At `87^(@)C`, the following equilibrium is established.
`H_(2)(g)+S(s) hArrH_(2)S(g), K_(c)=0.08`
If `0.3` mole hydrogen and 2 mole sulphur are heated to `87^(@)C` in a `2L` vessel, what will be concentration of `H_(2)S` at equilibrium ?

Options

A. `0.011M`
B. `0.022M`
C. `0.044M`
D. `0.08M`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction at equilibrium is:
\[ H_2(g) + S(s) \rightleftharpoons H_2S(g) \]

### Step 2: Identify the initial moles and volume
We are given:
- Initial moles of \( H_2 = 0.3 \) moles
- Initial moles of \( S = 2 \) moles (but it is a solid, so it does not affect the equilibrium expression)
- Volume of the vessel = \( 2 \) L

### Step 3: Calculate initial concentrations
The initial concentration of \( H_2 \) can be calculated as:
\[ \text{Concentration of } H_2 = \frac{\text{moles of } H_2}{\text{volume}} = \frac{0.3 \, \text{moles}}{2 \, \text{L}} = 0.15 \, \text{M} \]

The concentration of \( S \) is not included in the equilibrium expression since it is a solid.

### Step 4: Set up the change in moles at equilibrium
Let \( x \) be the moles of \( H_2S \) formed at equilibrium. Then, at equilibrium:
- Moles of \( H_2 = 0.3 - x \)
- Moles of \( H_2S = x \)

### Step 5: Write the equilibrium expression
The equilibrium constant \( K_c \) is given as \( 0.08 \). The expression for \( K_c \) is:
\[ K_c = \frac{[H_2S]}{[H_2]} \]
Since the concentration of \( S \) is not included, we have:
\[ K_c = \frac{x/2}{(0.3 - x)/2} \]
This simplifies to:
\[ K_c = \frac{x}{0.3 - x} \]

### Step 6: Substitute the value of \( K_c \)
Substituting the value of \( K_c \):
\[ 0.08 = \frac{x}{0.3 - x} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 0.08(0.3 - x) = x \]
Expanding this:
\[ 0.024 - 0.08x = x \]
Rearranging gives:
\[ 0.024 = x + 0.08x \]
\[ 0.024 = 1.08x \]
Now, solving for \( x \):
\[ x = \frac{0.024}{1.08} \approx 0.0222 \, \text{moles} \]

### Step 8: Calculate the concentration of \( H_2S \) at equilibrium
The concentration of \( H_2S \) at equilibrium is:
\[ [H_2S] = \frac{x}{\text{volume}} = \frac{0.0222}{2} \approx 0.0111 \, \text{M} \]

### Final Answer
The concentration of \( H_2S \) at equilibrium is approximately \( 0.011 \, \text{M} \).

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