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To solve the given problem, we need to follow these steps:

1. Calculate the initial molarity of each compound in the 250 mL flask.
2. Determine the changes in molarity for each compound using the stoichiometry of the reaction.
3. Find the equilibrium molarity for each compound.

Let's proceed step by step:

### Step 1: Calculate Initial Molarity

The initial moles of each gas are given:
- \( \text{NO}_2 = 1.1 \text{ mol} \)
- \( \text{NO} = 1.2 \text{ mol} \)

The volume of the flask is 250 mL, which is 0.250 L. Molarity is calculated as follows:

\[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume in liters}} \]

For \( \text{NO}_2 \):

\[ \text{Molarity of NO}_2 = \frac{1.1 \text{ mol}}{0.250 \text{ L}} = 4.4 \text{ M} \]

For \( \text{NO} \):

\[ \text{Molarity of NO} = \frac{1.2 \text{ mol}}{0.250 \text{ L}} = 4.8 \text{ M} \]

Oxygen (\( \text{O}_2 \)) starts with 0 M since it is not present initially.

### Step 2: Changes in Molarity

The balanced chemical equation is:

\[ 2\text{NO}_2 (g) \leftrightarrow 2\text{NO} (g) + \text{O}_2 (g) \]

Let \( x \) be the change in the molarity of \( \text{O}_2 \) produced. According to the stoichiometry of the reaction:
- For every \( x \) mol/L of \( \text{O}_2 \) produced, \( 2x \) mol/L of \( \text{NO}_2 \) is consumed and \( 2x \) mol/L of \( \text{NO} \) is produced.

Thus:
- Change in \( \text{NO}_2 \): \(-2x\)
- Change in \( \text{NO} \): \( +2x \)
- Change in \( \text{O}_2 \): \( +x \)

### Step 3: Equilibrium Molarity

The equilibrium molarity is calculated by adding the initial molarity and the change in molarity for each compound:

- \( \text{NO}_2 \) equilibrium molarity = \( 4.4 - 2x \)
- \( \text{NO} \) equilibrium molarity = \( 4.8 + 2x \)
- \( \text{O}_2 \) equilibrium molarity = \( x \)

### Filling the Table

Using the information above, we can fill out the table as follows:

|             | \(\text{NO}_2\) | \(\text{NO}\) | \(\text{O}_2\) |
|-------------|-----------------|---------------|----------------|
| **initial** | 4.4             | 4.8           | 0              |
| **change**  | \(-2x\)         | \(+2x\)       | \(+x\)         |
| **equilibrium** | \(4.4 - 2x\) | \(4.8 + 2x\) | \(x\)          |

Would you like more details or have any questions?

### Further Questions

1. How do you calculate the equilibrium constant (\( K_c \)) for this reaction?
2. What factors can affect the equilibrium position of this reaction?
3. How can you determine the value of \( x \) experimentally?
4. What happens to the equilibrium if the volume of the flask is changed?
5. How would adding an inert gas affect the equilibrium of this reaction?

### Tip

When dealing with chemical equilibria, always ensure to balance the chemical equation first and understand the stoichiometric relationships between reactants and products.

Suppose a 250 mL flask is filled with 1.1 mol of NO2 and 1.2 mol of NO. This reaction becomes possible:
2NO2 (g) ⇌ 2NO (g) + O2 (g)
Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O2.

Learn how to solve a chemical equilibrium problem involving the reaction 2NO2 (g) ⇌ 2NO (g) + O2 (g). This detailed solution covers the calculation of molarity changes, initial concentrations, and equilibrium concentrations. Suitable for high school students studying chemistry.

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