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This question is about the equilibrium 2 SO₂(g) + O₂(g) ⇌ 2 SO(g) Effect Explanation A 0.460 mol sample of SO₂ is mixed with a 0.250 mol sample of O₂ in a sealed container at a constant temperature - AQA - A-Level Chemistry - Question 5 - 2021 - Paper 1

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This question is about the equilibrium 2 SO₂(g) + O₂(g) ⇌ 2 SO(g) Effect Explanation A 0.460 mol sample of SO₂ is mixed with a 0.250 mol sample of O₂ in a sealed... show full transcript

Worked Solution & Example Answer:This question is about the equilibrium 2 SO₂(g) + O₂(g) ⇌ 2 SO(g) Effect Explanation A 0.460 mol sample of SO₂ is mixed with a 0.250 mol sample of O₂ in a sealed container at a constant temperature - AQA - A-Level Chemistry - Question 5 - 2021 - Paper 1

Step 1

Effect

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Answer

A decrease in overall pressure will decrease the yield of SO.

Step 2

Explanation

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Answer

According to Le Chatelier's Principle, a decrease in pressure shifts the equilibrium to the side with more moles of gas. In this case, the reaction produces 2 moles of SO for every 3 moles of reactants (2 SO₂ + O₂), so lowering the pressure shifts the equilibrium towards the reactants, which decreases the yield of SO.

Step 3

Calculate the partial pressure of SO₂ in this equilibrium mixture

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Answer

To find the partial pressure of SO₂, we first need to determine the moles of all gases at equilibrium. We have:

  • Total moles = 0.460 mol (SO₂) + 0.250 mol (O₂) - 0.180 mol (SO₂ at equilibrium)
  • Moles of SO at equilibrium = 0.180 mol

Using the ideal gas equation and knowing that the pressure is 215 kPa:

The total moles at equilibrium = (0.460 - 0.180) + 0.250 = 0.530 mol

Partial pressure of SO₂ can be calculated from:

P(SO₂) = rac{n(SO₂)}{n_{total}} imes P_{total}

where:

  • n(SO₂) = 0.180 mol
  • n(total) = 0.530 mol (assuming negligible other changes)

Thus, P(SO₂) = rac{0.180}{0.530} imes 215 ext{ kPa} = 61.1 ext{ kPa}

Step 4

Give an expression for the equilibrium constant (Kc) for this reaction

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Answer

The expression for the equilibrium constant (Kc) for the reaction is given by:

K_c = rac{[SO]^2}{[SO_2]^2[O_2]}

Step 5

Calculate the value of the equilibrium constant for this reaction and give its units

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Answer

From the equilibrium partial pressures given:

  • [SO][SO] = 1.85 x 10² kPa
  • [SO2][SO₂] = 1.67 x 10² kPa
  • [O2][O₂] = 1.02 x 10² kPa

Substituting back into the Kc expression gives:

K_c = rac{(1.85 imes 10^2)^2}{(1.67 imes 10^2)^2 imes (1.02 imes 10^2)}

Calculating this yields:

Kc=7.89K_c = 7.89

Units of Kc = kPa2^{-2}

Step 6

What is the effect on the value of Kc if the pressure of this equilibrium mixture is increased at a constant temperature?

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Answer

The value of Kc stays the same because Kc is only affected by temperature changes within the equilibrium system.

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