2020-2021學(xué)年高中化學(xué)新教材魯科版選擇性必修1學(xué)案：第1章 微專題1 蓋斯定律應(yīng)用的三類題型

1、微專題1蓋斯定律應(yīng)用的三類題型一、利用蓋斯定律確定反應(yīng)焓變的關(guān)系【例1】同溫同壓下，下列各組熱化學(xué)方程式中h1<h2的是()ac(s)o2(g)=co(g)h1；c(s)o2(g)=co2(g)h2bh2(g)cl2(g)=hcl(g)h1；h2(g)cl2(g)=2hcl(g)h2c2h2(g)o2(g)=2h2o(g)h1；2h2(g)o2(g)=2h2o(l)h2ds(g)o2(g)=so2(g)h1；s(s)o2(g)=so2(g) h2da項(xiàng)，前者為碳的不完全燃燒，放熱相對(duì)較少，則h1>h2，a項(xiàng)錯(cuò)誤；b項(xiàng)，前者放出的熱量是后者的一半，則h1>h2，b項(xiàng)錯(cuò)誤；c項(xiàng)，

2、前者生成能量高的氣態(tài)水，放出的熱量少，則h1>h2，c項(xiàng)錯(cuò)誤；d項(xiàng)，前者反應(yīng)物的能量高，放熱多，則h1<h2，d項(xiàng)正確。確定反應(yīng)熱關(guān)系的三大注意點(diǎn)(1)反應(yīng)物和反應(yīng)產(chǎn)物的狀態(tài)物質(zhì)的氣、液、固三態(tài)的變化與反應(yīng)熱的關(guān)系：(2)h的符號(hào)：比較反應(yīng)熱的大小時(shí)，不要只比較h數(shù)值的大小，還要考慮其符號(hào)。(3)參加反應(yīng)物質(zhì)的量：當(dāng)反應(yīng)物和反應(yīng)產(chǎn)物的狀態(tài)相同時(shí)，參加反應(yīng)物質(zhì)的量越多，放熱反應(yīng)的h越小，吸熱反應(yīng)的h越大。1室溫下，將1 mol的cuso4·5h2o(s)溶于水會(huì)使溶液溫度降低，熱效應(yīng)為h1，將1 mol的cuso4(s)溶于水會(huì)使溶液溫度升高，熱效應(yīng)為h2；cuso4

3、83;5h2o受熱分解的化學(xué)方程式為cuso4·5h2o(s)cuso4(s)5h2o(l)，熱效應(yīng)為h3。則下列判斷正確的是()ah2>h3bh1<h3ch1h3h2dh1h2>h3b由題給條件可知：cuso4·5h2o(s)=cu2(aq)so (aq)5h2o(l)h1>0；cuso4(s)=cu2(aq)so(aq)h2<0；根據(jù)蓋斯定律，由可得cuso4·5h2o(s)=cuso4(s)5h2o(l)h3h1h2>h1>0，選b。2已知：2h2(g)o2(g)=2h2o(g)h13h2(g)fe2o3(s)=2f

4、e(s)3h2o(g)h22fe(s)o2(g)=fe2o3(s)h32al(s)o2(g)=al2o3(s)h42al(s)fe2o3(s)=al2o3(s)2fe(s)h5下列關(guān)于上述反應(yīng)焓變的判斷正確的是()ah1<0，h3>0bh5<0，h4<h3ch1h2h3dh3h4h5b燃燒反應(yīng)都是放熱反應(yīng)，故h3<0，a錯(cuò)誤；將上述反應(yīng)分別編號(hào)為，反應(yīng)是鋁熱反應(yīng)，顯然是放熱反應(yīng)，h5<0，將反應(yīng)反應(yīng)可得反應(yīng)，即h5h4h3<0，b正確，d錯(cuò)誤；將反應(yīng)反應(yīng)可得反應(yīng)3h2(g)o2(g)=3h2o(g)，故h1(h2h3)，c錯(cuò)誤。3已知1 mol紅磷完全

