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:
What is the "mole" fraction of hydrogen gas in the equilibrium mixture in the gas phase?

Options

A. `1//4`
B. `3//4`
C. `1//8`
D. `4`

Explanation

To solve the problem step by step, we will analyze the decomposition reactions and calculate the mole fraction of hydrogen gas in the equilibrium mixture.

### Step 1: Understand the reactions and initial conditions
We have the following reactions:
1. \( \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \) with \( K_c = 8.91 \times 10^{-2} \, \text{M}^2 \)
2. \( 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \) with \( K_c = 3 \times 10^{-4} \, \text{M}^2 \)

Initially, we have 1 mole of \( \text{NH}_4\text{HS} \) in a 1 L container, which means that the initial concentration of \( \text{NH}_4\text{HS} \) is not relevant since it is a solid.

### Step 2: Set up the equilibrium for the first reaction
Let \( x \) be the number of moles of \( \text{NH}_3 \) produced at equilibrium. Then, the number of moles of \( \text{H}_2\text{S} \) produced will also be \( x \).

At equilibrium:
- Moles of \( \text{NH}_3 = x \)
- Moles of \( \text{H}_2\text{S} = x \)

Using the equilibrium constant expression for the first reaction:
\[ K_c = \frac{[\text{NH}_3][\text{H}_2\text{S}]}{1} = 8.91 \times 10^{-2} \]
Substituting the concentrations:
\[ 8.91 \times 10^{-2} = x \cdot x = x^2 \]
Thus, \( x = \sqrt{8.91 \times 10^{-2}} \approx 0.298 \).

### Step 3: Set up the equilibrium for the second reaction
From the problem, we know the ratio of moles of \( \text{NH}_3 \) to \( \text{H}_2 \) at equilibrium is 3:1. Let the moles of \( \text{H}_2 \) be \( A \). Then, the moles of \( \text{NH}_3 \) will be \( 3A \).

At equilibrium for the second reaction:
- Moles of \( \text{N}_2 = \frac{A}{3} \)
- Moles of \( \text{H}_2 = A \)
- Moles of \( \text{NH}_3 = 3A \)

Using the equilibrium constant expression for the second reaction:
\[ K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} = 3 \times 10^{-4} \]
Substituting the concentrations:
\[ 3 \times 10^{-4} = \frac{\left(\frac{A}{3}\right)(A)^3}{(3A)^2} \]
This simplifies to:
\[ 3 \times 10^{-4} = \frac{A^4}{27A^2} = \frac{A^2}{27} \]
Thus, \( A^2 = 3 \times 10^{-4} \times 27 = 8.1 \times 10^{-3} \) and \( A = \sqrt{8.1 \times 10^{-3}} \approx 0.090 \).

### Step 4: Calculate moles of other gases
Now we can find the moles of other gases:
- Moles of \( \text{NH}_3 = 3A = 3 \times 0.090 = 0.27 \)
- Moles of \( \text{H}_2\text{S} = 0.33 \) (from previous calculations)
- Moles of \( \text{N}_2 = \frac{A}{3} = \frac{0.090}{3} = 0.03 \)

### Step 5: Total moles and mole fraction of \( \text{H}_2 \)
Total moles of gas at equilibrium:
\[ \text{Total moles} = \text{Moles of } \text{H}_2 + \text{Moles of } \text{NH}_3 + \text{Moles of } \text{H}_2\text{S} + \text{Moles of } \text{N}_2 \]
\[ = 0.090 + 0.27 + 0.33 + 0.03 = 0.72 \]

Now, the mole fraction of \( \text{H}_2 \):
\[ \text{Mole fraction of } \text{H}_2 = \frac{\text{Moles of } \text{H}_2}{\text{Total moles}} = \frac{0.090}{0.72} \approx 0.125 \]

### Final Answer
The mole fraction of hydrogen gas in the equilibrium mixture in the gas phase is \( \frac{1}{8} \).