2 NOCl (g)  = 2NO (g) + Cl2 (g); Kc = 1.6 × 10-5 3.0 mol of NOCl (g) are added to a 2.5 L flask. What are the equilibrium concentrations of each component?

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The chemical equation for the dissociation of NOCl is:

2NOCl (g) --> 2NO (g) + Cl2 (g)

2 molecules of NOCl give 2 molecules of NO and one molecule of Cl2.

Kc for the equation is defined as ([NO]^2*[Cl2])/ [NOCl]^2

Kc is given to be 1.6*10^-5

If 3 mole of NOCl are added to a 2.5 L flask, the concentration of NOCl is [NOCl] = 3/2.5

([NO]^2*[Cl2])/ [NOCl]^2 = 1.6*10^-5

=> ([NO]^2*[Cl2])/ 1.2^2 = 1.6*10^-5

The concentration of NO is twice that of Cl2 as 2 NO are formed for each Cl2. If this is x

=> x^2*x = 1.6*10^-5*1.2^2

=> x^3 = 2.304*10^-5

=> x = (2.304*10^-5)^(1/3)

=> x = 0.0285

The equilibrium constants of NO is 0.0569 mole/L and the equilibrium constant of Cl2 is 0.0285 mole/L

Sources:

We’ve answered 318,945 questions. We can answer yours, too.

Ask a question