Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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# Solving Equilibrium

Dr. Dahm is attempting to store a sample of HI. Little does he realize that there exists an equilibrium between HI, I₂, and H₂. If he starts with an initial pressure of HI of 1.51 atm, H₂ of 0.45 atm and I₂ of 1.17 atm, find the pressures at equilibrium. The equilibrium constant (K) for this reaction is 1.34 x 10⁻². Calculate the pressure of I₂ and H₂ when the reaction comes to equilibrium.

### Reaction
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

### Initial Pressures
- \( \text{HI} = 1.51 \, \text{atm} \)
- \( \text{H}_2 = 0.45 \, \text{atm} \)
- \( \text{I}_2 = 1.17 \, \text{atm} \)

### Change in Pressures 
- Change for \( \text{HI} = -2p \)
- Change for \( \text{H}_2 = +p \)
- Change for \( \text{I}_2 = +p \)

### Equilibrium Pressures
- \( \text{HI} = 1.51 - 2p \)
- \( \text{H}_2 = 0.45 + p \)
- \( \text{I}_2 = 1.17 + p \)

### Calculations
- Use the equilibrium constant \( K = 1.34 \times 10^{-2} \) to solve for the pressures:

\[
K = \frac{(0.45 + p)(1.17 + p)}{(1.51 - 2p)^2}
\]

Solve for \( p \) and substitute back to find the equilibrium pressures for \( \text{H}_2 \) and \( \text{I}_2 \).

---

### Additional Problem

Dr. Dahm wants to make \( \text{N}_2\text{O} \) so he reacts \( \text{N}_2 \) with \( \text{O}_2 \) to give \( \text{N}_2\text{O} \). Unfortunately, this is an equilibrium process that has a K
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Transcribed Image Text:# Solving Equilibrium Dr. Dahm is attempting to store a sample of HI. Little does he realize that there exists an equilibrium between HI, I₂, and H₂. If he starts with an initial pressure of HI of 1.51 atm, H₂ of 0.45 atm and I₂ of 1.17 atm, find the pressures at equilibrium. The equilibrium constant (K) for this reaction is 1.34 x 10⁻². Calculate the pressure of I₂ and H₂ when the reaction comes to equilibrium. ### Reaction \[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \] ### Initial Pressures - \( \text{HI} = 1.51 \, \text{atm} \) - \( \text{H}_2 = 0.45 \, \text{atm} \) - \( \text{I}_2 = 1.17 \, \text{atm} \) ### Change in Pressures - Change for \( \text{HI} = -2p \) - Change for \( \text{H}_2 = +p \) - Change for \( \text{I}_2 = +p \) ### Equilibrium Pressures - \( \text{HI} = 1.51 - 2p \) - \( \text{H}_2 = 0.45 + p \) - \( \text{I}_2 = 1.17 + p \) ### Calculations - Use the equilibrium constant \( K = 1.34 \times 10^{-2} \) to solve for the pressures: \[ K = \frac{(0.45 + p)(1.17 + p)}{(1.51 - 2p)^2} \] Solve for \( p \) and substitute back to find the equilibrium pressures for \( \text{H}_2 \) and \( \text{I}_2 \). --- ### Additional Problem Dr. Dahm wants to make \( \text{N}_2\text{O} \) so he reacts \( \text{N}_2 \) with \( \text{O}_2 \) to give \( \text{N}_2\text{O} \). Unfortunately, this is an equilibrium process that has a K
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