Question

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentrations of all species at equilibrium for each of the following cases. (a) 1.3 g H2O and 2.2 g Cl2O are mixed in a 1.5-L flask.

Answer 1

Answer:

• [HOCl] = 0.00909 mol/liter

• [H₂O] = 0.03901 mol/liter

• [Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

• Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

• Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

• Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

• Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

            $H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

4. Equilibrium expression

       $K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

     $0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

            $x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

• [HOCl] = 0.00909 mol/liter

• [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

• [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter