Question

At a particular temperature, Kp = 0.25 for the following reaction.

N2O4(
g. 2 NO2(
g. (
a. A flask containing only N2O4 at an initial pressure of 5.3 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.
(
b. A flask containing only NO2 at an initial pressure of 7.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.

Answer 1

The reaction is
N₂O₄(g) <=> 2NO₂(g)

a. Initial pressure is 5.3 atm
We assume that the there is 1 mole at the start
N₂O₄(g) <=> 2NO₂(g)
1
-x 2x
---------------------------
1 - x 2x
The total moles is
1 + x
The mole fraction of N2O4 is
(1 - x)/(1 + x)
The mole fraction of NO2
2x/(1 + x)

Kp = [2x/(1 + x)]² / [(1 - x)/(1 + x)] = 0.25
Solve for x
And the total number of moles
The total pressure is solved by
(1 + x) (5.3)

The partial pressure of N2O4 is
5.3 (1 - x)
The partial pressure of NO4 is
5.3 (2x)

Do the same for the second problem.

Answer 2

See explanation below

Step-by-step explanation:

Let's write the equation:

N₂O₄(g) <------> 2NO₂(g) Kp = 0.25

The general expression for Kp for this reaction is:

Kp = pNO₂²/pN₂O₄ (1)

With these expression we will calculate the partial pressure of the gases. Now, let's solve this by parts:

a) When pN₂O₄ = 5.3 atm

In this case, we'll do an ICE chart of this reaction:

N₂O₄(g) <------> 2NO₂(g)

i) 5.3 0

c) -x +2x

e) 5.3 - x 2x

Replacing into (1) we have:

0.25 = (2x)² / (5.3 - x)

From here, we solve for x:

0.25(5.3-x) = 4x²

1.325 - 0.25x = 4x²

4x² + 0.25x - 1.325 = 0

Now, we use the general expression for calculate x in these conditions:

x = -b ±√b² - 4ac / 2a (2)

Replacing the data here: (a = 4; b = 0.25; c = -1.325)

x = -0.25 ±√(0.25)² - 4*4*(-1.325) / 2*4

x = -0.25 ±√21.2625 / 8

x = -0.25 ± 4.61 / 8

x1 = -0.25 + 4.61 / 8 = 0.55

x2 = -0.25 - 4.61 / 8 = -0.61

So, we take the positive value.

x = 0.55 atm

Therefore the partial pressure of the gases are:

pN₂O₄ = 5.3 - 0.55 = 4.75 atm

pNO₂ = 2 * 0.55 = 1.10 atm

b) When pNO₂ = 7 atm

In this part we do the same but with NO2 so:

N₂O₄(g) <------> 2NO₂(g)

i) 0 7

c) +x +2x

e) x 7 - 2x

replacing in (1):

0.25 = (7-2x)²/x

0.25x = 4x² - 28x + 49

4x² - 28.25x + 49 = 0

Using (2):

x = 28.25 ±√(28.25)² - 4*4*49 / 2*4

x = 28.25 ±√14.0625 / 8

x = 28.25 ± 3.75 / 8

x1 = 28.25 + 3.75 / 8 = 4

x2 = 28.25 - 3.75 / 8 = 3.0625

We can use either of the two values, but 2x is 2*4 = 8, it's higher than 7, and the result would be a negative number, so we use the smaller number, x2 will be the one:

pNO₂ = 7 - (2*3.0625) = 0.8744 atm

pN₂O₄ = 4.0625 atm