精修版高考化學(xué)復(fù)習(xí) 考點(diǎn)18 蓋斯定律反應(yīng)熱的計(jì)算與比較練習(xí)
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精修版高考化學(xué)復(fù)習(xí) 考點(diǎn)18 蓋斯定律反應(yīng)熱的計(jì)算與比較練習(xí)
精品文檔高考化學(xué)考點(diǎn)18蓋斯定律反應(yīng)熱的計(jì)算與比較1(2015·海南化學(xué),4)已知丙烷的燃燒熱H2 215 kJ·mol1。若一定量的丙烷完全燃燒后生成1.8 g水,則放出的熱量約為()A55 kJ B220 kJ C550 kJ D1 108 kJ2(2015·重慶理綜,6)黑火藥是中國(guó)古代的四大發(fā)明之一,其爆炸的熱化學(xué)方程式為:S(s)2KNO3(s)3C(s)=K2S(s)N2(g)3CO2(g)Hx kJ·mol1已知:碳的燃燒熱H1a kJ·mol1S(s)2K(s)=K2S(s)H2b kJ·mol12K(s)N2(g)3O2(g)=2KNO3(s)H3c kJ·mol1,則x為()A3abc Bc3abCabc Dcab32015·四川理綜,11(4)FeSO4可轉(zhuǎn)化為FeCO3,F(xiàn)eCO3在空氣中加熱反應(yīng)可制得鐵系氧化物材料。已知25 ,101 kPa時(shí):4Fe(s)3O2(g)=2Fe2O3(s)H1 648 kJ/molC(s)O2(g)=CO2(g)H393 kJ/mol2Fe(s)2C(s)3O2(g)=2FeCO3(s)H1 480 kJ/molFeCO3在空氣中加熱反應(yīng)生成Fe2O3的熱化學(xué)方程式是_。42015·廣東理綜,31(1)用O2將HCl轉(zhuǎn)化為Cl2,可提高效益,減少污染。傳統(tǒng)上該轉(zhuǎn)化通過(guò)如下圖所示的催化循環(huán)實(shí)現(xiàn)。其中,反應(yīng)為2HCl(g) CuO(s)H2O(g)CuCl2(s)H1反應(yīng)生成1 mol Cl2的反應(yīng)熱為H2,則總反應(yīng)的熱化學(xué)方程式為_(kāi),(反應(yīng)熱用H1和H2表示)。52015·課標(biāo)全國(guó),27(1)甲醇既是重要的化工原料,又可作為燃料,利用合成氣(主要成分為CO、CO2和H2)在催化劑作用下合成甲醇,發(fā)生的主要反應(yīng)如下:CO(g)2H2(g)CH3OH(g)H1CO2(g)3H2(g)CH3OH(g)H2O(g)H2CO2(g)H2(g)CO(g)H2O(g)H3回答下列問(wèn)題:已知反應(yīng)中相關(guān)的化學(xué)鍵鍵能數(shù)據(jù)如下:化學(xué)鍵HHCOHOCHE/(kJ·mol1)4363431 076465413由此計(jì)算H1_kJ·mol1;已知H258 kJ·mol1,則H3_kJ·mol1。62015·山東理綜,30(3)貯氫合金ThNi5可催化由CO、H2合成CH4的反應(yīng),溫度為T(mén)時(shí),該反應(yīng)的熱化學(xué)方程式為_(kāi)。已知溫度為T(mén)時(shí):CH4(g)2H2O(g)=CO2(g)4H2(g)H165 kJ·mol1CO(g)H2O(g)=CO2(g)H2(g)H41 kJ·mol172015·江蘇化學(xué),20(1)煙氣(主要污染物SO2、NOx)經(jīng)O3預(yù)處理后用CaSO3水懸浮液吸收,可減少煙氣中SO2、NOx的含量。O3氧化煙氣中SO2、NOx的主要反應(yīng)的熱化學(xué)方程式為:NO(g)O3(g)=NO2(g)O2(g)H200.9 kJ·mol1NO(g)O2(g)=NO2(g)H58.2 kJ·mol1SO2(g)O3(g)=SO3(g)O2(g)H241.6 kJ·mol1反應(yīng)3NO(g)O3(g)=3NO2(g)的H_kJ·mol1。