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

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

Given that the total pressure at equilibrium is \( 1.12 \, \text{atm} \) at \( 106^\circ C \).

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction involves the solid ammonium hydrosulfide \( NH_4HS \) decomposing into gaseous ammonia \( NH_3 \) and hydrogen sulfide \( H_2S \).

2. **Identify the Total Pressure**:
The total pressure of the gases at equilibrium is given as \( P_{total} = 1.12 \, \text{atm} \).

3. **Determine Partial Pressures**:
Since the stoichiometry of the reaction shows that one mole of \( NH_3 \) and one mole of \( H_2S \) are produced from one mole of \( NH_4HS \), we can denote the partial pressures of \( NH_3 \) and \( H_2S \) as equal:
\[ P_{NH_3} = P_{H_2S} = x \]
Therefore, the total pressure can be expressed as:
\[ P_{total} = P_{NH_3} + P_{H_2S} = x + x = 2x \]
Setting this equal to the total pressure:
\[ 2x = 1.12 \, \text{atm} \]
Solving for \( x \):
\[ x = \frac{1.12}{2} = 0.56 \, \text{atm} \]

4. **Calculate \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction is given by the expression:
\[ K_p = \frac{P_{NH_3} \cdot P_{H_2S}}{P_{NH_4HS}} \]
Since \( NH_4HS \) is a solid, its activity is considered to be 1, and thus:
\[ K_p = P_{NH_3} \cdot P_{H_2S} \]
Substituting the values of the partial pressures:
\[ K_p = (0.56 \, \text{atm}) \cdot (0.56 \, \text{atm}) = 0.56^2 = 0.3136 \, \text{atm}^2 \]

5. **Final Result**:
Rounding to three significant figures, we find:
\[ K_p \approx 0.314 \]

### Conclusion:
The value of \( K_p \) for the reaction at the given temperature is approximately \( 0.314 \).