`64 gm` of `CH_(4)` and `68gm "of" H_(2)S` was placed in an close container and heated up to `727^(@)C` following equilibrium is established in gaseous phase reaction is:
`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)`
The total pressure at equilibrium is `1.6` atm and partial pressure of `H_(2) "is" 0.8` atm, then

Options

A. Total moles at equilibrium `4.8`
B. `K_(P)=K_(C)(RT)^(2)`
C. Mole fraction `H_(2)` at equilibrium`=0.5`
D. On increasing moles of `H_(2)S` equilibrium constant increases.

Explanation

To solve the problem step by step, we will analyze the given information and apply the principles of chemical equilibrium.

### Step 1: Determine the moles of reactants
We start with the given masses of the reactants:
- Mass of CH₄ (methane) = 64 g
- Mass of H₂S (hydrogen sulfide) = 68 g

Next, we calculate the number of moles of each reactant using their molar masses:
- Molar mass of CH₄ = 12 (C) + 4 (H) = 16 g/mol
- Molar mass of H₂S = 2 (H) + 32 (S) = 34 g/mol

Calculating the moles:
- Moles of CH₄ = 64 g / 16 g/mol = 4 moles
- Moles of H₂S = 68 g / 34 g/mol = 2 moles

**Hint:** Remember to use the formula: Moles = Mass / Molar Mass.

### Step 2: Write the balanced chemical equation
The reaction is given as:
\[ CH₄(g) + 2 H₂S(g) \rightleftharpoons CS₂(g) + 4 H₂(g) \]

### Step 3: Set up the initial moles and changes at equilibrium
Let \( x \) be the moles of CH₄ that react at equilibrium. The changes in moles of each species can be represented as follows:
- Moles of CH₄ at equilibrium = \( 4 - x \)
- Moles of H₂S at equilibrium = \( 2 - 2x \)
- Moles of CS₂ at equilibrium = \( x \)
- Moles of H₂ at equilibrium = \( 4x \)

### Step 4: Calculate total moles at equilibrium
Total moles at equilibrium can be expressed as:
\[ \text{Total moles} = (4 - x) + (2 - 2x) + x + 4x = 6 - 2x \]

**Hint:** Keep track of the stoichiometry of the reaction when determining changes in moles.

### Step 5: Use the total pressure and partial pressure to find \( x \)
Given:
- Total pressure at equilibrium, \( P_{total} = 1.6 \) atm
- Partial pressure of H₂, \( P_{H₂} = 0.8 \) atm

Using the ideal gas law, we know:
\[ P_{total} = \frac{n_{total}RT}{V} \]
\[ P_{H₂} = \frac{n_{H₂}RT}{V} \]

From the partial pressure of H₂:
\[ P_{H₂} = \frac{4xRT}{V} \]
Setting this equal to 0.8 atm gives:
\[ 0.8 = \frac{4xRT}{V} \]

From the total pressure:
\[ 1.6 = \frac{(6 - 2x)RT}{V} \]

Dividing the total pressure equation by the partial pressure equation:
\[ \frac{1.6}{0.8} = \frac{(6 - 2x)}{(4x)} \]
This simplifies to:
\[ 2 = \frac{6 - 2x}{4x} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 2(4x) = 6 - 2x \]
\[ 8x + 2x = 6 \]
\[ 10x = 6 \]
\[ x = 0.6 \]

### Step 7: Calculate total moles at equilibrium
Substituting \( x \) back into the total moles equation:
\[ \text{Total moles} = 6 - 2(0.6) = 6 - 1.2 = 4.8 \]

### Step 8: Calculate mole fraction of H₂
Moles of H₂ at equilibrium:
\[ n_{H₂} = 4x = 4(0.6) = 2.4 \]

Mole fraction of H₂:
\[ \text{Mole fraction of H₂} = \frac{n_{H₂}}{n_{total}} = \frac{2.4}{4.8} = 0.5 \]

### Step 9: Determine the equilibrium constant \( K_p \)
Using the relationship:
\[ K_p = K_c (RT)^{\Delta n} \]
Where \( \Delta n = \text{moles of products} - \text{moles of reactants} = 5 - 3 = 2 \).

### Conclusion
All options provided in the question are correct based on the calculations:
1. Total moles at equilibrium is 4.8.
2. \( K_p = K_c (RT)^2 \).
3. Mole fraction of hydrogen at equilibrium is 0.5.
4. Increasing moles of H₂S increases the equilibrium constant.