N₂O4 (g) = 2NO₂(g) K = 6.14 x 10-3 @ 298 K A sealed 2.0 L container initially holds 5.0 moles N2O4 and 0.050 mole NO2. Determine the concentrations at equilibrium. What is the concentration of NO2 at equilibrium? [NO₂] = [?] M [NO₂], M Enter
N₂O4 (g) = 2NO₂(g) K = 6.14 x 10-3 @ 298 K A sealed 2.0 L container initially holds 5.0 moles N2O4 and 0.050 mole NO2. Determine the concentrations at equilibrium. What is the concentration of NO2 at equilibrium? [NO₂] = [?] M [NO₂], M Enter
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 Concentrations Calculation for NO₂**
**Chemical Reaction:**
\[ \text{N}_2\text{O}_4\text{ (g)} \rightleftharpoons 2\text{NO}_2\text{ (g)} \]
**Equilibrium Constant:**
\[ K = 6.14 \times 10^{-3} \text{ at } 298 \text{ K} \]
**Initial Conditions:**
A sealed 2.0 L container initially holds:
- 5.0 moles of \(\text{N}_2\text{O}_4\)
- 0.050 moles of \(\text{NO}_2\)
**Problem Statement:**
Determine the concentrations at equilibrium. What is the concentration of \(\text{NO}_2\) at equilibrium?
\[ [\text{NO}_2] = ? \text{ M} \]
**Diagram:**
An interactive element is present at the bottom of the image for inputting the equilibrium concentration of \(\text{NO}_2\) in molarity (M). It includes a text box labeled `[NO₂], M` and a blue 'Enter' button to submit the value.
--------------------------------------------------------------------------------------------------
To find the equilibrium concentrations, the following steps are usually performed:
1. **Initial Concentrations:**
- \(\text{N}_2\text{O}_4\): \(\frac{5.0 \text{ moles}}{2.0 \text{ L}} = 2.5 \text{ M}\)
- \(\text{NO}_2\): \(\frac{0.050 \text{ moles}}{2.0 \text{ L}} = 0.025 \text{ M}\)
2. **Set up the ICE Table (Initial, Change, Equilibrium):**
| | \( \text{N}_2\text{O}_4 \) | \(\text{NO}_2\) |
|---------------|--------------------|-------------------|
| Initial (M) | 2.5 | 0.025 |
| Change (M) | - x | + 2x |
| Equilibrium (M)| 2.5 - x | 0.025 + 2x |
3. **Expression for the Equ](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F02aa1c36-956b-4289-8507-865f3947cc14%2F7bd5506a-09a4-4ba8-9fde-f5d066dfdedc%2Fahzoas7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Concentrations Calculation for NO₂**
**Chemical Reaction:**
\[ \text{N}_2\text{O}_4\text{ (g)} \rightleftharpoons 2\text{NO}_2\text{ (g)} \]
**Equilibrium Constant:**
\[ K = 6.14 \times 10^{-3} \text{ at } 298 \text{ K} \]
**Initial Conditions:**
A sealed 2.0 L container initially holds:
- 5.0 moles of \(\text{N}_2\text{O}_4\)
- 0.050 moles of \(\text{NO}_2\)
**Problem Statement:**
Determine the concentrations at equilibrium. What is the concentration of \(\text{NO}_2\) at equilibrium?
\[ [\text{NO}_2] = ? \text{ M} \]
**Diagram:**
An interactive element is present at the bottom of the image for inputting the equilibrium concentration of \(\text{NO}_2\) in molarity (M). It includes a text box labeled `[NO₂], M` and a blue 'Enter' button to submit the value.
--------------------------------------------------------------------------------------------------
To find the equilibrium concentrations, the following steps are usually performed:
1. **Initial Concentrations:**
- \(\text{N}_2\text{O}_4\): \(\frac{5.0 \text{ moles}}{2.0 \text{ L}} = 2.5 \text{ M}\)
- \(\text{NO}_2\): \(\frac{0.050 \text{ moles}}{2.0 \text{ L}} = 0.025 \text{ M}\)
2. **Set up the ICE Table (Initial, Change, Equilibrium):**
| | \( \text{N}_2\text{O}_4 \) | \(\text{NO}_2\) |
|---------------|--------------------|-------------------|
| Initial (M) | 2.5 | 0.025 |
| Change (M) | - x | + 2x |
| Equilibrium (M)| 2.5 - x | 0.025 + 2x |
3. **Expression for the Equ
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