Question

In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (0.790 atm N₂, 0.210 atm O₂). What are the equilibrium partial pressures of N₂, O₂, and NO if $K_p$ = 0.0460 for the following reaction:
$N_2(g)+O_2(g) \rightleftharpoons 2NO(g)$

Answer 1

Answer:

N₂ = 0.7515atm

O₂ = 0.1715atm

NO = 0.0770atm

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g)

Where Kp is defined as:

$Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}$

Pressures in equilibrium are:

N₂ = 0.790atm - X

O₂ = 0.210atm - X

NO = 2X

Replacing in Kp:

0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]

0.0460 = 4X² / 0.1659 - X + X²

0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²

-3.954X² - 0.0460X + 7.6314x10⁻³ = 0

Solving for X:

X = - 0.050 → False answer. There is no negative concentrations.

X = 0.0385 atm → Right answer.

Replacing for pressures in equilibrium:

N₂ = 0.790atm - X = 0.7515atm

O₂ = 0.210atm - X = 0.1715atm

NO = 2X = 0.0770atm

Answer 2

Answer:

partial pressure N2 =  0.7515 atm

partial pressure O2 =  0.1715 atm

partial pressure NO =  0.077 atm

Explanation:

Step 1: Data given

Temperature = 2200 °C

Pressure of N2 = 0.790 atm

Pressure of O2 = 0.210 atm

Kp = 0.0460

Step 2: The balanced equation

N2(g) + O2(g) ⇆ 2NO(g)

Step 3: The pressure at equilibrium

pN2 = 0.790 - X atm

pO2 = 0.210 - X atm

pNO = 2X

Step 4: Define Kp and the partial pressures

Kp = (pNO)² / (pO2 * pN2)

0.0460 = 4X² / (0.210 - X)(0.790 - X)

X = 0.0385

pN2 = 0.790 - 0.0385 =  0.7515 atm

pO2 = 0.210 - 0.0385 = 0.1715 atm

pNO = 2*0.0385 = 0.077 atm