`NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)`
`The3 equilibrium pressure at `25^(@)C` is 0.660 atm . What is `K_(p)` for the reaction ?

Options

A. 0.109
B. `0.218`
C. `1.89`
D. `2.18`

Answer

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

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

with an equilibrium pressure of 0.660 atm at 25°C, we will follow these steps:

### Step 1: Identify the Reaction Components
The reaction involves solid ammonium hydrosulfide (\( \text{NH}_4\text{HS} \)) dissociating into gaseous ammonia (\( \text{NH}_3 \)) and hydrogen sulfide (\( \text{H}_2\text{S} \)). The solid does not appear in the equilibrium expression.

### Step 2: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by the partial pressures of the gaseous products:

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

### Step 3: Determine the Partial Pressures
At equilibrium, we know the total pressure \( P_t = 0.660 \, \text{atm} \). Since the reaction produces 1 mole of \( \text{NH}_3 \) and 1 mole of \( \text{H}_2\text{S} \) from 1 mole of \( \text{NH}_4\text{HS} \), we can denote the moles of \( \text{NH}_3 \) and \( \text{H}_2\text{S} \) as \( x \).

Thus, the total number of moles of gas at equilibrium is:

\[ \text{Total moles} = x + x = 2x \]

The mole fraction of each gas is:

\[ \text{Mole fraction of } \text{NH}_3 = \frac{x}{2x} = \frac{1}{2} \]
\[ \text{Mole fraction of } \text{H}_2\text{S} = \frac{x}{2x} = \frac{1}{2} \]

Using the total pressure to find the partial pressures:

\[ P_{\text{NH}_3} = \text{Mole fraction of } \text{NH}_3 \times P_t = \frac{1}{2} \times 0.660 \, \text{atm} = 0.330 \, \text{atm} \]
\[ P_{\text{H}_2\text{S}} = \text{Mole fraction of } \text{H}_2\text{S} \times P_t = \frac{1}{2} \times 0.660 \, \text{atm} = 0.330 \, \text{atm} \]

### Step 4: Calculate \( K_p \)
Now we can substitute the partial pressures into the \( K_p \) expression:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = 0.330 \, \text{atm} \times 0.330 \, \text{atm} = 0.1089 \, \text{atm}^2 \]

### Step 5: Round to Appropriate Significant Figures
Rounding \( 0.1089 \) to three significant figures gives:

\[ K_p \approx 0.109 \]

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

\[ \boxed{0.109} \]