December 11, 2023, 04:23:14 PM
Forum Rules: Read This Before Posting

### Topic: Equilibrium-looking for Kp  (Read 8325 times)

0 Members and 1 Guest are viewing this topic.

#### sweetdaisy186

• Guest
##### Equilibrium-looking for Kp
« on: July 29, 2005, 06:06:59 PM »
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

• Mr. pH
• Deity Member
• Posts: 27487
• Mole Snacks: +1788/-409
• Gender:
• I am known to be occasionally wrong.
##### Re:Equilibrium-looking for Kp
« Reply #1 on: July 29, 2005, 06:37:04 PM »
How many moles of ideal gas are there in the flask in the equilibrium? Is this information helpfull in finding x?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### sweetdaisy186

• Guest
##### Re:Equilibrium-looking for Kp
« Reply #2 on: July 29, 2005, 09:34:54 PM »
Hmmm, I don't know, that is the exact problem, word for word

#### Donaldson Tan

• Editor, New Asia Republic
• Retired Staff
• Sr. Member
• Posts: 3177
• Mole Snacks: +261/-13
• Gender:
##### Re:Equilibrium-looking for Kp
« Reply #3 on: July 29, 2005, 11:10:45 PM »
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.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### sdekivit

• Chemist
• Full Member
• Posts: 403
• Mole Snacks: +32/-3
• Gender:
• B.Sc Biomedical Sciences, Utrecht University
##### Re:Equilibrium-looking for Kp
« Reply #4 on: July 30, 2005, 04:52:29 PM »
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)
« Last Edit: July 30, 2005, 04:59:25 PM by sdekivit »

#### sweetdaisy186

• Guest
##### Re:Equilibrium-looking for Kp
« Reply #5 on: July 30, 2005, 10:10:37 PM »
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

• Guest
##### Re:Equilibrium-looking for Kp
« Reply #6 on: July 30, 2005, 10:13:28 PM »
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

• Chemist
• Full Member
• Posts: 403
• Mole Snacks: +32/-3
• Gender:
• B.Sc Biomedical Sciences, Utrecht University
##### Re:Equilibrium-looking for Kp
« Reply #7 on: July 31, 2005, 04:48:47 AM »
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)
« Last Edit: July 31, 2005, 05:11:09 AM by sdekivit »

#### sweetdaisy186

• Guest
##### Re:Equilibrium-looking for Kp
« Reply #8 on: July 31, 2005, 10:47:12 AM »
oooooo, okay, I see it now. You converted from atm to pascal? I see it now! Thanks for your *delete me*

#### Donaldson Tan

• Editor, New Asia Republic
• Retired Staff
• Sr. Member
• Posts: 3177
• Mole Snacks: +261/-13
• Gender:
##### Re:Equilibrium-looking for Kp
« Reply #9 on: July 31, 2005, 12:54:51 PM »
1atm = 1.013 x 105 Pa
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### sweetdaisy186

• Guest
##### Re:Equilibrium-looking for Kp
« Reply #10 on: July 31, 2005, 03:26:49 PM »
ahah! yup, that's what I did. Thanks!