Consider the following reaction.

Question

Consider the following reaction.

2NO2(g)⇌N2O4(g)

When the system is at equilibrium, it contains NO2 at a pressure of 0.724 atm , and N2O4 at a pressure of 0.0524 atm . The volume of the container is then reduced to half its original volume. What is the pressure of each gas after equilibrium is reestablished?

𝑃NO2 =
atm
𝑃N2O4=
atm

Answer 1

..................2NO2(g) ⇌ N2O4(g)

E............0.724 atm........0.0524 atm
Kp = p(N2O4)/p^2(NO2) = 0.0524/(0.724)^2 = 0.1 (rounded to 0.1)
Reducing the volume to 1/2 means pN2O4 becomes 0.0524 x 2 = 0.105 atm and pNO2 = 0.724 x 2 = 1.45 atm. Then, Le Chatelier's Principles tells us that when pressure is increased in a system at equilibrium that the equilibrium will shift to the side with fewer moles. So it will shift to the right producing more N2O4 at the expense of NO2.
.........................2NO2(g) ⇌ N2O4(g)
I.........................1.45...............0.105
C.......................-2x.....................+x
E.....................1.45-2x................0.105+x
Plug the E line into the Kp expression and solve for x, then evaluate 1.45 - 2x and 0.105 + x. Note that I have rounded here and there so you should start from the beginning and use your numbers and not mine.
Post your work if you get stuck.

Answer 2

Well, when you reduce the volume of a container, gases tend to feel a bit claustrophobic. So, they try to escape from the pressure by increasing it. It's like those mini stress balls you squeeze - the harder you squeeze, the more pressure you generate.

In this case, since the volume is reduced to half, the pressure of each gas will double. So, the pressure of NO2 will be 0.724 atm * 2 = 1.448 atm.

As for N2O4, its pressure will also double, so it will be 0.0524 atm * 2 = 0.1048 atm.

Remember, in the world of gases, a little squeeze can go a long way!

Answer 3

To determine the pressure of each gas after the volume of the container is reduced to half its original volume, we can use the principle of Le Chatelier's principle.

According to Le Chatelier's principle, when the volume of the container decreases, the system will shift in the direction that reduces the number of moles of gas.

In this case, the reaction 2NO2(g) ⇌ N2O4(g) consists of two moles of NO2 and one mole of N2O4. Therefore, reducing the volume of the container will favor the formation of the product N2O4 to reduce the number of moles of gas.

Let's denote the original pressure of NO2 as P1 and the original pressure of N2O4 as P2.

We are given:
Initial P1 = 0.724 atm
Initial P2 = 0.0524 atm

Since the volume is halved, the new volume is half of the original volume.

Now, to find the new pressures of NO2 and N2O4 after equilibrium is reestablished, we can use the following expression:

P1' = P1 x (V1'/V1)
P2' = P2 x (V2'/V2)

Where:
P1' = new pressure of NO2
P2' = new pressure of N2O4
V1' = new volume of the container (half of the original volume)
V1 = original volume of the container
V2' = new volume of the container (half of the original volume)
V2 = original volume of the container

Since both V1' and V2' are half of the original volume, the ratios of V1'/V1 and V2'/V2 are both 1/2.

Substituting the given values into the expressions, we get:

P1' = 0.724 atm x (1/2) = 0.362 atm
P2' = 0.0524 atm x (1/2) = 0.0262 atm

Therefore, the pressure of NO2 after equilibrium is reestablished is 0.362 atm, and the pressure of N2O4 after equilibrium is reestablished is 0.0262 atm.

Answer 4

To solve this question, we can use the concept of the ideal gas law and the relationship between pressure, volume, and moles of gas.

First, let's calculate the initial number of moles of NO2 and N2O4. We can use the ideal gas law equation:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature (which remains constant in this case).

Given:
Pressure of NO2 (PNO2) = 0.724 atm
Pressure of N2O4 (PN2O4) = 0.0524 atm

Since the moles of both gases are related by the balanced equation (2NO2 ⇌ N2O4), we can assume x moles of N2O4 and 2x moles of NO2 initially.

Using the ideal gas law, we can rearrange the equation to solve for n:

n = PV / RT

For NO2:
n(NO2) = (PNO2 * V) / RT

For N2O4:
n(N2O4) = (PN2O4 * V) / RT

Now, let's consider the change in volume. If the volume is reduced to half its original size, the new volume (V') will be V/2.

At equilibrium, the number of moles of NO2 and N2O4 remains constant. The equilibrium expression can be written as:

K = (PN2O4 * PNO2^2) / (PNO2^2)

where K is the equilibrium constant.

Now, let's rewrite the equation using the initial moles:

K = (n(N2O4) * n(NO2)^2) / (n(NO2)^2)

Since the number of moles remains constant at equilibrium, we can equate the two expressions for K:

(n(N2O4) * n(NO2)^2) / (n(NO2)^2) = (PN2O4 * PNO2^2) / (PNO2^2)

Substituting the expressions for n(N2O4) and n(NO2):

((PN2O4 * V) / RT) * (((PNO2 * V) / RT)^2) / (((PNO2 * V) / RT)^2) = (PN2O4 * PNO2^2) / (PNO2^2)

Canceling out the common terms and rearranging, we get:

(V / RT) * (((PNO2 * V) / RT)^2) = PN2O4

Simplifying further:

(PNO2 * V)^2 / (RT)^2 = PN2O4

Now, let's calculate the new pressures using the reduced volume (V/2) after equilibrium is reestablished.

(PNO2' * V/2)^2 / (RT)^2 = PN2O4'

(PNO2' * V^2) / (4 * (RT)^2) = PN2O4'

Substituting the initial pressures, volume, and the constant R:

(0.724 * V^2) / (4 * (RT)^2) = 0.0524

Since the volume remains constant when solving for the ratio of pressures, we can cancel out the V^2 term:

(0.724) / (4 * (RT)^2) = 0.0524

Now, let's solve for PNO2':

PNO2' = (0.0524 * 4 * (RT)^2) / 0.724

Similarly, for PN2O4':

PN2O4' = PN2O4

By plugging in the appropriate values for R and T, you can compute the new pressures PNO2' and PN2O4'.