第六章　微題型57　蓋斯定律與反應(yīng)熱的計算

微題型57蓋斯定律與反應(yīng)熱的計算1．已知H2(g)、C2H4(g)和C2H5OH(l)的燃燒熱分別是－285.8kJ·mol－1、－1411.0kJ·mol－1和－1366.8kJ·mol－1，則由C2H4(g)和H2O(l)反應(yīng)生成C2H5OH(l)的ΔH為()A．－44.2kJ·mol－1 B．＋44.2kJ·mol－1C．－330kJ·mol－1 D．＋330kJ·mol－1答案A解析由題干條件可得如下反應(yīng)：H2(g)＋eq\f(1,2)O2(g)=H2O(l)ΔH＝－285.8kJ·mol－1①C2H4(g)＋3O2(g)=2CO2(g)＋2H2O(l)ΔH＝－1411.0kJ·mol－1②C2H5OH(l)＋3O2(g)=2CO2(g)＋3H2O(l)ΔH＝－1366.8kJ·mol－1③根據(jù)蓋斯定律，②－③得：C2H4(g)＋H2O(l)=C2H5OH(l)ΔH＝－44.2kJ·mol－1。2．乙苯催化脫氫制苯乙烯的反應(yīng)：eq\o(,\s\up7(催化劑))＋H2(g)己知：化學(xué)鍵C—HC—CC=CH—H鍵能/(kJ·mol－1)412348612436根據(jù)表中數(shù)據(jù)計算上述反應(yīng)的ΔH(kJ·mol－1)為()A．－124 B．＋124C．＋1172 D．－1048答案B解析反應(yīng)熱＝反應(yīng)物總鍵能－生成物總鍵能，由有機物的結(jié)構(gòu)可知，應(yīng)是—CH2CH3中總鍵能與—CH=CH2、H2總鍵能之差，故ΔH＝(5×412＋348－3×412－612－436)kJ·mol－1＝＋124kJ·mol－1。3．在1200℃時，天然氣脫硫工藝中會發(fā)生下列反應(yīng)：①H2S(g)＋eq\f(3,2)O2(g)=SO2(g)＋H2O(g)ΔH1②2H2S(g)＋SO2(g)=eq\f(3,2)S2(g)＋2H2O(g)ΔH2③H2S(g)＋eq\f(1,2)O2(g)=S(g)＋H2O(g)ΔH3④2S(g)=S2(g)ΔH4則ΔH4的正確表達(dá)式為()A．ΔH4＝eq\f(2,3)(ΔH1＋ΔH2－3ΔH3)B．ΔH4＝eq\f(2,3)(3ΔH3－ΔH1－ΔH2)C．ΔH4＝eq\f(3,2)(ΔH1＋ΔH2－3ΔH3)D．ΔH4＝eq\f(3,2)(ΔH1－ΔH2－3ΔH3)答案A4．已知：Fe2O3(s)＋eq\f(3,2)C(s)=eq\f(3,2)CO2(g)＋2Fe(s)ΔH1＝＋234.1kJ·mol－1①C(s)＋O2(g)=CO2(g)ΔH2＝－393.5kJ·mol－1②則2Fe(s)＋eq\f(3,2)O2(g)=Fe2O3(s)的ΔH3是()A．－824.4kJ·mol－1 B．－627.6kJ·mol－1C．－744.7kJ·mol－1 D．－169.4kJ·mol－1答案A1)－(＋234.1kJ·mol－1)≈－824.4kJ·mol－1。5．已知：①2C(s)＋O2(g)=2CO(g)ΔH＝－221.0kJ·mol－1②2H2(g)＋O2(g)=2H2O(g)ΔH＝－483.6kJ·mol－1則制備水煤氣的反應(yīng)C(s)＋H2O(g)=CO(g)＋H2(g)的ΔH為()A．＋262.6kJ·mol－1 B．－131.3kJ·mol－1C．－352.3kJ·mol－1 D．＋131.3kJ·mol－1答案D－1。6．用H2O2和H2SO4的混合溶液可溶出廢舊印刷電路板上的銅。已知：①Cu(s)＋2H＋(aq)=Cu2＋(aq)＋H2(g)ΔH1＝＋64.39kJ·mol－1②2H2O2(l)=2H2O(l)＋O2(g)ΔH2＝－196.46kJ·mol－1③H2(g)＋eq\f(1,2)O2(g)=H2O(l)ΔH3＝－285.84kJ·mol－1在H2SO4溶液中，1molCu(s)與H2O2(l)反應(yīng)生成Cu2＋(aq)和H2O(l)的反應(yīng)熱ΔH等于()A．