M

Where X is reaction coordinate

a) Concentrations in equilibrium are:

[H₂O] = 0.0455M

[Cl₂O] = 0.0130M

[HOCl] = 0.0200M

b) Replacing in Kc:

Kc = [0.0200]² / [0.0455][0.0130] = 0.676

c) Initial concentration of HOCl is 1.0mol / 2.0L = 0.50M. In equilibrium concentrations are:

[H₂O] = X

[Cl₂O] = X

[HOCl] = 0.50M - 2X

Replacing in Kc formula:

0.676 = [0.50-2X]² / [X][X]

0.676X² = 4X² - 2X + 0.25

0 = 3.324X² - 2X + 0.25

Solving for X:

X = 0.177M

X = 0.425M → False answer, produce negative concentrations.

As X = [H₂O] = [Cl₂O]; equilibrium concentrations of both compounds is 0.177M