Given : C+2H2⟶CH4 ∆fH∘=−76.2kJmol−1......(i) C+O2⟶CO2 ∆fH∘=−349.8kJmol−1.....(ii) H2+
1
2
O2⟶H2O(l)
∆fH∘=−285.82kJmol−1.......(iii)
H2O(l)⟶H2O(g) ∆vH∘=44kJmol−1.....(iv) Oxidation of CH4 into its gaseous products is given as CH4+2O2⟶CO2+2H2O∆H=? To get the required equation, rewrite the above equation. CH4⟶C+2H2 ∆fH∘=+76.2kJmol−1.....(i) C+O2⟶CO2 ∆fH∘=−394.8kJmol−1.....(ii) 2H2+O2⟶2H2O(l) ∆fH∘=2×−285.82kJmol−1.......(iii) 2H2O(l)⟶H2O(g) ∆vH∘=2×44kJmol−1......(iv) given