For `NH_(4)HS(s)hArr NH_(3)(g)+H_(2)S(g)` If `K_(p)=64atm^(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 will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

### Step 2: Identify the equilibrium expression
For the reaction, the equilibrium constant \( K_p \) is defined as:
\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{P_{\text{NH}_4\text{HS}}} \]
Since NH4HS is a solid, it does not appear in the expression. Thus:
\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]

### Step 3: Set up the initial conditions
Assume we start with 1 mole of NH4HS. Initially:
- Moles of NH4HS = 1
- Moles of NH3 = 0
- Moles of H2S = 0

### Step 4: Define the change in moles at equilibrium
Let \( \alpha \) be the amount of NH4HS that dissociates. At equilibrium:
- Moles of NH4HS = \( 1 - \alpha \)
- Moles of NH3 = \( \alpha \)
- Moles of H2S = \( \alpha \)

### Step 5: Write the expression for \( K_p \)
At equilibrium, the partial pressures can be expressed as:
\[ P_{\text{NH}_3} = \alpha \]
\[ P_{\text{H}_2\text{S}} = \alpha \]
Thus, the expression for \( K_p \) becomes:
\[ K_p = \alpha \cdot \alpha = \alpha^2 \]

### Step 6: Substitute the value of \( K_p \)
We know that \( K_p = 64 \, \text{atm}^2 \). Therefore:
\[ \alpha^2 = 64 \]

### Step 7: Solve for \( \alpha \)
Taking the square root of both sides:
\[ \alpha = 8 \, \text{atm} \]

### Step 8: Calculate the total pressure at equilibrium
The total pressure \( P_{\text{total}} \) at equilibrium is the sum of the partial pressures of NH3 and H2S:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = \alpha + \alpha = 2\alpha \]
Substituting the value of \( \alpha \):
\[ P_{\text{total}} = 2 \times 8 = 16 \, \text{atm} \]

### Final Answer
The equilibrium pressure of the mixture is:
\[ P_{\text{total}} = 16 \, \text{atm} \]

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