Hydrogen iodide decomposes according to the reaction 2 HI(g) → H₂(g) + I₂(g) A sealed 1.50-L container initially holds 0.00623 mol of H₂, 0.00414 mol of I₂, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H₂(g) is 0.00467 M. What are the concentrations of HI(g) and I₂(g)?
Hydrogen iodide decomposes according to the reaction 2 HI(g) → H₂(g) + I₂(g) A sealed 1.50-L container initially holds 0.00623 mol of H₂, 0.00414 mol of I₂, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H₂(g) is 0.00467 M. What are the concentrations of HI(g) and I₂(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
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![**Chemical Equilibrium and Concentration Calculations**
**Problem Statement:**
Hydrogen iodide decomposes according to the reaction:
\[ 2 \, \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]
A sealed 1.50-L container initially holds 0.00623 mol of H₂, 0.00414 mol of I₂, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H₂(g) is 0.00467 M. What are the concentrations of HI(g) and I₂(g)?
**Given Data:**
- Initial moles of H₂: 0.00623 mol
- Initial moles of I₂: 0.00414 mol
- Initial moles of HI: 0.0244 mol
- Volume of container: 1.50 L
- Temperature: 703 K
- Equilibrium concentration of H₂(g): 0.00467 M
**Unknowns to Find:**
1. Equilibrium concentration of \(\text{I}_2(g)\) (\([\text{I}_2]_{\text{eq}}\))
2. Equilibrium concentration of \(\text{HI}(g)\) (\([\text{HI}]_{\text{eq}}\))
**Solution Approach:**
1. Calculate the initial concentrations of the substances.
2. Use the stoichiometry of the balanced chemical equation to establish changes in concentration.
3. Determine the equilibrium concentrations based on the given equilibrium data.
**Calculate Initial Concentrations:**
1. Initial concentration of H₂:
\[
[\text{H}_2]_{\text{initial}} = \frac{0.00623 \, \text{mol}}{1.50 \, \text{L}} = 0.00415 \, \text{M}
\]
2. Initial concentration of I₂:
\[
[\text{I}_2]_{\text{initial}} = \frac{0.00414 \, \text{mol}}{1.50 \, \text{L}} = 0.00276 \, \text{M}
\]
3. Initial concentration of HI:
\[
[\text{HI}]_{\text{initial}} = \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8fc0129-1b11-49f3-869a-8ad1c019dc55%2F8baa76e3-327b-4ebf-8b1c-d2aca474a89d%2Ftmky6hb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium and Concentration Calculations**
**Problem Statement:**
Hydrogen iodide decomposes according to the reaction:
\[ 2 \, \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]
A sealed 1.50-L container initially holds 0.00623 mol of H₂, 0.00414 mol of I₂, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H₂(g) is 0.00467 M. What are the concentrations of HI(g) and I₂(g)?
**Given Data:**
- Initial moles of H₂: 0.00623 mol
- Initial moles of I₂: 0.00414 mol
- Initial moles of HI: 0.0244 mol
- Volume of container: 1.50 L
- Temperature: 703 K
- Equilibrium concentration of H₂(g): 0.00467 M
**Unknowns to Find:**
1. Equilibrium concentration of \(\text{I}_2(g)\) (\([\text{I}_2]_{\text{eq}}\))
2. Equilibrium concentration of \(\text{HI}(g)\) (\([\text{HI}]_{\text{eq}}\))
**Solution Approach:**
1. Calculate the initial concentrations of the substances.
2. Use the stoichiometry of the balanced chemical equation to establish changes in concentration.
3. Determine the equilibrium concentrations based on the given equilibrium data.
**Calculate Initial Concentrations:**
1. Initial concentration of H₂:
\[
[\text{H}_2]_{\text{initial}} = \frac{0.00623 \, \text{mol}}{1.50 \, \text{L}} = 0.00415 \, \text{M}
\]
2. Initial concentration of I₂:
\[
[\text{I}_2]_{\text{initial}} = \frac{0.00414 \, \text{mol}}{1.50 \, \text{L}} = 0.00276 \, \text{M}
\]
3. Initial concentration of HI:
\[
[\text{HI}]_{\text{initial}} = \frac{
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