82015·海南化學(xué),20(2)已知:Fe2O3(s)3C(s)=2Fe(s)3CO(g)H494 kJ·mol1CO(g)O2(g)=CO2(g)H283 kJ·mol1C(s)O2(g)=CO(g)H110 kJ·mol1則反應(yīng)Fe2O3(s)3C(s)O2(g)=2Fe(s)3CO2(g)的H_ kJ·mol1。理論上反應(yīng)_放出的熱量足以供給反應(yīng)_所需要的熱量(填上述方程式序號(hào))。1(2014·江蘇化學(xué),10)已知:C(s)O2(g)=CO2(g)H1CO2(g)C(s)=2CO(g)H22CO(g)O2(g)=2CO2(g)H34Fe(s)3O2(g)=2Fe2O3(s)H43CO(g)Fe2O3(s)=3CO2(g)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<0 BH2>0,H4>0CH1H2H3 DH3H4H52(2014·重慶理綜,6)已知:C(s)H2O(g)=CO(g)H2(g)Ha kJ·mol12C(s)O2(g)=2CO(g)H220 kJ·mol1HH、O=O和OH鍵的鍵能分別為436、496和462 kJ·mol1,則a為()A332 B118 C350 D1303(2014·課標(biāo)全國(guó),13)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;CuSO4·5H2O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>H3 BH1<H3CH1H3H2 DH1H2>H34(2013·海南化學(xué),5)已知下列反應(yīng)的熱化學(xué)方程式:6C(s)5H2(g)3N2(g)9O2(g)=2C3H5(ONO2)3(l)H12H2(g)O2(g)=2H2O(g)H2C(s)O2(g)=CO2(g)H3則反應(yīng)4C3H5(ONO2)3(l)=12CO2(g)10H2O(g)O2(g)6N2(g)的H為()A12H35H22H1 B2H15H212H3C12H35H22H1 DH15H212H35(2013·課標(biāo)全國(guó),12)在1 200 時(shí),天然氣脫硫工藝中會(huì)發(fā)生下列反應(yīng)H2S(g)O2(g)=SO2(g)H2O(g)H12H2S(g)SO2(g)=S2(g)2H2O(g)H2H2S(g)O2(g)=S(g)H2O(g)H32S(g)=S2(g)H4則H4的正確表達(dá)式為()AH4(H1H23H3)BH4(3H3H1H2)CH4(H1H23H3)DH4(H1H23H3)62014·四川理綜,11(3)已知:25 、101 kPa時(shí),Mn(s)O2(g)=MnO2(s)H520 kJ·mol1S(s)O2(g)=SO2(g)H297 kJ·mol1Mn(s)S(s)2O2(g)=MnSO4(s)H1 065 kJ·mol1SO2與MnO2反應(yīng)生成無(wú)水MnSO4的熱化學(xué)方程式是_。7(2014·廣東理綜,31(1)用CaSO4代替O2與燃料CO反應(yīng),既可提高燃燒效率,又能得到高純CO2,是一種高效、清潔、經(jīng)濟(jì)的新型燃燒技術(shù),反應(yīng)為主反應(yīng),反應(yīng)和為副反應(yīng)。 1/4CaSO4(s)CO(g)1/4CaS(s)CO2(g)H147.3 kJ·mol1 CaSO4(s)CO(g)CaO(s)CO2(g) SO2(g)H2210.5 kJ·mol1 CO(g)1/2C(s)1/2CO2(g)H386.2 kJ·mol1反應(yīng)2CaSO4(s)7CO(g)CaS(s)CaO(s)6CO2(g)C(s)SO2(g)的H_(用H1、H2和H3表示) 82014·課標(biāo)全國(guó),28(2)已知:甲醇脫水反應(yīng)2CH3OH(g)=CH3OCH3(g)H2O(g)H123.9 kJ·mol1甲醇制烯烴反應(yīng)2CH3OH(g)=C2H4(g)2H2O(g)H229.1 kJ·mol1乙醇異構(gòu)化反應(yīng)C2H5OH(g)=CH3OCH3(g)H350.7 kJ·mol1則乙烯氣相直接水合反應(yīng)C2H4(g)H2O(g)=C2H5OH(g)的H_kJ·mol1。