At a certain temperature, the K, for the decomposition of H, S is 0.846. H,S(g) = H,(g)+S(g) Initially, only H, S is present at a pressure of 0.188 atm in a closed container. What is the total pressure in the container at equilibrium? Ptotal = atm
At a certain temperature, the K, for the decomposition of H, S is 0.846. H,S(g) = H,(g)+S(g) Initially, only H, S is present at a pressure of 0.188 atm in a closed container. What is the total pressure in the container at equilibrium? Ptotal = atm
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
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![### Decomposition of Hydrogen Sulfide (H₂S)
At a certain temperature, the equilibrium constant \( K_p \) for the decomposition of H₂S is 0.846.
\[ \text{H}_2\text{S(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{S(g)} \]
Initially, only H₂S is present at a pressure of 0.188 atm in a closed container. The task is to determine the total pressure in the container at equilibrium.
### Calculation of Total Pressure at Equilibrium
Given data:
- Initial pressure of H₂S: \( P_{\text{H}_2\text{S, initial}} = 0.188 \text{ atm} \)
- Equilibrium constant: \( K_p = 0.846 \)
At equilibrium, let the change in pressure of H₂S be \( -x \) atm. For each mole of H₂S that decomposes, 1 mole of H₂ and 1 mole of S are produced.
### Change in Pressures:
- H₂S: \( 0.188 - x \) atm
- H₂: \( x \) atm
- S: \( x \) atm
Using the equilibrium expression for \( K_p \):
\[ K_p = \frac{(P_{\text{H}_2})(P_{\text{S}})}{(P_{\text{H}_2\text{S}})} \]
Substitute the equilibrium pressures into the expression:
\[ 0.846 = \frac{(x)(x)}{0.188 - x} \]
\[ 0.846 = \frac{x^2}{0.188 - x} \]
Solve for \( x \) to find the equilibrium pressure change.
Finally, calculate the total pressure in the container at equilibrium:
\[ P_{\text{total}} = (0.188 - x) + x + x \]
\[ = 0.188 + x \]
Enter the value:
\[ P_{\text{total}} = \_\_\_\_ \text{ atm} \]
This interactive problem helps in understanding chemical equilibrium and partial pressures in gas-phase reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86546da6-58c7-4384-8a8e-1318a5c89f9f%2Fe1090e66-5bd0-47da-b39c-1f657f4e4289%2F5lrn45n_processed.png&w=3840&q=75)
Transcribed Image Text:### Decomposition of Hydrogen Sulfide (H₂S)
At a certain temperature, the equilibrium constant \( K_p \) for the decomposition of H₂S is 0.846.
\[ \text{H}_2\text{S(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{S(g)} \]
Initially, only H₂S is present at a pressure of 0.188 atm in a closed container. The task is to determine the total pressure in the container at equilibrium.
### Calculation of Total Pressure at Equilibrium
Given data:
- Initial pressure of H₂S: \( P_{\text{H}_2\text{S, initial}} = 0.188 \text{ atm} \)
- Equilibrium constant: \( K_p = 0.846 \)
At equilibrium, let the change in pressure of H₂S be \( -x \) atm. For each mole of H₂S that decomposes, 1 mole of H₂ and 1 mole of S are produced.
### Change in Pressures:
- H₂S: \( 0.188 - x \) atm
- H₂: \( x \) atm
- S: \( x \) atm
Using the equilibrium expression for \( K_p \):
\[ K_p = \frac{(P_{\text{H}_2})(P_{\text{S}})}{(P_{\text{H}_2\text{S}})} \]
Substitute the equilibrium pressures into the expression:
\[ 0.846 = \frac{(x)(x)}{0.188 - x} \]
\[ 0.846 = \frac{x^2}{0.188 - x} \]
Solve for \( x \) to find the equilibrium pressure change.
Finally, calculate the total pressure in the container at equilibrium:
\[ P_{\text{total}} = (0.188 - x) + x + x \]
\[ = 0.188 + x \]
Enter the value:
\[ P_{\text{total}} = \_\_\_\_ \text{ atm} \]
This interactive problem helps in understanding chemical equilibrium and partial pressures in gas-phase reactions.
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