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

For gaseous reactions, work done (w) at constant temperature is given by:

w = −Δn RT

Where,

Δn = Change in moles of gaseous products − moles of gaseous reactants

R = 8.314 J K⁻¹ mol⁻¹

T = 298 K

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

For 1 mole of SO₂:

Reactant moles = `1 + 1/2 = 3/2`

Product moles = 1

Δn = `1 - 3/2`

= `-1/2`

For 2 moles of SO₂:

Δn = `2 xx (-1/2)`

= −1

Work done (w) = −Δn RT

= −(−1) × 8.314 × 298

= 8.314 × 298

= 2477.6 J

= +2.48 × 10³ J