When sulphur ( in the form of `S_(8)` is heated at temperature `T`, at equilibrium , the pressure of `S_(8)` falls by `30%` from `1.0 atm`, because `S_(8)(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

Answer

To find the value of \( K_p \) for the reaction where \( S_8(g) \) is partially converted into \( S_2(g) \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced reaction for the conversion of sulfur is:
\[ S_8(g) \rightleftharpoons 4 S_2(g) \]

### Step 2: Determine the initial and equilibrium pressures
Initially, the pressure of \( S_8 \) is given as \( 1.0 \, \text{atm} \). Since the pressure of \( S_8 \) falls by \( 30\% \), we can calculate the equilibrium pressure of \( S_8 \):
\[ \text{Pressure of } S_8 \text{ at equilibrium} = 1.0 \, \text{atm} - 0.3 \, \text{atm} = 0.7 \, \text{atm} \]

### Step 3: Calculate the change in pressure of \( S_8 \)
The change in pressure of \( S_8 \) is:
\[ \Delta P_{S_8} = 1.0 \, \text{atm} - 0.7 \, \text{atm} = 0.3 \, \text{atm} \]

### Step 4: Determine the pressure of \( S_2 \) at equilibrium
According to the stoichiometry of the reaction, for every \( 1 \, \text{atm} \) of \( S_8 \) that reacts, \( 4 \, \text{atm} \) of \( S_2 \) are produced. Therefore, the pressure of \( S_2 \) at equilibrium can be calculated as:
\[ \text{Pressure of } S_2 = 4 \times \Delta P_{S_8} = 4 \times 0.3 \, \text{atm} = 1.2 \, \text{atm} \]

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{S_2})^4}{(P_{S_8})} \]

### Step 6: Substitute the equilibrium pressures into the \( K_p \) expression
Substituting the values we found:
\[ K_p = \frac{(1.2 \, \text{atm})^4}{(0.7 \, \text{atm})} \]

### Step 7: Calculate \( K_p \)
Calculating \( (1.2)^4 \):
\[ (1.2)^4 = 2.0736 \]
Now substituting this back into the expression for \( K_p \):
\[ K_p = \frac{2.0736 \, \text{atm}^4}{0.7 \, \text{atm}} = 2.0736 \div 0.7 \approx 2.96 \, \text{atm}^{3} \]

### Final Answer
Thus, the value of \( K_p \) for the reaction is:
\[ K_p \approx 2.96 \, \text{atm}^{3} \]

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