92013·四川理綜,11(5)焙燒產(chǎn)生的SO2可用于制硫酸。已知25 、101 kPa時(shí):2SO2(g)O2(g)2SO3(g)H1197 kJ/mol;H2O(g)=H2O(l)H244 kJ/mol;2SO2(g)O2(g)2H2O(g)=2H2SO4(l)H3545 kJ/mol。則SO3(g)與H2O(l)反應(yīng)的熱化學(xué)方程式是_。102013·廣東理綜,31(1)大氣中的部分碘源于O3對(duì)海水中I的氧化,將O3持續(xù)通入NaI溶液中進(jìn)行模擬研究。O3將I氧化成I2的過(guò)程由3步反應(yīng)組成:aI(aq)O3(g)=IO(aq)O2(g)H1bIO(aq)H(aq)HOI(aq)H2cHOI(aq)I(aq)H(aq)I2(aq)H2O(l)H3總反應(yīng)的化學(xué)方程式為_(kāi),其反應(yīng)熱H_。112013·課標(biāo)全國(guó),28(3)二甲醚(CH3OCH3)是無(wú)色氣體,可作為一種新型能源。由合成氣(組成為H2、CO和少量的CO2)直接制備二甲醚,其中的主要過(guò)程包括以下四個(gè)反應(yīng):甲醇合成反應(yīng):(i)CO(g)2H2(g)=CH3OH (g)H190.1 kJ·mol1(ii)CO2(g)3H2(g)=CH3OH(g)H2O(g)H249.0 kJ·mol1水煤氣變換反應(yīng):(iii)CO(g)H2O(g)=CO2(g)H2(g)H341.1 kJ·mol1二甲醚合成反應(yīng):(iv)2CH3OH(g)=CH3OCH3(g)H2O(g)H424.5 kJ·mol1由H2和CO直接制備二甲醚(另一產(chǎn)物為水蒸氣)的熱化學(xué)方程式為_(kāi)。1(2014·河南商丘期末,7)X、Y、Z、W有如圖所示的轉(zhuǎn)化關(guān)系,已知焓變:HH1H2,則X、Y可能是()C、COAlCl3、Al(OH)3Fe、Fe(NO3)2Na2CO3、NaHCO3A B C D2(2014·廣西南寧模擬,8)已知:2CH3OH(g)CH3OCH3(g)H2O(g)H24.5 kJ·mol1,則下列熱化學(xué)方程式中,焓變最大的是()CO2(g)3H2(g) CH3OH(g)H2O(g)H1CO2(g)3H2(g) CH3OH(g)H2O(l)H2CO2(g)3H2(g) CH3OCH3(g)H2O(g)H3CO2(g)3H2(g) CH3OCH3(g)H2O(l)H4AH1 BH2 CH3 DH43(2014·太原測(cè)評(píng),10)在同溫同壓下,下列各組熱化學(xué)方程式中,H2>H1的是()AS(s)O2(g)=SO2(g)H1;S(g)O2(g)=SO2(g)H2B2H2(g)O2(g)=2H2O(g)H1;2H2(g)O2(g)=2H2O(l)H2CH2(g)Cl2(g)=2HCl(g)H1;H2(g)Cl2(g)=HCl(g)H2DCO(g)O2(g)=CO2(g)H1;2CO(g)O2(g)=2CO2(g)H24(2015·山東青島期末,10)T 時(shí),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H表示生成1 mol產(chǎn)物的數(shù)據(jù)),下列說(shuō)法或熱化學(xué)方程式錯(cuò)誤的是()AP(s)Cl2(g)=PCl3(g)H310 kJ·mol1B2PCl5(g)=2P(s)5Cl2(g)H800 kJ·mol1CPCl3(g)Cl2(g)=PCl5(g)H90 kJ·mol1DP和Cl2分兩步反應(yīng)生成1 mol PCl5的焓變與P和Cl2一步反應(yīng)生成1 mol PCl5的焓變不相等5(2015·蘭州質(zhì)檢,13)使用石油熱裂解的副產(chǎn)物甲烷來(lái)制取氫氣,分兩步進(jìn)行,其反應(yīng)過(guò)程中的能量變化如圖所示:則甲烷和水蒸氣反應(yīng)生成二氧化碳和氫氣的熱化學(xué)方程式為()ACH4(g)H2O(g)=3H2(g)CO(g)H103.3 kJ·mol1BCH4(g)2H2O(g)=4H2(g)CO2(g)H70.1 kJ·mol1CCH4(g)2H2O(g)=4H2(g)CO2(g)H70.1 kJ·mol1DCH4(g)2H2O(g)=4H2(g)CO2(g)H136.5 kJ·mol16(2014·銀川調(diào)研,4)白錫和灰錫(以粉末狀存在)是錫的兩種同素異形體。