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 need to analyze the given reaction and the equilibrium constant \( K_p \):

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

**Given:**
\[ K_p = 64 \, \text{atm}^2 \]

**Step 1: Understand the Reaction**
- The solid ammonium hydrosulfide (\( \text{NH}_4\text{HS} \)) decomposes into ammonia (\( \text{NH}_3 \)) and hydrogen sulfide (\( \text{H}_2\text{S} \)).
- The solid does not affect the equilibrium expression since it is not in the gaseous phase.

**Step 2: Write the Expression for \( K_p \)**
- The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{\text{NH}_3} \times P_{\text{H}_2\text{S}}}{1} \]
- Since the solid does not appear in the equilibrium expression, we only consider the partial pressures of the gases.

**Step 3: Define Partial Pressures**
- Let the partial pressure of \( \text{NH}_3 \) at equilibrium be \( P_{\text{NH}_3} = x \) atm.
- Let the partial pressure of \( \text{H}_2\text{S} \) at equilibrium be \( P_{\text{H}_2\text{S}} = x \) atm.
- Therefore, the total pressure \( P \) at equilibrium will be:
\[ P = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = x + x = 2x \]

**Step 4: Substitute into the \( K_p \) Expression**
- Substitute \( P_{\text{NH}_3} \) and \( P_{\text{H}_2\text{S}} \) into the expression for \( K_p \):
\[ K_p = x \cdot x = x^2 \]
- Given that \( K_p = 64 \):
\[ x^2 = 64 \]

**Step 5: Solve for \( x \)**
- Taking the square root of both sides:
\[ x = \sqrt{64} = 8 \, \text{atm} \]

**Step 6: Calculate Total Equilibrium Pressure**
- The total pressure at equilibrium is:
\[ P = 2x = 2 \times 8 = 16 \, \text{atm} \]

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

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