（江苏专用）2022届高考化学二轮复习 压轴题热点练2 盖斯定律的应用

（江苏专用）2022届高考化学二轮复习 压轴题热点练2 盖斯定律的应用1联氨(又称联肼，N2H4，无色液体)是一种应用广泛的化工原料，可用作火箭燃料，回答下列问题：2O2(g)N2(g)=N2O4(l)H1N2(g)2H2(g)=N2H4(l)H2O2(g)2H2(g)=2H2O(g)H32N2H4(l)N2O4(l)=3N2(g)4H2O(g)H41048.9 kJ/mol上述反应热效应之间的关系式为H4_，联氨和N2O4可作为火箭推进剂的主要原因为_。答案2H32H2H1反应放热量大、产生大量气体解析根据盖斯定律，22即得2N2H4(l)N2O4(l)=3N2(g)4H2O(g)的H4，所以反应热效应之间的关系式为H42H32H2H1。联氨有强还原性，N2O4有强氧化性，两者在一起易发生氧化还原反应，反应放热量大、产生大量气体，所以联氨和N2O4可作为火箭推进剂。2已知：C(s，石墨)O2(g)=CO2(g)H1393.5 kJmol12H2(g)O2(g)=2H2O(l)H2571.6 kJmol12C2H2(g)5O2(g)=4CO2(g)2H2O(l)H32599 kJmol1根据盖斯定律，计算反应2C(s，石墨)H2(g)=C2H2(g)的H_。答案226.7 kJmol1解析C(s，石墨)O2(g)=CO2(g)H1393.5 kJmol1，2H2(g)O2(g)=2H2O(l)H2571.6 kJmol1，2C2H2(g)5O2(g)=4CO2(g)2H2O(l)H32599 kJmol1；根据盖斯定律计算(2)得2C(s，石墨)H2(g)=C2H2(g)H(393.5 kJmol1)2(571.6 kJmol1)(2599 kJmol1)226.7 kJmol1。3由金红石(TiO2)制取单质Ti，涉及的步骤为：TiO2TiCl4Ti已知：C(s)O2(g)=CO2(g)H1393.5 kJmol12CO(g)O2(g)=2CO2(g)H2566 kJmol1TiO2(s)2Cl2(g)=TiCl4(s)O2(g)H3141 kJmol1则TiO2(s)2Cl2(g)2C(s)=TiCl4(s)2CO(g)的H_。答案80 kJmol1解析2就可得TiO2(s)2Cl2(g)2C(s)=TiCl4(s)2CO(g)，则HH3H12H280 kJmol1。4甲醇质子交换膜燃料电池中将甲醇蒸气转化为氢气的两种反应的热化学方程式如下：CH3OH(g)H2O(g)=CO2(g)3H2(g)H49.0 kJmol1CH3OH(g)O2(g)=CO2(g)2H2(g)H192.9 kJmol1又知H2O(g)=H2O(l)H44 kJmol1，则甲醇蒸气燃烧生成液态水的热化学方程式为_。答案CH3OH(l)O2(g)=CO2(g)2H2O(l)H764.7 kJmol1解析根据盖斯定律计算(322)得：CH3OH(l)O2(g)=CO2(g)2H2O(l)H3(192.9 kJmol1)249.0 kJmol1(44 kJmol1)2764.7 kJmol1；则甲醇蒸气燃烧为液态水的热化学方程式为：CH3OH(l)O2(g)=CO2(g)2H2O(l)H764.7 kJmol1。5已知：H2的热值为142.9 kJg1(热值是表示单位质量的燃料完全燃烧生成稳定的化合物时所放出的热量)；N2(g)2O2(g)=2NO2(g)H133 kJmol1H2O(g)=H2O(l)H44 kJmol1催化剂存在下，H2还原NO2生成水蒸气和其他无毒物质的热化学方程式：_。答案4H2(g)2NO2(g)=N2(g)4H2O(g)H1100.2 kJmol1解析已知：H2的热值为142.9 kJg1，则H2(g)O2(g)=H2O(l)H285.8 kJmol1；N2(g)2O2(g)=2NO2(g)H133 kJmol1；H2O(g)=H2O(l)H44 kJmol1；根据盖斯定律由44可得4H2(g)2NO2(g)=4H2O(g)N2(g)H(285.8 kJmol1)4(133 kJmol1)(44 kJmol1)41100.2 kJmol1，故此反应的热化学方程式为4H2(g)2NO2(g)=N2(g)4H2O(g)H1100.2 kJmol1。6能源问题是人类社会面临的重大课题，H2、CO、CH3OH都是重要的能源物质，它们的燃烧热依次为285.8 kJmol1、282.5 kJmol1、726.7 kJmol1。已知CO和H2在一定条件下可以合成甲醇CO(g)2H2(g)=CH3OH(l)。则CO与H2反应合成甲醇的热化学方程式为_。答案CO(g)2H2(g)=CH3OH(l)H127.4 kJmol1解析根据目标反应与三种反应热的关系，利用盖斯定律，计算出目标反应的反应热H2(285.8 kJmol1)(282.5 kJmol1)(726.7 kJmol1)127.4 kJmol1。7已知：25 、101 kPa时，Mn(s)O2(g)=MnO2(s)H520 kJmol1S(s)O2(g)=SO2(g)H297 kJmol1Mn(s)S(s)2O2(g)=MnSO4(s)H1065 kJmol1则SO2与MnO2反应生成无水MnSO4的热化学方程式是_。答案MnO2(s)SO2(g)=MnSO4(s)H248 kJmol1解析将题给三个热化学方程式依次编号为，根据盖斯定律，由可得SO2(g)MnO2(s)=MnSO4(s)H(1065 kJmol1)(520 kJmol1)(297 kJmol1)248 kJmol1。8已知下列热化学方程式：Fe2O3(s)3CO(g)=2Fe(s)3CO2(g)H25 kJmol13Fe2O3(s)CO(g)=2Fe3O4(s)CO2(g)H47 kJmol1Fe3O4(s)CO(g)=3FeO(s)CO2(g)H19 kJmol1写出FeO(s)被CO还原成Fe和CO2的热化学方程式_。答案FeO(s)CO(g)=Fe(s)CO2(g)H11 kJmol1解析32就可得6FeO(s)6CO(g)=6Fe(s)6CO2(g)H66 kJmol1，即FeO(s)CO(g)=Fe(s)CO2(g)H11 kJmol1。9已知：2Cu2S(s)3O2(g)=2Cu2O(s)2SO2(g)H768.2 kJmol12Cu2O(s)Cu2S(s)=6Cu(s)SO2(g)H116.0 kJmol1则Cu2S(s)O2(g)=2Cu(s)SO2(g)H_。答案217.4 kJmol1解析根据盖斯定律，将方程式()得Cu2S(s)O2(g)=2Cu(s)SO2(g)H(768.2116.0) kJmol1217.4 kJmol1。