(1)A.1mol N2(g)与适量H2(g)起反应,生成2mol NH3(g),放出92.2kJ热量的热化学方程式为:N2(g)+3H2(g)?2NH3(g)△H=-92.2kJ/mol,故答案为:N2(g)+3H2(g)?2NH3(g)△H=-92.2kJ/mol;
B.1mol N2(g)与适量O2(g)起反应,生成2mol NO2(g),吸收68kJ热量的热化学方程式为N2(g)+2O2(g)=2NO2(g)△H=+68kJ/mol,故答案为:N2(g)+2O2(g)=2NO2(g)△H=+68kJ/mol;
(2)因燃烧热在某一温度和压强下lmol某物质B完全燃烧生成稳定化合物时所释放出的热量,而H2O的稳定化合物是液态水,H2O蒸汽转变为液态水需要放热,所以CH3OH的燃烧热大于192.9kJ/mol;
CH3OH(g)+H2O(g)=CO2(g)+3H2(g)△H=+49.0kJ/mol ①
CH3OH(g)+3/2O2(g)=CO2(g)+2H2O(g)△H=-192.9kJ/mol ②
②-①得 3H2(g)+3/2O2(g)=3H2O(g)△H=-241.9kJ/mol ③
H2O(l)=H2O(g)△H=+44kJ/mol ④
③-④×3 得 3H2(g)+3/2O2(g)=3H2O(g)△H=-373.9kJ/mol 即H2(g)+1/2O2(g)=H2O(l)△H=-124.6kJ/mol,
故答案为:大于;H2(g)+1/2O2(g)=H2O(l)△H=-124.6kJ/mol.