A reaction `S_(8)(g) Leftrightarrow 4S_(2)(g)` is carried out by taking 2 mol of `S_(g(g))` and 0.2 mol of `S_(2(g))` is a reaction vessel of 1 L and `K=6.30 xx 10^(-6)` then
(a) Reaction qutient is `8 xx 10^(-4)`
b) Reaction proceeds in backward direction
c) Reaction proceeds in forward direction
The correct options are

Options

A. a,b
B. b,c
C. a,c
D. All

Answer

To solve the problem, we need to analyze the reaction and calculate the reaction quotient (Q) and compare it with the equilibrium constant (K) to determine the direction of the reaction.

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The reaction given is:
\[ S_8(g) \leftrightarrow 4S_2(g) \]

2. **Identify Initial Moles:**
We are given:
- Moles of \( S_8 \) = 2 mol
- Moles of \( S_2 \) = 0.2 mol

3. **Calculate Concentrations:**
Since the volume of the reaction vessel is 1 L, the concentrations of the reactants and products can be calculated as follows:
- Concentration of \( S_8 \) = \(\frac{2 \text{ mol}}{1 \text{ L}} = 2 \text{ M}\)
- Concentration of \( S_2 \) = \(\frac{0.2 \text{ mol}}{1 \text{ L}} = 0.2 \text{ M}\)

4. **Calculate the Reaction Quotient (Q):**
The reaction quotient \( Q \) is calculated using the formula:
\[ Q = \frac{[S_2]^4}{[S_8]} \]
Substituting the values we calculated:
\[ Q = \frac{(0.2)^4}{(2)} = \frac{0.0016}{2} = 0.0008 = 8 \times 10^{-4} \]

5. **Compare Q with K:**
The equilibrium constant \( K \) is given as:
\[ K = 6.30 \times 10^{-6} \]
Now we compare \( Q \) and \( K \):
- \( Q = 8 \times 10^{-4} \)
- \( K = 6.30 \times 10^{-6} \)

Since \( Q > K \), according to Le Chatelier's principle, the reaction will shift to the left (backward direction) to reach equilibrium.

6. **Conclusion:**
- (a) The reaction quotient is \( 8 \times 10^{-4} \) (Correct)
- (b) The reaction proceeds in the backward direction (Correct)
- (c) The reaction proceeds in the forward direction (Incorrect)

### Final Answer:
The correct options are (a) and (b).

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