`NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)`
In the above reaction, if the pressure at equilibrium and at 300K is 100atm then what will be equilibrium constant `K_(p)` ?

Options

A. `2500 atm^(2)`
B. `50 atm^(2)`
C. `100 atm^(2)`
D. `200 atm^(2)`

Explanation

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

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

given that the total pressure at equilibrium is 100 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction involves a solid (NH₄HS) decomposing into two gaseous products (NH₃ and H₂S). In equilibrium expressions, the concentration or partial pressure of solids is not included.

### Step 2: Define Partial Pressures
Let the partial pressure of NH₃ at equilibrium be \( P \) atm and the partial pressure of H₂S also be \( P \) atm. Since both gases have the same stoichiometric coefficient (1), their partial pressures are equal.

### Step 3: Write the Total Pressure Equation
The total pressure at equilibrium is the sum of the partial pressures of the gaseous products:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = P + P = 2P \]

Given that the total pressure is 100 atm, we can set up the equation:

\[ 2P = 100 \text{ atm} \]

### Step 4: Solve for Partial Pressure
Now, solve for \( P \):

\[ P = \frac{100 \text{ atm}}{2} = 50 \text{ atm} \]

### Step 5: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed as:

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

Since the solid does not appear in the expression, we can write:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = P \cdot P = P^2 \]

### Step 6: Substitute the Value of \( P \)
Substituting the value of \( P \):

\[ K_p = (50 \text{ atm})^2 = 2500 \text{ atm}^2 \]

### Conclusion
Thus, the equilibrium constant \( K_p \) for the reaction at 300 K is:

\[ K_p = 2500 \text{ atm}^2 \]