An amount of solid ` NH_4 HS ` is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure . Ammonium hydrogen sulphide decomposses 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`

Explanation

To solve the problem, we need to analyze the decomposition reaction of ammonium hydrogen sulfide (NH₄HS) and calculate the equilibrium constant (K) based on the given pressures.

### Step-by-Step Solution:

1. **Write the decomposition reaction:**
\[ \text{NH}_4\text{HS (s)} \rightleftharpoons \text{NH}_3 (g) + \text{H}_2\text{S} (g) \]

2. **Identify initial conditions:**
- Initial pressure of NH₃ = 0.50 atm
- Initial pressure of H₂S = 0 atm (since it is produced from the decomposition of NH₄HS)
- Since NH₄HS is a solid, it does not contribute to the pressure.

3. **Define changes at equilibrium:**
- Let \( x \) be the change in pressure of H₂S at equilibrium.
- At equilibrium:
- Pressure of NH₃ = \( 0.50 + x \)
- Pressure of H₂S = \( x \)

4. **Total pressure at equilibrium:**
- Given that the total pressure at equilibrium is 0.84 atm:
\[ (0.50 + x) + x = 0.84 \]
- Simplifying this equation:
\[ 0.50 + 2x = 0.84 \]
\[ 2x = 0.84 - 0.50 \]
\[ 2x = 0.34 \]
\[ x = \frac{0.34}{2} = 0.17 \, \text{atm} \]

5. **Calculate equilibrium pressures:**
- Pressure of NH₃ at equilibrium:
\[ P_{\text{NH}_3} = 0.50 + x = 0.50 + 0.17 = 0.67 \, \text{atm} \]
- Pressure of H₂S at equilibrium:
\[ P_{\text{H}_2\text{S}} = x = 0.17 \, \text{atm} \]

6. **Calculate the equilibrium constant (K):**
- The equilibrium constant expression for the reaction is:
\[ K = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{1} = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]
- Substituting the equilibrium pressures:
\[ K = (0.67) \cdot (0.17) \]
- Performing the multiplication:
\[ K = 0.1139 \approx 0.11 \, \text{(to two decimal places)} \]

7. **Final answer:**
- The equilibrium constant \( K \) for the decomposition of NH₄HS at this temperature is approximately **0.11**.