`05` moles of `NH_(4)HS(s)` are taken in a container having air at `1` atm. On warming the closed container to `50^(@)C` the pressure attained a constant value of `1.5` atm, with some `NH_(4)HS(s)` remaining unreacted. The `K_(p)` of reaction
`NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g) "at" 50^@)C` is:

Options

A. `0.25`
B. `0.625`
C. `0.025`
D. `0.0625`

Explanation

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

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

### Step 1: Understanding the Initial Conditions
We start with 5 moles of \( NH_4HS(s) \) in a closed container that also contains air at a pressure of 1 atm. When the container is warmed to \( 50^\circ C \), the total pressure reaches a constant value of 1.5 atm.

### Step 2: Identify the Change in Pressure
The total pressure of the system after warming is 1.5 atm. However, since the container initially had air at 1 atm, the pressure contributed by the gases \( NH_3 \) and \( H_2S \) can be calculated as follows:

\[ P_{\text{NH}_3} + P_{\text{H}_2S} = P_{\text{total}} - P_{\text{air}} = 1.5 \, \text{atm} - 1 \, \text{atm} = 0.5 \, \text{atm} \]

### Step 3: Define Partial Pressures
Let \( P_{\text{NH}_3} = \alpha \) and \( P_{\text{H}_2S} = \alpha \) (since they are produced in a 1:1 ratio). Therefore, the total pressure due to these gases can be expressed as:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2S} = \alpha + \alpha = 2\alpha \]

### Step 4: Solve for \( \alpha \)
From the equation derived in Step 2, we have:

\[ 2\alpha = 0.5 \, \text{atm} \]

Now, solving for \( \alpha \):

\[ \alpha = \frac{0.5}{2} = 0.25 \, \text{atm} \]

### Step 5: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by the expression:

\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2S} \]

Substituting the values of the partial pressures:

\[ K_p = \alpha \times \alpha = \alpha^2 = (0.25)^2 = 0.0625 \, \text{atm}^2 \]

### Final Answer
Thus, the value of \( K_p \) at \( 50^\circ C \) is:

\[ \boxed{0.0625 \, \text{atm}^2} \]