`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 \( NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g) \) at a pressure of 100 atm and a temperature of 300 K, we can follow these steps:

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

### Step 2: 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 gaseous products:
\[ K_p = P_{NH_3} \times P_{H_2S} \]

### Step 3: Determine the Total Pressure and Partial Pressures
At equilibrium, the total pressure is given as 100 atm. The reaction produces one mole of \( NH_3 \) and one mole of \( H_2S \), so the total number of moles of gas at equilibrium is 2.

Assuming that the partial pressures of \( NH_3 \) and \( H_2S \) are equal (since they are produced in a 1:1 ratio), we can denote the partial pressure of each gas as \( P \). Therefore:
\[ P_{NH_3} + P_{H_2S} = 100 \, \text{atm} \]
\[ P + P = 100 \, \text{atm} \implies 2P = 100 \, \text{atm} \implies P = 50 \, \text{atm} \]

### Step 4: Calculate \( K_p \)
Now that we have the partial pressures:
\[ P_{NH_3} = 50 \, \text{atm} \]
\[ P_{H_2S} = 50 \, \text{atm} \]
Substituting these values into the expression for \( K_p \):
\[ K_p = P_{NH_3} \times P_{H_2S} = 50 \, \text{atm} \times 50 \, \text{atm} = 2500 \, \text{atm}^2 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is:
\[ K_p = 2500 \, \text{atm}^2 \]

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