A reaction `S_(8)(g) hArr 4S_(2)(g)` is carried out by taking `2` mol of `S_(8)(g)` and `0.2` mol of `S_(2)(g)` is a reaction vessel of `1 L`. Which one is not correct if `K_(c)=6.30xx10^(-6)`

Options

A. Reaction qutient is `8xx10^(-4)`
B. Reaction proceed in backward direction.
C. Reaction proceed is forward direction
D. `K_(p)=2.55 atm^(3)`

Explanation

To solve the problem, we need to analyze the given reaction and the provided data step by step.

### Step 1: Write the Reaction and Initial Conditions
The reaction given is:
\[ S_8(g) \rightleftharpoons 4 S_2(g) \]

We start with:
- 2 moles of \( S_8 \)
- 0.2 moles of \( S_2 \)
- Volume of the reaction vessel = 1 L

### Step 2: Calculate Initial Concentrations
Since the volume of the reaction vessel is 1 L, the initial concentrations are:
- Concentration of \( S_8 \) = \( \frac{2 \text{ moles}}{1 \text{ L}} = 2 \, \text{M} \)
- Concentration of \( S_2 \) = \( \frac{0.2 \text{ moles}}{1 \text{ L}} = 0.2 \, \text{M} \)

### Step 3: Calculate the Reaction Quotient \( Q \)
The reaction quotient \( Q \) is calculated using the formula:
\[ Q = \frac{[S_2]^4}{[S_8]} \]

Substituting the concentrations:
\[ Q = \frac{(0.2)^4}{2} \]

Calculating \( (0.2)^4 \):
\[ (0.2)^4 = 0.0016 \]

Now substituting this value into the equation for \( Q \):
\[ Q = \frac{0.0016}{2} = 0.0008 \]

Expressing \( Q \) in scientific notation:
\[ Q = 8.0 \times 10^{-4} \]

### Step 4: Compare \( Q \) with \( K_c \)
Given that \( K_c = 6.30 \times 10^{-6} \):
- Since \( Q (8.0 \times 10^{-4}) > K_c (6.30 \times 10^{-6}) \), the reaction will proceed in the backward direction.

### Step 5: Analyze the Options
1. **Reaction quotient is \( 8.0 \times 10^{-4} \)**: This is correct.
2. **Reaction proceeds in backward direction**: This is correct.
3. **Reaction proceeds in forward direction**: This is incorrect, as we established that the reaction proceeds backward.
4. **\( K_p \) is \( 2.55 \, \text{atm}^3 \)**: We need to check this.

### Step 6: Calculate \( K_p \)
The relationship between \( K_c \) and \( K_p \) is given by:
\[ K_p = K_c (RT)^{\Delta n} \]
Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 298 \, \text{K} \)
- \( \Delta n = \text{moles of products} - \text{moles of reactants} = 4 - 1 = 3 \)

Calculating \( K_p \):
\[ K_p = (6.30 \times 10^{-6}) \times (0.0821 \times 298)^3 \]

Calculating \( (0.0821 \times 298) \):
\[ 0.0821 \times 298 \approx 24.4658 \]

Now calculating \( (24.4658)^3 \):
\[ (24.4658)^3 \approx 14606.4 \]

Now substituting back:
\[ K_p = (6.30 \times 10^{-6}) \times 14606.4 \]
\[ K_p \approx 9.22 \times 10^{-2} \, \text{atm}^3 \]

### Conclusion
The calculated \( K_p \) does not match the given \( 2.55 \, \text{atm}^3 \), making this option incorrect.

### Final Answer
The options that are not correct are:
- **3. Reaction proceeds in forward direction.**
- **4. \( K_p \) is \( 2.55 \, \text{atm}^3 \).**