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

Answer

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

\[ 2H_{2}S(g) \rightleftharpoons 2H_{2}(g) + S_{2}(g) \]

given the amounts of each species in a 2-liter flask, we will follow these steps:

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

The equilibrium constant \( K_c \) for the reaction can be expressed as:

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

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

### Step 2: Calculate the concentrations of each species

We are given the number of moles of each species and the volume of the flask (2 liters). The concentration is calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

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

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

- For \( S_2 \):
\[ [S_2] = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ mol/L} \]

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

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

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

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

Calculating the values:

\[ K_c = \frac{(0.01)(0.4)}{(0.25)} = \frac{0.004}{0.25} = 0.016 \]

### Final Answer

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

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

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