已知:Sn(s、白)2HCl(aq)=SnCl2(aq)H2(g)H1Sn(s、灰)2HCl(aq)=SnCl2(aq)H2(g)H2Sn(s、灰)Sn(s、白)H3m kJ·mol1若H2>H1,則下列說(shuō)法正確的是()Am<0B錫在常溫下以白錫形式存在C灰錫轉(zhuǎn)化為白錫的反應(yīng)是放熱反應(yīng)D要延長(zhǎng)錫制器皿的使用時(shí)間,盡量保持在13.2 以下7(2015·烏魯木齊一次診斷,12)常溫下,0.01 mol·L1MOH溶液的pH為10。已知:2MOH(aq)H2SO4(aq)=M2SO4(aq)2H2O(l)H124.2 kJ·mol1;H(aq)OH(aq)=H2O(l)H257.3 kJ·mol1。則MOH在水溶液中電離的H為()A33.1 kJ·mol1 B45.2 kJ·mol1C81.5 kJ·mol1 D33.1 kJ·mol18(2014·??谡{(diào)研,6)已知CH4(g)和CO(g)的燃燒熱分別是890.3 kJ·mol1和283.0 kJ·mol1,則由CH4(g)不完全燃燒生成1 mol CO(g)和H2O(l)的H為()A607.3 kJ·mol1 B607.3 kJ·mol1C324.3 kJ·mol1 D324.3 kJ·mol19(2014·山東濰坊期末)已知3.6 g碳在6.4 g氧氣中燃燒,至反應(yīng)物耗盡,測(cè)得放出熱量a kJ。又知道12.0 g碳完全燃燒,放出熱量為b kJ,則熱化學(xué)方程式C(s)O2(g)=CO(g)H,則H等于()A(ab) kJ·mol1B(ab) kJ·mol1C(5a0.5b) kJ·mol1D(10ab) kJ·mol110(2015·江蘇鹽城調(diào)研,12)已知:4NH3(g)5O2(g)=4NO(g)6H2O(l)Hx kJ·mol1。蒸發(fā)1 mol H2O(l)需要吸收的熱量為44 kJ,其他相關(guān)數(shù)據(jù)如下表:NH3(g)O2(g)NO(g)H2O(g)1 mol分子中的化學(xué)鍵斷裂時(shí)需要吸收的能量/kJabzd則表中z(用x、a、b、d表示)的大小為()A. B.C. D.11(2014·沈陽(yáng)質(zhì)檢,9)已知反應(yīng):C2H2(g)H2(g)C2H4(g)2CH4(g)C2H4(g)2H2(g)在降低溫度時(shí)式平衡向右移動(dòng),式平衡向左移動(dòng),則下列三個(gè)反應(yīng):C(s)2H2(g)=CH4(g)HQ1C(s)H2(g)=C2H2(g)HQ2C(s)H2(g)=C2H4(g)HQ3若Q1、Q2、Q3均為正值,則下列對(duì)“Q值”大小比較正確的是()AQ1>Q3>Q2 BQ1>Q2>Q3CQ2>Q1>Q3 DQ3>Q1>Q212(2014·石家莊聯(lián)考)將氧化鐵還原為鐵的技術(shù)在人類(lèi)文明的進(jìn)步中占有十分重要的地位。煉鐵高爐中發(fā)生的關(guān)鍵反應(yīng)如下:C(s)O2(g)=CO2(g)H393.5 kJ·mol1CO2(g)C(s)=2CO(g)H172.46 kJ·mol1Fe2O3COFeCO2若已知:2Fe(s)O2(g)=Fe2O3(s)H824.21 kJ·mol1根據(jù)上面三個(gè)熱化學(xué)反應(yīng)方程式,回答下列問(wèn)題:(1)寫(xiě)出CO的燃燒熱的熱化學(xué)反應(yīng)方程式_。