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

N2O4(

g. 2 NO2(

g. (

AA 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.

(

BA 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.

Verified Answer :

MissPhiladelphia

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:

See explanation below

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 - 44(-1.325) / 24

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 - 4449 / 24

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 24 = 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

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