Question

Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitrogen x oxygens") are serious contributors to air pollution. They can often be interconverted, sometimes by reaction with oxygen or ozone (O3) in the air.

An atmospheric scientist decides to study the reaction between nitrogen monoxide and oxygen that produces nitrogen dioxide. He fills a stainless steel reaction chamber with 4.9atm of nitrogen monoxide gas and 5.1atm of oxygen gas and raises the temperature considerably. At equilibrium he measures the mole fraction of nitrogen dioxide to be 0.52.

Calculate the pressure equilibrium constant Kp for the equilibrium between nitrogen monoxide, oxygen, and nitrogen dioxide at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer 1

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm    5.1 atm    0       Initial

-2x             -x           +2x    Reacts (stoichiometry is 2:1:2)

4.9-2x      5.1-x        2x      Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

Answer 2

The value of pressure equilibrium constant for the reaction has been 9.2.

The balanced chemical equation for the reaction has been:

$\rm 2\;NO\;+\;O_2\;\leftrightharpoons\;2\;NO_2$

The balanced chemical equation has been used in the determination of the partial pressure of each gas for the estimation of equilibrium constant.

Computation of Equilibrium constant

The partial pressure of a gas has been given as the product of mole fraction to the total pressure of the mixture.

The pressure of each gas has been given in the ICE table attached.

The mole fraction of nitrogen dioxide has been measured as 0.52. The value of x from mole fraction has been given as:

$\rm P_{NO_2}=Mole\;fraction\;\times\;P$

Where, the mole fraction of nitrogen dioxide, 0.52

The partial pressure of nitrogen dioxide at equilibrium, $\rm P_{NO_2}=2x$

The total pressure of the reaction has been given as, P:

$\rm P=4.9 - 2x + 5.1 -x + 2x\\P=10-x$

Substituting the values for the value of x:

$\rm 2x=0.52\;\times\;10-x\\2.52x=5.2\\x=2.06$

The value of x has been 2.06.

The partial pressure of the gas has been found as:

• Partial pressure of NO:

$\rm P_{NO}=4.9-2x\\P_{NO}=4.9-2\;\times\;2.06\\P_{NO}=0.773$

The partial pressure of NO has been 0.773 atm.

• Partial pressure of Oxygen:

$\rm P_{O_2}=5.1-x\\P_{O_2}=5.1-2.06\\P_{O_2}=3.04$

The partial pressure of oxygen has been 3.04atm.

• Partial pressure of Nitrogen dioxide:

$\rm P_{NO_2}=2x\\P_{NO_2}=2\;\times2.06\\P_{NO_2}=4.12$

The partial pressure of oxygen has been 4.12 atm.

The equilibrium constant (Kp) for the reaction has been given as:

$Kp=\rm\dfrac{[P_{NO_2}]^2}{[P_N_O]^2[P_{O_2}]}$

Substituting the values in the equation for the equilibrium constant,

$Kp=\dfrac{(4.12)^2}{(0.773)^2\;(3.04)} \\K_p=\dfrac{16.9744}{1.849536} \\K_p=9.2$

The value of equilibrium constant for the reaction has been 9.2.

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