(2)高爐內(nèi)Fe2O3被CO還原為Fe的熱化學(xué)反應(yīng)方程式為_(kāi)。(3)煉制1 t含鐵96%的生鐵所需焦炭的理論用量是_t,實(shí)際生產(chǎn)中所需焦炭遠(yuǎn)高于理論用量,其原因是_。13(2014·洛陽(yáng)模擬)參考下列圖表和有關(guān)要求回答問(wèn)題:(1)如圖是1 mol NO2(g)和1 mol CO(g)反應(yīng)生成CO2和NO過(guò)程中能量變化示意圖,若在反應(yīng)體系中加入催化劑,反應(yīng)速率增大,E1的變化是_,H的變化是_(填“增大”、“減小”或“不變”)。請(qǐng)寫(xiě)出NO2和CO反應(yīng)的熱化學(xué)方程式:_。(2)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)原理是:CH3OH(g)H2O(g)=CO2(g)3H2(g)H49.0 kJ·mol1CH3OH(g)1/2O2(g)=CO2(g)2H2(g)H192.9 kJ·mol1又知H2O(g)=H2O(l)H44 kJ·mol1則甲醇燃燒生成液態(tài)水的熱化學(xué)方程式為_(kāi)。14(2014·南昌質(zhì)檢,18)(1)將CH4與H2O(g)通入聚焦太陽(yáng)能反應(yīng)器,可發(fā)生反應(yīng):CH4(g)H2O(g)CO(g)3H2(g),該反應(yīng)的H206 kJ·mol1已知:CH4(g)2O2(g)=CO2(g)2H2O(g)H802 kJ·mol1,寫(xiě)出由CO2和H2O(g)生成CO的熱化學(xué)方程式_。(2)汽車(chē)尾氣中主要含有CO、NO2、SO2、CO2,目前采用的是在汽車(chē)排氣裝置中安裝一個(gè)凈化器可以有效地將尾氣中的有害氣體轉(zhuǎn)化。如:CO(g)NO2(g)=NO(g)2CO2(g)Ha kJ·mol1(a>0)2CO(g)2NO(g)=N2(g)2CO2(g)Hb kJ·mol1(b>0)則用標(biāo)準(zhǔn)狀況下的3.36 L CO還原NO2至N2(CO完全反應(yīng))的整個(gè)過(guò)程中放出的熱量為_(kāi)kJ(用含有a和b的代數(shù)式表示)??键c(diǎn)專練18蓋斯定律反應(yīng)熱的計(jì)算與比較【三年高考真題演練】2015年高考真題1A由丙烷的燃燒熱H2 215 kJ·mol1,可寫(xiě)出其燃燒的熱化學(xué)方程式C3H8(g)5O2(g)=3CO2(g)4H2O(l)H2 215 kJ·mol1,丙烷完全燃燒產(chǎn)生1.8 g水,n(H2O)m÷M1.8 g÷18 g/mol0.1 mol,所以反應(yīng)放出的熱量是Q(2 215 kJ÷4 mol)×0.155.4 kJ,A選項(xiàng)正確。2A由碳的燃燒熱H1a kJ·mol1,得C(s) O2(g)=CO2(g)H1a kJ·mol1,目標(biāo)反應(yīng)可由×3得到,所以H3H1H2H3,即x3abc。3解析對(duì)三個(gè)已知的化學(xué)方程式由上到下依次標(biāo)記為、,對(duì)照反應(yīng)4FeCO3(s)O2(g)2Fe2O3(s)4CO2(g),根據(jù)蓋斯定律,由4×2×可得所求的熱化學(xué)方程式,其反應(yīng)熱H1 648 kJ/mol 4×(393 kJ/mol)2×(1 480 kJ/mol)260 kJ/mol。答案4FeCO3(s)O2(g)=2Fe2O3(s)4CO2(g)H260 kJ/mol4解析根據(jù)題中催化循環(huán)圖示得出:CuCl2(s)O2(g)CuO(s)Cl2(g)H2,然后利用蓋斯定律將兩個(gè)熱化學(xué)方程式相加即可得出2HCl(g)O2(g) H2O(g)Cl2(g)HH1H2。答案2HCl(g)O2(g)H2O(g)Cl2(g)HH1H25解析根據(jù)反應(yīng),H1E(CO)2E(HH)3E(CH)E(CO)E(HO)1 076 kJ·mol12×436 kJ·mol13×413 kJ·mol1343 kJ·mol1465 kJ·mol1 99 kJ·mol1;根據(jù)蓋斯定律,可得反應(yīng),H3H2H158 kJ·mol1(99 kJ·mol1)41 kJ·mol1。