What is the work done during oxidation of 4 moles of \(\mathrm{SO}_{2(\mathrm{~g})}\) to \(\mathrm{SO}_{3(\mathrm{~g})}\) at \(27^{\circ} \mathrm{C}\) ? \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)

For oxidation of 4 moles of \(\mathrm{SO}_2\), the reaction is given as follows:
\(\begin{aligned}
& 4 \mathrm{SO}_{2(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 4 \mathrm{SO}_{3(\mathrm{~g})} \\
& \Delta \mathrm{n}_{\mathrm{g}}=\text { (moles of product gases) }
\\& \text { - (moles of reactant gases) } \\
& \Delta \mathrm{n}_{\mathrm{g}}=4-6=-2 \mathrm{~mol} \\
& \mathrm{~W}=-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
& =-(-2 \mathrm{~mol}) \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \\
& =+4988.4 \mathrm{~J} \\
& =+4.988 \mathrm{~kJ} \\
&
\end{aligned}\)
[Note: For the given question, none of the provided options is the correct answer.]