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.
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.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1067c048-27d4-462b-825c-b686bc07047b%2Fe649b221-bbc6-4026-a47d-dfac46b019f4%2Fqmhwgm.png&w=3840&q=75)
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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