An amount of solid `NH_4HS` is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure.Ammonium hydrogen sulphide decomposes to yield `NH_3` and `H_2S` gases in the flask.When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for `NH_4HS` decomposition at this temperature is :

Options

A. 0.11
B. 0.17
C. 0.18
D. `0.30`

Answer

To solve the problem, we need to analyze the decomposition of ammonium hydrogen sulfide (NH₄HS) into ammonia (NH₃) and hydrogen sulfide (H₂S) gases. The reaction can be represented as follows:

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

### Step 1: Initial Conditions
We know the initial pressure of ammonia (NH₃) in the flask is 0.50 atm. Since NH₄HS is a solid, it does not contribute to the pressure. Therefore, the initial pressures are:
- \( P_{\text{NH}_3} = 0.50 \, \text{atm} \)
- \( P_{\text{H}_2\text{S}} = 0 \, \text{atm} \)

### Step 2: Change in Pressure at Equilibrium
Let \( P \) be the increase in pressure of NH₃ and H₂S due to the decomposition of NH₄HS. According to the stoichiometry of the reaction, for every mole of NH₄HS that decomposes, one mole of NH₃ and one mole of H₂S is produced.

At equilibrium:
- The pressure of NH₃ will be \( 0.50 + P \)
- The pressure of H₂S will be \( P \)

### Step 3: Total Pressure at Equilibrium
The total pressure at equilibrium is given as 0.84 atm. Therefore, we can write the equation:
\[ (0.50 + P) + P = 0.84 \]
This simplifies to:
\[ 0.50 + 2P = 0.84 \]

### Step 4: Solve for P
Now, we can solve for \( P \):
\[ 2P = 0.84 - 0.50 \]
\[ 2P = 0.34 \]
\[ P = \frac{0.34}{2} = 0.17 \, \text{atm} \]

### Step 5: Equilibrium Pressures
Now we can find the equilibrium pressures:
- \( P_{\text{NH}_3} = 0.50 + 0.17 = 0.67 \, \text{atm} \)
- \( P_{\text{H}_2\text{S}} = 0.17 \, \text{atm} \)

### Step 6: Calculate the Equilibrium Constant (Kp)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} \]
Substituting the equilibrium pressures:
\[ K_p = (0.67) \times (0.17) \]

### Step 7: Perform the Calculation
Calculating \( K_p \):
\[ K_p = 0.67 \times 0.17 = 0.1139 \approx 0.11 \, \text{atm}^2 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) for the decomposition of NH₄HS at this temperature is approximately:
\[ K_p = 0.11 \, \text{atm}^2 \]