At a particular temperature, Kp = 0.490 for the reaction

Question

N2O4(g) = 2NO2(g)

A flask containing only NO2(g) at an initial pressure of 10.00 atm is allowed to reach equilibrium.
a) Calculate the total pressure in this flask at equilibrium.
answer: 5.724 atm
b) With no change in the amount of material in the flask, the volume of the container in question is decreased to 0.100 times the original volume. Assuming constant temperature, calculate the (new) total pressure, at equilibrium.
answer: ?

Answer 1

I........................0....................10
C.....................+x...................-2x
E.....................+x...................10-2x

Kp = 0.490 = (NO2)^2/(N2O4)
0.490 = (10-2x)^2/(x)
Solve for x and evaluate NO2 and N2O4, then rewrite the equation as above and knowing the volume decreased 0.100 as much so you know the pressure increased by a factor of 10. Calculate the new concentrations of NO2 and N2O4 for the E line, make those nujmbers the I line, and reaclculate x at that point.
Post your work if you get stuck. I'll help get you started.
From part a you know x = 4.276 so E line is
(NO2) = 10-2x = (10-2*4.276) atm and (NO2) = 4.276 atm. Plug those into the I line (initial) and go through the same calculation.

Answer 2

I understand a) but I'm stuck getting started on how take into account the 0.100 times the original volume. Im not sure how to rework the equation. Do I times the E line concentrations by 10? then that will be my new initial and then proceed to find the new x?

Answer 3

Yes, that's what you do. Here is what I wrote in my first response that summarizes what you do.
I'll help get you started.
From part a you know x = 4.276 so E line is
(NO2) = 10-2x = (10-2*4.276) atm and (NO2) = 4.276 atm. Plug those into the I line (initial) and go through the same calculation.