Calculate Kp?

Question

when a 2:1 mixture of SO2(g) and O2(g) at a total pressure of 5atm is passed over a catalyst at 425 degrees, the partial pressure of SO3(g) at equilibruim is found to be 3 atm, calculate the value of Kp

Answer 1

The mixture is 2 SO2:1 O2, so if p is the initial partial pressure of O2, then 2p is the initial partial pressure of SO2
The total initial pressure is P0= P0(SO2)+P0(O2) =2p+p= 5 =>3p=5 => p=5/3 = 1.67
Thus
P0(O2)=p =1.67 atm
P0(SO2)= 5-P0(O2)= 5-1.67= 3.33 atm

.. .. .. .. .. .. 2SO2 + O2<=>2SO3
Initial .. .. .. 3.33 ...1.67
React .. .. .. 2x .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. 2x
At Equil. 3.33-2x. 1.67-x .. 2x

Peq(SO3)=2x
Also Peq(SO3)=3
Thus from these two equations 2x=3 =>x=1.5 atm

Kp= Peq(SO3)^2 / (Peq(SO2)^2 *Peq(O2)) =>
Kp =(2x)^2 / ( (3.33-2x)^2(1.67-x) ) =>
Kp = 9 / ((0.33^2)*0.17) = 486