Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this: 3 NO₂(g) + H₂O(1)→ 2 HNO3(aq) + NO(g) At a certain temperature, a chemist finds that a 4.0 L reaction vessel containing a mixture of nitrogen dioxide, water, nitric acid, and nitrogen monoxide at equilibrium has the following composition: compound amount NO₂ 18.3 g H₂O 73.9 g HNO3 5.0 g NO 14.7 g Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits. K x10

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**Equilibrium in Chemical Reactions**

In this lesson, we will explore how to determine the equilibrium constant for a chemical reaction. As an example, consider the following reaction where nitrogen dioxide and water react to form nitric acid and nitrogen monoxide:

\[ 3 \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2 \text{HNO}_3(aq) + \text{NO}(g) \]

At a certain temperature, a chemist observes a 4.0 L reaction vessel containing a mixture of these substances at equilibrium with the following composition:

| Compound | Amount |
|----------|--------|
| NO\(_2\)   | 18.3 g |
| H\(_2\)O      | 73.9 g |
| HNO\(_3\)| 5.0 g   |
| NO           | 14.7 g |

To calculate the equilibrium constant (\(K_c\)) for this reaction, follow these steps:

1. **Convert Mass to Moles**:
   Calculate the number of moles for each compound using their molar mass.
   - Molar mass of NO\(_2\) = 14 (N) + 2×16 (O) = 46 g/mol
   - Molar mass of H\(_2\)O = 2×1 (H) + 16 (O) = 18 g/mol
   - Molar mass of HNO\(_3\) = 1 (H) + 14 (N) + 3×16 (O) = 63 g/mol
   - Molar mass of NO = 14 (N) + 16 (O) = 30 g/mol

2. **Calculate Concentrations (M)**:
   Convert moles to molarity by dividing by the volume of the reaction vessel (4.0 L).

3. **Determine \(K_c\) Using the Equilibrium Expression**:
   The equilibrium constant expression for the reaction is:

   \[ K_c = \frac{[\text{HNO}_3]^2 [\text{NO}]}{[\text{NO}_2]^3 [\text{H}_2\text{O}]} \]

4. **Insert Concentrations and Solve**:
   Use the concentrations computed in the previous step to find \(K_c\).
Transcribed Image Text:**Equilibrium in Chemical Reactions** In this lesson, we will explore how to determine the equilibrium constant for a chemical reaction. As an example, consider the following reaction where nitrogen dioxide and water react to form nitric acid and nitrogen monoxide: \[ 3 \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2 \text{HNO}_3(aq) + \text{NO}(g) \] At a certain temperature, a chemist observes a 4.0 L reaction vessel containing a mixture of these substances at equilibrium with the following composition: | Compound | Amount | |----------|--------| | NO\(_2\) | 18.3 g | | H\(_2\)O | 73.9 g | | HNO\(_3\)| 5.0 g | | NO | 14.7 g | To calculate the equilibrium constant (\(K_c\)) for this reaction, follow these steps: 1. **Convert Mass to Moles**: Calculate the number of moles for each compound using their molar mass. - Molar mass of NO\(_2\) = 14 (N) + 2×16 (O) = 46 g/mol - Molar mass of H\(_2\)O = 2×1 (H) + 16 (O) = 18 g/mol - Molar mass of HNO\(_3\) = 1 (H) + 14 (N) + 3×16 (O) = 63 g/mol - Molar mass of NO = 14 (N) + 16 (O) = 30 g/mol 2. **Calculate Concentrations (M)**: Convert moles to molarity by dividing by the volume of the reaction vessel (4.0 L). 3. **Determine \(K_c\) Using the Equilibrium Expression**: The equilibrium constant expression for the reaction is: \[ K_c = \frac{[\text{HNO}_3]^2 [\text{NO}]}{[\text{NO}_2]^3 [\text{H}_2\text{O}]} \] 4. **Insert Concentrations and Solve**: Use the concentrations computed in the previous step to find \(K_c\).
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