When sulphur is heated at 900 K, `S_(8)` is converted to `S_(2)`. What will be the equilibrim constant for the reaction if initial pressure of 1 atm falls by `25%` at equilibrium ?

Options

A. (a) `0.75 atm^(3)`
B. (b) `2.55atm^(3)`
C. (c) `25.0 atm^(3)`
D. (d) `1.33 atm^(3)`

Explanation

To solve the problem step by step, we will analyze the reaction and the changes in pressure to find the equilibrium constant \( K_p \).

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ S_8 \rightleftharpoons 4 S_2 \]

### Step 2: Determine initial and equilibrium pressures
We are given that the initial pressure of \( S_8 \) is 1 atm. The problem states that this pressure falls by 25% at equilibrium.

- Initial pressure \( P_{initial} = 1 \, \text{atm} \)
- Pressure drop = 25% of 1 atm = \( 0.25 \, \text{atm} \)
- Equilibrium pressure \( P_{equilibrium} = P_{initial} - \text{Pressure drop} = 1 - 0.25 = 0.75 \, \text{atm} \)

### Step 3: Calculate the change in pressure for \( S_2 \)
Since 1 mole of \( S_8 \) produces 4 moles of \( S_2 \), the change in pressure for \( S_2 \) can be calculated as follows:

- Change in pressure for \( S_8 \) = -0.25 atm (since it decreases)
- Therefore, the pressure of \( S_2 \) produced will be:
\[ P_{S_2} = 4 \times \text{change in pressure of } S_8 = 4 \times 0.25 = 1 \, \text{atm} \]

### Step 4: Write the expression for the equilibrium constant \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed as:
\[ K_p = \frac{(P_{S_2})^4}{(P_{S_8})} \]

### Step 5: Substitute the equilibrium pressures into the expression
From our calculations:
- \( P_{S_2} = 1 \, \text{atm} \)
- \( P_{S_8} = 0.75 \, \text{atm} \)

Substituting these values into the expression for \( K_p \):
\[ K_p = \frac{(1)^4}{(0.75)} = \frac{1}{0.75} = \frac{4}{3} \approx 1.33 \, \text{atm}^3 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) for the reaction at 900 K is approximately:
\[ K_p \approx 1.33 \, \text{atm}^3 \]

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