Consider the decomposition of solid `NH_(4)HS` in a flask containing `NH_(3)(g)` at a pressure of `2` atm. What will be the partial pressure of `NH_(3)(g) "and" H_(2)S(g)` after the equilibrium has been attained?
`K_(p)` for the reaction is `3`.

Options

A. `P_(NH_(3))=4 "atm", P_(H_(2))s=2` atm
B. `P_(MH_(3))=1.732 "atm", P_(h_2)s=1.732` atm
C. `P_(NH_(3))=3 "atm", P_(H_(2))s=1` atm
D. `P_(NH_(3))=1 "atm", P_(H_(2))s=1` atm

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of solid ammonium hydrosulfide (NH₄HS) can be represented by the following equation:
\[ \text{NH}_4\text{HS (s)} \rightleftharpoons \text{NH}_3\text{(g)} + \text{H}_2\text{S(g)} \]

### Step 2: Identify initial conditions
Initially, we have:
- Partial pressure of NH₃ = 2 atm
- Partial pressure of H₂S = 0 atm

### Step 3: Define changes at equilibrium
Let \( P \) be the change in pressure of H₂S at equilibrium. Therefore, at equilibrium:
- Partial pressure of NH₃ = \( 2 + P \)
- Partial pressure of H₂S = \( P \)

### Step 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{1} \]
Since the solid does not appear in the expression, we can simplify it to:
\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]

### Step 5: Substitute known values into the \( K_p \) expression
Given that \( K_p = 3 \):
\[ 3 = (2 + P) \cdot P \]

### Step 6: Solve for \( P \)
Expanding the equation:
\[ 3 = 2P + P^2 \]
Rearranging gives us:
\[ P^2 + 2P - 3 = 0 \]

Now, we can factor or use the quadratic formula to solve for \( P \):
\[ (P + 3)(P - 1) = 0 \]
Thus, \( P = -3 \) (not physically meaningful) or \( P = 1 \).

### Step 7: Calculate partial pressures at equilibrium
Now substituting \( P = 1 \) back into our expressions for partial pressures:
- Partial pressure of NH₃ = \( 2 + P = 2 + 1 = 3 \) atm
- Partial pressure of H₂S = \( P = 1 \) atm

### Final Answer
- The partial pressure of NH₃ at equilibrium is **3 atm**.
- The partial pressure of H₂S at equilibrium is **1 atm**.

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