The equilibrium constant Kp for the reaction `NH_(4)HS_((s)) Leftrightarrow NH_(3(g))+H_(2)S_((g))` is

Options

A. `K_(P) =P_(NH_(3) xx P_(H_(2)S))/(P_(NH_(4)HS))`
B. `K_(P)=(P_(NH_(4) HS))/(P_(NH_(3)) xx P_(H_(2)S))`
C. `K_(P)=P_(NH_(4)HS)`
D. `K_(P)=P_(NH_(3)) xx P_(H_(2)S)`

Answer

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

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

we will follow these steps:

### Step 1: Identify the Reaction Components
The reaction involves one solid reactant, \( NH_4HS \), and two gaseous products, \( NH_3 \) and \( H_2S \).

### Step 2: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the gaseous products over the reactants. The general formula for \( K_p \) is:

\[ K_p = \frac{(P_{products})^{\text{stoichiometric coefficient}}}{(P_{reactants})^{\text{stoichiometric coefficient}}} \]

### Step 3: Apply the Formula to Our Reaction
For our specific reaction, we have:

- Products: \( NH_3 \) and \( H_2S \)
- Reactant: \( NH_4HS \) (which is a solid)

The expression for \( K_p \) will be:

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

### Step 4: Consider the Role of Solids in Equilibrium
In equilibrium expressions, the activities of pure solids and liquids are considered to be 1. Therefore, the partial pressure of the solid \( NH_4HS \) does not appear in the expression for \( K_p \).

### Step 5: Simplify the Expression
Since the partial pressure of the solid \( NH_4HS \) is considered to be 1, we can simplify the expression for \( K_p \):

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

### Final Answer
Thus, the equilibrium constant \( K_p \) for the reaction is:

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

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