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?

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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{{(\
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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