Chemistry - Thermochemistry Question with Solution | TestHub

Enthalpy change is the standard enthalpy of formation, in any general chemical reaction, the reactants undergo chemical changes and combine to give products. It can be represented by the following equation:

$\mathrm{Reactant}\to \mathrm{Product}$

From the given equations,

$2{\mathrm{Fe}}_{\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{Fe}}_2{\mathrm{O}}_{3( \mathrm{s})},\Delta {\mathrm{H}}^{\mathrm{o}}=-822 \mathrm{kJ}/\mathrm{mol}$ .........(i)

${\mathrm{C}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{CO}}_{\left(\mathrm{g}\right)},\mathrm{\Delta H}°=-110 \mathrm{kJ}/\mathrm{mol}$....(ii)

$3{\mathrm{C}}_{\left(\mathrm{s}\right)}+{\mathrm{Fe}}_2{\mathrm{O}}_{3( \mathrm{s})}\to 2{\mathrm{Fe}}_{\left(\mathrm{s}\right)}+3{\mathrm{CO}}_{\left(\mathrm{g}\right)},{\mathrm{\Delta H}}_3=?$

The value of ${\mathrm{\Delta H}}_3$ can be calculated as,

${\mathrm{\Delta H}}_3=3\times {\mathrm{\Delta H}}_2-{\mathrm{\Delta H}}_1$

$=3(-110)+822$

$=492 \mathrm{kJ}/\mathrm{mole}$