解 CHOH()+1/202(g)-HCHO(g)+H2O()?? HCH0g)+02g)=C02(g)+H200 △H81=-563.58k.mo1 CH30H0+3/202(g)=C02(g)+2H20(0 入H9 =-776 64kI.mol-1 得： CH,OH(0+1/202g)=HCHO()+H20(0 4,H9m=-726.64-(-563.58)=-163.06kJ-mo1 解 CH3OH(l) +1/2O2 (g) → HCHO(g) + H2O(l) ??? HCHO(g) + O2 (g) = CO2 (g) + H2O(l) rH 1 = −563.58kJmol-1 CH3OH(l) +3/2O2 (g) =CO2 (g) +2H2O (l) rH 2 = −726.64 kJmol-1 - 得 ： CH3OH(l) + 1/2O2 (g) = HCHO(g) + H2O (l) rH m = −726.64 − (−563.58) = −163.06 kJmol-1