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:

N

_{2}O

_{4} (g)

2 NO

_{2} (g) Kp1 = 0.134

N

_{2}O

_{3} (g)

NO

_{2} (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 + 0.0509P

_{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 = 120.27x

^{2} + 1.116x + 0.00259

0 = 120.27x

^{2} + 2.587x - 0.0179

0 = x

^{2} + 0.0215x - 0.000149

x = [-0.0215 + (0.0215

^{2} + 4 · 0.000149)

^{1/2}]/2

=

0.0055This 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