JEE Main 2017 — Thermodynamics (C) Question with Solution

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.

The formation reaction of methane is 

$\mathrm{C}\left(\mathrm{graphite}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)$

The heat of combustion data for $\mathrm{C}\left(\mathrm{graphite}\right), {\mathrm{H}}_2\left(\mathrm{g}\right) \text{and} {\mathrm{CH}}_4\left(\mathrm{g}\right)$

$∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}=\sum _{}∆{\mathrm{H}}_{\mathrm{Combustion}}\left(\mathrm{reactants}\right)-\underset{}{\sum ∆{\mathrm{H}}_{\mathrm{Combustion}}\left(\mathrm{products}\right)}$$∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{CH}}_4\right)=∆{\mathrm{H}}_{\mathrm{C}}^{\mathrm{o}}\left(\mathrm{C}\left(\mathrm{graphite}\right)\right)+2\times ∆{\mathrm{H}}_{\mathrm{C}}^{\mathrm{o}}\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)-∆{\mathrm{H}}_{\mathrm{C}}^{\mathrm{o}}\left({\mathrm{CH}}_4\left(\mathrm{g}\right)\right)$

$∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{CH}}_4\right)=-393.5-2\times 285.8+890.3$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\mathrm{o}}$ of ${\mathrm{CH}}_{4 }\left(\mathrm{g}\right)=-74.8 \mathrm{kJ}/\mathrm{mol}$