Calculate the % age dissociation of `H_2S(g)` if 0.1 mole of `H_2S` is kept in a 0.4 L vessel at 1000 K . The value of `K_c` for the reaction, `2H_2S(g) hArr 2H_2(g) +S_2(g)` is `1.0 xx 10^-3` .

Options

A. `2.8%`
B. `2.2%`
C. `3.1%`
D. `3.8%`

Explanation

To calculate the percentage dissociation of \( H_2S(g) \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of hydrogen sulfide can be represented by the following equation:
\[ 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \]

### Step 2: Calculate the initial concentration of \( H_2S \)
Given:
- Number of moles of \( H_2S = 0.1 \) moles
- Volume of the vessel = \( 0.4 \) L

The concentration \( C \) is calculated using the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume}} = \frac{0.1 \, \text{moles}}{0.4 \, \text{L}} = 0.25 \, \text{mol/L} \]

### Step 3: Set up the expression for equilibrium concentrations
Let \( x \) be the amount of \( H_2S \) that dissociates at equilibrium. Therefore, at equilibrium:
- Concentration of \( H_2S = 0.25 - 2x \)
- Concentration of \( H_2 = 2x \)
- Concentration of \( S_2 = x \)

### Step 4: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[H_2]^2[S_2]}{[H_2S]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2 (x)}{(0.25 - 2x)^2} \]
Given \( K_c = 1.0 \times 10^{-3} \), we can write:
\[ 1.0 \times 10^{-3} = \frac{4x^2 \cdot x}{(0.25 - 2x)^2} \]
or
\[ 1.0 \times 10^{-3} = \frac{4x^3}{(0.25 - 2x)^2} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ 4x^3 = 1.0 \times 10^{-3} (0.25 - 2x)^2 \]
Expanding and rearranging this equation will allow us to solve for \( x \).

Assuming \( x \) is small compared to \( 0.25 \), we can approximate \( 0.25 - 2x \approx 0.25 \):
\[ 4x^3 = 1.0 \times 10^{-3} (0.25)^2 \]
Calculating \( (0.25)^2 = 0.0625 \):
\[ 4x^3 = 1.0 \times 10^{-3} \cdot 0.0625 = 6.25 \times 10^{-5} \]
Thus,
\[ x^3 = \frac{6.25 \times 10^{-5}}{4} = 1.5625 \times 10^{-5} \]
Taking the cube root:
\[ x \approx (1.5625 \times 10^{-5})^{1/3} \approx 0.022 \, \text{mol/L} \]

### Step 6: Calculate the percentage dissociation
The percentage dissociation of \( H_2S \) is given by:
\[ \text{Percentage dissociation} = \left( \frac{x}{\text{initial concentration}} \right) \times 100 = \left( \frac{0.022}{0.25} \right) \times 100 \approx 8.8\% \]

### Final Answer
The percentage dissociation of \( H_2S \) is approximately **8.8%**.