Calculate `K_(p)` for the equilibrium,
`NH_(4)HS_((s))hArrNH_(3(g))+H_(2)S_((g))`
if the total pressure inside reaction vessel s 1.12 atm at `105.^(@)C`.

Options

A. 0.56
B. 1.25
C. 0.31
D. 0.63

Explanation

To calculate \( K_p \) for the equilibrium reaction:

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

given that the total pressure inside the reaction vessel is 1.12 atm at \( 105^\circ C \), 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 gaseous ammonia (\( \text{NH}_3 \)) and hydrogen sulfide (\( \text{H}_2\text{S} \)). Since \( \text{NH}_4\text{HS} \) is a solid, it does not appear in the equilibrium expression.

### Step 2: Write the Equilibrium Expression
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gaseous products:

\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{1} \]

### Step 3: Define Partial Pressures
Let:
- \( P_{\text{NH}_3} = x \) (partial pressure of ammonia)
- \( P_{\text{H}_2\text{S}} = y \) (partial pressure of hydrogen sulfide)

From the stoichiometry of the reaction, we know that:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = x + y \]

Given that the total pressure \( P_{\text{total}} = 1.12 \, \text{atm} \), we can express \( y \) in terms of \( x \):

\[ y = P_{\text{total}} - x = 1.12 - x \]

### Step 4: Substitute into the \( K_p \) Expression
Substituting \( y \) into the equilibrium expression gives:

\[ K_p = x \cdot (1.12 - x) \]

### Step 5: Solve for \( K_p \)
To find \( K_p \), we need to consider the total number of moles of gas produced. Since one mole of \( \text{NH}_4\text{HS} \) produces one mole of \( \text{NH}_3 \) and one mole of \( \text{H}_2\text{S} \), the total number of moles of gas is 2.

Assuming ideal gas behavior, we can relate the total pressure and the partial pressures:

\[ P_{\text{NH}_3} = \frac{P_{\text{total}}}{2} = \frac{1.12}{2} = 0.56 \, \text{atm} \]
\[ P_{\text{H}_2\text{S}} = \frac{P_{\text{total}}}{2} = 0.56 \, \text{atm} \]

Now substituting these values back into the \( K_p \) expression:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = 0.56 \cdot 0.56 = 0.3136 \, \text{atm}^2 \]

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

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