An equilibrium mixture of the reaction `2H_(2)S(g)hArr2H_(2)(g) + S_(2)(g)` had `0.5` mole `H_(2)S`, `0.10` mole `H_(2)` and `0.4` mole `S_(2)` in one litre vessel. The value of equilibrium constants (K) in mole `litre^(-1)` is

Options

A. `0.0004`
B. `0.008`
C. `0.016`
D. `0.160`

Answer

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

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

given the equilibrium concentrations of the reactants and products, we can follow these steps:

### Step 1: Identify the equilibrium concentrations
We are given the following amounts in a 1-liter vessel:
- \( [H_2S] = 0.5 \) moles
- \( [H_2] = 0.1 \) moles
- \( [S_2] = 0.4 \) moles

Since the volume of the vessel is 1 liter, the concentrations in moles per liter are:
- \( [H_2S] = 0.5 \, \text{mol/L} \)
- \( [H_2] = 0.1 \, \text{mol/L} \)
- \( [S_2] = 0.4 \, \text{mol/L} \)

### Step 2: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:

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

### Step 3: Substitute the equilibrium concentrations into the expression
Now we can substitute the concentrations we found into the expression for \( K_c \):

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

### Step 4: Calculate the values
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 these values back into the equation for \( K_c \):

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

### Step 5: Perform the final calculation
Now, performing the division:

\[ K_c = 0.016 \]

### Conclusion
The equilibrium constant \( K_c \) for the reaction is

\[ K_c = 0.016 \, \text{mol/L} \]