• Support PF! Buy your school textbooks, materials and every day products Here!

Equilibrium pressure of a gaseous system

  • #1

Homework Statement



2NO2(g) -> N2O4(g)

  • Calculate ΔG at 298.15 K - complete
  • Calculate KP at 298.15 K
  • Calculate equilibrium pressure of a system with initial conditions of 1 mol N2O4, confined to 24.46 liters fixed volume, at 298.15 K.

Homework Equations



ΔG = -RTlnK
KP = (Pproducts)n / (Preactants)n

The Attempt at a Solution



ΔG = -RTlnK
ΔG = ΔGN2O4 - 2ΔGNO2
ΔG = (97.9 kJ/mol) - 2(51.3 kJ/mol)
ΔG = -4.73 kJ/mol

KP = PN2O4 / (PNO2)2

How do you determine the partial pressures of either NO2 or N2O4?
 

Answers and Replies

  • #2
Borek
Mentor
28,448
2,843
Calculate K first.
 
  • #4
Even if K = KP, I will still have 2 unknowns in KP = PN2O4 / (PNO2)2

Please explain.
 
  • #5
Borek
Mentor
28,448
2,843
Amounts of both gases are linked through reaction stoichiometry.
 
  • #6
So I'm assuming by that you mean PNO2 = 2PN2O4 ?

i.e. KP = P / (2P)2 ?
 
  • #7
Borek
Mentor
28,448
2,843
No. You started with 1 mole of N2O4. If x moles of N2O4 reacted - how many moles of N2O4 were left? How many moles of NO2 were produced?
 
  • #8
No. You started with 1 mole of N2O4. If x moles of N2O4 reacted - how many moles of N2O4 were left? How many moles of NO2 were produced?
2

Here's where I am right now:

I calculated P (x) in the above equation to be 0.0371

Using PV = nRT, I calculated the initial pressure of N2O4 to be 101,348 Pa:

P = [(1 mol)(8.3145J K-1 mol-1)(298.15 K)] / (0.02446 m3)
P = 101,348 Pa

NO2
initial: 0
change: +2x
equil.: 2x

N2O4
initial: 101,348
change: -x
equil.: 101,348 - x

Seems wrong.
 
Last edited:
  • #9
Borek
Mentor
28,448
2,843
Doesn't look wrong to me. Just enter these values into reaction quotient and solve for x. Use this x to calculate pressure - and you are done.

Note: I am not sure about K/Kp thing, black hole between my ears :(
 
  • #10
Doesn't look wrong to me. Just enter these values into reaction quotient and solve for x. Use this x to calculate pressure - and you are done.
I already did this and got 0.0371 (see my previous post). What does 0.0371 refer to?

Right now I am doing this, which is clearly wrong:

NO2
initial: 0
change: +2x
equil.: 2x = 2(0.0371) = 0.0742

N2O4
initial: 101,348
change: -x
equil.: 101,348 - x = 101,348 - 0.0371 = 101,347.96

Please clarify.

EDIT: If it would help, just let me know and I can type out the example given in the textbook. It's long, but it might help you see exactly what I should be doing (for some reason, I can't seem to be able to translate the example into a solution for this specific problem).
 
Last edited:
  • #11
Borek
Mentor
28,448
2,843
Final pressure is sum of both pressures - that of N2O4 and that of NO2. Use x to calculate both pressures (or number of moles, then pressures - math will be identical, as pressure is directly proportional to the number of moles, just numbers will be different).
 
  • #12
I still do not understand what the above calculated x value represents. Is it a proportion of Ptotal? Or is it an actual pressure value as I have calculated?

i.e. are my above numbers correct? The change in contributed pressures seems much too small to be the correct answer (with NO2 contributing only 0.00007% to the total pressure, which strikes me as ridiculously low - the K value is nowhere near high enough to have such a bias favouring the product, N2O4).
 
Last edited:
  • #13
Borek
Mentor
28,448
2,843
It becomes hard to follow.

What value of Kp have you used?

