The `K_(p)` of the reaction is `NH_(4)HS_((s)) Leftrightarrow NH_(3(g))+H_(2)S_((g))`. If the total pressure at equilibrium is 30 atm.

Options

A. `15 atm^(2)`
B. `225 atm^(2)`
C. `30 atm^(2)`
D. 15 atm

Explanation

To solve the problem of finding the \( K_p \) for the reaction

\[ NH_4HS_{(s)} \leftrightarrow NH_3_{(g)} + H_2S_{(g)} \]

given that the total pressure at equilibrium is 30 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction involves the dissociation of solid ammonium hydrosulfide (\( NH_4HS \)) into gaseous ammonia (\( NH_3 \)) and hydrogen sulfide (\( H_2S \)). Since \( NH_4HS \) is a solid, it does not appear in the equilibrium expression.

### Step 2: Define the Variables
Let \( p \) be the partial pressure of \( NH_3 \) and \( H_2S \) at equilibrium. Since both gases are produced in a 1:1 ratio, we can express the total pressure at equilibrium as:

\[ P_{total} = P_{NH_3} + P_{H_2S} = p + p = 2p \]

### Step 3: Use the Total Pressure to Find \( p \)
We are given that the total pressure at equilibrium is 30 atm:

\[ 2p = 30 \text{ atm} \]

Now, solving for \( p \):

\[ p = \frac{30}{2} = 15 \text{ atm} \]

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the products:

\[ K_p = \frac{P_{NH_3} \cdot P_{H_2S}}{P_{NH_4HS}} \]

Since \( NH_4HS \) is a solid, it does not appear in the expression. Therefore:

\[ K_p = P_{NH_3} \cdot P_{H_2S} \]

### Step 5: Substitute the Values
Now substituting the values of the partial pressures:

\[ K_p = p \cdot p = 15 \text{ atm} \cdot 15 \text{ atm} = 225 \text{ atm}^2 \]

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

\[ K_p = 225 \text{ atm}^2 \]

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