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.

[HOCl] = 0.00909 mol/liter [H₂O] = 0.03901 mol/liter[Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 molMolarity, 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 molMolarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

           

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

4. Equilibrium expression

       

     

5. Solve:

           

Use the quadatic formula:

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

ok, will do. Hope you hit 3k soon.