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
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
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e5f0bee-bbde-421f-8347-e8b63f453811%2Ff85def1e-6b8d-4164-a36e-e995c515fe40%2Fv8hakn5_processed.jpeg&w=3840&q=75)
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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