A sample of N2O4(g) is placed in an empty cylinder at a certain temperature. It is allowed to decompose and form NO2(g). At equilibrium, the flask has a total pressure of 1.000 atm. If the KP at this temperature is 0.85, calculate the partial pressure (in atm) of NO2(g) at equilibrium. g

Based on the total pressure at equilibrium, the partial pressure NO₂ of at equilibrium is 0.590 atm.

From the equation of the reaction, the equilibrium is determined:

At equilibrium, N₂O₄ and NO₂ exist in the ratio 1 : 2.

N₂O₄ at x pressure releases y pressure to form NO₂ .

Thus at equilibrium:

N₂O₄ → x - y

NO₂ → 2y

Ptotal = 1 atm

Thus:

(x-y) + 2y = 1 atm

x + y = 1 atm

Also, Kp = 0.85

From the equation of the reaction:

Kp =  [Partial pressure NO₂]² / [ Partial pressure N₂O₄]

Kp = (2y)² / (x-y)

Kp = 4y² / (x-y)

Solving for x from the first equation:

x + y = 1 atm

x = 1 atm - y

Substitute x = 1 - y in the Kp expression

0.85 = 4y² / ( 1 - y - y)

0.85 = 4y² / 1 -2y

0.85 - 1.7y - 4y² = 0

Solving the quadractic equation:

where a = -4, b = -1.7 c = 0.85

y = (-b +- √(b² - 4ac)) / (2a)

y =(1.7 +-√((-1.7)² - 4 (-4) × 0.85) / 2 × (-4)

y = 0.295 or y = -0.720

We take positive value of y only.

Since partial pressure of NO₂ in equilibrium is 2y

Partial pressure of NO₂ = 0.295 × 2

Partial pressure of NO₂ = 0.590 atm

Therefore, the partial pressure of at equilibrium is 0.590 atm.

Learn more about about equilibrium partial pressure at: link

Part. press. NO₂ in equilibrium is 0.590 atm

Explanation:

First of all, we determine the equilibrium:

N₂O₄ (g) ⇄ 2NO₂ (g)

These is a system of two unknown values.

In the begining we have x pressure of N₂O₄ and no value for NO₂.

During the reaction, y pressure has been released from N₂O₄. As ratio is 1:2, 2y will be the value for the pressure of NO₂. So in the equilibrium we have:

N₂O₄ → x - y

NO₂ → 2y

Data from the excersise states that the total pressure is 1 atm so we know that the sum of partial pressures in a mixture, will be the total one. In the equilibrium, total pressure will be:

(x-y) + 2y = 1 atm

x + y = 1 atm

Let's make the expression for Kp

Kp =  [Partial pressure NO₂]² / [ Partial pressure N₂O₄]

Kp = (2y)² / (x-y)

Kp = 4y² / (x-y)

We split the x value in the first equation:

x + y = 1 atm

x = 1 atm - y

x = 1 - y → we put this in the Kp expression

0.85 = 4y² / ( 1 - y - y)

0.85 = 4y² / 1 -2y

This is a quadractic equation

0.85 - 1.7y - 4y² = 0 where (a = -4, b = -1.7 c = 0.85)

(-b +- √(b² - 4ac)) / (2a)

(1.7 +-√((-1.7)² - 4 (-4) . 0.85) / 2 .(-4) → 0.295 = y

As [ Part. press. NO₂] in equilibrium is 2y → 0.295 . 2 = 0.590 atm

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