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

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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:

(2.00x)+(2.00x2)+x=3.40(2.00 - x) + (2.00 - \frac{x}{2}) + x = 3.40

Simplifying this:

4.00x+xx2=3.404.00 - x + x - \frac{x}{2} = 3.40

This simplifies to:

4.00x2=3.404.00 - \frac{x}{2} = 3.40

Isolating x gives:

x2=0.60    x=1.20\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:

Mole fraction of SO3=Moles of SO3Total moles=1.203.40\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:

Mole fraction of SO3=1.203.400.353\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.

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