`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)`

Answer

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

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

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

### Step 1: Understand the Reaction
The reaction involves the decomposition of solid ammonium hydrosulfide (\( \text{NH}_4\text{HS} \)) into gaseous ammonia (\( \text{NH}_3 \)) and hydrogen sulfide (\( \text{H}_2\text{S} \)).

### Step 2: Identify the Components
At equilibrium, we have:
- \( P_{\text{NH}_3} \): partial pressure of ammonia
- \( P_{\text{H}_2\text{S}} \): partial pressure of hydrogen sulfide

### Step 3: Total Pressure
The total pressure at equilibrium is given as 100 atm. Since there are two moles of gas produced (1 mole of \( \text{NH}_3 \) and 1 mole of \( \text{H}_2\text{S} \)), we can express the total pressure as:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = 100 \text{ atm} \]

### Step 4: Assume Equal Partial Pressures
Assuming that the partial pressures of \( \text{NH}_3 \) and \( \text{H}_2\text{S} \) are equal, we can denote them as \( P \):

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

Thus, we have:

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

### Step 5: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by the expression:

\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} \]

Substituting the values we found:

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

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

\[ \boxed{2500} \]

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