1、1第三節(jié)化學(xué)反應(yīng)熱的計(jì)算A組1.已知:CH4(g)2O2(g)CO2(g)2H2O(l)ΔH1=890.3kJmol1H2(g)O2(g)H2O(l)ΔH2=12285.8kJmol1。CO2與H2反應(yīng)生成甲烷氣體與液態(tài)水的熱化學(xué)方程式為CO2(g)4H2(g)CH4(g)2H2O(l)ΔH3則ΔH3為()A.252.9kJmol1B.252.9kJmol1C.604.5kJmol1D.604.5kJmol1解析:由蓋斯定律可知ΔH3=

2、4ΔH2ΔH1=(285.8kJmol1)4890.3kJmol1=252.9kJmol1。答案:A2.已知熱化學(xué)方程式:①C2H2(g)O2(g)2CO2(g)H2O(l)ΔH1=1301.0kJmol152②C(s)O2(g)CO2(g)ΔH2=393.5kJmol1③H2(g)O2(g)H2O(l)ΔH3=285.8kJmol112則反應(yīng)④2C(s)H2(g)C2H2(g)的ΔH為()A.228.2kJmol1B.228.2kJm

3、ol1C.1301.0kJmol1D.621.7kJmol1解析:熱化學(xué)方程式①②③和④之間存在如下關(guān)系:2②③①=④。所以2ΔH2ΔH3ΔH1=ΔH=2393.5kJmol1285.8kJmol11301.0kJmol1=228.2kJmol1。答案:A3.已知:2CO(g)O2(g)2CO2(g)ΔH=566kJmol1Na2O2(s)CO2(g)Na2CO3(s)O2(g)12ΔH=226kJmol1。根據(jù)以上熱化學(xué)方程式判斷下列

4、說(shuō)法正確的是()A.CO的燃燒熱為283kJB.如圖所示可表示由CO生成CO2的反應(yīng)過(guò)程和能量關(guān)系C.2Na2O2(s)2CO2(s)2Na2CO3(s)O2(g)ΔH452kJmol1D.CO(g)與Na2O2(s)反應(yīng)放出509kJ熱量時(shí)電子轉(zhuǎn)移數(shù)為6.021023解析:A選項(xiàng)中燃燒熱單位錯(cuò)誤應(yīng)為kJmol1B項(xiàng)圖中沒(méi)有表示出ΔH=566kJmol1時(shí)對(duì)應(yīng)的反應(yīng)物與生成物物質(zhì)的量的關(guān)系及聚集狀態(tài)B項(xiàng)錯(cuò)由于2Na2O2(s)2CO2(

5、g)2Na2CO3(s)O2(g)ΔH=452kJmol1固態(tài)CO2轉(zhuǎn)化為氣態(tài)CO2需吸收熱量所以2Na2O2(s)2CO2(s)2Na2CO3(s)O2(g)ΔH452kJmol1C項(xiàng)正確根據(jù)蓋斯定律將已知的兩個(gè)熱化學(xué)方程式分別編號(hào)①②并作①②運(yùn)算得:CO(g)Na2O2(s)Na2CO3(s)ΔH=509kJmol1可知當(dāng)放出50912kJ熱量時(shí)消耗1molCO(g)電子轉(zhuǎn)移數(shù)為26.021023D項(xiàng)錯(cuò)誤。答案:C4.已知:2SO2

6、(g)O2(g)2SO3(g)ΔH=196.6kJmol1實(shí)驗(yàn)室測(cè)得4molSO2發(fā)生上述化學(xué)反應(yīng)時(shí)放出314.3kJ熱量參加反應(yīng)的SO2的物質(zhì)的量為()A.1.6molB.2.0molC.3.2molD.3.6mol解析:參加反應(yīng)的SO2為≈3.2mol。314.32196.63C.H—H鍵的鍵能為436kJD.a=200解析:A項(xiàng)2HBr(g)H2(g)Br2(g)應(yīng)為吸熱反應(yīng)ΔH0B項(xiàng)液體變成氣體應(yīng)吸收熱量ΔH0故B項(xiàng)錯(cuò)誤C項(xiàng)單位

7、錯(cuò)H—H鍵的鍵能應(yīng)為436kJmol1故C項(xiàng)錯(cuò)誤D項(xiàng)已知H2(g)Br2(g)2HBr(g)ΔH=72kJmol1蒸發(fā)1molBr2(l)需要吸收的能量為30kJ則有H2(g)Br2(g)2HBr(g)ΔH=102kJmol1反應(yīng)熱等于反應(yīng)物的總鍵能減去生成物的總鍵能則有436kJa2369kJ=102kJa=200kJ故D項(xiàng)正確。故選D。答案:D4.廢舊印刷電路板的回收利用可實(shí)現(xiàn)資源再生并減少污染。廢舊印刷電路板經(jīng)粉碎分離能得到非金屬

8、粉末和金屬粉末。用H2O2和H2SO4的混合溶液可溶出印刷電路板金屬粉末中的銅。已知:Cu(s)2H(aq)Cu2(aq)H2(g)ΔH=64.39kJmol12H2O2(l)2H2O(l)O2(g)ΔH=196.46kJmol1H2(g)O2(g)H2O(l)ΔH=285.84kJmol112在H2SO4溶液中Cu與H2O2反應(yīng)生成Cu2和H2O的熱化學(xué)方程式為。解析:已知熱化學(xué)方程式中①②③得:12Cu(s)2H(aq)H2O2(l

9、)Cu2(aq)2H2O(l)ΔH=319.68kJmol1ΔH=ΔH1ΔH2ΔH3=64.39kJmol1(196.46kJmol1)(285.84kJmol1)=319.681212kJmol1答案:Cu(s)H2O2(l)2H(aq)Cu2(aq)2H2O(l)ΔH=319.68kJmol15.紅磷P(s)和Cl2(g)發(fā)生反應(yīng)生成PCl3(g)和PCl5(g)。反應(yīng)過(guò)程和能量關(guān)系如圖所示(圖中的ΔH表示生成1mol產(chǎn)物的數(shù)據(jù))。

10、根據(jù)圖回答下列問(wèn)題:(1)P和Cl2反應(yīng)生成PCl3的熱化學(xué)方程式為。(2)PCl5分解成PCl3和Cl2的熱化學(xué)方程式為。(3)P和Cl2分兩步反應(yīng)生成1molPCl5的ΔH3=P和Cl2一步反應(yīng)生成1molPCl5的ΔH4(填“”“”或“=”)ΔH3。解析:(1)由題圖看出1molP在Cl2中燃燒生成PCl3(g)放出的熱量為306kJ所以P與Cl2反應(yīng)生成PCl3的熱化學(xué)方程式為P(s)Cl2(g)PCl3(g)ΔH=306kJm

11、ol1。(2)中間產(chǎn)物PCl3和未完32全反應(yīng)的Cl2的總能量高于最終產(chǎn)物PCl5的能量其ΔH=93kJmol1所以PCl5(g)PCl3(g)Cl2(g)ΔH=93kJmol1。(3)由蓋斯定律可知一步生成PCl5和兩步生成PCl5的熱效應(yīng)相等即ΔH3=ΔH4=ΔH1ΔH2=399kJmol1。答案:(1)P(s)Cl2(g)PCl3(g)ΔH=306kJmol132(2)PCl5(g)PCl3(g)Cl2(g)ΔH=93kJmol1