Question

Lanthanum oxalate decomposes when heated to lanthanum oxide, CO, and CO2

La2(C2O4)3 (s) AE La2O3 (s) + 3 CO (g) + 3 CO2 (g)

a) If, at equilibrium, the total pressure in a 10.0L flask is 0.200 stm, what is the value of Kp?

b)Suppose 0.100 mol of La2(C2O4)3 was originally placed in the 10.0L flask. What quantity of La2(C2O4)3 remains unreacted at equilibrium?

Answer 1

a) there are two gaseous products, partial pressure of CO= partial pressure of CO2

partial pressure of CO= mole fraction* Total pressure= 3/(3+3)*0.2= 0.1 atm

Similarly partial pressure of CO2= 0.1 atm

Kp = [CO]³ [CO2]₃ = (0.1)6 =1*10-6

b) concentration of La2(CO4)3 =0.1 moles

let x= moles of La2(CO4)3 decomposed,

3x= moles of CO and CO2 each formed

total moles = 6x

mole fraction =0.5 for CO and 0.5 for CO2. Temperature is assumed to be 373 K

Whichi is not given

number of moles at Equilibrium = PV/RT= 0.2*10/(0.08206*373)=0.0653 moles

6x= 0.0653, x = 0.001083

Moles of La2(C2O4)3 at equilibrium= 0.1-0.001083=0.098917

Partial pressure = 0.5*P for CO and 0.5P for CO2

KP= (0.5P)6= 1*10-6

P= 0.2 atm