equilibrium partial pressures?

Question

At room temperature the following reaction has Kp = 9.48E-10:
2SO2 (g) + O2 (g) (equilibrium arrows/yields) 2SO3 (g)

A closed reaction vessel is initially charged with the following partial pressures: 2.91 atm SO2 and 2.97 atm O2. After the reaction has come to equilibrium, what are the equilibrium partial pressures of all reactants and products?
p(SO2) = ? atm
p(O2) = ? atm
p(SO3) = ? atm

The answers are:
p(SO2) = 2.91 atm
p(O2) = 2.97 atm
p(SO3) = .000154 atm, but I don't understand how?

Answer 1

Set up the following table

.. .. .. .. .. .. 2SO2 + O2 <=> 2SO3
Initial .. .. .. 2.91 .. 2.97
React .. .. .. .. 2x .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. 2x
At Equil. 2.91-2x . 2.97-x .. .. 2x

Kp = P(SO3)^2 / ( P(SO2)^2 * P(O2)) =>
Kp = (2x)^2 / ( (2.91-2x)^2*(2.97-x) ) = 9.48*10^-10

If you do the calculations this boils down to a cubic equation.
Since Kp is very small and the initial pressures are quite big, we can avoid solving the difficult cubic equation by doing the following approximation;

Let's assume that x<<2.9 so that 2.91-2x=2.91 and 2.97-x=2.97. Then the equation is simplified to

4x^2/(2.91^2 *2.97) = 9.48*10^-10 =>
x= squareroot ( ((2.91^2)*2.97*9.48/4)*10^-10) = 7.7*10^-5

so
P(SO2)= 2.91 atm (remember this was part of our assumption)
P(O2) = 2.97 atm (remember this was part of our assumption)
P(SO3) =2x =2*7.7*10^-5 = 1.54*10^-4 atm