Question

Calculate the work done in oxidation of 2 moles of SO₂ at 298 K, if \[\ce{SO2_{(g)} + \frac{1}{2}O2_{(g)} -> SO3_{(g)}}\]

[Given: R = 8.314 J/K/mol]

Answer 1

The given equation for the oxidation of 1 mole of SO₂ is:

\[\ce{SO2_{(g)} + \frac{1}{2}O2_{(g)} -> SO3_{(g)}}\]

∴ For 2 moles, the balanced equation becomes,

\[\ce{2SO2_{(g)} + O2_{(g)} -> 2SO3_{(g)}}\]

Δn_g = (moles of product gases) − (moles of reactant gases)

= 2 − 3

= −1

Work done (w) = −Δn_g RT

= −(−1) × 8.314 × 298

= 8.314 × 298

= 2477.6 J

= +2.478 kJ