Use the References to access important values if needed for this question. The equilibrium constant, Ke, for the following reaction is 0.00650 at 298 K. 2NOBr(g)2NO(g) + Br₂ (9) If an equilibrium mixture of the three gases in a 17.8 L container at 298 K contains 0.378 mol of NOBr(g) and 0.461 mol of NO, the equilibrium concentration of Br2 Is M. Submit Answer Retry Entire Group 9 more group attempts remaining

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### Equilibrium Constant Calculation for a Gas Reaction

**Problem Statement:**

The equilibrium constant, \( K_c \), for the following reaction is 0.00650 at 298 K:

\[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \]

If an equilibrium mixture of the three gases in a 17.8 L container at 298 K contains 0.378 mol of \(\text{NOBr}(g)\) and 0.461 mol of \(\text{NO}\), calculate the equilibrium concentration of \(\text{Br}_2\) in molarity.

**Solution Steps:**

1. **Define the Reaction:**
   \[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \]

2. **Given Data:**
   - Volume of container = 17.8 L
   - Moles of \(\text{NOBr}\) = 0.378 mol
   - Moles of \(\text{NO}\) = 0.461 mol
   - \( K_c = 0.00650 \) at 298 K

3. **Calculate Concentrations:**
   \[
   [\text{NOBr}] = \frac{0.378 \text{ mol}}{17.8 \text{ L}} = \text{Concentration of NOBr}
   \]
   \[
   [\text{NO}] = \frac{0.461 \text{ mol}}{17.8 \text{ L}} = \text{Concentration of NO}
   \]

4. **Equilibrium Constant Expression:**
   \[
   K_c = \frac{{[\text{NO}]^2 [\text{Br}_2]}}{{[\text{NOBr}]^2}}
   \]

5. **Solve for \([\text{Br}_2]\):**
   - Rearrange the equation and solve for the unknown concentration of \(\text{Br}_2\).

6. **Submit Answer:**
   - Enter the calculated equilibrium concentration of \(\text{Br}_2\) in the text box provided.

**Note:** More group attempts remaining for recalculating or retrying if needed.
Transcribed Image Text:### Equilibrium Constant Calculation for a Gas Reaction **Problem Statement:** The equilibrium constant, \( K_c \), for the following reaction is 0.00650 at 298 K: \[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \] If an equilibrium mixture of the three gases in a 17.8 L container at 298 K contains 0.378 mol of \(\text{NOBr}(g)\) and 0.461 mol of \(\text{NO}\), calculate the equilibrium concentration of \(\text{Br}_2\) in molarity. **Solution Steps:** 1. **Define the Reaction:** \[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \] 2. **Given Data:** - Volume of container = 17.8 L - Moles of \(\text{NOBr}\) = 0.378 mol - Moles of \(\text{NO}\) = 0.461 mol - \( K_c = 0.00650 \) at 298 K 3. **Calculate Concentrations:** \[ [\text{NOBr}] = \frac{0.378 \text{ mol}}{17.8 \text{ L}} = \text{Concentration of NOBr} \] \[ [\text{NO}] = \frac{0.461 \text{ mol}}{17.8 \text{ L}} = \text{Concentration of NO} \] 4. **Equilibrium Constant Expression:** \[ K_c = \frac{{[\text{NO}]^2 [\text{Br}_2]}}{{[\text{NOBr}]^2}} \] 5. **Solve for \([\text{Br}_2]\):** - Rearrange the equation and solve for the unknown concentration of \(\text{Br}_2\). 6. **Submit Answer:** - Enter the calculated equilibrium concentration of \(\text{Br}_2\) in the text box provided. **Note:** More group attempts remaining for recalculating or retrying if needed.
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