Suppose a 250. mL flask is filled with 1.1 mol of NO3 and 2.0 mol of NO. The following reaction becomes possible: NO3(g) + NO(g) → 2NO,(g) The equilibrium constant K for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

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### Equilibrium Concentration Calculation

#### Problem Statement:
Suppose a 250. mL flask is filled with 1.1 mol of \( NO_3 \) and 2.0 mol of \( NO \). The following reaction becomes possible:

\[ NO_3 (g) + NO (g) \rightleftharpoons 2NO_2 (g) \]

The equilibrium constant \( K \) for this reaction is 8.88 at the temperature of the flask. 

Calculate the equilibrium molarity of \( NO \). Round your answer to two decimal places.

#### Given:
- Volume of the flask = 250 mL (or 0.250 L)
- Initial moles of \( NO_3 \) = 1.1 mol
- Initial moles of \( NO \) = 2.0 mol
- Equilibrium constant \( K = 8.88 \)

#### Required:
- Equilibrium molarity of \( NO \).

#### Solution Approach:
1. **Set up the ICE (Initial, Change, Equilibrium) table:**

   | Species | Initial (M) | Change (M) | Equilibrium (M) |
   |---------|-------------|------------|------------------|
   | \(NO_3\) | \( \frac{1.1}{0.250} \) | -x | \( \frac{1.1}{0.250} - x \) |
   | \(NO\)  | \( \frac{2.0}{0.250} \) | -x | \( \frac{2.0}{0.250} - x \) |
   | \(NO_2\) | 0 | +2x | 2x |

2. **Write the equilibrium expression for \( K \):**
   \[ K = \frac{[NO_2]^2}{[NO_3] \cdot [NO]} = 8.88 \]

3. **Substitute equilibrium concentrations into the expression:**
   \[ K = \frac{(2x)^2}{\left( \frac{1.1}{0.250} - x \right) \left( \frac{2.0}{0.250} - x \right)} = 8.88 \]

4. **Solve the equation for \( x \), which represents the change in concentration from the initial to
Transcribed Image Text:### Equilibrium Concentration Calculation #### Problem Statement: Suppose a 250. mL flask is filled with 1.1 mol of \( NO_3 \) and 2.0 mol of \( NO \). The following reaction becomes possible: \[ NO_3 (g) + NO (g) \rightleftharpoons 2NO_2 (g) \] The equilibrium constant \( K \) for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of \( NO \). Round your answer to two decimal places. #### Given: - Volume of the flask = 250 mL (or 0.250 L) - Initial moles of \( NO_3 \) = 1.1 mol - Initial moles of \( NO \) = 2.0 mol - Equilibrium constant \( K = 8.88 \) #### Required: - Equilibrium molarity of \( NO \). #### Solution Approach: 1. **Set up the ICE (Initial, Change, Equilibrium) table:** | Species | Initial (M) | Change (M) | Equilibrium (M) | |---------|-------------|------------|------------------| | \(NO_3\) | \( \frac{1.1}{0.250} \) | -x | \( \frac{1.1}{0.250} - x \) | | \(NO\) | \( \frac{2.0}{0.250} \) | -x | \( \frac{2.0}{0.250} - x \) | | \(NO_2\) | 0 | +2x | 2x | 2. **Write the equilibrium expression for \( K \):** \[ K = \frac{[NO_2]^2}{[NO_3] \cdot [NO]} = 8.88 \] 3. **Substitute equilibrium concentrations into the expression:** \[ K = \frac{(2x)^2}{\left( \frac{1.1}{0.250} - x \right) \left( \frac{2.0}{0.250} - x \right)} = 8.88 \] 4. **Solve the equation for \( x \), which represents the change in concentration from the initial to
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