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

\[ \text{NH}_4\text{HS}_{(s)} \rightleftharpoons \text{NH}_3_{(g)} + \text{H}_2\text{S}_{(g)} \]

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 a solid reactant (NH₄HS) that decomposes into two gaseous products (NH₃ and H₂S). In equilibrium expressions, we only consider the gaseous components.

### Step 2: Set Up the Expression for \( K_p \)
The equilibrium constant \( K_p \) is defined as:

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

Since NH₄HS is a solid, its partial pressure is considered to be 1, thus:

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

### Step 3: Define Variables
Let \( x \) be the amount of NH₄HS that decomposes at equilibrium. At equilibrium:
- Moles of NH₃ = \( x \)
- Moles of H₂S = \( x \)
- Total moles of gas = \( x + x = 2x \)

### Step 4: Calculate Partial Pressures
The partial pressure of each gas can be calculated using the mole fraction and total pressure:

\[ P_{\text{NH}_3} = \frac{x}{2x} \cdot P_{\text{total}} = \frac{1}{2} P_{\text{total}} \]

\[ P_{\text{H}_2\text{S}} = \frac{x}{2x} \cdot P_{\text{total}} = \frac{1}{2} P_{\text{total}} \]

### Step 5: Substitute into the \( K_p \) Expression
Substituting the partial pressures into the \( K_p \) expression:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = \left(\frac{1}{2} P_{\text{total}}\right) \cdot \left(\frac{1}{2} P_{\text{total}}\right) = \frac{1}{4} P_{\text{total}}^2 \]

### Step 6: Plug in the Total Pressure
Given \( P_{\text{total}} = 1.12 \, \text{atm} \):

\[ K_p = \frac{1}{4} (1.12)^2 \]

### Step 7: Calculate \( K_p \)
Calculating \( (1.12)^2 \):

\[ (1.12)^2 = 1.2544 \]

Now substituting back into the equation for \( K_p \):

\[ K_p = \frac{1.2544}{4} = 0.3136 \]

### Final Answer
Thus, \( K_p \approx 0.31 \).

---