2023高考化学一轮复习章末排查练6化学反应与能量转化含解析鲁科版202303101131

章末排查练(六)　化学反应与能量转化(本栏目内容，在学生用书中以独立形式分册装订！)排查点一　热化学方程式的再书写1．沼气是一种能源，它的主要成分是CH4，常温下，0.5molCH4完全燃烧生成CO2(g)和液态水时，放出445kJ热量，则热化学方程式为________________________________________________________________________________________________________________________________________________。答案：　CH4(g)＋2O2(g)===CO2(g)＋2H2O(l)　ΔH＝－890kJ·mol－12．已知H2S完全燃烧生成SO2(g)和H2O(l)，H2S的燃烧热为akJ·mol－1，写出H2S的燃烧热化学方程式________________________________________________________________________________________________________________________________________________。答案：　2H2S(g)＋3O2(g)===2SO2(g)＋2H2O(l)　ΔH＝－2akJ·mol－13．已知H—H键能436kJ·mol－1，H—N键能391kJ·mol－1，NN键能945.6kJ·mol－1，试写出N2和H2反应生成NH3的热化学方程式________________________________________________________________________________________________________________________________________________________________________________________________________________________。答案：　N2(g)＋3H2(g)===2NH3(g)　ΔH＝－92.4kJ·mol－14．已知N2(g)＋H2(g)===N(g)＋3H(g)　ΔH1＝＋akJ·mol－1N(g)＋3H(g)===NH3(g)　ΔH2＝－bkJ·mol－1NH3(g)===NH3(l)　ΔH3＝－ckJ·mol－1写出N2(g)和H2(g)反应生成液氨的热化学方程式________________________________________________________________________________________________________________________________________________。答案：　N2(g)＋3H2(g)===2NH3(l)　ΔH＝－2(b＋c－a)kJ·mol－15．已知：①HF(aq)＋OH－(aq)===F－(aq)＋H2O(l)　ΔH＝－67.7kJ·mol－1-8-\n②H＋(aq)＋OH－(aq)===H2O(l)ΔH＝－57.3kJ·mol－1试写出HF电离的热化学方程式________________________________________________________________________________________________________________________________________________。答案：　HF(aq)⇌F－(aq)＋H＋(aq)　ΔH＝－10.4kJ·mol－16．SF6是一种优良的绝缘气体，分子结构中，只存在S—F键，已知1molS(s)转化为气态硫原子吸收能量280kJ，F—F键能为160kJ·mol－1，S—F键能为330kJ·mol－1，试写出S(s)和F2(g)反应生成SF6(g)的热化学方程式________________________________________________________________________________________________________________________________________________。答案：　S(s)＋3F2(g)===SF6(g)　ΔH＝－1220kJ·mol－1排查点二　电池电极反应式或总反应式的再书写1．用惰性电极电解下列溶液(1)NaCl溶液阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。(2)CuSO4溶液阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。答案：　(1)2H＋＋2e－===H2↑2Cl－－2e－===Cl2↑2NaCl＋2H2O2NaOH＋H2↑＋Cl2↑(2)2Cu2＋＋4e－===2Cu-8-\n4OH－－4e－===2H2O＋O2↑2CuSO4＋2H2O2Cu＋2H2SO4＋O2↑2．用铜作电极电解下列溶液(1)H2O阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。(2)H2SO4溶液阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。(3)NaOH溶液阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。答案：　(1)2H＋＋2e－===H2↑Cu－2e－===Cu2＋Cu＋2H2OCu(OH)2＋H2↑(2)2H＋＋2e－===H2↑Cu－2e－===Cu2＋Cu＋H2SO4CuSO4＋H2↑(3)2H2O＋2e－===H2↑＋2OH－Cu－2e－＋2OH－===Cu(OH)2-8-\nCu＋2H2OCu(OH)2＋H2↑3．用Al作电极电解下列溶液(1)H2SO4溶液阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。(2)NaOH溶液阴极：________________________________________________________________________；阳极：________________________________________________________________________；总反应式：________________________________________________________________________。答案：　(1)6H＋＋6e－===3H2↑2Al－6e－===2Al3＋2Al＋3H2SO4Al2(SO4)3＋3H2↑(2)6H2O＋6e－===3H2↑＋6OH－2Al－6e－＋8OH－===2[Al(OH)4]－2Al＋6H2O＋2NaOH2Na[Al(OH)4]＋3H2↑4．铁镍电池(负极—Fe，正极—NiO2，电解液—KOH溶液)已知Fe＋NiO2＋2H2OFe(OH)2＋Ni(OH)2，则：负极：________________________________________________________________________；正极：________________________________________________________________________。阴极：________________________________________________________________________；阳极：________________________________________________________________________。-8-\n答案：　Fe－2e－＋2OH－===Fe(OH)2NiO2＋2H2O＋2e－===Ni(OH)2＋2OH－Fe(OH)2＋2e－===Fe＋2OH－Ni(OH)2－2e－＋2OH－===NiO2＋2H2O5．LiFePO4电池(正极—LiFePO4，负极—Li，含Li＋导电固体为电解质)已知FePO4＋LiLiFePO4，则负极：________________________________________________________________________；正极：________________________________________________________________________。