Be sure to answer all parts. Kp = 6.5 x 104 at 308 K for the following reaction: P 2 NO (g) + Cl₂(g) ⇒ 2 NOCI(g) At equilibrium, PNO = 0.35 atm and Pcl₂ = 0.20 atm What is the equilibrium partial pressure of NOCI(g)? PNOCI= X 10 atm
Be sure to answer all parts. Kp = 6.5 x 104 at 308 K for the following reaction: P 2 NO (g) + Cl₂(g) ⇒ 2 NOCI(g) At equilibrium, PNO = 0.35 atm and Pcl₂ = 0.20 atm What is the equilibrium partial pressure of NOCI(g)? PNOCI= X 10 atm
Chemistry
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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 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8fc0129-1b11-49f3-869a-8ad1c019dc55%2Fe4c547d1-5b85-4edc-b4aa-0f4b85e2ddf6%2F3usiinr_processed.jpeg&w=3840&q=75)
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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