A vessel of `10L` was filled with `6` mole of `Sb_(2)S_(3)` and `6` mole of `H_(2)` to attain the equilibrium at `440^(@)C` as:
`Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)`
After equilibrium the `H_(2)S` formed was analysed by dissolving it in water and treating with excess of `Pb^(2+)` to give `708 g "of" PbS` as precipitate. What is value of `K_(c)` of the reaction at `440^(@)C`?(At. weight of `Pb=206)`.

Options

A. `0.08`
B. `0.8`
C. `0.4`
D. `0.04`

Explanation

To find the equilibrium constant \( K_c \) for the reaction at \( 440^\circ C \), we will follow these steps:

### Step 1: Determine the number of moles of \( H_2S \) formed
The reaction is:
\[ Sb_2S_3(s) + 3H_2(g) \rightleftharpoons 2Sb(s) + 3H_2S(g) \]
From the problem, we know that \( 708 \, g \) of \( PbS \) was formed. The molar mass of \( PbS \) is \( 206 \, g/mol \).

Using the formula for the number of moles:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
we can calculate the number of moles of \( PbS \):
\[ n_{PbS} = \frac{708 \, g}{206 \, g/mol} = 3.43 \, mol \]

Since the formation of \( PbS \) is directly related to the amount of \( H_2S \) produced, we can say that:
\[ n_{H_2S} = 3.43 \, mol \]

### Step 2: Set up the initial and equilibrium concentrations
Initially, we have:
- \( n_{Sb_2S_3} = 6 \, mol \)
- \( n_{H_2} = 6 \, mol \)
- \( n_{H_2S} = 0 \, mol \)
- \( n_{Sb} = 0 \, mol \)

At equilibrium, let \( x \) be the amount of \( Sb_2S_3 \) that reacts. The changes in moles will be:
- \( Sb_2S_3 \): \( 6 - x \)
- \( H_2 \): \( 6 - 3x \)
- \( H_2S \): \( 0 + 3x \)
- \( Sb \): \( 0 + 2x \)

Given that \( 3x = 3.43 \), we can find \( x \):
\[ x = 1.14 \]

Now substituting \( x \) back into the equilibrium concentrations:
- \( n_{Sb_2S_3} = 6 - 1.14 = 4.86 \, mol \)
- \( n_{H_2} = 6 - 3(1.14) = 2.58 \, mol \)
- \( n_{H_2S} = 3(1.14) = 3.43 \, mol \)
- \( n_{Sb} = 2(1.14) = 2.28 \, mol \)

### Step 3: Calculate the equilibrium concentrations
Since the volume of the vessel is \( 10 \, L \), we can calculate the concentrations:
\[ [C] = \frac{n}{V} \]
- \( [Sb_2S_3] = \frac{4.86}{10} = 0.486 \, M \)
- \( [H_2] = \frac{2.58}{10} = 0.258 \, M \)
- \( [H_2S] = \frac{3.43}{10} = 0.343 \, M \)
- \( [Sb] = \frac{2.28}{10} = 0.228 \, M \)

### Step 4: Write the expression for \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[Sb]^2 [H_2S]^3}{[Sb_2S_3][H_2]^3} \]

Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{(0.228)^2 (0.343)^3}{(0.486)(0.258)^3} \]

### Step 5: Calculate \( K_c \)
Calculating each part:
- \( (0.228)^2 = 0.051984 \)
- \( (0.343)^3 = 0.040244 \)
- \( (0.258)^3 = 0.017188 \)

Now substituting these values back into the \( K_c \) expression:
\[ K_c = \frac{0.051984 \times 0.040244}{0.486 \times 0.017188} \]
Calculating the numerator:
\[ 0.051984 \times 0.040244 = 0.002094 \]
Calculating the denominator:
\[ 0.486 \times 0.017188 = 0.008345 \]
Finally, calculating \( K_c \):
\[ K_c = \frac{0.002094}{0.008345} \approx 0.251 \]

### Final Answer
Thus, the value of \( K_c \) at \( 440^\circ C \) is approximately \( 0.251 \).