Calculate enthalpy of formation of methane (CH4) from the following data:
(i) C(s) + O2(g) → CO2(g), ΔrH° = −393.5 kJ mol−1
(ii) H2(g) + 1/2O2(g) → H2O(l), ΔrH° = −285.8 kJ mol−1
(iii) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔrH° = −890.3 kJ mol−1
We aim at : C(s) + 2H2(g) → CH4(g); ΔfH° = ?
Multiplying equation (ii) with 2, adding to equation (i) and then subtracting equation (iii) from the sum, i.e., operating equation (i) + 2 × equation (ii) − equation (iii), we get
C(s) + 2H2(g) − CH4(g) → 0; ΔrH° = −393.5 + 2(−285.8) − (−890.3) = −74.8 kJ mol−1
or C(s) + 2H2(g) → CH4(g); ΔfH° = −74.8 kJ mol−1
Hence, enthalpy of formation of methane is : ΔfH° = − 74.8 kJ mol−1