例3-4：已知Sn(s)+ Cl2 (g)=SnCl (s) △Hm=-349.8kJmo1 SnCL(s)+Cl(g)=SnCla ( 4,Hn=-195.4kJmo1 求：Sns)+2Cl2g)=SnCl,(0的反应热4，Hm 解：Sn(s)+Cl2(g)=SnCl2(s)4,H (1) SnCl2(s)+Cl(g)=SnCIa (l)4H2 (2) (1)+(2)Sn(s)+2Cl2(g)=SnCl() 4H9 。 4Hm=4,H81+4H2 =-349.8+(-195.4)=-545.2 kJ.mol-1 例3-4 ： 已知 Sn (s) + Cl2 (g) = SnCl2 (s) rH m = −349.8 kJmol-1 SnCl2 (s) + Cl2 (g) = SnCl4 (l) rH m = −195.4 kJmol-1 求: Sn(s)+ 2Cl2 (g) = SnCl4 (l) 的反应热rH m 解：Sn (s) + Cl2 (g) = SnCl2 (s) rH 1 (1) SnCl2 (s) +Cl2 (g) = SnCl4 (l) rH 2 (2) (1) + (2) 得 Sn(s) + 2Cl2 (g) = SnCl4 (l) rH m ∴ rH m = rH 1 + rH 2 = −349.8+ (−195.4) = −545.2 kJmol-1