Respuesta :
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.
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 - 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
