1、 全国卷 (2)(2019全国卷节选)Deacon直接氧化法可按下列催化过程进行：CuCl2(s)=CuCl(s)Cl2(g)H183 kJmol1CuCl(s)O2(g)=CuO(s)Cl2(g)H220 kJmol1CuO(s)2HCl(g)=CuCl2(s)H2O(g)H3121 kJmol1则4HCl(g)O2(g)=2Cl2(g)2H2O(g)的H_kJmol1。解析(1)根据盖斯定律，由反应反应得反应，则H3H1H2(100.311.0 kJmol189.3 kJmol1。(2)将已知热化学方程式依次编号为、，根据盖斯定律，由()2得4HCl(g)O2(g)=2Cl2(g)2H2O
2、(g)H116 kJmol1。答案(1)89.3(2)1162(1)(2018全国卷节选)SiHCl3在催化剂作用下发生反应：2SiHCl3(g)=SiH2Cl2(g)SiCl4(g)H148 kJmol13SiH2Cl2(g)=SiH4(g)2SiHCl3(g)H230 kJmol1则反应4SiHCl3(g)=SiH4(g)3SiCl4(g)的H为_kJmol1。(2)(2018全国卷节选)CH4CO2催化重整反应为：CH4(g)CO2(g)=2CO(g)2H2(g)。已知：C(s)2H2(g)=CH4(g)H75 kJmol1C(s)O2(g)=CO2(g)H394 kJmol1C(s)O
3、2(g)=CO(g)H111 kJmol1该催化重整反应的H_kJmol1。解析(1)将题给两个热化学方程式依次编号为、，根据盖斯定律，由3可得：4SiHCl3(g)=SiH4(g)3SiCl4(g)，则有H3H1H2348 kJmol1(30 kJmol1)114 kJmol1。(2)将题给三个已知热化学方程式依次编号为，由2可得CH4(g)CO2(g)=2CO(g)2H2(g)；根据盖斯定律，则该反应的H(111 kJmol1)2(75 kJmol1)(394 kJmol1)247 kJmol1。答案(1)114(2)2473(2017全国卷)下图是通过热化学循环在较低温度下由水或硫化氢分
4、解制备氢气的反应系统原理。通过计算，可知系统()和系统()制氢的热化学方程式分别为_、_，制得等量H2所需能量较少的是_。解析将题干中的四个热化学方程式分别编号为，根据盖斯定律，将可得系统()中的热化学方程式：H2O(l)=H2(g)O2(g)HH1H2H3327 kJmol1151 kJmol1110 kJmol1286 kJmol1同理，将可得系统()中的热化学方程式：H2S(g)=H2(g)S(s)HH2H3H4151 kJmol1110 kJmol161 kJmol120 kJmol1由所得两热化学方程式可知，制得等量H2所需能量较少的是系统()。答案H2O(l)=H2(g)O2(g)
5、H286 kJmol1H2S(g)=H2(g)S(s)H20 kJmol1系统()省市卷1(2019天津卷)硅粉与HCl在300时反应生成1 mol SiHCl3气体和H2，放出225 kJ热量，该反应的热化学方程式为_。答案Si(s)3HCl(g)=SiHCl3(g)H2(g)H225 kJmol12(2018江苏卷)NOx(主要指NO和NO2)是大气主要污染物之一。有效去除大气中的NOx是环境保护的重要课题。用水吸收NOx的相关热化学方程式如下：2NO2(g)H2O(l)=HNO3(aq)HNO2(aq)H116.1 kJmol13HNO2(aq)=HNO3(aq)2NO(g)H2O(l)
6、H75.9 kJmol1反应3NO2(g)H2O(l)=2HNO3(aq)NO(g)的H_kJmol1。解析将题给三个热化学方程式依次编号为、和，根据盖斯定律可知，(3)/2，则H(116.1 kJmol1375.9 kJmol1)/2136.2 kJmol1。答案136.23(2017浙江卷)已知：2Al2O3(s)=4Al(g)3O2(g)H13351 kJmol12C(s)O2(g)=2CO(g)H2221 kJmol12Al(g)N2(g)=2AlN(s)H3318 kJmol1碳热还原Al2O3合成AlN的总热化学方程式是_。答案Al2O3(s)3C(s)N2(g)=3CO(g)2AlN(s)H1026 kJmol14(2017海南卷节选)碳酸钠是一种重要的化工原料，主要采用氨碱法生产。回答下列问题：已知：2NaOH(s)CO2(g)=Na2CO3(s)H2O(g)H1127.4 kJmol1NaOH(s)CO2(g)=NaHCO3(s)H2131.5 kJmol1反应2NaHCO3(s)=Na2CO3(s)H2O(g)CO2(g)的H_kJmol1。解析根据盖斯定律，NaHCO3(s)分解反应可由2得到，故HH12H2127.4 kJmol1(131.5 kJmol1)2135.6 kJmol1。答案135.6
