5 mole of `NH_(4)HS(s)` start to decompose at a particular temperature in a closed vessel. If pressure of `NH_(3)(g)` in the vessel is 2 atm, then `K_(p)` for the reaction, `NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)`, will be

Options

A. 2
B. 4
C. `0.4`
D. `0.8`

Explanation

To solve the problem, we need to determine the equilibrium constant \( K_p \) for the decomposition reaction of ammonium hydrosulfide (\( NH_4HS \)) given by:

\[ NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g) \]

### Step 1: Understand the Reaction
The reaction shows that solid ammonium hydrosulfide decomposes into ammonia gas and hydrogen sulfide gas. The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the gaseous products.

### Step 2: Identify Given Information
We are given:
- The pressure of \( NH_3(g) \) at equilibrium is \( 2 \, \text{atm} \).
- The reaction involves solid \( NH_4HS \), which does not contribute to the equilibrium expression.

### Step 3: Write the Expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) for the reaction is given by:

\[ K_p = \frac{P_{NH_3} \cdot P_{H_2S}}{P_{NH_4HS}} \]

Since \( NH_4HS \) is a solid, its pressure is not included in the expression. Thus, the equation simplifies to:

\[ K_p = P_{NH_3} \cdot P_{H_2S} \]

### Step 4: Determine Partial Pressures
At equilibrium, we know:
- \( P_{NH_3} = 2 \, \text{atm} \)

Since the stoichiometry of the reaction shows that 1 mole of \( NH_3 \) and 1 mole of \( H_2S \) are produced, the partial pressure of \( H_2S \) will also be equal to the partial pressure of \( NH_3 \) at equilibrium:

\[ P_{H_2S} = 2 \, \text{atm} \]

### Step 5: Calculate \( K_p \)
Now we can substitute the values into the expression for \( K_p \):

\[ K_p = P_{NH_3} \cdot P_{H_2S} = (2 \, \text{atm}) \cdot (2 \, \text{atm}) = 4 \, \text{atm}^2 \]

### Conclusion
Thus, the value of \( K_p \) for the reaction is:

\[ \boxed{4 \, \text{atm}^2} \]