答案99416解析CO、H2合成CH4的反應(yīng)為CO(g)3H2(g)=CH4(g)H2O(g)H,將已知的兩個(gè)熱化學(xué)方程式依次編號(hào)為、,即得所求的反應(yīng),根據(jù)蓋斯定律有:H41 kJ·mol1(165 kJ·mol1)206 kJ·mol1。答案CO(g)3H2(g)=CH4(g)H2O(g)H206 kJ·mol17解析對(duì)所給的三個(gè)熱化學(xué)方程式由上到下依次標(biāo)記為、,由反應(yīng)和可知O2是中間產(chǎn)物,×2消去O2,可得目標(biāo)反應(yīng)的H200.9 kJ·mol1(58.2 kJ·mol1)×2317.3 kJ·mol1。答案317.38解析由題中已知的三個(gè)熱化學(xué)方程式:Fe2O3(s)3C(s)=2Fe(s)3CO(g)H494 kJ·mol1CO(g)O2(g)=CO2(g)H283 kJ·mol1C(s)O2(g)=CO(g)H110 kJ·mol1根據(jù)蓋斯定律:×3得反應(yīng):Fe2O3(s)3C(s)O2(g)=2Fe(s)3CO2(g)的H355 kJ·mol1。理論上反應(yīng)放出的熱量足以供給反應(yīng)所需的熱量。答案355兩年經(jīng)典高考真題1C設(shè)題中反應(yīng)由上到下分別為、,反應(yīng)為碳的燃燒,是放熱反應(yīng),H1<0,反應(yīng)為吸熱反應(yīng),H2>0,反應(yīng)為CO的燃燒,是放熱反應(yīng),H3<0,反應(yīng)為鐵的氧化反應(yīng)(化合反應(yīng)),是放熱反應(yīng),H4<0,A、B錯(cuò)誤;C項(xiàng),由于反應(yīng)反應(yīng)反應(yīng),所以H1H2H3,正確;D項(xiàng),反應(yīng)(反應(yīng)反應(yīng)×2)/3,所以H3,錯(cuò)誤。2DC(s)H2O(g)=CO(g)H2(g)Ha kJ·mol12C(s)O2(g)=2CO(g)H220 kJ·mol1根據(jù)蓋斯定律×2得2H2O(g)=2H2(g)O2(g)H(2a220) kJ·mol1根據(jù)HH、O=O和OH鍵的鍵能數(shù)據(jù),得4×462 kJ·mol12×436 kJ·mol1496 kJ·mol1(2a220) kJ·mol1,解得a130,A、B、C錯(cuò)誤;D正確。3B根據(jù)題意,可知CuSO4·5H2O(s)=Cu2(aq)SO(aq)5H2O(l)H1>0,CuSO4(s)=Cu2(aq)SO(aq)H2<0,CuSO4·5H2O(s)CuSO4(s)5H2O(l)H3,根據(jù)蓋斯定律有,則H1H2H3,即有H3H1H2>0。A項(xiàng),由以上分析知,H2<0,H3>0,故有H3>H2,錯(cuò)誤;B項(xiàng),由于H1>0,H2<0,故有H1<H1H2H3,正確;由以上分析H3H1H2,故C、D錯(cuò)誤。4A設(shè)三個(gè)熱化學(xué)方程式依次是、,根據(jù)蓋斯定律,×12×5×2得:4C3H5(ONO2)3(l)=12CO2(g)10H2O(g)O2(g)6N2(g)H12H35H22H1。5A式中含有S2(g)、式中含有S(g),根據(jù)蓋斯定律可知,××2得2S(g)SO2(g)H2O(g)=H2S(g)S2(g)O2(g),然后再加×得2S(g)=S2(g),所以H4H2×H1×H3×2。6解析將題中三個(gè)熱化學(xué)方程式由上到下依次編號(hào)為、;根據(jù)蓋斯定律可知,得MnO2(s)SO2(g)=MnSO4(s)H1 065 kJ·mol1520 kJ·mol1297 kJ·mol1248 kJ·mol1答案MnO2(s)SO2(g)=MnSO4(s)H248 kJ·mol17解析根據(jù)蓋斯定律,可知×4×2得H4H1H22H3。