The reaction for the oxidation of sulfur dioxide, SO₂, is shown below - VCE - SSCE Chemistry - Question 8 - 2021 - Paper 1

To calculate the equilibrium constant Kc, we first determine the concentration of SO₃ at equilibrium:

1. Calculate the number of moles of SO₃: Given that 20.0 g of SO₃ is present: $\text{Molecular weight of SO₃} = 32.07 + (3 \times 16.00) = 80.07 \text{ g mol}^{-1}$ $\text{Moles of SO₃} = \frac{20.0 \text{ g}}{80.07 \text{ g mol}^{-1}} \approx 0.2498 \text{ mol}$

2. Calculate the change in moles: From the stoichiometry of the reaction: 2 moles of SO₂ produce 2 moles of SO₃. Therefore, for every mole of SO₃ produced, a mole of SO₂ is consumed. Initial moles:

   • $n(SO₂)_{initial} = 1.00$ mol
   • $n(O₂)_{initial} = 1.00$ mol
   • $n(SO₃)_{initial} = 0.00$ mol

   Changes at equilibrium:

   • $n(SO₂)_{change} = -x$ (where $x$ is moles of SO₂ reacted)
   • $n(O₂)_{change} = -\frac{x}{2}$
   • $n(SO₃)_{change} = +x$

   With $x = 0.25$ mol (since 0.25 mol of SO₃ is produced):

   • $n(SO₂)_{final} = 1.00 - 0.25 = 0.75$ mol
   • $n(O₂)_{final} = 1.00 - 0.125 = 0.875$ mol
   • $n(SO₃)_{final} = 0.25$ mol

3. Convert moles to concentrations (in 3.00 L):

   • $[SO₂] = \frac{0.75}{3.00} = 0.25 \text{ M}$
   • $[O₂] = \frac{0.875}{3.00} \approx 0.292 \text{ M}$
   • $[SO₃] = \frac{0.25}{3.00} \approx 0.083 \text{ M}$

4. Plug values into the equilibrium constant expression: $K_c = \frac{[SO₃]^2}{[SO₂]^2[O₂]}$ $K_c = \frac{(0.083)^2}{(0.25)^2(0.292)} \approx 0.38 \text{ M}^{-1}$