5、轉(zhuǎn)化為白磷，吸收18.39 kj熱量。4p(紅，s)5o2(g)=2p2o5(s)；h1p4(白，s)5o2(g)=2p2o5(s)；h2則h1與h2的關(guān)系正確的是()ah1h2 bh1>h2ch1<h2d無(wú)法確定b根據(jù)題供信息，由反應(yīng)減去反應(yīng)可得，4p(紅，s)=p4(白，s)；hh1h218.39 kj·mol1×473.56 kj·mol1>0，故h1>h2，b正確。二、利用蓋斯定律書寫熱化學(xué)方程式【例2】下圖是通過熱化學(xué)循環(huán)在較低溫度下由水或硫化氫分解制備氫氣的反應(yīng)系統(tǒng)原理。通過計(jì)算，可知系統(tǒng)()和系統(tǒng)()制氫的熱化學(xué)方程式分別為_

6、、_，制得等量h2所需能量較少的是_。解析根據(jù)蓋斯定律，把系統(tǒng)()和系統(tǒng)()的三個(gè)熱化學(xué)方程式分別相加，可以得到制氫的熱化學(xué)方程式，分別為h2o(l)=h2(g)o2(g)h286 kj·mol1、h2s(g)=h2(g)s(s)h20 kj·mol1，制得等量h2所需能量較少的是系統(tǒng)()。答案h2o(l)=h2(g)o2(g)h286 kj·mol1h2s(g)=h2(g)s(s)h20 kj·mol1系統(tǒng)()1(1)高爐冶鐵過程中，甲烷在催化反應(yīng)室中產(chǎn)生水煤氣(co和h2)還原氧化鐵，有關(guān)反應(yīng)為ch4(g)co2(g)=2co(g)2h2(g)h26

7、0 kj·mol1 2co(g)o2(g)=2co2(g)h566 kj·mol1則ch4與o2反應(yīng)生成co和h2的熱化學(xué)方程式為_。(2)已知na2co3·10h2o(s)=na2co3(s)10h2o(g)h532.36 kj·mol1na2co3·10h2o(s)=na2co3·h2o(s)9h2o(g)h473.63 kj·mol1寫出na2co3·h2o脫水反應(yīng)的熱化學(xué)方程式：_。解析(1)反應(yīng)×2得：2ch4(g)o2(g)=2co(g)4h2(g)h46 kj·mol1。(2)反應(yīng)

8、得：na2co3·h2o(s)=na2co3(s)h2o(g)h58.73 kj·mol1。答案(1)2ch4(g)o2(g)=2co(g)4h2(g)h46 kj·mol1(2)na2co3·h2o(s)=na2co3(s)h2o(g)h58.73 kj·mol12(1)工業(yè)上制取硝酸銨的流程圖如下所示：請(qǐng)回答下列問題：已知：4no(g)4nh3(g)o2(g)=4n2(g)6h2o(g)h1 745.2 kj·mol16no(g)4nh3(g)=5n2(g)6h2o(g)h1 925.2 kj·mol1則反應(yīng)的熱化學(xué)方程

9、式可表示為_。(2)蓋斯定律在生產(chǎn)和科學(xué)研究中有很重要的意義。有些反應(yīng)的反應(yīng)熱雖然無(wú)法直接測(cè)得，但可通過間接的方法測(cè)定?，F(xiàn)根據(jù)下列3個(gè)熱化學(xué)反應(yīng)方程式：fe2o3(s)3co(g)=2fe(s)3co2(g)h24.8 kj·mol13fe2o3(s)co(g)=2fe3o4(s)co2(g)h47.2 kj·mol1fe3o4(s)co(g)=3feo(s)co2(g)h640.5 kj·mol1寫出co氣體還原feo固體得到fe固體和co2氣體的熱化學(xué)反應(yīng)方程式：_。解析(1)將已知的兩個(gè)熱化學(xué)方程式從上到下依次標(biāo)記為和，根據(jù)蓋斯定律由×5×

