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1、（江蘇專用）2022屆高考化學(xué)二輪復(fù)習(xí) 壓軸題熱點(diǎn)練2 蓋斯定律的應(yīng)用1聯(lián)氨(又稱聯(lián)肼，N2H4，無(wú)色液體)是一種應(yīng)用廣泛的化工原料，可用作火箭燃料，回答下列問(wèn)題：2O2(g)N2(g)=N2O4(l)H1N2(g)2H2(g)=N2H4(l)H2O2(g)2H2(g)=2H2O(g)H32N2H4(l)N2O4(l)=3N2(g)4H2O(g)H41048.9 kJ/mol上述反應(yīng)熱效應(yīng)之間的關(guān)系式為H4_，聯(lián)氨和N2O4可作為火箭推進(jìn)劑的主要原因?yàn)開(kāi)。答案2H32H2H1反應(yīng)放熱量大、產(chǎn)生大量氣體解析根據(jù)蓋斯定律，22即得2N2H4(l)N2O4(l)=3N2(g)4H2O(g)的H4，所

2、以反應(yīng)熱效應(yīng)之間的關(guān)系式為H42H32H2H1。聯(lián)氨有強(qiáng)還原性，N2O4有強(qiáng)氧化性，兩者在一起易發(fā)生氧化還原反應(yīng)，反應(yīng)放熱量大、產(chǎn)生大量氣體，所以聯(lián)氨和N2O4可作為火箭推進(jìn)劑。2已知：C(s，石墨)O2(g)=CO2(g)H1393.5 kJmol12H2(g)O2(g)=2H2O(l)H2571.6 kJmol12C2H2(g)5O2(g)=4CO2(g)2H2O(l)H32599 kJmol1根據(jù)蓋斯定律，計(jì)算反應(yīng)2C(s，石墨)H2(g)=C2H2(g)的H_。答案226.7 kJmol1解析C(s，石墨)O2(g)=CO2(g)H1393.5 kJmol1，2H2(g)O2(g)=2

3、H2O(l)H2571.6 kJmol1，2C2H2(g)5O2(g)=4CO2(g)2H2O(l)H32599 kJmol1；根據(jù)蓋斯定律計(jì)算(2)得2C(s，石墨)H2(g)=C2H2(g)H(393.5 kJmol1)2(571.6 kJmol1)(2599 kJmol1)226.7 kJmol1。3由金紅石(TiO2)制取單質(zhì)Ti，涉及的步驟為：TiO2TiCl4Ti已知：C(s)O2(g)=CO2(g)H1393.5 kJmol12CO(g)O2(g)=2CO2(g)H2566 kJmol1TiO2(s)2Cl2(g)=TiCl4(s)O2(g)H3141 kJmol1則TiO2(s

4、)2Cl2(g)2C(s)=TiCl4(s)2CO(g)的H_。答案80 kJmol1解析2就可得TiO2(s)2Cl2(g)2C(s)=TiCl4(s)2CO(g)，則HH3H12H280 kJmol1。4甲醇質(zhì)子交換膜燃料電池中將甲醇蒸氣轉(zhuǎn)化為氫氣的兩種反應(yīng)的熱化學(xué)方程式如下：CH3OH(g)H2O(g)=CO2(g)3H2(g)H49.0 kJmol1CH3OH(g)O2(g)=CO2(g)2H2(g)H192.9 kJmol1又知H2O(g)=H2O(l)H44 kJmol1，則甲醇蒸氣燃燒生成液態(tài)水的熱化學(xué)方程式為_(kāi)。答案CH3OH(l)O2(g)=CO2(g)2H2O(l)H764

