[Rutherford]

The cock between the two vessels in the figure is closed. Opening it the two gases mix to form NO₂. A part of this NO₂ will dimerize to N₂O₄. After the system has reached equilibrium and the initial temperature appears again, the differences in height of the mercury columns in the attached manometers amount to 7.1 cm instead of 10 cm in the beginning. Calculate the percentage of NO₂ that has dimerized to N₂O₄.
Assumptions: N₂O₄ is totally gaseous. The vapor pressure of mercury in the closed ends of the manometers can be neglected, you may assume vacuum there. The changes in height of the mercury columns have no influence on the total volume, it can be regarded as constant.

Solving:
NO+0.5O₂ NO₂
 3       1            0
 -2     -1           +2
=1     =0          =2

NO₂ 0.5N₂O₄
  2                 0
 -x                +0.5x
=2-x             =0.5x
x needs to be calculated, but I am not sure how to use the manometer data. I need some help.

[Borek]

Tells you about pressure changes.

[Rutherford]

Okay, but how to use the 7.1cm? When the cock is open, how will the gasses distribute?

By the way the answer should be 0.16.

[Borek]

After the cock is opened, gases mix, so the pressure is identical in both tanks (it was also identical before, as both manometers showed the same 10 cm). Change in the height tells you by how much the pressure changed from the initial, which is related to the stoichiometry of both reactions.

[Rutherford]

How so? Didn't one manometer show 10cm higher than the other? By doing the way you proposed:
4:10=V:7.1
V=2.84l
V=1+1-x+0.5x=2-0.5x
2-0.5x=2.84 not a positive result .

[Borek]

Think about it this way:

1. You have initially 1L of O₂ and 3 L of NO, both at 100 mmHg.

2. They react, and the reaction goes to completion. What is the pressure now?

3. NO₂ starts to dimerize, and pressure changes from the one calculated in 2, to 71 mmHg.

[Rutherford]

Found the mistake. The initial expression I wrote for the first reaction volume change was wrong. The final volume after the first reaction would be 3l, then:
4:V=10:7.1
V=2.84l
3-0.5x=2.84
x=0.32
α=0.32/2=0.16
Thank you for your help.