A mixture of 0.590 M H 2 O , 0.340 M Cl 2 O , and 0.747 M HClO are enclosed in a vessel at 25 ° C . H 2 O ( g ) + Cl 2 O ( g ) − ⇀ ↽ − 2 HOCl ( g ) K c = 0.0900 at 25 ° C Calculate the equilibrium concentrations of each gas at 25 ° C .

Answer : The concentrations of , and at equilibrium is, 0.215 M, 0.856 M and 0.606 M respectively.

Explanation :

The given chemical reaction is:

Initial conc.     0.590       0.340            0.747

At eqm.       (0.590-x)       (0.340-x)     (0.747+2x)

As we are given:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

x = -0.520 and x = -0.266

We are neglecting the value of x = -0.520 because equilibrium concentration can not be more than initial concentration.

Thus, the value of x will be, -0.266

The concentrations of at equilibrium = (0.747+2x) = (0.747+2(-0.266)) = 0.215 M

The concentrations of at equilibrium = (0.590-x) = (0.590+0.266) = 0.856 M

The concentrations of at equilibrium = (0.340-x) = (0.340+0.266) = 0.606 M

- lead + phosphoric acid = hydrogen + lead(ii) phosphate