已知下列热化学方程式:①Fe2O3(s)+3CO(g)═2Fe(s)+3CO2(g)△H1=-26.7kJ?mol-1②3Fe2O3(s)+CO

2025-04-06 07:20:30
推荐回答(2个)
回答1:

由①×3-②-③×2得:6FeO(s)+6CO(g)═6Fe(s)+6CO2(g) ④
再由④÷6得:FeO(s)+CO(g)═Fe(s)+CO2(g)
故△H=

3△H1?△H2?2△H3
6
=
3×(?26.7kJ?mol?1)?(?50.75kJ?mol?1)?2×(?36.5kJ?mol?1)
6
=+7.28 kJ?mol-1
故选A.

回答2:

(1)*3-(2):2Fe3O4
+
8CO
===
6Fe
+
8CO2
(1)*3-(2)-(3)*2:6FeO
+
6CO
===
6Fe
+
6CO2
FeO(s)
+
CO(g)
===
Fe(s)
+
CO2(g)的焓变为:[(-27.6)*3-(-50.75)-(-36.5)*2]/6=+6.825