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

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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.
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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