[sweetdaisy186]

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!

[Borek]

How many moles of ideal gas are there in the flask in the equilibrium? Is this information helpfull in finding x?

[sweetdaisy186]

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

[Donaldson Tan]

1. assume the process is isobaric.
2. the intial pressure of the system is the sum of partial pressure at equilibrium.
3. use PV = nRT to find the partial pressure of each species at equilibrium.

[sdekivit]

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)

[sweetdaisy186]

Hmm, I am a little confused. The equation is nrt/v and you divided by the R and the T. I don't understand why. Also, where did the 1,41855 x 10^5 and 1,0 x 10^-3 come from? Sorry, I'm just very confused. Thanks for your *delete me*  

[sweetdaisy186]

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

[sdekivit]

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

sorry, i'm confident with the unbit Pascal of pressure, so that number is the pressure in Pa equivalent to 1,4 atm.

That's why my gas constant has also another value: 8,314 J/(Kmol)

[sweetdaisy186]

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

[Donaldson Tan]

1atm = 1.013 x 10⁵ Pa

[sweetdaisy186]

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