For the reaction,`Fe(s) + S(s) rarr FeS(s)` the expression for equilibrium constant is

Options

A. `[FeS]/([S][Fe])`
B. `[Fe][S]/[FeS]`
C. `[Fe][S][FeS]`
D. None of these

Explanation

To derive the expression for the equilibrium constant for the reaction \( \text{Fe}(s) + \text{S}(s) \rightleftharpoons \text{FeS}(s) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
The reaction given is:
\[ \text{Fe}(s) + \text{S}(s) \rightleftharpoons \text{FeS}(s) \]
Here, we have iron (Fe) and sulfur (S) as reactants, and iron sulfide (FeS) as the product.

2. **Understand the Equilibrium Constant Expression**:
The equilibrium constant \( K_c \) for a general reaction:
\[ aA + bB \rightleftharpoons cC + dD \]
is expressed as:
\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
where \([X]\) denotes the concentration of species \(X\).

3. **Identify the States of Matter**:
In the given reaction, all species (Fe, S, and FeS) are in the solid state.

4. **Consider the Concentration of Solids**:
The concentration of pure solids is defined as 1 (unity) in equilibrium expressions. Therefore, for the reactants and products in solid form:
\[ [\text{Fe}] = 1, \quad [\text{S}] = 1, \quad [\text{FeS}] = 1 \]

5. **Write the Equilibrium Constant Expression**:
Since all components are solids, the equilibrium constant expression becomes:
\[ K_c = \frac{[\text{FeS}]}{[\text{Fe}][\text{S}]} = \frac{1}{1 \times 1} = 1 \]

6. **Conclusion**:
Since the equilibrium constant expression results in 1, we conclude that:
\[ K_c = 1 \]

### Final Answer:
The expression for the equilibrium constant \( K_c \) for the reaction \( \text{Fe}(s) + \text{S}(s) \rightleftharpoons \text{FeS}(s) \) is:
\[ K_c = 1 \]