If the following reaction at 698 K has equilibrium concentration values of 0.584 M HI, 0.0793 M H₂, and 0.0793 M 12, what is the value of the equilibrium constant? 2H1 H₂(g) + 1₂ (g) 2 (0.584m) → 0.0793 M + 0.0793 M L -3

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
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
icon
Concept explainers
Question
100%
Does this look right
**Equilibrium Constant Calculation for the Reaction 2HI ⇌ H₂ + I₂**

**Problem Statement:**
At 698 K, the equilibrium concentrations for the reaction 2HI (g) ⇌ H₂ (g) + I₂ (g) are given as:
- [HI] = 0.584 M
- [H₂] = 0.0793 M
- [I₂] = 0.0793 M

**Objective:**
Calculate the equilibrium constant (Kc) for this reaction.

**Solution:**

1. **Balanced Reaction:**
   \[
   2HI (g) \rightarrow H₂ (g) + I₂ (g)
   \]

2. **Expression for the Equilibrium Constant (Kc):**
   \[
   Kc = \frac{[H₂][I₂]}{[HI]^2}
   \]

3. **Substitute the Equilibrium Concentrations into the Expression:**
   - [HI] = 0.584 M
   - [H₂] = 0.0793 M
   - [I₂] = 0.0793 M
   
   \[
   Kc = \frac{(0.0793)(0.0793)}{(0.584)^2}
   \]

4. **Calculate:**
   \[
   Kc = \frac{0.0793 \times 0.0793}{0.584^2}
   \]
   \[
   Kc = \frac{0.00628049}{0.341056} \approx 5.80 \times 10^{-3}
   \]

**Conclusion:**
The equilibrium constant \( Kc \) for the reaction at 698 K is approximately \( 5.80 \times 10^{-3} \).
Transcribed Image Text:**Equilibrium Constant Calculation for the Reaction 2HI ⇌ H₂ + I₂** **Problem Statement:** At 698 K, the equilibrium concentrations for the reaction 2HI (g) ⇌ H₂ (g) + I₂ (g) are given as: - [HI] = 0.584 M - [H₂] = 0.0793 M - [I₂] = 0.0793 M **Objective:** Calculate the equilibrium constant (Kc) for this reaction. **Solution:** 1. **Balanced Reaction:** \[ 2HI (g) \rightarrow H₂ (g) + I₂ (g) \] 2. **Expression for the Equilibrium Constant (Kc):** \[ Kc = \frac{[H₂][I₂]}{[HI]^2} \] 3. **Substitute the Equilibrium Concentrations into the Expression:** - [HI] = 0.584 M - [H₂] = 0.0793 M - [I₂] = 0.0793 M \[ Kc = \frac{(0.0793)(0.0793)}{(0.584)^2} \] 4. **Calculate:** \[ Kc = \frac{0.0793 \times 0.0793}{0.584^2} \] \[ Kc = \frac{0.00628049}{0.341056} \approx 5.80 \times 10^{-3} \] **Conclusion:** The equilibrium constant \( Kc \) for the reaction at 698 K is approximately \( 5.80 \times 10^{-3} \).
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Knowledge Booster
Ionic Equilibrium
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY