# Solving EquilibriumDr. 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 ProblemDr. 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

Note- Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.” Here we to determine equilibrium pressure of H2 ,I2 and total equilibrium pressure in the following given equilibrium reaction.

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Pproducts PReactants PH₂ PIZ hp- hp= (PHI)2 1.34-10 = (0.45 +0) (1.17+p) t1.51-212 1.34-10--/151-701² (1 Dr. Dahm is attempting to store a sample of HI. Little does he realize that there exists an equilibrium between HI, I, 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. K for this reaction is 1.34 10¹ calculate the pressure of I, and H, when the reaction comes to equilibrium. 2 HI 1.51 Initial Change-2p Equilibrium-1.51-2p - Solving Equilibrium atm H₂ + 1₂ 0.45 alm +p 045+p 1.17 QAm +P 1.17-p

Pproducts PReactants PH₂ PIZ hp- hp= (PHI)2 1.34-10 = (0.45 +0) (1.17+p) t1.51-212 1.34-10--/151-701² (1 Dr. Dahm is attempting to store a sample of HI. Little does he realize that there exists an equilibrium between HI, I, 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. K for this reaction is 1.34 10¹ calculate the pressure of I, and H, when the reaction comes to equilibrium. 2 HI 1.51 Initial Change-2p Equilibrium-1.51-2p - Solving Equilibrium atm H₂ + 1₂ 0.45 alm +p 045+p 1.17 QAm +P 1.17-p

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter17: Equilibrium

Section: Chapter Questions

Problem 10ALQ: . Consider an equilibrium mixture consisting of H2O(g), CO(g). H2(g), and CO2(g) reacting in a...

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Question

# 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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