Calculate the per cent dissociation of `H_2S(g)` if 0.1 mole of `H_2S` is kept in a 0.4-litre vessel at 1000 K.
For the reaction
`2H_2S (g) iff 2H_2(g) + S_2(g)`
the value of `K_c` is `1.0 xx 10^(-6)`

Answer

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:
\[ 2 H_2S(g) \rightleftharpoons 2 H_2(g) + S_2(g) \]

### Step 2: Set up the initial concentrations
Initially, we have 0.1 moles of \( H_2S \) in a 0.4-liter vessel. Therefore, the initial concentration of \( H_2S \) can be calculated as:
\[ \text{Initial concentration of } H_2S = \frac{0.1 \text{ moles}}{0.4 \text{ L}} = 0.25 \text{ M} \]

### Step 3: Define the change in concentration
Let \( x \) be the amount of \( H_2S \) that dissociates. According to the stoichiometry of the reaction:
- The change in concentration of \( H_2S \) will be \( -x \).
- The change in concentration of \( H_2 \) will be \( +\frac{x}{2} \) (since 2 moles of \( H_2S \) produce 2 moles of \( H_2 \)).
- The change in concentration of \( S_2 \) will be \( +\frac{x}{2} \).

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \( [H_2S] = 0.25 - x \)
- \( [H_2] = \frac{x}{2} \)
- \( [S_2] = \frac{x}{2} \)

### Step 5: 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 into the expression:
\[ K_c = \frac{\left(\frac{x}{2}\right)^2 \left(\frac{x}{2}\right)}{(0.25 - x)^2} \]

### Step 6: Substitute the known value of \( K_c \)
Given that \( K_c = 1.0 \times 10^{-6} \), we can write:
\[ 1.0 \times 10^{-6} = \frac{\left(\frac{x}{2}\right)^3}{(0.25 - x)^2} \]

### Step 7: Simplify the equation
This simplifies to:
\[ 1.0 \times 10^{-6} = \frac{x^3}{8(0.25 - x)^2} \]

### Step 8: Make an assumption
Assuming \( x \) is small compared to 0.25 (which is valid since \( K_c \) is very small), we can approximate \( 0.25 - x \approx 0.25 \):
\[ 1.0 \times 10^{-6} = \frac{x^3}{8(0.25)^2} \]

### Step 9: Solve for \( x \)
Calculating \( (0.25)^2 = 0.0625 \):
\[ 1.0 \times 10^{-6} = \frac{x^3}{8 \times 0.0625} = \frac{x^3}{0.5} \]
Thus:
\[ x^3 = 1.0 \times 10^{-6} \times 0.5 = 5.0 \times 10^{-7} \]
Taking the cube root:
\[ x = (5.0 \times 10^{-7})^{1/3} \approx 8.0 \times 10^{-3} \text{ moles} \]

### Step 10: Calculate the percent dissociation
Percent dissociation is given by:
\[ \text{Percent dissociation} = \left(\frac{x}{\text{Initial moles of } H_2S}\right) \times 100 = \left(\frac{8.0 \times 10^{-3}}{0.1}\right) \times 100 = 8.0\% \]

### Final Answer
The percent dissociation of \( H_2S \) is approximately **8.0%**.