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.

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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
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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