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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. $$ ext{2SO}_2(g) + ext{O}_2(g) \rightleftharpoons \text{2SO}_3(g)$$ In an experiment, 2.... 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 of the reaction

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Answer

Let the change in moles of SO2 that reacts be denoted as x. The balanced equation shows that for every 2 moles of SO2 reacted, 2 moles of SO3 are produced.

Starting with:

  • SO2: 2.00 mol
  • O2: 2.00 mol
  • SO3: 0 mol

At equilibrium, we have:

  • SO2: 2.00 - x mol
  • O2: 2.00 - (0.5)x mol (since 1 mole of O2 reacts with 2 moles of SO2)
  • SO3: 0 + x mol (it produces the same amount as consumed)

The total moles at equilibrium is: extTotal=(2x)+(20.5x)+x=3.40 ext{Total} = (2 - x) + (2 - 0.5x) + x = 3.40

Simplifying this: extTotal=40.5x=3.40 ext{Total} = 4 - 0.5x = 3.40

Solving for x, we find:

ightarrow x = 1.20$$

Step 2

Determine the equilibrium moles of sulfur trioxide

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Answer

Using the value of x found, the moles of SO3 at equilibrium is: extSO3=x=1.20 extmol ext{SO3} = x = 1.20\ ext{mol}

Step 3

Calculate the total moles at equilibrium

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Answer

The total moles at equilibrium is 3.40 mol (given).

Step 4

Determine the mole fraction of sulfur trioxide

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Answer

The mole fraction (X) of a component is calculated as: X=moles of componenttotal molesX = \frac{\text{moles of component}}{\text{total moles}}

For sulfur trioxide: XSO3=1.203.400.353X_{SO3} = \frac{1.20}{3.40} \approx 0.353

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