For `NH_4HS(s)hArrNH_3(g)+H_2S(g)` , if `K_p = 64 atm^2` , equilibrium pressure of mixture is

Options

A. 8 atm
B. 16 atm
C. 64 atm
D. 4 atm

Answer

To solve the problem, we need to find the equilibrium pressure of the mixture for the reaction:

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

Given that the equilibrium constant \( K_p = 64 \, \text{atm}^2 \), we can follow these steps:

### Step 1: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by the formula:

\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{P_{\text{NH}_4\text{HS}}} \]

Since \( \text{NH}_4\text{HS} \) is a solid, its partial pressure is not included in the expression. Therefore, we can simplify the expression to:

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

### Step 2: Define the partial pressures
Let the partial pressure of \( \text{NH}_3 \) at equilibrium be \( P \) and the partial pressure of \( \text{H}_2\text{S} \) also be \( P \) (since they are produced in a 1:1 ratio). Thus, we have:

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

### Step 3: Substitute into the \( K_p \) expression
Now substituting these values into the \( K_p \) expression:

\[ K_p = P \cdot P = P^2 \]

### Step 4: Set up the equation
We know that \( K_p = 64 \, \text{atm}^2 \), so we can set up the equation:

\[ P^2 = 64 \]

### Step 5: Solve for \( P \)
Taking the square root of both sides gives:

\[ P = \sqrt{64} = 8 \, \text{atm} \]

### Step 6: Calculate the total pressure
The total pressure \( P_{\text{total}} \) at equilibrium is the sum of the partial pressures of \( \text{NH}_3 \) and \( \text{H}_2\text{S} \):

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

Substituting the value of \( P \):

\[ P_{\text{total}} = 2 \times 8 = 16 \, \text{atm} \]

### Final Answer
The equilibrium pressure of the mixture is \( 16 \, \text{atm} \).

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