Find the value of `K_(p)` for the reaction : `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`, if the partial pressures of `SO_(2), O_(2)`, and `SO_(3)` are 0.559 atm, 0.101 atm and 0.331 atm respectively. What will be the partial pressure of `O_(2)` gas if at equilibrium, equal amounts (in moles) of `SO_(2) and SO_(3)` are observed?

Answer

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

Given the partial pressures:
- \( P_{SO_2} = 0.559 \, \text{atm} \)
- \( P_{O_2} = 0.101 \, \text{atm} \)
- \( P_{SO_3} = 0.331 \, \text{atm} \)

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) for the reaction is given by the formula:

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

### Step 2: Substitute the given values into the \( K_p \) expression

Substituting the values into the equation:

\[ K_p = \frac{(0.331)^2}{(0.559)^2 \cdot (0.101)} \]

### Step 3: Calculate \( K_p \)

Calculating the numerator:

\[ (0.331)^2 = 0.109561 \]

Calculating the denominator:

\[ (0.559)^2 = 0.312481 \quad \text{and} \quad (0.101) = 0.101 \]

Now, multiplying the denominator:

\[ 0.312481 \cdot 0.101 = 0.031558 \]

Now substituting back into the \( K_p \) expression:

\[ K_p = \frac{0.109561}{0.031558} \approx 3.47 \]

### Step 4: Find the partial pressure of \( O_2 \) at equilibrium

Now, we need to find the partial pressure of \( O_2 \) when equal amounts of \( SO_2 \) and \( SO_3 \) are observed at equilibrium. Let’s assume \( x \) moles of \( SO_2 \) react, then:

- The change in \( SO_2 \) will be \( -2x \)
- The change in \( O_2 \) will be \( -x \)
- The change in \( SO_3 \) will be \( +2x \)

At equilibrium, we have:

\[ P_{SO_2} = 0.559 - 2x \]
\[ P_{O_2} = 0.101 - x \]
\[ P_{SO_3} = 0.331 + 2x \]

Since \( P_{SO_2} = P_{SO_3} \) at equilibrium, we can set up the equation:

\[ 0.559 - 2x = 0.331 + 2x \]

### Step 5: Solve for \( x \)

Rearranging gives:

\[ 0.559 - 0.331 = 4x \]
\[ 0.228 = 4x \]
\[ x = 0.057 \]

### Step 6: Calculate the new partial pressure of \( O_2 \)

Substituting \( x \) back into the equation for \( P_{O_2} \):

\[ P_{O_2} = 0.101 - x = 0.101 - 0.057 = 0.044 \, \text{atm} \]

### Final Answers

1. The value of \( K_p \) is approximately **3.47**.
2. The partial pressure of \( O_2 \) at equilibrium is **0.044 atm**.