Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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### Equilibrium Partial Pressure Calculation for a Chemical Reaction

**Problem Statement:**

Given the equilibrium constant \( K_p = 6.5 \times 10^4 \) at 308 K for the following reaction:

\[ 2 \text{NO} (g) + \text{Cl}_2 (g) \rightleftharpoons 2 \text{NOCl} (g) \]

At equilibrium, the partial pressures are given as:

\[ P_{\text{NO}} = 0.35 \text{ atm} \]
\[ P_{\text{Cl}_2} = 0.20 \text{ atm} \]

**Question:**

What is the equilibrium partial pressure of NOCl(g)?

**Solution Approach:**

1. **Write the Expression for \( K_p \):**

\[ K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot (P_{\text{Cl}_2})} \]

2. **Substitute the Known Values:**

\[ 6.5 \times 10^4 = \frac{(P_{\text{NOCl}})^2}{(0.35)^2 \cdot 0.20} \]

3. **Solve for \( P_{\text{NOCl}} \):**

\[ 6.5 \times 10^4 = \frac{(P_{\text{NOCl}})^2}{0.01225} \]

\[ P_{\text{NOCl}}^2 = 6.5 \times 10^4 \times 0.01225 \]

\[ P_{\text{NOCl}}^2 = 796.25 \]

\[ P_{\text{NOCl}} = \sqrt{796.25} \]

\[ P_{\text{NOCl}} \approx 28.20 \text{ atm} \]

**Answer:**

The equilibrium partial pressure of NOCl(g) (\( P_{\text{NOCl}} \)) is:

\[ P_{\text{NOCl}} = 2.8 \times 10^1 \text{ atm} \]

**Graph/Diagram Explanation:**

There are no graphs or diagrams in the provided image. The problem is purely text-based and involves algebraic manipulation to solve for the equilibrium partial pressure.
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Transcribed Image Text:### Equilibrium Partial Pressure Calculation for a Chemical Reaction **Problem Statement:** Given the equilibrium constant \( K_p = 6.5 \times 10^4 \) at 308 K for the following reaction: \[ 2 \text{NO} (g) + \text{Cl}_2 (g) \rightleftharpoons 2 \text{NOCl} (g) \] At equilibrium, the partial pressures are given as: \[ P_{\text{NO}} = 0.35 \text{ atm} \] \[ P_{\text{Cl}_2} = 0.20 \text{ atm} \] **Question:** What is the equilibrium partial pressure of NOCl(g)? **Solution Approach:** 1. **Write the Expression for \( K_p \):** \[ K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot (P_{\text{Cl}_2})} \] 2. **Substitute the Known Values:** \[ 6.5 \times 10^4 = \frac{(P_{\text{NOCl}})^2}{(0.35)^2 \cdot 0.20} \] 3. **Solve for \( P_{\text{NOCl}} \):** \[ 6.5 \times 10^4 = \frac{(P_{\text{NOCl}})^2}{0.01225} \] \[ P_{\text{NOCl}}^2 = 6.5 \times 10^4 \times 0.01225 \] \[ P_{\text{NOCl}}^2 = 796.25 \] \[ P_{\text{NOCl}} = \sqrt{796.25} \] \[ P_{\text{NOCl}} \approx 28.20 \text{ atm} \] **Answer:** The equilibrium partial pressure of NOCl(g) (\( P_{\text{NOCl}} \)) is: \[ P_{\text{NOCl}} = 2.8 \times 10^1 \text{ atm} \] **Graph/Diagram Explanation:** There are no graphs or diagrams in the provided image. The problem is purely text-based and involves algebraic manipulation to solve for the equilibrium partial pressure.
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