At 800 K, hydrogen iodide can decompose into hydrogen and iodine gases. 2HI(9) 2(9) +H2 (g) At this temperature, K = 0.0169. What are the partial pressures at equilibrium of the hydrogen and iodine if initially a sealed flask at 800 K contains only HI at a pressure of 0.226 atm? Partial pressure of hydrogen = atm Partial pressure of iodine = atm
At 800 K, hydrogen iodide can decompose into hydrogen and iodine gases. 2HI(9) 2(9) +H2 (g) At this temperature, K = 0.0169. What are the partial pressures at equilibrium of the hydrogen and iodine if initially a sealed flask at 800 K contains only HI at a pressure of 0.226 atm? Partial pressure of hydrogen = atm Partial pressure of iodine = atm
Chemistry
10th Edition
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
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![### Decomposition of Hydrogen Iodide at 800 K
At 800 K, hydrogen iodide (HI) can decompose into hydrogen (H₂) and iodine (I₂) gases according to the equilibrium reaction:
\[ 2HI(g) \rightleftharpoons I_2(g) + H_2(g) \]
At this temperature, the equilibrium constant, \( K \), is 0.0169. Given this information, we are asked to determine the partial pressures of hydrogen and iodine at equilibrium if initially a sealed flask at 800 K contains only HI at a pressure of 0.226 atm.
The problem can be broken down into the following steps:
1. **Initial Condition:**
- Initial pressure of HI, \( P_{HI} \): 0.226 atm.
- Initial pressure of H₂ and I₂: 0 atm (since initially only HI is present).
2. **Change at Equilibrium:**
- Let the change in pressure of HI be \( -2x \) (since 2 moles of HI produce 1 mole each of H₂ and I₂).
- The change in pressure of both H₂ and I₂ will be \( +x \).
3. **Equilibrium Condition:**
- Equilibrium pressure of HI: \( 0.226 - 2x \) atm.
- Equilibrium pressure of H₂: \( x \) atm.
- Equilibrium pressure of I₂: \( x \) atm.
4. **Equilibrium Expression:**
\[ K = \frac{(P_{I2})(P_{H2})}{(P_{HI})^2} \]
Substituting the equilibrium conditions:
\[ 0.0169 = \frac{(x)(x)}{(0.226 - 2x)^2} \]
5. **Solving for \( x \):**
This equation needs to be solved to find the value of \( x \) which will give us the partial pressures at equilibrium.
6. **Substitute \( x \) into Equilibrium Pressures:**
- Partial pressure of hydrogen \( P_{H₂} = x \) atm.
- Partial pressure of iodine \( P_{I₂} = x \) atm.
### User Interaction
**Question:**
What are the partial pressures at equilibrium of the hydrogen and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F848b3b37-0e47-4626-8c62-5eaa520ec29b%2Fccfa248a-c551-4b96-bdf1-9f9f1db6375a%2Fy4wiw5n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Decomposition of Hydrogen Iodide at 800 K
At 800 K, hydrogen iodide (HI) can decompose into hydrogen (H₂) and iodine (I₂) gases according to the equilibrium reaction:
\[ 2HI(g) \rightleftharpoons I_2(g) + H_2(g) \]
At this temperature, the equilibrium constant, \( K \), is 0.0169. Given this information, we are asked to determine the partial pressures of hydrogen and iodine at equilibrium if initially a sealed flask at 800 K contains only HI at a pressure of 0.226 atm.
The problem can be broken down into the following steps:
1. **Initial Condition:**
- Initial pressure of HI, \( P_{HI} \): 0.226 atm.
- Initial pressure of H₂ and I₂: 0 atm (since initially only HI is present).
2. **Change at Equilibrium:**
- Let the change in pressure of HI be \( -2x \) (since 2 moles of HI produce 1 mole each of H₂ and I₂).
- The change in pressure of both H₂ and I₂ will be \( +x \).
3. **Equilibrium Condition:**
- Equilibrium pressure of HI: \( 0.226 - 2x \) atm.
- Equilibrium pressure of H₂: \( x \) atm.
- Equilibrium pressure of I₂: \( x \) atm.
4. **Equilibrium Expression:**
\[ K = \frac{(P_{I2})(P_{H2})}{(P_{HI})^2} \]
Substituting the equilibrium conditions:
\[ 0.0169 = \frac{(x)(x)}{(0.226 - 2x)^2} \]
5. **Solving for \( x \):**
This equation needs to be solved to find the value of \( x \) which will give us the partial pressures at equilibrium.
6. **Substitute \( x \) into Equilibrium Pressures:**
- Partial pressure of hydrogen \( P_{H₂} = x \) atm.
- Partial pressure of iodine \( P_{I₂} = x \) atm.
### User Interaction
**Question:**
What are the partial pressures at equilibrium of the hydrogen and
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