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

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.

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}

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

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

From the equilibrium partial pressures given:

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

$K_c = 7.89$

Units of Kc = kPa$^{-2}$

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