2017年高考化学三轮冲刺热点题型-填空题五、用盖斯定律计算反应热专练.doc

五、用盖斯定律计算反应热专练1O3氧化烟气中SO2、NOx的主要反应的热化学方程式：NO(g)O3(g)=NO2(g)O2(g)H200.9 kJmol1NO(g)O2(g)=NO2(g)H58.2 kJmol1SO2(g)O3(g)=SO3(g)O2(g)H241.6 kJmol1则反应3NO(g)O3(g)=3NO2(g)的H_kJmol1答案317.3解析利用盖斯定律将2得：3NO(g)O3(g)=3NO2(g)H200.9 kJmol158.2 kJmol12317.3 kJmol1。2已知下列两个热化学方程式：H2(g)O2(g)=H2O(l)H285 kJmol1C3H8(g)5O2(g)=3CO2(g)4H2O(l)H2 220.0 kJmol1已知：H2O(l)=H2O(g)H44.0 kJmol1，写出丙烷燃烧生成CO2和气态水的热化学方程式：_。答案C3H8(g)5O2(g)=3CO2(g)4H2O(g)H2 044.0 kJmol1解析C3H8(g)5O2(g)=3CO2(g)4H2O(l)H2 220.0 kJmol1，H2O(l)=H2O(g)H44.0 kJmol1，用前式4后式得：C3H8(g)5O2(g)=3CO2(g)4H2O(g)H2 220.0 kJmol1444.0 kJmol12 044.0 kJmol1。3已知下列反应的热化学方程式：6C(s)5H2(g)3N2(g)9O2(g)=2C3H5(ONO2)3(l)H12H2(g)O2(g)=2H2O(g)H2C(s)O2(g)=CO2(g)H3则反应4C3H5(ONO2)3(l)=12CO2(g)10H2O(g)O2(g)6N2(g)的H为_(用H1、H2、H3表示)。答案12H35H22H1解析根据盖斯定律，12第三个热化学方程式5第二个热化学方程式2第一个热化学方程式得：4C3H5(ONO2)3(l)=12CO2(g)10H2O(g)O2(g)6N2(g)H12H35H22H1。4(1)用CaSO4代替O2与燃料CO反应，既可提高燃烧效率，又能得到高纯CO2，是一种高效、清洁、经济的新型燃烧技术，反应为主反应，反应和为副反应。CaSO4(s)CO(g)=CaS(s)CO2(g)H147.3 kJmol1CaSO4(s)CO(g)=CaO(s)CO2(g)SO2(g)H2210.5 kJmol1CO(g)=C(s)CO2(g)H386.2 kJmol1反应2CaSO4(s)7CO(g)=CaS(s)CaO(s)6CO2(g)C(s)SO2(g)的H_(用H1、H2和H3)。(2)已知：甲醇脱水反应2CH3OH(g)=CH3OCH3(g)H2O(g)H123.9 kJmol1甲醇制烯烃的反应2CH3OH(g)=C2H4(g)2H2O(g)H229.1 kJmol1乙醇异构化反应C2H5OH(g)=CH3OCH3(g)H350.7 kJmol1则乙烯气相直接水合反应C2H4(g)H2O(g)=C2H5OH(g)的H_ kJmol1。(3)已知：25 、101 kPa时，Mn(s)O2(g)=MnO2(s)H520 kJmol1，S(s)O2(g)=SO2(g)H297 kJmol1，Mn(s)S(s)2O2(g)=MnSO4(s)H1 065 kJmol1。SO2与MnO2反应生成无水MnSO4的热化学方程式是_。答案(1)4H1H22H3(2)45.5(3)MnO2(s)SO2(g)=MnSO4(s)H248 kJmol1解析(1)根据盖斯定律，可知42得H4H1H22H3。(2)将三个热化学方程式依次编号为，根据盖斯定律，由得：C2H4(g)H2O(g)=C2H5OH(g)H45.5 kJmol1。(3)将题中三个热化学方程式由上到下依次编号为、，根据盖斯定律，得：MnO2(s)SO2(g)=MnSO4(s)H1 065 kJmol1520 kJmol1297 kJmol1248 kJmol1。5(2016九江模拟)已知在常温常压下：2CH3OH(l)3O2(g)=2CO2(g)4H2O(g)H1 275.6 kJmol12CO(g)O2(g)=2CO2(g)H566.0 kJmol1H2O(g)=H2O(l)H44.0 kJmol1则甲醇不完全燃烧生成一氧化碳和液态水的热化学方程式为_。答案CH3OH(l)O2(g)=CO(g)2H2O(l)H442.8 kJmol1解析根据盖斯定律，(4)2可得：CH3OH(l)O2(g)=CO(g)2H2O(l)，则：H1 275.6 kJmol1(566.0 kJmol1)4(44.0 kJmol1)2442.8 kJmol1。6铁及其化合物在日常生活、生产中应用广泛，研究铁及其化合物的应用意义重大。回答下列问题：已知高炉炼铁过程中会发生如下反应：FeO(s)CO(g)=Fe(s)CO2(g)H1Fe2O3(s)CO(g)=Fe3O4(s)CO2(g)H2Fe3O4(s)CO(g)=3Fe(s)CO2(g)H3Fe2O3(s)CO(g)=2Fe(s)3CO2(g)H4则H4的表达式为_(用含H1、H2、H3的代数式表示)。答案H2H3解析根据盖斯定律，将所给的反应可得：Fe2O3(s)3CO(g)=2Fe(s)3CO2(g)，故H4H2H3。7金属锰可用铝热法制得。已知Al的燃烧热为c kJmol1，其余相关热化学方程式：3MnO2(s)=Mn3O4(s)O2(g)H1a kJmol13Mn3O4(s)8Al(s)=9Mn(s)4Al2O3(s)H2b kJmol1则3MnO2(s)4Al(s)=3Mn(s)2Al2O3(s)H_ kJmol1(用含a、b、c的代数式表示)。答案abc解析根据盖斯定律，可得：3MnO2(s)4Al(s)=3Mn(s)2Al2O3(s)，则H(abc)kJmol1。8肼可作为火箭发动机的燃料，与氧化剂N2O4反应生成N2和水蒸气。已知：N2(g)2O2(g)=N2O4(l)H119.5 kJmol1N2H4(l)O2(g)=N2(g)2H2O(g)H2534.2 kJmol1写出肼和N2O4反应的热化学方程式_。答案2N2H4(l)N2O4(l)=3N2(g)4H2O(g)H1 048.9 kJmol1解析肼可作为火箭发动机的燃料，与氧化剂N2H4反应生成N2和水蒸气，将方程式2得肼和N2H4反应的热化学方程式2N2H4(l)N2O4(l)=3N2(g)4H2O(g)H1 048.9 kJmol1。9金红石(TiO2)制取单质Ti，涉及的步骤：TiO2TiCl4Ti已知：C(s)O2(g)=CO2(g)H393.5 kJmol12CO(g)O2(g)=2CO2(g)H556 kJmol1TiO2(s)2Cl2(g)=TiCl4(s)O2(g)H141 kJmol1则：TiO2(s)2Cl2(g)2C(s)=TiCl4(s)2CO(g)H_。答案80 kJmol1解析2得到TiO2(s)2Cl2(g)2C(s)=TiCl4(s)2CO(g)，所以H141 kJmol1(393.5 kJmol1)2566 kJmol180 kJmol1。10Bunsen热化学循环制氢工艺由下列三个反应组成：SO2(g)I2(g)2H2O(g)=2HI(g)H2SO4(l)Ha kJ