5、.7 kJmol1解析根據(jù)蓋斯定律計(jì)算(322)得：CH3OH(l)O2(g)=CO2(g)2H2O(l)H3(192.9 kJmol1)249.0 kJmol1(44 kJmol1)2764.7 kJmol1；則甲醇蒸氣燃燒為液態(tài)水的熱化學(xué)方程式為：CH3OH(l)O2(g)=CO2(g)2H2O(l)H764.7 kJmol1。5已知：H2的熱值為142.9 kJg1(熱值是表示單位質(zhì)量的燃料完全燃燒生成穩(wěn)定的化合物時(shí)所放出的熱量)；N2(g)2O2(g)=2NO2(g)H133 kJmol1H2O(g)=H2O(l)H44 kJmol1催化劑存在下，H2還原NO2生成水蒸氣和其他無(wú)毒物質(zhì)

6、的熱化學(xué)方程式：_。答案4H2(g)2NO2(g)=N2(g)4H2O(g)H1100.2 kJmol1解析已知：H2的熱值為142.9 kJg1，則H2(g)O2(g)=H2O(l)H285.8 kJmol1；N2(g)2O2(g)=2NO2(g)H133 kJmol1；H2O(g)=H2O(l)H44 kJmol1；根據(jù)蓋斯定律由44可得4H2(g)2NO2(g)=4H2O(g)N2(g)H(285.8 kJmol1)4(133 kJmol1)(44 kJmol1)41100.2 kJmol1，故此反應(yīng)的熱化學(xué)方程式為4H2(g)2NO2(g)=N2(g)4H2O(g)H1100.2 kJ

7、mol1。6能源問(wèn)題是人類社會(huì)面臨的重大課題，H2、CO、CH3OH都是重要的能源物質(zhì)，它們的燃燒熱依次為285.8 kJmol1、282.5 kJmol1、726.7 kJmol1。已知CO和H2在一定條件下可以合成甲醇CO(g)2H2(g)=CH3OH(l)。則CO與H2反應(yīng)合成甲醇的熱化學(xué)方程式為_(kāi)。答案CO(g)2H2(g)=CH3OH(l)H127.4 kJmol1解析根據(jù)目標(biāo)反應(yīng)與三種反應(yīng)熱的關(guān)系，利用蓋斯定律，計(jì)算出目標(biāo)反應(yīng)的反應(yīng)熱H2(285.8 kJmol1)(282.5 kJmol1)(726.7 kJmol1)127.4 kJmol1。7已知：25 、101 kPa時(shí)，M

8、n(s)O2(g)=MnO2(s)H520 kJmol1S(s)O2(g)=SO2(g)H297 kJmol1Mn(s)S(s)2O2(g)=MnSO4(s)H1065 kJmol1則SO2與MnO2反應(yīng)生成無(wú)水MnSO4的熱化學(xué)方程式是_。答案MnO2(s)SO2(g)=MnSO4(s)H248 kJmol1解析將題給三個(gè)熱化學(xué)方程式依次編號(hào)為，根據(jù)蓋斯定律，由可得SO2(g)MnO2(s)=MnSO4(s)H(1065 kJmol1)(520 kJmol1)(297 kJmol1)248 kJmol1。8已知下列熱化學(xué)方程式：Fe2O3(s)3CO(g)=2Fe(s)3CO2(g)H25

9、kJmol13Fe2O3(s)CO(g)=2Fe3O4(s)CO2(g)H47 kJmol1Fe3O4(s)CO(g)=3FeO(s)CO2(g)H19 kJmol1寫(xiě)出FeO(s)被CO還原成Fe和CO2的熱化學(xué)方程式_。答案FeO(s)CO(g)=Fe(s)CO2(g)H11 kJmol1解析32就可得6FeO(s)6CO(g)=6Fe(s)6CO2(g)H66 kJmol1，即FeO(s)CO(g)=Fe(s)CO2(g)H11 kJmol1。9已知：2Cu2S(s)3O2(g)=2Cu2O(s)2SO2(g)H768.2 kJmol12Cu2O(s)Cu2S(s)=6Cu(s)SO2(g)H116.0 kJmol1則Cu2S(s)O2(g)=2Cu(s)SO2(g)H_。答案217.4 kJmol1解析根據(jù)蓋斯定律，將方程式()得Cu2S(s)O2(g)=2Cu(s)SO2(g)H(768.2116.0) kJmol1217.4 kJmol1。