A vessel of `250` litre was filled with `0.01` mole of `Sb_(2)S_(3)` and `0.01` 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 dissolved it in water and treating with execess of `Pb^(2+)` to give `1.19` g of PbS as precipitate. What is the value of `K_(c)` at `440^(@)C` ?

Options

A. `1`
B. `2`
C. `4`
D. `8`

Answer

To solve the problem step by step, we will follow the given data and the equilibrium reaction:

### Step 1: Write the balanced chemical equation.
The reaction given is:
\[ \text{Sb}_2\text{S}_3(s) + 3\text{H}_2(g) \rightleftharpoons 2\text{Sb}(s) + 3\text{H}_2\text{S}(g) \]

### Step 2: Identify the initial moles of reactants.
We have:
- Moles of \(\text{Sb}_2\text{S}_3 = 0.01\) moles
- Moles of \(\text{H}_2 = 0.01\) moles

### Step 3: Set up the change in moles at equilibrium.
Let \(x\) be the amount of \(\text{Sb}_2\text{S}_3\) that reacts. Then:
- Moles of \(\text{Sb}_2\text{S}_3\) at equilibrium = \(0.01 - x\)
- Moles of \(\text{H}_2\) at equilibrium = \(0.01 - 3x\)
- Moles of \(\text{H}_2\text{S}\) at equilibrium = \(3x\)

### Step 4: Analyze the formation of \(\text{H}_2\text{S}\).
After equilibrium, the \(\text{H}_2\text{S}\) formed was treated with excess \(\text{Pb}^{2+}\) to form \(\text{PbS}\). The mass of \(\text{PbS}\) precipitate formed is given as \(1.19\) g.

### Step 5: Calculate the moles of \(\text{PbS}\).
The molar mass of \(\text{PbS}\) is approximately \(238 \, \text{g/mol}\).
\[ \text{Moles of PbS} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.19 \, \text{g}}{238 \, \text{g/mol}} \approx 0.005 \, \text{moles} \]

### Step 6: Relate moles of \(\text{H}_2\text{S}\) to moles of \(\text{PbS}\).
Since \(1\) mole of \(\text{H}_2\text{S}\) produces \(1\) mole of \(\text{PbS}\), the moles of \(\text{H}_2\text{S}\) formed at equilibrium is also \(0.005\) moles.

### Step 7: Set up the equation for \(x\).
From the equilibrium expression, we have:
\[ 3x = 0.005 \implies x = \frac{0.005}{3} \approx 0.00167 \, \text{moles} \]

### Step 8: Calculate the equilibrium moles of \(\text{H}_2\).
Substituting \(x\) back into the equilibrium expression for \(\text{H}_2\):
\[ \text{Moles of H}_2 = 0.01 - 3(0.00167) \approx 0.01 - 0.005 = 0.005 \, \text{moles} \]

### Step 9: Calculate the concentrations.
The volume of the vessel is \(250 \, \text{liters}\).
\[ [\text{H}_2\text{S}] = \frac{0.005}{250} = 2 \times 10^{-5} \, \text{mol/L} \]
\[ [\text{H}_2] = \frac{0.005}{250} = 2 \times 10^{-5} \, \text{mol/L} \]

### Step 10: Write the expression for \(K_c\).
The equilibrium constant \(K_c\) is given by:
\[ K_c = \frac{[\text{H}_2\text{S}]^3}{[\text{H}_2]^3} \]

### Step 11: Substitute the concentrations into the \(K_c\) expression.
\[ K_c = \frac{(2 \times 10^{-5})^3}{(2 \times 10^{-5})^3} = 1 \]

### Final Answer:
The value of \(K_c\) at \(440^\circ C\) is \(1\).

---