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.
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.
Chemistry
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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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbca33481-84f0-4d0e-9458-2fdcb949c0ec%2F78cecbf1-303f-46f5-bf96-35010068fc71%2Fefbalyo_processed.jpeg&w=3840&q=75)
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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