Calculate the percent dissociation of `H_(2)S(g)` if `0.1 mol` of `H_(2)S` is kept in `0.4 L` vessel at `1000 K`. For the reaction:
`2H_(2)S(g) hArr 2H_(2)(g)+S(g)`
The value of `K_(c )` is `1.0xx10^(-6)`

Explanation

To calculate the percent dissociation of \( H_2S(g) \) in the given reaction, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the dissociation of hydrogen sulfide is:
\[ 2 H_2S(g) \rightleftharpoons 2 H_2(g) + S(g) \]

### Step 2: Set up the initial conditions
We start with \( 0.1 \) moles of \( H_2S \) in a \( 0.4 \, L \) vessel. The initial concentrations can be calculated as follows:
\[ \text{Initial concentration of } H_2S = \frac{0.1 \, \text{mol}}{0.4 \, \text{L}} = 0.25 \, \text{M} \]
The initial concentrations of \( H_2 \) and \( S \) are both \( 0 \).

### Step 3: Define the change in concentration at equilibrium
Let \( x \) be the amount of \( H_2S \) that dissociates at equilibrium. Therefore, at equilibrium:
- Concentration of \( H_2S = 0.25 - \frac{x}{2} \)
- Concentration of \( H_2 = \frac{x}{2} \)
- Concentration of \( S = \frac{x}{2} \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[H_2]^2[S]}{[H_2S]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{x}{2}\right)^2 \left(\frac{x}{2}\right)}{\left(0.25 - x\right)^2} \]

### Step 5: Substitute the value of \( K_c \)
Given \( K_c = 1.0 \times 10^{-6} \), we can set up the equation:
\[ 1.0 \times 10^{-6} = \frac{\left(\frac{x}{2}\right)^2 \left(\frac{x}{2}\right)}{\left(0.25 - x\right)^2} \]
This simplifies to:
\[ 1.0 \times 10^{-6} = \frac{\frac{x^3}{8}}{(0.25 - x)^2} \]

### Step 6: Simplify and solve for \( x \)
Assuming \( x \) is very small compared to \( 0.25 \), we can approximate \( 0.25 - x \approx 0.25 \):
\[ 1.0 \times 10^{-6} = \frac{\frac{x^3}{8}}{(0.25)^2} \]
\[ 1.0 \times 10^{-6} = \frac{x^3}{8 \times 0.0625} \]
\[ 1.0 \times 10^{-6} = \frac{x^3}{0.5} \]
\[ x^3 = 1.0 \times 10^{-6} \times 0.5 = 5.0 \times 10^{-7} \]
\[ x = (5.0 \times 10^{-7})^{1/3} \approx 0.008 \]

### Step 7: Calculate percent dissociation
The percent dissociation is given by:
\[ \text{Percent dissociation} = \left(\frac{x}{\text{initial moles of } H_2S}\right) \times 100 \]
\[ \text{Percent dissociation} = \left(\frac{0.008}{0.1}\right) \times 100 = 8\% \]

### Final Answer
The percent dissociation of \( H_2S \) is approximately **8%**.

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