For the following reaction at a certain temperature H2(g) + F2(g) = 2HF(g) the equilibrium concentrations in a 10.00-L rigid container are [H2] = 0.0600 M, [F2] = 0.0200 M, and [HF] = 0.620 M. If 0.220 moles of F2 is added to the equilibrium mixture, calculate the concentration of H2 after equilibrium is reestablished.
For the following reaction at a certain temperature H2(g) + F2(g) = 2HF(g) the equilibrium concentrations in a 10.00-L rigid container are [H2] = 0.0600 M, [F2] = 0.0200 M, and [HF] = 0.620 M. If 0.220 moles of F2 is added to the equilibrium mixture, calculate the concentration of H2 after equilibrium is reestablished.
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 Calculation
**Reaction:**
\[ \text{H}_2(g) + \text{F}_2(g) \rightleftharpoons 2\text{HF}(g) \]
**Initial Equilibrium Concentrations in a 10.00-L Rigid Container:**
- \([ \text{H}_2 ] = 0.0600 \, \text{M}\)
- \([ \text{F}_2 ] = 0.0200 \, \text{M}\)
- \([ \text{HF} ] = 0.620 \, \text{M}\)
**Problem:**
If 0.220 moles of \(\text{F}_2\) are added to the equilibrium mixture, calculate the concentration of \(\text{H}_2\) after equilibrium is reestablished.
**Detailed Explanation:**
1. **Calculate the new concentration of \(\text{F}_2\) after addition:**
- Additional moles of \(\text{F}_2 = 0.220 \, \text{moles}\)
- Volume of the container = 10.00 L
New concentration of \(\text{F}_2\):
\[
[\text{F}_2]_{\text{initial}} + \left(\frac{0.220 \, \text{moles}}{10.00 \, \text{L}}\right) = 0.0200 \, \text{M} + 0.0220 \, \text{M} = 0.0420 \, \text{M}
\]
2. **Using the equilibrium constant (Kc):**
- Establish the expression for the equilibrium constant, \(K_c\):
\[
K_c = \frac{[ \text{HF} ]^2}{[ \text{H}_2 ][ \text{F}_2 ]}
\]
Using the initial equilibrium concentrations:
\[
K_c = \frac{(0.620)^2}{(0.0600)(0.0200)} = \frac{0.3844}{0.0012} = 320.33
\]
3. **Set up the ICE table (Initial, Change, Equilibrium):**
- Let \(x\) be the change in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F911c9e4a-8e4d-4d6f-a5f5-4c91474dfa08%2Fcf433af0-7d02-406e-99d0-2ffb9cab96f9%2F5v5kt2n.png&w=3840&q=75)
Transcribed Image Text:### Chemical Equilibrium Calculation
**Reaction:**
\[ \text{H}_2(g) + \text{F}_2(g) \rightleftharpoons 2\text{HF}(g) \]
**Initial Equilibrium Concentrations in a 10.00-L Rigid Container:**
- \([ \text{H}_2 ] = 0.0600 \, \text{M}\)
- \([ \text{F}_2 ] = 0.0200 \, \text{M}\)
- \([ \text{HF} ] = 0.620 \, \text{M}\)
**Problem:**
If 0.220 moles of \(\text{F}_2\) are added to the equilibrium mixture, calculate the concentration of \(\text{H}_2\) after equilibrium is reestablished.
**Detailed Explanation:**
1. **Calculate the new concentration of \(\text{F}_2\) after addition:**
- Additional moles of \(\text{F}_2 = 0.220 \, \text{moles}\)
- Volume of the container = 10.00 L
New concentration of \(\text{F}_2\):
\[
[\text{F}_2]_{\text{initial}} + \left(\frac{0.220 \, \text{moles}}{10.00 \, \text{L}}\right) = 0.0200 \, \text{M} + 0.0220 \, \text{M} = 0.0420 \, \text{M}
\]
2. **Using the equilibrium constant (Kc):**
- Establish the expression for the equilibrium constant, \(K_c\):
\[
K_c = \frac{[ \text{HF} ]^2}{[ \text{H}_2 ][ \text{F}_2 ]}
\]
Using the initial equilibrium concentrations:
\[
K_c = \frac{(0.620)^2}{(0.0600)(0.0200)} = \frac{0.3844}{0.0012} = 320.33
\]
3. **Set up the ICE table (Initial, Change, Equilibrium):**
- Let \(x\) be the change in
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