Equilibrium pressure of a gaseous system

[anisotropic]

Homework Statement

2NO_2(g) -> N₂O_4(g)

• Calculate ΔG at 298.15 K - complete
• Calculate K_P 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
K_P = (P_products)ⁿ / (P_reactants)ⁿ

The Attempt at a Solution

ΔG = -RTlnK
ΔG = ΔG_N2O4 - 2ΔG_NO2
ΔG = (97.9 kJ/mol) - 2(51.3 kJ/mol)
ΔG = -4.73 kJ/mol

K_P = P_N2O4 / (P_NO2)²

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

 

[Borek]

Calculate K first.

 

[anisotropic]

K = K_p ?

 

[anisotropic]

Even if K = K_P, I will still have 2 unknowns in K_P = P_N2O4 / (P_NO2)²

Please explain.

 

[Borek]

Amounts of both gases are linked through reaction stoichiometry.

 

[anisotropic]

So I'm assuming by that you mean P_NO2 = 2P_N2O4 ?

i.e. K_P = P / (2P)² ?

 

[Borek]

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?

 

[anisotropic]

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.

 

[Borek]

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 :(

 

[anisotropic]

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).

 

[Borek]

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).

 

[anisotropic]

I still do not understand what the above calculated x value represents. Is it a proportion of P_total? 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).

 

[Borek]

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)?

 

[anisotropic]

= K_P
= 6.74
= P_N2O4 / (2P_NO2)²
= P_N2O4 / 4P²_NO2
= x / 4x²

 

[Borek]

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.

 

[anisotropic]

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 P_initial, 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 P_initial - 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?

 

[Borek]

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.

 

[anisotropic]

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.

 

[Borek]

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.

 

[anisotropic]

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 = K_P
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 P_initial of the system (N₂O₄) as follows:

P_initial = (nRT)/V
P_initial = [(1 mol)(8.3145 J K^-1 mol^-1)(298.15 K)]/(0.02446 m³)
P_initial = 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?

 

[Borek]

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.

 

[anisotropic]

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
P_total > P⁰_total : 140,873 Pa > 101,348 Pa

Is this correct? Yes or no?

 

[Borek]

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

 

[anisotropic]

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.

 

[cristo]

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 realize 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.