Question

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 1

Answer : The value of
`K_p` for the equilibrium is
`1.0* 10^(-6)`

Explanation : Given,

Total pressure = 0.200 atm

The balanced equilibrium reaction is,

`La_2(C_2O_4)_3(s)\rightleftharpoons La_2O3(s)+3CO_2(g)+3CO(g)`

First we have to calculate the partial pressure of
`CO_2` and
`CO`.

From the balanced reaction we conclude that,

The moles of
`CO_2` is equal to the moles of
`CO`. So,

Total number of moles = 2

Partial pressure of
`CO_2` =
`\frac{\text{Moles of }CO_2}{\text{Total moles}}* \text{Total pressure}=(1)/(2)* 0.2atm=0.1atm`

and,

Partial pressure of
`CO` =
`\frac{\text{Moles of }CO}{\text{Total moles}}* \text{Total pressure}=(1)/(2)* 0.2atm=0.1atm`

Now we have to calculate the value of equilibrium constant.

The expression of equilibrium constant
`K_p` for the reaction will be:

`K_p=(p_(CO_2))^3* (p_(CO))^3`

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

`K_p=(0.1)^3* (0.1)^3`

`K_p=1.0* 10^(-6)`

Therefore, the value of
`K_p` for the equilibrium is
`1.0* 10^(-6)`