The enthalpy of combustion of methane, graphite and dihydrogen at 298
Question:

The enthalpy of combustion of methane, graphite and dihydrogen at $298 \mathrm{~K}$ are, $-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$, and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Enthalpy of formation of $\mathrm{CH}_{4(g)}$ will be

(i) $-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $-52.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $+74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iv) $+52.26 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:

According to the question,

Thus, the desired equation is the one that represents the formation of CH4 (g) i.e.,

$\therefore$ Enthalpy of formation of $\mathrm{CH}_{4(g)}=-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Hence, alternative (i) is correct.