阴极：________________________________________________________________________；阳极：________________________________________________________________________。答案：　Li－e－===Li＋FePO4＋Li＋＋e－===LiFePO4Li＋＋e－===LiLiFePO4－e－===FePO4＋Li＋6．高铁电池(负极—Zn，正极—石墨，电解质为浸湿的固态碱性物质)已知：3Zn＋2K2FeO4＋8H2O3Zn(OH)2＋2Fe(OH)3＋4KOH，则：负极：________________________________________________________________________；正极：________________________________________________________________________。阴极：________________________________________________________________________；阳极：________________________________________________________________________。答案：　3Zn－6e－＋6OH－===3Zn(OH)22FeO＋6e－＋8H2O===2Fe(OH)3＋10OH－3Zn(OH)2＋6e－===3Zn＋6OH－2Fe(OH)3－6e－＋10OH－===2FeO＋8H2O7．氢氧燃料电池-8-\n(1)电解质是KOH溶液(碱性电解质)负极：________________________________________________________________________；正极：________________________________________________________________________；总反应方程式：________________________________________________________________________。(2)电解质是H2SO4溶液(酸性电解质)负极：________________________________________________________________________；正极：________________________________________________________________________；总反应方程式：________________________________________________________________________。(3)电解质是NaCl溶液(中性电解质)负极：________________________________________________________________________；正极：________________________________________________________________________；总反应方程式：________________________________________________________________________。答案：　(1)2H2－4e－＋4OH－===4H2OO2＋2H2O＋4e－===4OH－2H2＋O2===2H2O(2)2H2－4e－===4H＋O2＋4H＋＋4e－===2H2O2H2＋O2===2H2O(3)2H2－4e－===4H＋O2＋2H2O＋4e－===4OH－2H2＋O2===2H2O8．甲烷燃料电池(铂为两极，正极通入O2和CO2，负极通入甲烷，电解液有三种)(1)电解质是熔融碳酸盐(K2CO3或Na2CO3)正极：________________________________________________________________________；-8-\n负极：________________________________________________________________________；总反应方程式：________________________________________________________________________。(2)酸性电解质(电解液为H2SO4溶液)正极：________________________________________________________________________；负极：________________________________________________________________________；总反应方程式：________________________________________________________________________。(3)碱性电解质(电解液为KOH溶液)正极：________________________________________________________________________；负极：________________________________________________________________________；总反应方程式：________________________________________________________________________。答案：　(1)2O2＋8e－＋4CO2===4COCH4－8e－＋4CO===5CO2＋2H2OCH4＋2O2===CO2＋2H2O(2)2O2＋8e－＋8H＋===4H2OCH4－8e－＋2H2O===CO2＋8H＋CH4＋2O2===CO2＋2H2O(3)2O2＋8e－＋4H2O===8OH－CH4－8e－＋10OH－===CO＋7H2OCH4＋2O2＋2OH－===CO＋3H2O9．甲醇燃料电池(1)碱性电解质(铂为两极，电解液为KOH溶液)正极：________________________________________________________________________；负极：________________________________________________________________________；总反应方程式：-8-\n________________________________________________________________________。(2)酸性电解质(铂为两极，电解液为H2SO4溶液)正极：________________________________________________________________________；负极：________________________________________________________________________；总反应方程式：________________________________________________________________________。答案：　(1)3O2＋12e－＋6H2O===12OH－2CH3OH－12e－＋16OH－===2CO＋12H2O2CH3OH＋3O2＋4KOH===2K2CO3＋6H2O(2)3O2＋12e－＋12H＋===6H2O2CH3OH－12e－＋2H2O===12H＋＋2CO22CH3OH＋3O2===2CO2＋4H2O10．CO燃料电池(总反应方程式均为2CO＋O2===2CO2)(1)熔融盐(铂为两极，Li2CO3和Na2CO3的熔融盐作电解质，CO为负极燃气，空气与CO2的混合气为正极助燃气)正极：________________________________________________________________________；负极：________________________________________________________________________。(2)酸性电解质(铂为两极，电解液为H2SO4溶液)正极：________________________________________________________________________；负极：________________________________________________________________________。答案：　(1)O2＋4e－＋2CO2===2CO2CO＋2CO－4e－===4CO2(2)O2＋4e－＋4H＋===2H2O2CO－4e－＋2H2O===2CO2＋4H＋-8-