Hey guys!

Here is my question:

Starting with 0.25 mol pure N₂O₄ in a 1.0 L flask at 35 degrees C, what is the value of Kp if the final pressure of NO₂ is 1.4 atm at equilibrium?

At first, I wanted to use ICE, and then I realized, that I didn't know what X is. So then, I thought about using the Kp=Kc/RT, and then I realized that I didn't know Kp or Kc. So, now I am stumped about what to do, because I don't know Kp, Kc, or the x when I use the ICE. A hint would be greatly appreciated. Thanks!

Hmmm, I don't know, that is the exact problem, word for word

here is my solution:

we now the partial pressure of NO2 in equilibrium. Assuming the gasses behave as ideal gasses, p(NO2) is given by:

pNO2 = nRT/V

--> n(nO2) in equilibrium = (1,41855 x 10^5 * 1,0 x 10^-3) / (8,314 * 308) = 0,05540 mol

We started with 0,25 mol N2O4 so 0,05540/2 = 0,02770 mol N2O4 has reacted.

now we thus have, in equilibrium, 0,25 - 0,02770 = 0,2230 mol N2O4

--> p(N2O4) = 0,2230 * 8,314 * 308 / 1,0 x 10^-3 = 5,711 x 10^5 Pa.

This looks nice, since the Kp for this equilibrium at T = 298 K is 1,48 x 10^4 Pa.

Now Kp = p(NO2)^2 / p(N2O4)

--> Kp = (1,41855 x 10^5)^2 / 5,711 x 10^5 = 3,5 x 10^4 Pa.
( In this case, the relation between Kp and Kc:

Kp = Kc * RT

--> delta n = 2-1 = 1)

oo, wait, I get it now, we are solving for n and I get where the 1.00 times ten to the 3rd came from. But what about 1,41855 x 10^5 ? Shouldn't we use ATM? Thanks

oooooo, okay, I see it now. You converted from atm to pascal? I see it now! Thanks for your *delete me*

ahah! yup, that's what I did. Thanks!