5.52×10 3kJ
首先依据燃烧热概念写出:
①CH 4(g)+2O 2(g)
CO 2(g)+2H 2O(l)ΔH 1="-890.3" kJ·mol -1,
②H 2(g)+O 2(g)
H 2O(l)ΔH 2="-285.8" kJ·mol -1,
③CO(g)+O 2(g)
CO 2(g)ΔH 3="-283.0" kJ·mol -1;
再写出CH 4(g)+CO 2(g)
2CO(g)+2H 2(g),
①式-②式×2-③式×2即得CH 4(g)+CO 2(g)
2CO(g)+2H 2(g),
所以ΔH=ΔH 1-2ΔH 2-2ΔH 3="(-890.3+285.8×2+283.0×2)" kJ·mol -1="+247.3" kJ·mol -1,
生成1 m 3CO所需热量为×247.3 kJ·mol -1×≈5 520 kJ。