An equilibrium mixture for the reaction
`2H_(2)S(g) hArr 2H_(2)(g) + S_(2)(g)`
had 1 mole of `H_(2)S, 0.2` mole of `H_(2)` and 0.8 mole of `S_(2)` in a 2 litre flask. The value of `K_(c)` in mol `L^(-1)` is

Options

A. `0.004`
B. `0.08`
C. `0.016`
D. `0.0016`

Answer

To find the equilibrium constant \( K_c \) for the reaction

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

given the equilibrium concentrations, we can follow these steps:

### Step 1: Calculate the concentrations of each species

We have the following moles in a 2-liter flask:

- Moles of \( H_2S = 1 \) mole
- Moles of \( H_2 = 0.2 \) mole
- Moles of \( S_2 = 0.8 \) mole

To find the concentration of each species, we use the formula:

\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \]

#### Concentration of \( H_2S \):

\[ \text{Concentration of } H_2S = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ mol/L} \]

#### Concentration of \( H_2 \):

\[ \text{Concentration of } H_2 = \frac{0.2 \text{ moles}}{2 \text{ L}} = 0.1 \text{ mol/L} \]

#### Concentration of \( S_2 \):

\[ \text{Concentration of } S_2 = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ mol/L} \]

### Step 2: Write the expression for \( K_c \)

The expression for the equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[H_2]^2 \cdot [S_2]}{[H_2S]^2} \]

### Step 3: Substitute the concentrations into the \( K_c \) expression

Now we can substitute the calculated concentrations into the \( K_c \) expression:

\[ K_c = \frac{(0.1)^2 \cdot (0.4)}{(0.5)^2} \]

### Step 4: Calculate \( K_c \)

Calculating the numerator:

\[ (0.1)^2 = 0.01 \]
\[ 0.01 \cdot 0.4 = 0.004 \]

Calculating the denominator:

\[ (0.5)^2 = 0.25 \]

Now substituting back into the equation for \( K_c \):

\[ K_c = \frac{0.004}{0.25} = 0.016 \]

### Final Answer

Thus, the value of \( K_c \) is:

\[ K_c = 0.016 \text{ mol L}^{-1} \]

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