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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**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
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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