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

Using the stoichiometry of the reaction:

Initial moles:

• SO2 = 1.00 mol
• O2 = 1.00 mol

From the reaction, we see that for every 2 moles of SO2 reacting, 1 mole of O2 and 2 moles of SO3 are produced.

We denote changes as follows:

• Change in SO2 = -2x
• Change in O2 = -x
• Change in SO3 = +2x

We know that 0.25 mol of SO3 corresponds to x = 0.125 mol. Therefore, changes for SO2 and O2 are:

• Change in SO2 = -2(0.125) = -0.25 mol
• Change in O2 = -0.125 mol.

Calculating the final moles:

• Final moles of SO2 = 1.00 - 0.25 = 0.75 mol
• Final moles of O2 = 1.00 - 0.125 = 0.875 mol
• Final moles of SO3 = 0.25 mol

We can now calculate the equilibrium concentrations:

$[SO2]_{eq} = \frac{0.75 \, \text{mol}}{3.00 \, \text{L}} = 0.25 \, \text{mol/L}$
$[O2]_{eq} = \frac{0.875 \, \text{mol}}{3.00 \, \text{L}} \approx 0.292 \, \text{mol/L}$
$[SO3]_{eq} = \frac{0.25 \, \text{mol}}{3.00 \, \text{L}} \approx 0.083 \, \text{mol/L}$

Now, we can calculate the equilibrium constant Kc using the equilibrium concentrations:

$K_c = \frac{[SO3]^2}{[SO2]^2[O2]}$

Substituting the values:

$K_c = \frac{(0.083)^2}{(0.25)^2 \times 0.292} \approx 0.38 \, \text{M}^{-1}$

On the graph provided, the Maxwell-Boltzmann distribution curve for SO3 at a significantly lower temperature should be sketched.

• The new peak should be to the left of the original peak, indicating lower average kinetic energy.
• The new peak should be higher than the original peak, representing an increase in the number of particles at lower energies.

Make sure to label the axes clearly: 'Number of particles' on the y-axis and 'Kinetic energy' on the x-axis.