For `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)`, the observed, pressure for reaction mixture in equilibrium is `1.12` atm at `106^(@)C`. What is the value of `K_(p)` for the reaction?

Explanation

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

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

given that the total pressure at equilibrium is \( 1.12 \, \text{atm} \) at \( 106^\circ C \), we can follow these steps:

### Step 1: Understand the Reaction
The reaction involves the decomposition of solid ammonium hydrosulfide (\( \text{NH}_4\text{HS} \)) into ammonia gas (\( \text{NH}_3 \)) and hydrogen sulfide gas (\( \text{H}_2\text{S} \)). Since \( \text{NH}_4\text{HS} \) is a solid, it does not contribute to the pressure in the equilibrium expression.

### Step 2: Define the Variables
Let \( P \) be the partial pressure of \( \text{NH}_3 \) and \( \text{H}_2\text{S} \) at equilibrium. Since both gases are produced in a 1:1 ratio, the total pressure at equilibrium can be expressed as:

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

### Step 3: Calculate the Partial Pressure
Given that the total pressure at equilibrium is \( 1.12 \, \text{atm} \):

\[ 2P = 1.12 \]

To find \( P \):

\[ P = \frac{1.12}{2} = 0.56 \, \text{atm} \]

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the gaseous products:

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

Since \( \text{NH}_4\text{HS} \) is a solid, its activity is considered to be 1. Therefore, we can simplify the expression to:

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

### Step 5: Substitute the Values
Substituting the values we found:

\[ K_p = P \cdot P = (0.56) \cdot (0.56) = 0.3136 \, \text{atm}^2 \]

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

\[ K_p = 0.3136 \, \text{atm}^2 \]

---