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

you need to give more information

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.