At a certain temperature, the K, for the decomposition of H,S is 0.873. H,S(g) = H,(g)+S(g) Initially, only H,S is present at a pressure of 0.118 bar in a closed container. What is the total pressure in the container at equilibrium? bar Protal =

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### Equilibrium Calculation for Decomposition of H₂S

At a certain temperature, the equilibrium constant \( K_p \) for the decomposition of hydrogen sulfide (H₂S) is 0.873.

\[ H_2S(g) \rightleftharpoons H_2(g) + S(g) \]

Initially, only H₂S is present at a pressure of 0.118 bar in a closed container. What is the total pressure in the container at equilibrium?

**Equilibrium Calculation:**

\[ P_{total} = \] _______ bar

---

Explanation:

1. **Chemical Equation:** The reaction involves the decomposition of hydrogen sulfide gas (H₂S) into hydrogen gas (H₂) and sulfur gas (S). This process is reversible and reaches equilibrium.

2. **Given Value:** The equilibrium constant \( K_p \) for this reaction is provided as 0.873.

3. **Initial Condition:** The initial pressure of H₂S in the container is 0.118 bar.

4. **Objective:** Determine the total pressure in the container once the system reaches equilibrium.

The linear equation implies that as H₂S decomposes, the partial pressures of the products (H₂ and S) and reactant (H₂S) will adjust until the equilibrium constant expression is satisfied. Use the initial pressure and the equilibrium constant to solve for the equilibrium pressures and subsequently the total pressure.

### Step-by-Step Approach:

1. **Write the expression for \( K_p \):**

\[ K_p = \frac{P_{H_2} \cdot P_S}{P_{H_2S}} \]

2. **Assign variables** for changes in partial pressures at equilibrium:

Let the change in pressure of H₂S be \( -x \) bar (it decreases as it decomposes), then the pressures of H₂ and S will both be \( +x \) bar (they increase as they are produced).

3. **Set up the equilibrium pressures:**

\[ P_{H_2S} = 0.118 - x \]
\[ P_{H_2} = x \]
\[ P_S = x \]

4. **Substitute these expressions into the \( K_p \) equation:**

\[ 0.873 = \frac{x \cdot x}{0.118 - x} \]

5. **
Transcribed Image Text:Certainly! Here is the transcription for an Educational website: --- ### Equilibrium Calculation for Decomposition of H₂S At a certain temperature, the equilibrium constant \( K_p \) for the decomposition of hydrogen sulfide (H₂S) is 0.873. \[ H_2S(g) \rightleftharpoons H_2(g) + S(g) \] Initially, only H₂S is present at a pressure of 0.118 bar in a closed container. What is the total pressure in the container at equilibrium? **Equilibrium Calculation:** \[ P_{total} = \] _______ bar --- Explanation: 1. **Chemical Equation:** The reaction involves the decomposition of hydrogen sulfide gas (H₂S) into hydrogen gas (H₂) and sulfur gas (S). This process is reversible and reaches equilibrium. 2. **Given Value:** The equilibrium constant \( K_p \) for this reaction is provided as 0.873. 3. **Initial Condition:** The initial pressure of H₂S in the container is 0.118 bar. 4. **Objective:** Determine the total pressure in the container once the system reaches equilibrium. The linear equation implies that as H₂S decomposes, the partial pressures of the products (H₂ and S) and reactant (H₂S) will adjust until the equilibrium constant expression is satisfied. Use the initial pressure and the equilibrium constant to solve for the equilibrium pressures and subsequently the total pressure. ### Step-by-Step Approach: 1. **Write the expression for \( K_p \):** \[ K_p = \frac{P_{H_2} \cdot P_S}{P_{H_2S}} \] 2. **Assign variables** for changes in partial pressures at equilibrium: Let the change in pressure of H₂S be \( -x \) bar (it decreases as it decomposes), then the pressures of H₂ and S will both be \( +x \) bar (they increase as they are produced). 3. **Set up the equilibrium pressures:** \[ P_{H_2S} = 0.118 - x \] \[ P_{H_2} = x \] \[ P_S = x \] 4. **Substitute these expressions into the \( K_p \) equation:** \[ 0.873 = \frac{x \cdot x}{0.118 - x} \] 5. **
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