Question

Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called “NOx”) that are of interest to atmospheric chemistry. It can react with itself to form another form of NOX, dinitrogen tetroxide.

A chemical engineer studying this reaction fills a2.0 L flask at 17. °C with 0.71 atm of nitrogen dioxide gas. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 0.29 atm of nitrogen dioxide gas.

The engineer then adds another 0.24 atm of nitrogen dioxide and allows the mixture to come to equilibrium again. Calculate the pressure of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer 1

Reaction is:

2NO₂ (g) N₂O₄(g)

I(atm) 0.71 0

E(atm) 0.29 0.21

Thus, change which takes place is :

C(atm) . 0.42 0.21

Kp = pN₂O₄ / (pNO₂)²

Kp = 0.21/ (0.29)² = 2.497

now, added NO₂ = 0.24 atm

Added(atm) 0.24 0

Total (atm) 0.53 0.21

change that takes place after the addition of NO₂ is:

C(atm) -2x x

E(atm) 0.53-2x 0.21+x

Now, Kp = pN₂O₄ / (pNO₂)²

2.497 = (0.21+x )/ (0.53-2x )²

2.497 (0.2809 +4x² - 2.12x) = 0.21 + x

0.7014 + 9.988 x² - 5.294 x = 0.21 +x

9.988 x² - 6.294 x + 0.4914 = 0

x = 0.0913

Thus, at equilibrium, pN₂O₄ = 0.21+x =0.21+0.0913 = 0.30 atm

pNO₂ = 0.53 - 2(0.0913) = 0.35 atm