5. At 355 K, a 10.00 L reaction flask is charged with 1.44 atm each of hydrogen and fluorine gas. The following reaction takes place: H2 (8) + F2 (g) = 2 HF (g) Kp = 176 Based on building an ICE table, determine the partial pressure of H2, F2 and HF once equilibrium has been established.

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### Equilibrium in Chemical Reactions

**Problem Description:**

**Context:**
At 355 K, a 10.00 L reaction flask is charged with 1.44 atm each of hydrogen (H₂) and fluorine gas (F₂).

**Chemical Reaction:**
The following reaction takes place:
\[ \text{H}_2 \text{(g)} + \text{F}_2 \text{(g)} \rightleftharpoons 2 \text{HF} \text{(g)} \]

**Equilibrium Constant:**
Given \( K_p = 176 \)

**Task:**
Based on building an ICE (Initial, Change, Equilibrium) table, determine the partial pressure of H₂, F₂, and HF once equilibrium has been established.

**Steps to Solution:**

1. **Set Up the ICE Table:**

   - **Initial:**  
     - \(P_{\text{H}_2} = 1.44\) atm
     - \(P_{\text{F}_2} = 1.44\) atm
     - \(P_{\text{HF}} = 0\) atm
     
   - **Change:**  
     - Let the change in pressure for H₂ and F₂ be \( -x \) atm.
     - The change in pressure for HF will be \( +2x \) atm (since 2 moles of HF are produced for every mole of H₂ and F₂ reacted).
   
   - **Equilibrium:**  
     - \(P_{\text{H}_2} = 1.44 - x\) atm
     - \(P_{\text{F}_2} = 1.44 - x\)
     - \(P_{\text{HF}} = 2x\)

2. **Express the Equilibrium Constant \( K_p \):**

   \[ K_p = \frac{(P_{\text{HF}})^2}{(P_{\text{H}_2})(P_{\text{F}_2})} \]
   
   Substitute the equilibrium expressions into this equation:
   
   \[ 176 = \frac{(2x)^2}{(1.44 - x)(1.44 - x)} \]
   
   This simplifies to:
   
   \[ 176 = \frac{4x^2}{(1.44 - x)^2} \]

3
Transcribed Image Text:### Equilibrium in Chemical Reactions **Problem Description:** **Context:** At 355 K, a 10.00 L reaction flask is charged with 1.44 atm each of hydrogen (H₂) and fluorine gas (F₂). **Chemical Reaction:** The following reaction takes place: \[ \text{H}_2 \text{(g)} + \text{F}_2 \text{(g)} \rightleftharpoons 2 \text{HF} \text{(g)} \] **Equilibrium Constant:** Given \( K_p = 176 \) **Task:** Based on building an ICE (Initial, Change, Equilibrium) table, determine the partial pressure of H₂, F₂, and HF once equilibrium has been established. **Steps to Solution:** 1. **Set Up the ICE Table:** - **Initial:** - \(P_{\text{H}_2} = 1.44\) atm - \(P_{\text{F}_2} = 1.44\) atm - \(P_{\text{HF}} = 0\) atm - **Change:** - Let the change in pressure for H₂ and F₂ be \( -x \) atm. - The change in pressure for HF will be \( +2x \) atm (since 2 moles of HF are produced for every mole of H₂ and F₂ reacted). - **Equilibrium:** - \(P_{\text{H}_2} = 1.44 - x\) atm - \(P_{\text{F}_2} = 1.44 - x\) - \(P_{\text{HF}} = 2x\) 2. **Express the Equilibrium Constant \( K_p \):** \[ K_p = \frac{(P_{\text{HF}})^2}{(P_{\text{H}_2})(P_{\text{F}_2})} \] Substitute the equilibrium expressions into this equation: \[ 176 = \frac{(2x)^2}{(1.44 - x)(1.44 - x)} \] This simplifies to: \[ 176 = \frac{4x^2}{(1.44 - x)^2} \] 3
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