What is the reaction quotient in terms of x (the equation that you have solved)?
 
  • #14
= KP
= 6.74
= PN2O4 / (2PNO2)2
= PN2O4 / 4P2NO2
= x / 4x2
 
  • #15
Borek
Mentor
28,448
2,843
x/4x2 is just 1/4x. You have prepared ICE table, but you are not using E line. The way it is written now it looks like both gases came from nowehere.
 
  • #16
You have prepared ICE table, but you are not using E line. The way it is written now it looks like both gases came from nowehere.
Why?

I calculated Pinitial, which applies only to N2O4, as its the only form of gas initially present. According to the E line, the change in N2O4 is -x (so Pinitial - x), while the change in NO2 is +2x (so 0 + 2x):

NO2
equil.: 2x = 2(0.0371) = 0.0742

N2O4
equil.: 101,348 - x = 101,348 - 0.0371 = 101,347.96
Can you please clarify what you mean by "came from nowhere"?

And is x = 0.037 in Pa or in bar?
 
Last edited:
  • #17
Borek
Mentor
28,448
2,843
2NO2 <-> N2O4
I: 0 1
C: +2x -x
E: +2x 1-x

K = (1-x)/(2x)2

Take a look at numerator - it is 1-x, that means initial 1 minus x that reacted; denominator is 2x produced from what is missing in the numerator. When it was x/4x2 both numerator and denominator look like they contain x produced from nowhere - nothing is consumed.
 
  • #18
So in other words, per 1 mol of N2O4 originally present, 2 x 0.0371 mol of N2O4 is converted to NO2. This would answer my earlier question of what the x value represents. I was under the impression that x was the pressure value itself.
 
  • #19
Borek
Mentor
28,448
2,843
This is a constant volume system, so the total pressure can change. In the case of constant pressure systems, n and partial pressures are directly protportional, so you can use whichever you want.
 
Last edited:
  • #20
Having returned to this problem after a brief absence, I am at a loss as to how I originally calculated the value of K as being 6.74. The value I am coming up with now is 1.002.

I am going ahead with the problem assuming the latter value to be the correct one, unless someone here can correct me.

K = KP
K = e-ΔG/RT
K = e-(-4.74 kJ mol-1)/(8.3145 J K-1 mol-1)(298.15 K)
K = 1.002

* Value previously calculated in post #1

equilibrium values (from ICE table):
E (2NO2): 2x
E (N2O4): 1 - x

K = 1.002 = (1 - x)/(2x)2
K = 1.002 = (1 - x)/4x2
4.008x2 + x - 1 = 0

Using quadratic formula, solve for x...

x = 0.390, -0.640

Discard negative value, as molar value cannot be negative.

x = 0.390

Thus, per 1 mol of N2O4, 2 x 0.390 = 0.780 mol of it is converted into NO2 at equilibrium.

-or-

Per 1 unit of pressure initially exerted by N2O4, 2 x 0.390 = 0.780 units of pressure will be exerted by NO2 at equilibrium.

Is this correct? Yes or no?

Assuming it is correct, then, calculating Pinitial of the system (N2O4) as follows:

Pinitial = (nRT)/V
Pinitial = [(1 mol)(8.3145 J K-1 mol-1)(298.15 K)]/(0.02446 m3)
Pinitial = 101,348 Pa

And based on the initial calculations above, the equilibrium pressure would be:

2NO2: 79,051 Pa
N2O4: 22,297 Pa

Is this correct? Yes or no?
 
Last edited:
  • #21
Borek
Mentor
28,448
2,843
Having returned to this problem after a brief absence, I am at a loss as to how I originally calculated the value of K as being 6.74. The value I am coming up with now is 1.002.
Perhaps you forgot (then or now) about volume change.

2NO2: 79,051 Pa
N2O4: 22,297 Pa

Is this correct? Yes or no?
No, and it is obvious even without looking at your calculations. You have a fixed volume, constant temperature, you have a reaction in which number of total moles change, but miraculously total pressure stays identical.
 
