For the equilibrium `2SO_(2)+O_(2)hArr 2SO_(3)` we start with 2 moles of `SO_(2)` and 1 mole of `O_(2)` at 3 atm. When equilibrium is attained, pressure changes to 2.5atm. Hence, `K_(p)` is `:`

Options

A. `3 atm^(-1)`
B. `2.5 atm^(-1)`
C. `2 atm ^(-1)`
D. `0.5 atm^(-1)`

Explanation

To find the equilibrium constant \( K_p \) for the reaction

\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]

we start with the following information:

- Initial moles of \( SO_2 \) = 2 moles
- Initial moles of \( O_2 \) = 1 mole
- Initial total pressure = 3 atm
- Equilibrium pressure = 2.5 atm

### Step 1: Determine the change in moles at equilibrium

Let \( x \) be the amount of \( SO_2 \) that reacts at equilibrium. According to the stoichiometry of the reaction:

- Moles of \( SO_2 \) at equilibrium = \( 2 - 2x \)
- Moles of \( O_2 \) at equilibrium = \( 1 - x \)
- Moles of \( SO_3 \) at equilibrium = \( 2x \)

### Step 2: Calculate total moles at equilibrium

The total moles at equilibrium can be expressed as:

\[ \text{Total moles} = (2 - 2x) + (1 - x) + (2x) = 3 - x \]

### Step 3: Relate the total pressure to the moles at equilibrium

Using the ideal gas law, we know that pressure is directly proportional to the number of moles. Therefore, we can set up the following relationship:

\[ \frac{P_{eq}}{P_{initial}} = \frac{n_{eq}}{n_{initial}} \]

Where:
- \( P_{eq} = 2.5 \, \text{atm} \)
- \( P_{initial} = 3 \, \text{atm} \)
- \( n_{initial} = 3 \) moles (2 moles of \( SO_2 \) + 1 mole of \( O_2 \))

Thus,

\[ \frac{2.5}{3} = \frac{3 - x}{3} \]

### Step 4: Solve for \( x \)

Cross-multiplying gives:

\[ 2.5 \times 3 = 3(3 - x) \]

\[ 7.5 = 9 - 3x \]

Rearranging gives:

\[ 3x = 9 - 7.5 \]

\[ 3x = 1.5 \implies x = 0.5 \]

### Step 5: Calculate moles at equilibrium

Now substituting \( x \) back into the expressions for moles at equilibrium:

- Moles of \( SO_2 \) at equilibrium = \( 2 - 2(0.5) = 1 \)
- Moles of \( O_2 \) at equilibrium = \( 1 - 0.5 = 0.5 \)
- Moles of \( SO_3 \) at equilibrium = \( 2(0.5) = 1 \)

### Step 6: Calculate partial pressures

Using the total pressure at equilibrium (2.5 atm), we can find the partial pressures:

- Partial pressure of \( SO_2 \):

\[ P_{SO_2} = P_{total} \times \frac{n_{SO_2}}{n_{total}} = 2.5 \times \frac{1}{2.5} = 1 \, \text{atm} \]

- Partial pressure of \( O_2 \):

\[ P_{O_2} = P_{total} \times \frac{n_{O_2}}{n_{total}} = 2.5 \times \frac{0.5}{2.5} = 0.5 \, \text{atm} \]

- Partial pressure of \( SO_3 \):

\[ P_{SO_3} = P_{total} \times \frac{n_{SO_3}}{n_{total}} = 2.5 \times \frac{1}{2.5} = 1 \, \text{atm} \]

### Step 7: Calculate \( K_p \)

The expression for \( K_p \) is given by:

\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

Substituting the values:

\[ K_p = \frac{(1)^2}{(1)^2 \cdot (0.5)} = \frac{1}{0.5} = 2 \]

Thus, the equilibrium constant \( K_p \) is:

\[ \boxed{2} \]