－319.68kJ·mol－1B．－417.91kJ·mol－1C．－448.46kJ·mol－1D．＋546.69kJ·mol－1答案A解析根據(jù)蓋斯定律，①＋eq\f(1,2)×②＋③得：Cu(s)＋H2O2(l)＋2H＋(aq)=Cu2＋(aq)＋2H2O(l)ΔH＝－319.68kJ·mol－1。7．環(huán)戊二烯()是重要的有機化工原料，廣泛用于農(nóng)藥、橡膠、塑料等生產(chǎn)?；卮鹣铝袉栴}：已知：①(g)=(g)＋H2(g)ΔH1＝＋100.3kJ·mol－1②H2(g)＋I(xiàn)2(g)=2HI(g)ΔH2＝－11.0kJ·mol－1對于反應(yīng)：③(g)＋I(xiàn)2(g)=(g)＋2HI(g)ΔH3＝________kJ·mol－1。答案＋89.3解析根據(jù)蓋斯定律，①＋②可得反應(yīng)③，則ΔH3＝(＋100.3kJ·mol－1)＋(－11.0kJ·mol－1)＝＋89.3kJ·mol－1。8．已知：①As(s)＋eq\f(3,2)H2(g)＋2O2(g)=H3AsO4(s)ΔH1②H2(g)＋eq\f(1,2)O2(g)=H2O(l)ΔH2③2As(s)＋eq\f(5,2)O2(g)=As2O5(s)ΔH3則反應(yīng)As2O5(s)＋3H2O(l)=2H3AsO4(s)的ΔH＝________________________。答案2ΔH1－3ΔH2－ΔH3解析待求解反應(yīng)可由反應(yīng)①×2－反應(yīng)②×3－反應(yīng)③得到，則ΔH＝2ΔH1－3ΔH2－ΔH3。9．Cu2O是一種重要的工業(yè)原料，廣泛用作催化劑。在100℃的Cu(NO3)2溶液中加入一定體積的NaOH溶液，攪拌均勻，再逐滴加入肼(N2H4)產(chǎn)生紅色沉淀，抽濾、洗滌、干燥，得到Cu2O。已知：N2H4(l)＋O2(g)=N2(g)＋2H2O(l)ΔH＝－akJ·mol－1Cu(OH)2(s)=CuO(s)＋H2O(l)ΔH＝＋bkJ·mol－14CuO(s)=2Cu2O(s)＋O2(g)ΔH＝＋ckJ·mol－1則由N2H4(l)和Cu(OH)2(s)反應(yīng)制備Cu2O(s)的熱化學(xué)方程式為____________________________________________________________________________________________________。答案N2H4(l)＋4Cu(OH)2(s)=N2(g)＋2Cu2O(s)＋6H2O(l)ΔH＝(4b＋c－a)kJ·mol－1解析(1)反應(yīng)由上到下排序為①②③，根據(jù)蓋斯定律，①＋②×4＋③得N2H4(l)＋4Cu(OH)2(s)=N2(g)＋2Cu2O(s)＋6H2O(l)ΔH＝(4b＋c－a)kJ·mol－1。10．多晶硅是制作光伏電池的關(guān)鍵材料。回答下列問題：(1)硅粉與HCl在300℃時反應(yīng)生成1molSiHCl3氣體和H2，放出225kJ熱量，該反應(yīng)的熱化學(xué)方程式為___________________________________________________________________________________________________________________________________________________。(2)將SiCl4氫化為SiHCl3有三種方法，對應(yīng)的反應(yīng)依次為①SiCl4(g)＋H2(g)SiHCl3(g)＋HCl(g)ΔH1>0②3SiCl4(g)＋2H2(g)＋Si(s)4SiHCl3(g)ΔH2<0③2SiCl4(g)＋H2(g)＋Si(s)＋HCl(g)3SiHCl3(g)ΔH3反應(yīng)③的ΔH3＝________(用ΔH1、ΔH2表示)。答案(1)Si(s)＋3HCl(g)=SiHCl3(g)＋H2(g)ΔH＝－225kJ·mol－1(