One "mole" of `NH_(4)HS(s)` was allowed to decompose in a `1-L` container at `200^(@)C`. It decomposes reversibly to `NH_(3)(g)` and `H_(2)S(g). NH_(3)(g)` further undergoes decomposition to form `N_(2)(g)` and `H_(2)(g)`. Finally, when equilibrium was set up, the ratio between the number of moles of `NH_(3)(g)` and `H_(2)(g)` was found to be `3`.
`NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2)`
`2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2)`
Answer the following:
To attain equilibrium, how much `%` by weight of folid `NH_(4)HS` got dissociated?

Options

A. `19%`
B. `30%`
C. `33%`
D. `15%`

Explanation

To solve the problem, we need to analyze the decomposition of `NH4HS` and the subsequent reactions involving `NH3`. Here’s a step-by-step breakdown of the solution:

### Step 1: Set Up the Initial Conditions
1 mole of `NH4HS(s)` is placed in a 1-L container. Initially, we have:
- Moles of `NH4HS = 1`
- Moles of `NH3 = 0`
- Moles of `H2S = 0`
- Moles of `N2 = 0`
- Moles of `H2 = 0`

### Step 2: Define the Degree of Dissociation
Let `α1` be the degree of dissociation of `NH4HS`. At equilibrium:
- Moles of `NH4HS = 1 - α1`
- Moles of `NH3 = α1`
- Moles of `H2S = α1`

### Step 3: Consider the Decomposition of `NH3`
The reaction for the decomposition of `NH3` is:
\[ 2NH3(g) \rightleftharpoons N2(g) + 3H2(g) \]
Let `α2` be the degree of dissociation of `NH3`. At equilibrium:
- Moles of `NH3 = α1(1 - α2)`
- Moles of `N2 = 0.5α1α2`
- Moles of `H2 = 1.5α1α2`

### Step 4: Use the Given Ratio
We are given that the ratio of moles of `NH3` to `H2` at equilibrium is 3:
\[ \frac{NH3}{H2} = 3 \implies \frac{α1(1 - α2)}{1.5α1α2} = 3 \]
This simplifies to:
\[ \frac{(1 - α2)}{1.5α2} = 3 \]
Cross-multiplying gives:
\[ 1 - α2 = 4.5α2 \implies 1 = 5.5α2 \implies α2 = \frac{1}{5.5} = \frac{2}{11} \]

### Step 5: Calculate the Equilibrium Concentration for `NH3`
Substituting `α2` back into the expression for `NH3`:
\[ NH3 = α1(1 - \frac{2}{11}) = α1 \cdot \frac{9}{11} \]

### Step 6: Calculate the Equilibrium Constant for `NH4HS`
The equilibrium constant `Kc` for the first reaction is given by:
\[ Kc = [NH3][H2S] = \frac{(α1 \cdot \frac{9}{11}) \cdot (α1)}{1} = α1^2 \cdot \frac{9}{11} \]
Given that `Kc = 8.91 \times 10^{-2}`:
\[ α1^2 \cdot \frac{9}{11} = 8.91 \times 10^{-2} \]
Solving for `α1`:
\[ α1^2 = \frac{8.91 \times 10^{-2} \cdot 11}{9} \implies α1^2 = \frac{0.9801}{9} \implies α1^2 = 0.1089 \]
\[ α1 = \sqrt{0.1089} \approx 0.33 \]

### Step 7: Calculate the Percentage Dissociation
The percentage by weight of `NH4HS` that got dissociated is given by:
\[ \text{Percentage dissociation} = α1 \times 100 = 0.33 \times 100 \approx 33\% \]

### Final Answer
Thus, the percentage by weight of solid `NH4HS` that got dissociated is approximately **33%**.