(1)A.1mol N 2(g)与适量H 2(g)起反应,生成2mol NH 3(g),放出92.2kJ热量的热化学方程式为:N 2(g)+3H 2(g)⇌2NH 3(g)△H=-92.2kJ/mol,故答案为:N 2(g)+3H 2(g)⇌2NH 3(g)△H=-92.2kJ/mol;
B.1mol N 2(g)与适量O 2(g)起反应,生成2mol NO 2(g),吸收68kJ热量的热化学方程式为N 2(g)+2O 2(g)=2NO 2(g)△H=+68kJ/mol,故答案为:N 2(g)+2O 2(g)=2NO 2(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H 2O的稳定化合物是液态水,H 2O蒸汽转变为液态水需要放热,所以CH 3OH的燃烧热大于192.9kJ/mol;
CH 3OH(g)+H 2O(g)=CO 2(g)+3H 2(g)△H=+49.0kJ/mol①
CH 3OH(g)+3/2O 2(g)=CO 2(g)+2H 2O(g)△H=-192.9kJ/mol ②
②-①得3H 2(g)+3/2O 2(g)=3H 2O(g)△H=-241.9kJ/mol③
H 2O(l)=H 2O(g)△H=+44kJ/mol④
③-④×3 得 3H 2(g)+3/2O 2(g)=3H 2O(g)△H=-373.9kJ/mol 即H 2(g)+1/2O 2(g)=H 2O(l)△H=-124.6kJ/mol,
故答案为:大于;H 2(g)+1/2O 2(g)=H 2O(l)△H=-124.6kJ/mol.