When heated, lanthanum(iii) oxalate decomposes as follows: la2(c2o4)3(s) < ===> la2o3(s) + 3 co2(g) + 3 co(g) starting with just the oxalate in a 10.0 l flask, at equilibrium the total pressure observed is 0.200 atm. what is the value of kp for the equilibrium? (dalton’s law of partial pressure! )

Answer : The value of for the equilibrium is

Explanation : Given,

Total pressure = 0.200 atm

The balanced equilibrium reaction is,

First we have to calculate the partial pressure of and .

From the balanced reaction we conclude that,

The moles of is equal to the moles of . So,

Total number of moles = 2

Partial pressure of =

and,

Partial pressure of =

Now we have to calculate the value of equilibrium constant.

The expression of equilibrium constant for the reaction will be:

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

Therefore, the value of for the equilibrium is