[XeLa.]

Hi,
I recently came across this question in an engineer's chemistry textbook (see attached).

I can do part (a) with relative ease; I obtain a Kp of 0.134, which is the right answer. However, I seem to be stuck on part (b). Since we must consider the effect of nitrogen dioxide acting on both equilibria, I take this into account as follows:

Equations:

N₂O₄ (g) 2 NO₂ (g)  Kp1 = 0.134
N₂O₃ (g) NO₂ (g) + NO (g)  Kp2 = ?

At equilibrium:

n_N2O4 = 0.0139 - x
n_NO = 0.0167 - y
n_N2O3 = y
n_NO2 = 2x + y (since both reactions contribute nitrogen dioxide)

n_tot = 0.0139 - x + 0.0167 - y + y + 2x + y = 0.0139 + x + y ...1
n_tot = PV/RT = (39.20 kPa · 2.2296)/(8.314 · 298) = 0.0353 ...2

Subbing (1) into (2)...

0.0139 + x + y = 0.0353
                    y = 0.0046 - x ...3

Now,

Kp1 = (P_NO2)²/P_N2O4 ...4

Subbing in (3)...

P_NO2 = P'_NO2/P° = [(2x + y) · 8.314 · 298]/(2.2296 · 101.325) = 10.97(2x + y) = 10.97(2x + 0.0046 - x) = 10.97x + 0.0509

P_N2O4 = P'_N2O4/P° = [(0.0139 - x) · 8.314 · 298]/(2.2296 · 101.325) = 10.97(0.0139 - x) = 0.153 - 10.97x


Now, let us substitute these into (4), and solve for x...

0.314 = (10.97x + 0.0509)²/(0.153 - 10.97x)

0.0205 - 1.471x = (10.97x + 0.0509)²

0.0205 - 1.471x = 120.27x² + 1.116x + 0.00259
                     0 = 120.27x² + 2.587x - 0.0179
                     0 = x² + 0.0215x - 0.000149

                     x = [-0.0215 + (0.0215² + 4 · 0.000149)^1/2]/2
                        = 0.0055

This x value, of course, results in a negative y-value since 0.0055 > 0.0046...

This is not possible. I have checked my numbers, and assumptions, and I ensured not to round until the final step. I thought that I had made the typical assumptions when solving such a multiple equilibria question, but the figures seem to suggest that such assumptions are invalid.

Any help to identify where I went wrong would be appreciated!

Thanks,
XeLa

[mjc123]

Haven't been through the whole thing, but immediately noticed

nNO₂ = 2x + y (since both reactions contribute nitrogen dioxide)

Should be 2x - y. NO reacts with NO₂ to give N₂O₃.
What do you get now?

[mjc123]

I get K2 = 1.30 atm; does the textbook agree?

[XeLa.]

I get K2 = 1.30 atm; does the textbook agree?

Thank you, mjc! That makes a lot of sense. Well, I didn't have to change my calculations much; just had to change the y-value in (3) to -y...

After doing so, I get an x-value of 0.0055 mol, and a y-value of 0.000862 mol.

From there, I get the following equilibrium mole quantities of NO₂, N₂O₃, and NO...

n_NO2 = 2x - y = 2(0.0055 - 0.000862) = 0.0101 mol
n_N2O3 = y = 0.000862 mol
n_NO = 0.0167 - y = 0.0167 - 0.000826 = 0.0159 mol

Then, finding the respective partial pressures and dividing by atmospheric pressure (101.325 kPa/1.00 atm)...

P_NO2 = 0.111
P_N2O3 = 0.009
P_NO = 0.174

Kp2 = P_NO · P_NO2/P_N2O3 = 0.111 · 0.174/0.009 = 2.05

The textbook gets an answer of 2.07, which is 1% off the answer I achieved. I think that has to do with their using 101.3 kPa to represent 1.00 atm, instead of 101.325 kPa; as I used.

NOTE: The original question comes from "Physical Chemistry for Engineering and Applied Sciences" by F. Foulkes (p.291).

[mjc123]

I did it hurriedly and may have made a mistake. I'm glad you get an answer so close to the textbook.
Just be careful:

nNO2 = 2x - y = 2(0.0055 - 0.000862) = 0.0101 mol

2x - y is not 2(x - y). However, you seem to have got the calculation right, just written the brackets wrongly. But take care over things like that, you might cause yourself problems.

[mjc123]

I did make a mistake, in solving the quadratic equation. Now I agree with you.
It's worth noting that if x = 0.0055 and y = x - 0.0046, you have quite a large relative uncertainty in y, unless you are using more decimal places than you have quoted (0.0055 - 0.0046 ≠ 0.000862). It might be helpful in a case like this to work in millimoles, and use more sig figs (e.g. 5.5132 mmol, or whatever) - as long as you take care to get the conversion factors right!

[XeLa.]

I did it hurriedly and may have made a mistake. I'm glad you get an answer so close to the textbook.
Just be careful:

nNO2 = 2x - y = 2(0.0055 - 0.000862) = 0.0101 mol

2x - y is not 2(x - y). However, you seem to have got the calculation right, just written the brackets wrongly. But take care over things like that, you might cause yourself problems.

Yes, I must have had an oversight when I was copying my calculations from paper onto screen. :/ Thanks for pointing it out, though!

I did make a mistake, in solving the quadratic equation. Now I agree with you.
It's worth noting that if x = 0.0055 and y = x - 0.0046, you have quite a large relative uncertainty in y, unless you are using more decimal places than you have quoted (0.0055 - 0.0046 ≠ 0.000862). It might be helpful in a case like this to work in millimoles, and use more sig figs (e.g. 5.5132 mmol, or whatever) - as long as you take care to get the conversion factors right!

I typically tend to round for my final answer; I save all of my exact numbers on my calculator. However, I do agree that I tend to neglect sig figs when writing my steps out. I'll try to work on that in future.

Thanks for the help, mjc!