  • #22
I appreciate the help, but...

Why don't you just tell me the exact step where my logic is wrong, instead of giving me unspecific and haphazard answers? I have clearly outlined, step by step, my thought process, in order to make it easy for you to critique it, yet you continue to give me fragments of information that serve only to confuse me further.

It's not as if I haven't put in more than my share of effort. My work is all there, in a nicely formatted and presented form. Surely this warrants a little more directed help?

On that note, how's this?:

At equilibrium:
2NO2: 79,051 Pa
N2O4: 61,822 Pa
Ptotal > P0total : 140,873 Pa > 101,348 Pa

Is this correct? Yes or no?
 
Last edited:
  • #23
Borek
Mentor
28,448
2,843
Why don't you just tell me the exact step where my logic is wrong, instead of giving me unspecific and haphazard answers? I have clearly outlined, step by step, my thought process, in order to make it easy for you to critique it, yet you continue to give me fragments of information that serve only to confuse me further.
Because I have no time to babysit you, so if I am seeing at first sight that something is wrong, I am not going to spend my time looking for a mistake. And asking me in PM to check this thread again and again you are not helping your case.

Numbers presented this time look much better. Doesn't mean they are correct, but I can't see anything wrong about them (assuming K is OK).

EOT
 
  • #24
Because I have no time to babysit you, so if I am seeing at first sight that something is wrong, I am not going to spend my time looking for a mistake. And asking me in PM to check this thread again and again you are not helping your case.
The only babysitting that needs to be done here is those of your manners.

Either help properly, or don't help at all. Take some time from your oh-so-busy schedule and make up your mind as to which you want to do. And I say "oh-so-busy" facetiously, as you seem to be on these boards far too often, with far too high a post count, to really be able to claim any sort of busy schedule worthy of relevant note.

And a lesson for you, since you seem to be lacking in basic netiquette: the amount of help provided to a given user should be directly proportional to the effort that user puts in organizing, clarifying, and presenting a given problem; or in other words, if I've put an honest effort into solving the problem independently, and have presented the problem and all information as clearly as possible, then at the very least, I deserve an equally worthy response.

Instead, what we have here is an honest effort on my part, answered only with consistently curt and altogether ineffective responses on your part. If you're going to do something (in this case, help someone), do it right, or don't do it at all.

And asking me in PM to check this thread again and again you are not helping your case.
Icing on the cake of fail here. You bring up a non-issue, all in an effort to make it sound as though I am harassing you in PM, when all I have done is send you, on occasion (i.e. no more than thrice over the course of two weeks), short, polite messages, all in an effort to make sure you have not overlooked a thread which I very much depend on.

And at the end of the day, it's a PM for a reason. Next time, try not harping on about it in public chat in a childish effort to make me look like some sort of criminal. Really now, is this the sort of behaviour a senior PF member like you should be exhibiting? For shame.
 
  • #25
cristo
Staff Emeritus
Science Advisor
8,107
73
Either help properly, or don't help at all. Take some time from your oh-so-busy schedule and make up your mind as to which you want to do. And I say "oh-so-busy" facetiously, as you seem to be on these boards far too often, with far too high a post count, to really be able to claim any sort of busy schedule worthy of relevant note.
That's enough. Our homework helpers put in hours of their own time on a voluntary basis and do not have to put up with comments like this. You should also not send one PM bugging someone to check your thread, let alone three. If every student acted like this, our homework helpers would be swamped with hundreds of PMs per week.

if I've put an honest effort into solving the problem independently, and have presented the problem and all information as clearly as possible, then at the very least, I deserve an equally worthy response.
Again, you don't seem to realise that all help you are given here is on a voluntary basis. You do not "deserve" anything! Furthermore, you do not get to dictate to anyone when or how they should help you with your own homework.


I've seen enough. This thread is done.
 

Related Threads on Equilibrium pressure of a gaseous system

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
14
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
19
Views
2K
Replies
3
Views
910
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
2
Views
1K
Top