高考化學(xué)一輪復(fù)習(xí)章末排查練6化學(xué)反應(yīng)與能量含解析新人教版

章末排查練(六)化學(xué)反應(yīng)與能量排查點(diǎn)一熱化學(xué)方程式的再書寫1．沼氣是一種能源，它的主要成分是CH4，常溫下，0.5molCH4完全燃燒生成CO2(g)和液態(tài)水時(shí)，放出445kJ熱量，則熱化學(xué)方程式為________________________________________________________________________________________________________________________________________________。答案：CH4(g)＋2O2(g)=CO2(g)＋2H2O(l)ΔH＝－890kJ·mol－12．已知H2S完全燃燒生成SO2(g)和H2O(l)，H2S的燃燒熱為akJ·mol－1，寫出H2S的燃燒熱化學(xué)方程式________________________________________________________________________________________________________________________________________________。答案：2H2S(g)＋3O2(g)=2SO2(g)＋2H2O(l)ΔH＝－2akJ·mol－13．已知H—H鍵能436kJ·mol－1，H—N鍵能391kJ·mol－1，NN鍵能945.6kJ·mol－1，試寫出N2和H2反應(yīng)生成NH3的熱化學(xué)方程式________________________________________________________________________________________________________________________________________________________________________________________________________________________。答案：N2(g)＋3H2(g)=2NH3(g)ΔH＝－92.4kJ·mol－14．已知eq\f(1,2)N2(g)＋eq\f(3,2)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寫出N2(g)和H2(g)反應(yīng)生成液氨的熱化學(xué)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－1②H＋(aq)＋OH－(aq)=H2O(l)ΔH＝－57.3kJ·mol－1試寫出HF電離的熱化學(xué)方程式________________________________________________________________________________________________________________________________________________。答案：HF(aq)?F－(aq)＋H＋(aq)ΔH＝－10.4kJ·mol－16．SF6是一種優(yōu)良的絕緣氣體，分子結(jié)構(gòu)中，只存在S—F鍵，已知1molS(s)轉(zhuǎn)化為氣態(tài)硫原子吸收能量280kJ，F(xiàn)—F鍵能為160kJ·mol－1，S—F鍵能為330kJ·mol－1，試寫出S(s)和F2(g)反應(yīng)生成SF6(g)的熱化學(xué)方程式________________________________________________________________________________________________________________________________________________。答案：S(s)＋3F2(g)=SF6(g)ΔH＝－1220kJ·mol－1排查點(diǎn)二電池電極反應(yīng)式或總反應(yīng)式的再書寫1．用惰性電極電解下列溶液(1)NaCl溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)CuSO4溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)2H＋＋2e－=H2↑2Cl－－2e－=Cl2↑2NaCl＋2H2Oeq\o(=,\s\up7(電解))2NaOH＋H2↑＋Cl2↑(2)2Cu2＋＋4e－=2Cu4OH－－4e－=2H2O＋O2↑2CuSO4＋2H2Oeq\o(=,\s\up7(電解))2Cu＋2H2SO4＋O2↑2．用銅作電極電解下列溶液(1)H2O陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)H2SO4溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(3)NaOH溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)2H＋＋2e－=H2↑Cu－2e－=Cu2＋Cu＋2H2Oeq\o(=,\s\up7(電解))Cu(OH)2＋H2↑(2)2H＋＋2e－=H2↑Cu－2e－=Cu2＋Cu＋H2SO4eq\o(=,\s\up7(電解))CuSO4＋H2↑(3)2H2O＋2e－=H2↑＋2OH－Cu－2e－＋2OH－=Cu(OH)2Cu＋2H2Oeq\o(=,\s\up7(電解))Cu(OH)2＋H2↑3．用Al作電極電解下列溶液(1)H2SO4溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)NaOH溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)6H＋＋6e－=3H2↑2Al－6e－=2Al3＋2Al＋3H2SO4eq\o(=,\s\up7(電解))Al2(SO4)3＋3H2↑(2)6H2O＋6e－=3H2↑＋6OH－2Al－6e－＋8OH－=2AlOeq\o\al(－,2)＋4H2O2Al＋2H2O＋2NaOHeq\o(=,\s\up7(電解))2NaAlO2＋3H2↑4．鐵鎳電池(負(fù)極—Fe，正極—NiO2，電解液—KOH溶液)已知Fe＋NiO2＋2H2Oeq\o(,\s\up7(放電),\s\do5(充電))Fe(OH)2＋Ni(OH)2，則：負(fù)極：________________________________________________________________________；正極：________________________________________________________________________。陰極：________________________________________________________________________；陽極：________________________________________________________________________。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電池(正極—LiFePO4，負(fù)極—Li，含Li＋導(dǎo)電固體為電解質(zhì))已知FePO4＋Lieq\o(,\s\up7(放電),\s\do5(充電))LiFePO4，則負(fù)極：________________________________________________________________________；正極：________________________________________________________________________。陰極：________________________________________________________________________；陽極：________________________________________________________________________。答案：Li－e－=Li＋FePO4＋Li＋＋e－=LiFePO4Li＋＋e－=LiLiFePO4－e－=FePO4＋Li＋6．高鐵電池(負(fù)極—Zn，正極—石墨，電解質(zhì)為浸濕的固態(tài)堿性物質(zhì))已知：3Zn＋2K2FeO4＋8H2Oeq\o(?,\s\up7(放電),\s\do5(充電))3Zn(OH)2＋2Fe(OH)3＋4KOH，則：負(fù)極：________________________________________________________________________；正極：________________________________________________________________________。陰極：________________________________________________________________________；陽極：________________________________________________________________________。答案：3Zn－6e－＋6OH－=3Zn(OH)22FeOeq\o\al(2－,4)＋6e－＋8H2O=2Fe(OH)3＋10OH－3Zn(OH)2＋6e－=3Zn＋6OH－2Fe(OH)3－6e－＋10OH－=2FeOeq\o\al(2－,4)＋8H2O7．氫氧燃料電池(1)電解質(zhì)是KOH溶液(堿性電解質(zhì))負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(2)電解質(zhì)是H2SO4溶液(酸性電解質(zhì))負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(3)電解質(zhì)是NaCl溶液(中性電解質(zhì))負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；總反應(yīng)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．甲烷燃料電池(鉑為兩極，正極通入O2和CO2，負(fù)極通入甲烷，電解液有三種)(1)電解質(zhì)是熔融碳酸鹽(K2CO3或Na2CO3)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(2)酸性電解質(zhì)(電解液為H2SO4溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(3)堿性電解質(zhì)(電解液為KOH溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。答案：(1)2O2＋8e－＋4CO2=4COeq\o\al(2－,3)CH4－8e－＋4COeq\o\al(2－,3)=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－=COeq\o\al(2－,3)＋7H2OCH4＋2O2＋2OH－=COeq\o\al(2－,3)＋3H2O9．甲醇燃料電池(1)堿性電解質(zhì)(鉑為兩極，電解液為KOH溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(2)酸性電解質(zhì)(鉑為兩極，電解液為H2SO4溶液)正極：_______________________________________