答案4H1H22H38解析2CH3OH(g)=CH3OCH3(g)H2O(g)H123.9 kJ·mol12CH3OH(g)=C2H4(g)2H2O(g)H229.1 kJ·mol1C2H5OH(g)=CH3OCH3(g)H350.7 kJ·mol1根據(jù)蓋斯定律,由得:C2H4(g)H2O(g)=C2H5OH(g)H45.5 kJ·mol1。答案45.59解析根據(jù)蓋斯定律,得出SO3(g)H2O(l)=H2SO4(l)H4只須按下式進(jìn)行即可:H(H3H12H2)130 kJ·mol1熱化學(xué)方程式應(yīng)為:SO3(g)H2O(l)=H2SO4(l)H130 kJ·mol1答案SO3(g)H2O(l)=H2SO4(l)H130 kJ·mol1102I(aq)O3(g)2H(aq)=I2(aq)O2(g)H2O(l)H1H2H311解析篩選用熱化學(xué)方程式為()和(),用()×2()即可得目標(biāo)熱化學(xué)方程式:2CO(g)4H2(g)=CH3OCH3(g)H2O(g)H204.7 kJ·mol1答案2CO(g)4H2(g)=CH3OCH3(g)H2O(g)H204.7 kJ·mol1【兩年模擬試題精練】1A2A根據(jù)蓋斯定律,由得:H2O(l)=H2O(g)HH1H2,由于液態(tài)水變?yōu)闅鈶B(tài)水要吸熱,所以H1H2>0,即H1>H2。同理可得:H3>H4。設(shè)題目中的熱化學(xué)方程式的編號(hào)為,由于×,所以H3H1(24.5 kJ·mol1)×<H1。綜上,H1最大。3C分析四個(gè)選項(xiàng)知所涉及反應(yīng)均是放熱反應(yīng),H均小于0,所以反應(yīng)放出熱量越多,對(duì)應(yīng)的H反而越小。A項(xiàng),已知S(g)S(s)需要放熱,所以前一反應(yīng)放出的熱量少,H1>H2,錯(cuò)誤;B項(xiàng),H2O(g)H2O(l)需要放出熱量,所以前一反應(yīng)放出熱量少,H1>H2,錯(cuò)誤;C項(xiàng),由熱化學(xué)方程式化學(xué)計(jì)量數(shù)知H12H2<0,所以H2>H1,正確;D項(xiàng),由化學(xué)計(jì)量數(shù)知H1H2<0,所以H1>H2,錯(cuò)誤。4D由能量圖知A、B、C項(xiàng)正確;P和Cl2分兩步反應(yīng)生成1 mol PCl5的焓變與P和Cl2一步反應(yīng)生成1 mol PCl5的焓變相等,D項(xiàng)錯(cuò)誤。5D第步反應(yīng)對(duì)應(yīng)的熱化學(xué)方程式為CH4(g)H2O(g)=3H2(g)CO(g)H103.3 kJ·mol1,第步反應(yīng)對(duì)應(yīng)的熱化學(xué)方程式為CO(g)H2O(g)=H2(g)CO2(g)H33.2 kJ·mol1,根據(jù)蓋斯定律,將兩式相加即可得所求熱化學(xué)方程式,本題選D。6B根據(jù)蓋斯定律,由可得,故H3H2H1,因H2>H1,故H2H1>0,即H3m kJ·mol1>0,的正反應(yīng)為吸熱反應(yīng),A、C項(xiàng)錯(cuò)誤。根據(jù)可知,常溫下,灰錫會(huì)向白錫轉(zhuǎn)化,故常溫下錫以白錫形式存在,故B項(xiàng)正確。當(dāng)錫制器皿長(zhǎng)期處于13.2 以下的環(huán)境中時(shí),會(huì)轉(zhuǎn)化為灰錫,而灰錫以粉末狀存在,故D項(xiàng)錯(cuò)誤。7B根據(jù)題中0.01 mol/L MOH溶液的pH10,知MOH為弱堿,MOH溶液與硫酸的中和反應(yīng)可以看做兩個(gè)過(guò)程:MOH(aq)M(aq)OH(aq)H、H(aq)OH(aq)=H2O(l)H2,根據(jù)蓋斯定律知H12(HH2),則HH1H2(24.2 kJ·mol1)×(57.3 kJ·mol1)45.2 kJ·mol1,B項(xiàng)正確。