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.

Solutions

To answer these questions, we need to use the equilibrium constant expression Kp and the given equilibrium constant value.

(a) For the reaction N2O4(g) ⇌ 2 NO2(g), the equilibrium constant expression is:

Kp = [NO2]^2 / [N2O4]

Given that Kp = 0.25, we can use this value and the initial pressure of N2O4 (5.3 atm) to calculate the equilibrium partial pressures of the gases.

Let's assume the equilibrium partial pressures of N2O4 and NO2 are x atm and y atm respectively. At equilibrium:

[N2O4] = x atm
[NO2] = 2y atm

Now, substituting these values into the equilibrium constant expression:

Kp = [NO2]^2 / [N2O4]
0.25 = (2y)^2 / x
0.25 = 4y^2 / x
1 = 16y^2 / 4x
16y^2 = x

We also know that the total pressure is the sum of the partial pressures:

Total pressure = [N2O4] + [NO2]
5.3 atm = x + 2y

Since we have two equations and two unknowns, we can solve for x and y simultaneously.

Substituting the value of x from the first equation into the second equation:

5.3 atm = 16y^2 + 2y

Now, we need to solve this quadratic equation. Rearranging the equation:

16y^2 + 2y - 5.3 = 0

Solving this quadratic equation using factoring, completing the square, or the quadratic formula, we find that:

y = 0.353 atm or y = -0.387 atm

Since partial pressures cannot be negative, the only valid solution is y = 0.353 atm. Substituting this value back into the equation for x:

x = 16(0.353)^2
x = 2.0 atm

Therefore, at equilibrium, the partial pressure of N2O4 is 2.0 atm and the partial pressure of NO2 is 0.706 atm.

(b) For this part, we are given the initial pressure of NO2 (7.0 atm) and we need to calculate the equilibrium partial pressures of the gases.

Following a similar approach as in part (a), let's assume the equilibrium partial pressures of N2O4 and NO2 are x atm and y atm respectively.

At equilibrium:
[N2O4] = x atm
[NO2] = 2y atm

Substituting these values into the equilibrium constant expression:

Kp = [NO2]^2 / [N2O4]
0.25 = (2y)^2 / x
0.25 = 4y^2 / x
1 = 16y^2 / 4x
4y^2 = x

Using the total pressure equation:

Total pressure = [N2O4] + [NO2]
7.0 atm = x + 2y

Now, we have two equations and two unknowns. We can solve for x and y simultaneously.

Substituting the value of x from the first equation into the second equation:

7.0 atm = 4y^2 + 2y

Rearranging the equation:

4y^2 + 2y - 7.0 = 0

Solving this quadratic equation using factoring, completing the square, or the quadratic formula, we find that:

y = 1.0 atm or y = -1.75 atm

Since partial pressures cannot be negative, the only valid solution is y = 1.0 atm. Substituting this value back into the equation for x:

x = 4(1.0)^2
x = 4.0 atm

Therefore, at equilibrium, the partial pressure of N2O4 is 4.0 atm and the partial pressure of NO2 is 2.0 atm.

To determine the equilibrium partial pressures of the gases, we need to use the equilibrium constant expression (Kp) and the given initial pressures. The equilibrium constant expression for this reaction is:

Kp = [NO2]^2 / [N2O4]

(a) In this case, we are starting with only N2O4 and no NO2. Let's assume that at equilibrium, the partial pressure of N2O4 is x and the partial pressure of NO2 is y. Since we have no NO2 initially, its initial partial pressure is 0.

Using the given equilibrium constant (Kp = 0.25), we can write the equation as follows:

0.25 = y^2 / x

We also know that the initial pressure of N2O4 is 5.3 atm, which is equal to the partial pressure at equilibrium (x). Therefore, we can substitute x = 5.3 in the equation:

0.25 = y^2 / 5.3

Now we solve for y:

y^2 = 0.25 * 5.3
y^2 = 1.325
y ≈ √(1.325)
y ≈ 1.15

So, the equilibrium partial pressure of NO2 is approximately 1.15 atm, and the equilibrium partial pressure of N2O4 is 5.3 atm.

(b) In this case, we are starting with only NO2 and no N2O4. Let's assume that at equilibrium, the partial pressure of NO2 is x and the partial pressure of N2O4 is y. Since we have no N2O4 initially, its initial partial pressure is 0.

Using the given equilibrium constant (Kp = 0.25), we can write the equation as follows:

0.25 = x^2 / y

We also know that the initial pressure of NO2 is 7.0 atm, which is equal to the partial pressure at equilibrium (x). Therefore, we can substitute x = 7.0 in the equation:

0.25 = 7.0^2 / y

Now we solve for y:

y = (7.0^2) / 0.25
y = 49 / 0.25
y = 196

So, the equilibrium partial pressure of NO2 is 7.0 atm, and the equilibrium partial pressure of N2O4 is approximately 196 atm.

You should get into the habit of using an arrow; otherwise we must guess which are products and which are reactants.

..........N2O4 ==> 2NO2
initial....5.3atm....0
change.....-p.....2p
equil..... 5.3-p....2p

Kp = 0.25 = (NO2O^2/(N2O4)
Substitute from the ICE chart into Kp expression and solve for p.
Then 2p = PNO2 and 5.3-p = PN2O4.

For #2,
...........N2O4 ==> 2NO2
initial......0........7
change......+p........-p
equil........p.......7-p
Substitute into Ka and solve for p.