Q.5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol-1  –393.5 kJ mol-1, and –285.8 kJ mol-1 respectively. Enthalpy of formation of CH4g will be
(i) –74.8 kJ mol-1

(ii) –52.27 kJ mol-1

(iii) +74.8 kJ mol-1

(iv) +52.26 kJ mol-1

NEETprep Answer:
According to the question,
iCH4g+2O2gCO2g+2H2Og
H=-890.3kJmol-1
iiCs+O2gCO2g
H=-393.5kJmol-1
iii2H2g+O2g2H2Og
H=-285.8kJmol-1
Thus, the desired equation is the one that represents the formation of CH4g i.e.,
Cs+2H2gCH4gfHCH4=cHc+2cHH2-cHCO2=-393.5+2-285.8--890.3kJmol-1=-74.8kJmol-1
Enthalpy of formation of CH4g=-74.8kJmol-1 Hence, alternative (i) is correct.