When sulphur ( in the form of `S_(B))` is heated at temperature `T`, at equilibrium , the pressure of `S_(B)` falls by `30%` from `1.0 atm`, because `S_(B)(g)` in partially converted into `S_(2)(g)`.
Find the value of `K_(P)` for this reaction.

Options

A. `2.96`
B. `6.14`
C. `204.8`
D. None of these

Explanation

To solve the problem, we need to find the equilibrium constant \( K_P \) for the reaction where sulfur in the form of \( S_B \) (gaseous sulfur) is partially converted into \( S_2 \) (gaseous sulfur).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial pressure of \( S_B \) = 1.0 atm
- Initial pressure of \( S_2 \) = 0 atm

2. **Determine Change in Pressure:**
- The pressure of \( S_B \) falls by 30%.
- Therefore, the change in pressure of \( S_B \) = 30% of 1.0 atm = 0.3 atm.
- Thus, the pressure of \( S_B \) at equilibrium = \( 1.0 \, \text{atm} - 0.3 \, \text{atm} = 0.7 \, \text{atm} \).

3. **Calculate Pressure of \( S_2 \):**
- Since \( S_B \) is converted into \( S_2 \), the increase in pressure of \( S_2 \) will be equal to the decrease in pressure of \( S_B \).
- Therefore, the pressure of \( S_2 \) at equilibrium = 0.3 atm.

4. **Write the Equilibrium Expression:**
- The reaction can be represented as:
\[ 2 S_B(g) \rightleftharpoons S_2(g) \]
- The equilibrium constant \( K_P \) expression for this reaction is:
\[ K_P = \frac{P_{S_2}}{(P_{S_B})^2} \]

5. **Substitute the Values into the Expression:**
- Substitute \( P_{S_2} = 0.3 \, \text{atm} \) and \( P_{S_B} = 0.7 \, \text{atm} \):
\[ K_P = \frac{0.3}{(0.7)^2} \]

6. **Calculate \( K_P \):**
- Calculate \( (0.7)^2 = 0.49 \).
- Therefore,
\[ K_P = \frac{0.3}{0.49} \approx 0.6122 \]

### Final Answer:
The value of \( K_P \) for the reaction is approximately \( 0.6122 \, \text{atm}^{-1} \).