10、;4得：4nh3(g)5o2(g)=4no(g)6h2o(g)h1 025.2 kj·mol1。(2)co氣體還原feo固體生成fe固體和co2氣體的化學(xué)方程式為co(g)feo(s)=fe(s)co2(g)；應(yīng)用蓋斯定律，消去fe2o3(s)、fe3o4(s)，將×3×2÷6得，co(g)feo(s)=fe(s)co2(g)h(24.8 kj·mol1×347.2 kj·mol1640.5 kj·mol1×2)÷6218.0 kj·mol1，則熱化學(xué)方程式為co(g)feo(s)=f

11、e(s)co2(g)h218.0 kj·mol1。答案(1)4nh3(g)5o2(g)=4no(g)6h2o(g)h1 025.2 kj·mol1(2)co(g)feo(s)=fe(s)co2(g)h218.0 kj·mol1三、利用蓋斯定律計(jì)算焓變【例3】(1)已知(g)=(g)h2(g)h1100.3 kj·mol1h2(g)i2(g)=2hi(g)h211.0 kj·mol1對(duì)于反應(yīng)：(g)i2(g)=(g)2hi(g)h3_ kj·mol1。(2)deacon直接氧化法可按下列催化過程進(jìn)行：cucl2(s)=cucl(s)cl

12、2(g)h183 kj·mol1cucl(s)o2(g)=cuo(s)cl2(g)h220 kj·mol1cuo(s)2hcl(g)=cucl2(s)h2o(g)h3121 kj·mol1則4hcl(g)o2(g)=2cl2(g)2h2o(g)的h_kj·mol1。解析(1)反應(yīng)可得反應(yīng)：則h3h1h2100.3 kj·mol1(11.0 kj·mol1)89.3 kj·mol1。(2)利用蓋斯定律解答本題。cucl2(s)=cucl(s)cl2(g)h183 kj·mol1 cucl(s)o2(g)=cuo(s)

13、cl2(g)h220 kj·mol1 cuo(s)2hcl(g)=cucl2(s)h2o(g)h3121 kj·mol1 則4hcl(g)o2(g)=2cl2(g)2h2o(g)可由×2×2×2得到，所以其h83 kj·mol1×2(20 kj·mol1)×2(121 kj·mol1)×2116 kj·mol1。答案(1)89.3(2)1161由蓋斯定律結(jié)合下述反應(yīng)方程式，回答問題：(1)已知：c(s)o2(g)=co2(g)h1；2co(g)o2(g)=2co2(g)h2；

14、tio2(g)2cl2(g)=ticl4(s)o2(g)h3。則tio2(g)2cl2(g)2c(s)=ticl4(s)2co(g)的h_。(列出關(guān)于h1、h2、h3的表達(dá)式)(2)已知：溫度過高時(shí)，wo2(s)轉(zhuǎn)變?yōu)閣o2(g)：wo2(s)2h2(g)=w(s)2h2o(g)h166.0 kj·mol1wo2(g)2h2(g)=w(s)2h2o(g)h2137.9 kj·mol1則wo2(s)轉(zhuǎn)變?yōu)閣o2(g)的h_。解析(1)將熱化學(xué)方程式×2得：h2h1h2h3。(2)由可得wo2(s)=wo2(g)hh1h2(66.0 kj·mol1)(137

15、.9 kj·mol1)203.9 kj·mol1。答案(1)2h1h2h3(2)203.9 kj·mol12(1)已知：2n2o5(g)=2n2o4(g)o2(g)h14.4 kj·mol12no2(g)=n2o4(g)h255.3 kj·mol1則反應(yīng)n2o5(g)=2no2(g)o2(g)的h_ kj·mol1。(2)ch4­co2催化重整不僅可以得到合成氣(co和h2)，還對(duì)溫室氣體的減排具有重要意義?；卮鹣铝袉栴}：ch4­co2催化重整反應(yīng)為ch4(g)co2(g)=2co(g)2h2(g)。已知：c(s)2h2(g)=ch4(g)h75 kj·mol1c(s)o2(g)=co2(g)h394 kj·mol1c(s)o2