The equation for the reaction between sulfur dioxide and oxygen is shown. $$2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)$$ In an experiment, 2.00 mol of sulfur dioxide are mixed with 2.00 mol of oxygen. The total amount of the three gases at equilibrium is 3.40 mol. What is the mole fraction of sulfur trioxide in the equilibrium mixture?

A-Level AQA Chemistry Chemical Equilibria, Le Chateliers Principle & Kc: The equation for the reaction be

The equation for the reaction between sulfur dioxide and oxygen is shown - AQA - A-Level Chemistry - Question 11 - 2019 - Paper 3

Question 11

The equation for the reaction between sulfur dioxide and oxygen is shown. $$2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)$$ In an experiment, 2.00 mol of sul... show full transcript

Worked Solution & Example Answer:The equation for the reaction between sulfur dioxide and oxygen is shown - AQA - A-Level Chemistry - Question 11 - 2019 - Paper 3

Step 1

Calculate the Change in Moles

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Answer

Initially, we have 2.00 mol of sulfur dioxide and 2.00 mol of oxygen. The balanced equation indicates that 2 moles of sulfur dioxide react with 1 mole of oxygen to produce 2 moles of sulfur trioxide.

Let 'x' be the amount of sulfur dioxide reacted. The stoichiometric ratios tell us that:

2 moles of SO₂ produce 2 moles of SO₃, so the amount of SO₃ will be 'x'.
The amount of O₂ consumed will be 'x/2' (since 1 mol of O₂ reacts with 2 mol of SO₂).

At equilibrium, we know:

Total amount of gases = 3.40 mol

Let’s represent the amounts at equilibrium:

Moles of SO₂ left = 2.00 - x
Moles of O₂ left = 2.00 - x/2
Moles of SO₃ produced = x

Step 2

Write the Total Mole Equation

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Answer

Now we write the total in terms of x:

$(2.00 - x) + (2.00 - \frac{x}{2}) + x = 3.40$

Simplifying this:

$4.00 - x + x - \frac{x}{2} = 3.40$

This simplifies to:

$4.00 - \frac{x}{2} = 3.40$

Isolating x gives:

$\frac{x}{2} = 0.60 \implies x = 1.20$

Step 3

Determine Moles of SO₃

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Answer

From our calculation, the moles of sulfur trioxide at equilibrium = 1.20 mol.

Now we can find the mole fraction of SO₃:

$\text{Mole fraction of } SO₃ = \frac{\text{Moles of } SO₃}{\text{Total moles}} = \frac{1.20}{3.40}$

Step 4

Calculate the Mole Fraction

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Answer

Calculating this gives:

$\text{Mole fraction of } SO₃ = \frac{1.20}{3.40} \approx 0.353$

Thus, the mole fraction of sulfur trioxide in the equilibrium mixture is approximately 0.353.

The equation for the reaction between sulfur dioxide and oxygen is shown - AQA - A-Level Chemistry - Question 11 - 2019 - Paper 3

Worked Solution & Example Answer:The equation for the reaction between sulfur dioxide and oxygen is shown - AQA - A-Level Chemistry - Question 11 - 2019 - Paper 3

Calculate the Change in Moles

Write the Total Mole Equation

Determine Moles of SO₃

Calculate the Mole Fraction

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