A sample of N2O4 gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, 20.0% of the N₂O4 has been converted to NO₂ gas. (a) Calculate Kp. (b) If the original pressure of N₂O4 is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier's principle?
A sample of N2O4 gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, 20.0% of the N₂O4 has been converted to NO₂ gas. (a) Calculate Kp. (b) If the original pressure of N₂O4 is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier's principle?
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter15: Principles Of Chemical Reactivity: Equilibria
Section: Chapter Questions
Problem 40GQ: Consider the following equilibrium: COBr2(g) CO(g) + Br2(g)Kc = 0.190 at 73 C (a) A 0.50 mol sample...
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![### Equilibrium and Le Chatelier’s Principle: Sample Problem
**Problem Statement:**
A sample of \( \text{N}_2\text{O}_4 \) gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, 20.0% of the \( \text{N}_2\text{O}_4 \) has been converted to \( \text{NO}_2 \) gas.
#### Questions:
(a) Calculate \( K_p \).
(b) If the original pressure of \( \text{N}_2\text{O}_4 \) is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier’s principle?
---
**Detailed Explanation:**
There are no graphs or diagrams in the provided image, only text. Therefore, we'll proceed directly with the text transcription suitable for an educational website.
### Solution:
#### (a) Calculate \( K_p \).
1. **Initial Information:**
- Initial pressure of \( \text{N}_2\text{O}_4 \): 1.00 atm
- Percent dissociation to \( \text{NO}_2 \): 20.0%
2. **Write the dissociation reaction:**
\[
\text{N}_2\text{O}_4 (g) \leftrightarrow 2 \text{NO}_2 (g)
\]
3. **Establish equilibrium pressures:**
Since 20.0% of \( \text{N}_2\text{O}_4 \) dissociates:
- Pressure of \( \text{N}_2\text{O}_4 \) at equilibrium: \( 1.00 \text{ atm} \times (1 - 0.20) = 0.80 \text{ atm} \)
- Pressure of \( \text{NO}_2 \) at equilibrium: \( 1.00 \text{ atm} \times 0.20 \times 2 = 0.40 \text{ atm} \)
\[
\text{NO}_2 = 2 \times (\text{0.20 atm}) = 0.40 \text{ atm}
\]
4. **Calculate \( K_p \):**
\[
K_p = \frac{{(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f1e4f7e-3ee3-4328-9071-c5dd7af83341%2Fcf1c04cf-dfc4-429f-8858-c29ce8d106d8%2Ffssem2u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Equilibrium and Le Chatelier’s Principle: Sample Problem
**Problem Statement:**
A sample of \( \text{N}_2\text{O}_4 \) gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, 20.0% of the \( \text{N}_2\text{O}_4 \) has been converted to \( \text{NO}_2 \) gas.
#### Questions:
(a) Calculate \( K_p \).
(b) If the original pressure of \( \text{N}_2\text{O}_4 \) is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier’s principle?
---
**Detailed Explanation:**
There are no graphs or diagrams in the provided image, only text. Therefore, we'll proceed directly with the text transcription suitable for an educational website.
### Solution:
#### (a) Calculate \( K_p \).
1. **Initial Information:**
- Initial pressure of \( \text{N}_2\text{O}_4 \): 1.00 atm
- Percent dissociation to \( \text{NO}_2 \): 20.0%
2. **Write the dissociation reaction:**
\[
\text{N}_2\text{O}_4 (g) \leftrightarrow 2 \text{NO}_2 (g)
\]
3. **Establish equilibrium pressures:**
Since 20.0% of \( \text{N}_2\text{O}_4 \) dissociates:
- Pressure of \( \text{N}_2\text{O}_4 \) at equilibrium: \( 1.00 \text{ atm} \times (1 - 0.20) = 0.80 \text{ atm} \)
- Pressure of \( \text{NO}_2 \) at equilibrium: \( 1.00 \text{ atm} \times 0.20 \times 2 = 0.40 \text{ atm} \)
\[
\text{NO}_2 = 2 \times (\text{0.20 atm}) = 0.40 \text{ atm}
\]
4. **Calculate \( K_p \):**
\[
K_p = \frac{{(\
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