For `NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)` reaction started only with `NH_(4)HS(s)`, the observed pressure for reaction mixture in equilibrium is `1.2` atm at `106^(@)C`. What is the value of `K_(p)` for the reaction?

Options

A. `1.44atm^(2)`
B. `0.36atm^(2)`
C. `0.16atm^(2)`
D. `3.6atm^(2)`

Explanation

To find the value of \( K_p \) for the reaction

\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

we will follow these steps:

### Step 1: Understand the Reaction
The reaction involves the decomposition of solid ammonium hydrosulfide (\( \text{NH}_4\text{HS} \)) into gaseous ammonia (\( \text{NH}_3 \)) and hydrogen sulfide (\( \text{H}_2\text{S} \)). Since solids do not contribute to the equilibrium expression, we will only consider the gaseous products.

### Step 2: Set Up the Equilibrium Expression
The equilibrium constant \( K_p \) for this reaction can be expressed as:

\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{P_{\text{NH}_4\text{HS}}} \]

Since \( \text{NH}_4\text{HS} \) is a solid, its partial pressure is considered to be 1 (unity). Thus, the expression simplifies to:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]

### Step 3: Determine the Partial Pressures
At equilibrium, let the partial pressure of \( \text{NH}_3 \) be \( p \) and the partial pressure of \( \text{H}_2\text{S} \) also be \( p \). Therefore, the total pressure at equilibrium is:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = p + p = 2p \]

According to the problem, the total pressure at equilibrium is given as 1.2 atm. Thus, we can write:

\[ 2p = 1.2 \, \text{atm} \]

### Step 4: Solve for \( p \)
Now, solving for \( p \):

\[ p = \frac{1.2}{2} = 0.6 \, \text{atm} \]

### Step 5: Calculate \( K_p \)
Now that we have the partial pressures, we can substitute them into the expression for \( K_p \):

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = p \cdot p = 0.6 \cdot 0.6 \]

Calculating this gives:

\[ K_p = 0.36 \, \text{atm}^2 \]

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

\[ \boxed{0.36 \, \text{atm}^2} \]

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