2) Keq=0.0183 for the reaction: 2HI e)2 H, (g) + I, (g) (g) If 3.0 moles of HI are placed in a 5.00L vessel and allowed to reach equilibrium, what is the equilibrium concentration of H,?
2) Keq=0.0183 for the reaction: 2HI e)2 H, (g) + I, (g) (g) If 3.0 moles of HI are placed in a 5.00L vessel and allowed to reach equilibrium, what is the equilibrium concentration of H,?
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![**Equilibrium Concentration Calculation: Understanding the Reaction**
**Problem Statement:**
2) For the reaction:
\[ 2HI_{(g)} \leftrightarrow H_2_{(g)} + I_2_{(g)} \]
where \( K_{eq} = 0.0183 \),
Given that 3.0 moles of HI are placed in a 5.00 L vessel and allowed to reach equilibrium, what is the equilibrium concentration of \( H_2 \)?
**Explanation and Steps to Solve:**
1. **Initial Concentrations:**
- Initial moles of \( HI \) are 3.0 moles.
- Volume of the vessel = 5.00 L.
- Therefore, the initial concentration of \( HI \) is \( \frac{3.0 \text{ moles}}{5.00 \text{ L}} = 0.60 \text{ M} \).
- Initial concentrations of \( H_2 \) and \( I_2 \) are both 0 M since they have not been formed yet.
2. **Change in Concentration:**
- Let \( x \) be the concentration of \( H_2 \) formed at equilibrium.
- According to the stoichiometry of the reaction, the concentration of \( I_2 \) formed will also be \( x \) M.
- The change in concentration of \( HI \) will be \( -2x \) M (since 2 moles of \( HI \) form 1 mole each of \( H_2 \) and \( I_2 \)).
3. **Equilibrium Concentrations:**
- At equilibrium:
- Concentration of \( HI \) = \( 0.60 - 2x \) M.
- Concentration of \( H_2 \) = \( x \) M.
- Concentration of \( I_2 \) = \( x \) M.
4. **Expression for \( K_{eq} \):**
- \( K_{eq} = \frac{[H_2][I_2]}{[HI]^2} \)
- Plugging the equilibrium concentrations into this expression:
\[ K_{eq} = \frac{x \cdot x}{(0.60 - 2x)^2} = 0.018](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63a27622-a493-4bf8-aa5e-97adfc7108ed%2F6ebb6fed-256e-4ed6-8519-471db8ff06d5%2Fh89zp55_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Concentration Calculation: Understanding the Reaction**
**Problem Statement:**
2) For the reaction:
\[ 2HI_{(g)} \leftrightarrow H_2_{(g)} + I_2_{(g)} \]
where \( K_{eq} = 0.0183 \),
Given that 3.0 moles of HI are placed in a 5.00 L vessel and allowed to reach equilibrium, what is the equilibrium concentration of \( H_2 \)?
**Explanation and Steps to Solve:**
1. **Initial Concentrations:**
- Initial moles of \( HI \) are 3.0 moles.
- Volume of the vessel = 5.00 L.
- Therefore, the initial concentration of \( HI \) is \( \frac{3.0 \text{ moles}}{5.00 \text{ L}} = 0.60 \text{ M} \).
- Initial concentrations of \( H_2 \) and \( I_2 \) are both 0 M since they have not been formed yet.
2. **Change in Concentration:**
- Let \( x \) be the concentration of \( H_2 \) formed at equilibrium.
- According to the stoichiometry of the reaction, the concentration of \( I_2 \) formed will also be \( x \) M.
- The change in concentration of \( HI \) will be \( -2x \) M (since 2 moles of \( HI \) form 1 mole each of \( H_2 \) and \( I_2 \)).
3. **Equilibrium Concentrations:**
- At equilibrium:
- Concentration of \( HI \) = \( 0.60 - 2x \) M.
- Concentration of \( H_2 \) = \( x \) M.
- Concentration of \( I_2 \) = \( x \) M.
4. **Expression for \( K_{eq} \):**
- \( K_{eq} = \frac{[H_2][I_2]}{[HI]^2} \)
- Plugging the equilibrium concentrations into this expression:
\[ K_{eq} = \frac{x \cdot x}{(0.60 - 2x)^2} = 0.018
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