(1)化学反应过程中一定伴随着能量的变化,反应焓变主要是指反应前后的热量变化;对于化学反应A+B=C+D,若H(A)+H(B)>H(C)+H(D):△H=H(终态)-H(始态),判断可知△H<0;反应是放热反应; 故答案为:热能;小于; 放热; (2)已知:H 2 (g)+Cl 2 (g)=2HCl(g)△H=-185kJ?mol -1 ,△H(H 2 )=436kJ?mol -1 ,△H(Cl 2 )=247kJ?mol -1 ;依据△H=H(终态)-H(始态), △H=2△H(HCl)-△H(H 2 )-△H(Cl 2 )=-185kJ?mol -1 ;则△H(HCl)=434 kJ?mol -1 ;故答案为:434 kJ?mol -1 ; (3)依据盖斯定律 ①Fe 2 O 3 (s)+3CO(g)=2Fe(s)+3CO 2 (g)△H=-25kJ?mol -1 ②3Fe 2 O 3 (s)+CO(g)=2Fe 3 O 4 (s)+CO 2 (g)△H=-47kJ?mol -1 ③Fe 3 O 4 (s)+CO(g)=3FeO(s)+CO 2 (g)△H=19kJ?mol -1 ①×3-③×2+②得到:6FeO(s)+6CO(g)═6Fe(s)+6CO 2 (g)△H=-66 kJ?mol -1 得到热化学方程式为:FeO(s)+CO(g)═Fe(s)+CO 2 (g)△H=-11 kJ?mol -1 故答案为:FeO(s)+CO(g)═Fe(s)+CO 2 (g)△H=-11 kJ?mol -1 |