5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equilibrium , it was foumnd that `60%` of `SO_(2)` is used up .If the pressure of the equilibrium mixture is one aatmosphere, the parital pressure of `O_(2)` is :

Options

A. `0.52 atm`
B. `0.21atm`
C. `0.41 atm`
D. `0.82 atm`

Answer

To solve the problem step by step, we will follow the reaction and the changes in moles of the reactants and products at equilibrium.

### Step 1: Write the balanced chemical equation.
The reaction between sulfur dioxide (SO₂) and oxygen (O₂) to form sulfur trioxide (SO₃) can be represented as:
\[ 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \]

### Step 2: Determine initial moles of reactants.
Initially, we have:
- Moles of SO₂ = 5 moles
- Moles of O₂ = 5 moles
- Moles of SO₃ = 0 moles

### Step 3: Calculate the amount of SO₂ used.
It is given that 60% of SO₂ is used up at equilibrium. Therefore, the amount of SO₂ used is:
\[ \text{Amount of SO}_2 \text{ used} = 60\% \text{ of } 5 = 0.6 \times 5 = 3 \text{ moles} \]

### Step 4: Calculate the change in moles of O₂ and SO₃.
From the balanced equation, for every 2 moles of SO₂ that react, 1 mole of O₂ reacts and 2 moles of SO₃ are formed.

Since 3 moles of SO₂ are used:
- Moles of O₂ reacted = \( \frac{3}{2} = 1.5 \) moles
- Moles of SO₃ formed = \( 3 \) moles (since 2 moles of SO₂ produce 2 moles of SO₃)

### Step 5: Calculate moles at equilibrium.
At equilibrium, the moles of each substance are:
- Moles of SO₂ remaining = \( 5 - 3 = 2 \) moles
- Moles of O₂ remaining = \( 5 - 1.5 = 3.5 \) moles
- Moles of SO₃ formed = \( 3 \) moles

### Step 6: Calculate the total moles at equilibrium.
Total moles at equilibrium = Moles of SO₂ + Moles of O₂ + Moles of SO₃
\[ = 2 + 3.5 + 3 = 8.5 \text{ moles} \]

### Step 7: Calculate the mole fraction of O₂.
The mole fraction of O₂ is given by:
\[ \text{Mole fraction of O}_2 = \frac{\text{Moles of O}_2}{\text{Total moles}} = \frac{3.5}{8.5} \]

### Step 8: Calculate the partial pressure of O₂.
The partial pressure of a gas can be calculated using the formula:
\[ \text{Partial pressure of O}_2 = \text{Mole fraction of O}_2 \times \text{Total pressure} \]
Given that the total pressure at equilibrium is 1 atmosphere:
\[ \text{Partial pressure of O}_2 = \frac{3.5}{8.5} \times 1 \]

### Step 9: Perform the calculation.
Calculating the mole fraction:
\[ \frac{3.5}{8.5} \approx 0.4118 \]
Thus, the partial pressure of O₂ is approximately:
\[ \text{Partial pressure of O}_2 \approx 0.4118 \text{ atm} \]

### Final Answer:
The partial pressure of O₂ at equilibrium is approximately **0.41 atm**.

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