8A由CH4、CO的燃燒熱可以寫(xiě)出兩個(gè)熱化學(xué)方程式:CH4(g)2O2(g)=CO2(g)2H2O(l)H1890.3 kJ/mol,CO(g)O2(g)=CO2(g)H2283.0 kJ/mol,CH4不完全燃燒的熱化學(xué)方程式為CH4(g)O2(g)=CO(g)2H2O(l)H,該熱化學(xué)方程式可由熱化學(xué)方程式熱化學(xué)方程式得到,根據(jù)蓋斯定律HH1H2890.3 kJ/mol283.0 kJ/mol607.3 kJ/mol,A項(xiàng)正確。9C根據(jù)題意得兩個(gè)熱化學(xué)方程式:3C(s)2O2(g)=2CO(g)CO2(g)H10a kJ·mol1;C(s)O2(g)=CO2(g)Hb kJ·mol1;則()得C(s)O2(g)=CO(g)H(5a0.5b)kJ·mol1。10A由題給條件可得6H2O(l)=6H2O(g)H264 kJ·mol1,根據(jù)蓋斯定律,將此式與題干中熱化學(xué)方程式同向相加得4NH3(g)5O2(g)=4NO(g)6H2O(g)H(x264) kJ·mol1。則有4a5b(4z6d)(x264),解得z,A項(xiàng)正確。11A降低溫度時(shí),式分別向右、左移動(dòng),則式H<0,式H>0;由式×2式×2得式,即有2H()2H()<0,可得出Q3>Q2;再由式×2式×2得式,有2H()2H()>0,可得出Q1>Q3,故Q1>Q3>Q2,A正確。12解析由題意:C(s)O2(g)=CO2(g)H1393.5 kJ·mol1CO2(g)C(s)=2CO(g)H2172.46 kJ·mol12Fe(s)O2(g)=Fe2O3(s)H3824.21 kJ·mol1CO(g)O2(g)=CO2(g)H4Fe2O3(s)3CO(g)=2Fe(s)3CO2(g)H5根據(jù)蓋斯定律:(1)(),所以H4(H1H2)×(393.5 kJ·mol1172.46 kJ·mol1)282.98 kJ·mol1。(2)××,所以H5H1×H3H2×393.5 kJ·mol1×(824.21 kJ·mol1)172.46 kJ·mol1×24.73 kJ·mol1。(3)根據(jù)化學(xué)方程式計(jì)算理論值,由方程式得:x0.31 t。答案(1)CO(g)O2(g)=CO2(g)H282.98 kJ·mol1(2)Fe2O3(s)3CO(g)=2Fe(s)3CO2(g)H24.73 kJ·mol1(3)0.31焦炭沒(méi)有被充分利用(或其他合理說(shuō)法)13解析(1)觀察圖形,E1應(yīng)為反應(yīng)的活化能,加入催化劑反應(yīng)的活化能降低,但是H不變;1 mol NO2(g)和1 mol CO(g)反應(yīng)生成CO2和NO的反應(yīng)熱數(shù)值即反應(yīng)物和生成物的能量差,因此該熱化學(xué)方程式為NO2(g)CO(g)=CO2(g)NO(g)H234 kJ·mol1。(2)熱化學(xué)方程式:3××2×2得熱化學(xué)方程式:CH3OH(g)3/2O2(g)=CO2(g)2H2O(l)H3×(192.9 kJ·mol1)2×49.0 kJ·mol1(44 kJ·mol1)×2764.7 kJ·mol1。答案(1)減小不變NO2(g)CO(g)=CO2(g)NO(g)H234 kJ·mol1(2)CH3OH(g)3/2O2(g)=CO2(g)2H2O(l)H764.7 kJ·mol114解析(1)分析兩式子,把式式可得CO2(g)3H2O(g)=2O2(g)CO(g)3H2(g)H206 kJ·mol1(802 kJ·mol1)1 008 kJ·mol1。(2)CO(g)NO2(g)=NO(g)CO2(g)Ha kJ·mol1乘以2得2CO(g)2NO2(g)=2NO(g)2CO2(g)H2a kJ·mol1與2CO(g)2NO(g)=N2(g)2CO2(g)Hb kJ·mol1相加即得4CO(g)2NO2(g)=N2(g)4CO2(g)H(2ab) kJ·mol1,n(CO)0.15 mol,所以放出的熱量為×(2ab) kJ。答案(1)CO2(g)3H2O(g)=2O2(g)CO(g)3H2(g)H1 008 kJ·mol1(2)