An equilibrium mixture for the reaction `2H_2S (g) hArr 2H_2(g)+S_2(g)` had 1 mole of hydrogen sulphide, 0.2 mole of H, and 0.3 mole of `S_2` in 2 litre vessel. The value of `K_c` is

Options

A. 0.004
B. 0.08
C. 0.016
D. `0.160`

Explanation

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

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

given the amounts of substances in a 2-liter vessel, we can follow these steps:

### Step 1: Write the expression for \( K_c \)

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

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

where \([H_2]\), \([S_2]\), and \([H_2S]\) are the concentrations of hydrogen, sulfur, and hydrogen sulfide, respectively.

### Step 2: Calculate the concentrations

To find the concentrations, we need to divide the number of moles of each gas by the volume of the vessel (2 liters).

- For \( H_2S \):
\[ [H_2S] = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \]

- For \( H_2 \):
\[ [H_2] = \frac{0.2 \text{ moles}}{2 \text{ L}} = 0.1 \text{ M} \]

- For \( S_2 \):
\[ [S_2] = \frac{0.3 \text{ moles}}{2 \text{ L}} = 0.15 \text{ M} \]

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

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

\[ K_c = \frac{(0.1)^2 (0.15)}{(0.5)^2} \]

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

Now we will calculate \( K_c \):

1. Calculate the numerator:
\[ (0.1)^2 = 0.01 \]
\[ 0.01 \times 0.15 = 0.0015 \]

2. Calculate the denominator:
\[ (0.5)^2 = 0.25 \]

3. Now substitute back into the \( K_c \) expression:
\[ K_c = \frac{0.0015}{0.25} = 0.006 \]

### Final Answer

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

\[ K_c = 0.006 \]

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