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 \( K_p \) is defined as:

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

### 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:

\[ (0.559)^2 = 0.312481 \quad \text{and} \quad (0.312481) \cdot (0.101) = 0.031563 \]

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

\[ K_p = \frac{0.109561}{0.031563} \approx 3.472 \]

### Step 4: Find the partial pressure of \( O_2 \) at equilibrium when \( SO_2 \) and \( SO_3 \) are equal

Let \( x \) be the change in moles at equilibrium. Since equal amounts of \( SO_2 \) and \( SO_3 \) are observed, we can set:

\[ P_{SO_2} = P_{SO_3} \]

At equilibrium, we have:

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

Setting them equal:

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

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

Rearranging gives:

\[ 0.559 - 0.331 = 4x \]
\[ 0.228 = 4x \quad \Rightarrow \quad x = \frac{0.228}{4} = 0.057 \]

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

Now substituting \( x \) back into the expression for \( P_{O_2} \):

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

### Final Results

Thus, the equilibrium constant \( K_p \) is approximately \( 3.472 \) and the partial pressure of \( O_2 \) at equilibrium is \( 0.044 \, \text{atm} \).

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