H2(g)+1/202(g)→H20(g) △Hm°(298.15)=-241.82KJ.mol→△fHm C(s+02(g)→C02(g)△Hm°(g,C02)=-393.5 2H2(g)+O2(g)-→2H20(g) Hm°(298.15)=-483.64kJ.mol1 H2(g)+1/202(g)-→2H20(I) ΛHm°(L,298.15)=-285.83kJ.mol H2 (g)+1/2 O2(g) → H2O (g) Hm  (298.15)=-241.82 KJ.mol-1→ fHm  C (s)+O2(g) → CO2 (g) fHm  (g,CO2)=-393.5 2H2 (g) +O2(g)→ 2H2O (g) Hm  (298.15)=-483.64 kJ.mol-1 H2 (g) +1/2O2(g)→ 2H2O (l) fHm  (l,298.15)=-285.83 kJ.mol-1