### Equilibrium Constant \( K_p \) Calculation**Problem Statement:**A reaction vessel initially contains only 3.5 atm of \( \text{N}_2\text{O}_4 \). When the system reaches equilibrium according to the reaction below, there is 1.7 atm of \( \text{NO}_2 \) in the vessel. What is \( K_p \) for this reaction?\[ 2 \text{NO}_2 (g) \rightleftharpoons \text{N}_2\text{O}_4 (g) \]**Solution:**To calculate the equilibrium constant \( K_p \), follow these steps:1. **Determine Changes in Pressure:** - Initial partial pressure of \( \text{N}_2\text{O}_4 \) = 3.5 atm - Changes due to equilibrium: - The formation of \( \text{NO}_2 \) increases its pressure by 1.7 atm. - According to stoichiometry, for every 2 moles of \( \text{NO}_2 \) formed, 1 mole of \( \text{N}_2\text{O}_4 \) dissociates.2. **Equilibrium Pressures:** - The change in pressure of \( \text{N}_2\text{O}_4 \) = \(- \frac{1.7}{2} \) atm due to the stoichiometry. - Equilibrium pressure of \( \text{N}_2\text{O}_4 \) = Initial pressure - Change = \( 3.5 - \frac{1.7}{2} \) atm3. **Calculate \( K_p \):** - \( K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \) - Substitute the equilibrium pressures into the expression for \( K_p \) to find the value.This setup allows you to calculate the equilibrium constant \( K_p \) for the given reaction based on pressure changes in a closed system.

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A reaction vessel initially contains only 3.5 atm of N,O4. When the system reaches equilibrium according to the reaction below, there is 1.7 atm of NO, in the vessel. What is Kp for this reaction? 2 NO2 (g) = N204 (g)

A reaction vessel initially contains only 3.5 atm of N,O4. When the system reaches equilibrium according to the reaction below, there is 1.7 atm of NO, in the vessel. What is Kp for this reaction? 2 NO2 (g) = N204 (g)

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

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Question

### Equilibrium Constant \( K_p \) Calculation

**Problem Statement:**

A reaction vessel initially contains only 3.5 atm of \( \text{N}_2\text{O}_4 \). When the system reaches equilibrium according to the reaction below, there is 1.7 atm of \( \text{NO}_2 \) in the vessel. What is \( K_p \) for this reaction?

\[ 2 \text{NO}_2 (g) \rightleftharpoons \text{N}_2\text{O}_4 (g) \]

**Solution:**

To calculate the equilibrium constant \( K_p \), follow these steps:

1. **Determine Changes in Pressure:**
- Initial partial pressure of \( \text{N}_2\text{O}_4 \) = 3.5 atm
- Changes due to equilibrium:
- The formation of \( \text{NO}_2 \) increases its pressure by 1.7 atm.
- According to stoichiometry, for every 2 moles of \( \text{NO}_2 \) formed, 1 mole of \( \text{N}_2\text{O}_4 \) dissociates.

2. **Equilibrium Pressures:**
- The change in pressure of \( \text{N}_2\text{O}_4 \) = \(- \frac{1.7}{2} \) atm due to the stoichiometry.
- Equilibrium pressure of \( \text{N}_2\text{O}_4 \) = Initial pressure - Change = \( 3.5 - \frac{1.7}{2} \) atm

3. **Calculate \( K_p \):**
- \( K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \)
- Substitute the equilibrium pressures into the expression for \( K_p \) to find the value.

This setup allows you to calculate the equilibrium constant \( K_p \) for the given reaction based on pressure changes in a closed system.

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