2019高考化學(xué)大一輪復(fù)習(xí) 第6章 化學(xué)能與熱能 6-2 蓋斯定律及其應(yīng)用練習(xí) 新人教版.doc

6-2 蓋斯定律及其應(yīng)用板塊三 限時(shí)規(guī)范特訓(xùn)時(shí)間：45分鐘滿分：100分 一、選擇題(每題7分，共77分)12017河北武邑中學(xué)調(diào)研已知反應(yīng)：H2(g)O2(g)=H2O(g)H1N2(g)O2(g)=NO2(g)H2N2(g)H2(g)=NH3(g)H3則反應(yīng)2NH3(g)O2(g)=2NO2(g)3H2O(g)的H為()A2H12H22H3 BH1H2H3C3H12H22H3 D3H12H22H3答案C解析根據(jù)蓋斯定律，由322可得2NH3(g)O2(g)=2NO2(g)3H2O(g)則有H3H12H22H3，C項(xiàng)正確。2已知：NH3H2O(aq)與H2SO4(aq)反應(yīng)生成1 mol正鹽的H24.2 kJmol1，強(qiáng)酸、強(qiáng)堿稀溶液反應(yīng)的中和熱為H57.3 kJmol1，則NH3H2O在水溶液中電離的H等于()A69.4 kJmol1 B45.2 kJmol1C69.4 kJmol1 D45.2 kJmol1答案D解析由題給條件可先寫出NH3H2O(aq)H(aq)=NH(aq)H2O(l)H12.1 kJmol1，H(aq)OH(aq)=H2O(l)H57.3 kJmol1，根據(jù)蓋斯定律，由即可得到NH3H2O(aq)NH(aq)OH(aq)H45.2 kJmol1，D正確。32017河南洛陽期中已知：Fe2O3(s)C(s)=CO2(g)2Fe(s)H1234.1 kJmol1C(s)O2(g)=CO2(g)H2393.5 kJmol1則2Fe(s)O2(g)=Fe2O3(s)的H是()A824.4 kJmol1 B627.6 kJmol1C744.7 kJmol1 D159.4 kJmol1答案A解析分析題給兩個(gè)熱化學(xué)方程式，根據(jù)蓋斯定律，由可得：2Fe(s)O2(g)=Fe2O3(s)，則有HH2H1(393.5 kJmol1)(234.1 kJmol1)824.4 kJmol1。4已知下列四個(gè)熱化學(xué)方程式(H的單位均為kJ/mol)：NH4Cl(s)=NH3(g)HCl(g)H1a()Ba(OH)28H2O(s)=BaO(s)9H2O(l)H2b()2NH4Cl(s)Ba(OH)28H2O(s)=BaCl2(aq)2NH3(g)10H2O(l)H3c()BaO(s)2HCl(aq)=BaCl2(aq)H2O(l)H4d()其中a、b、c均大于0，HCl氣體溶解于水，溶液溫度升高。由此可知下列判斷一定正確的是()Ad<0 Bc<2abdCc2abd Dc>2abd答案B解析通過給出的信息無法判斷反應(yīng)()是放熱反應(yīng)還是吸熱反應(yīng)，A錯(cuò)誤；根據(jù)題目提示信息可得：HCl(g)=HCl(aq)H5e，由蓋斯定律可確定c2abd2e，因e<0，故c(2abd)<0，即c<2abd，B正確，C、D錯(cuò)誤。5Mg2Ni是一種儲(chǔ)氫合金，已知：Mg(s)H2(g)=MgH2(s)H174.5 kJmol1Mg2Ni(s)2H2(g)=Mg2NiH4(s)H264.4 kJmol1Mg2Ni(s)2MgH2(s)=2Mg(s)Mg2NiH4(s)H3則H3等于()A84.6 kJmol1 B84.6 kJmol1C138.9 kJmol1 D138.9 kJmol1答案B解析根據(jù)蓋斯定律，由2可得Mg2Ni(s)2MgH2(s)=2Mg(s)Mg2NiH4(s)，則有H3H22H1(64.4 kJmol1)(74.5 kJmol1)284.6 kJmol1。6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 DH3H4H5答案C解析C與O2生成CO2的反應(yīng)是放熱反應(yīng)，H1<0，CO2與C生成CO的反應(yīng)是吸熱反應(yīng)，H2>0，CO與O2生成CO2的反應(yīng)是放熱反應(yīng)，H3<0，鐵與氧氣的反應(yīng)是放熱反應(yīng)，H4<0，A、B項(xiàng)錯(cuò)誤；前兩個(gè)方程式相減得：2CO(g)O2(g)=2CO2(g)H3H1H2，即H1H2H3，C項(xiàng)正確；由4Fe(s)3O2(g)=2Fe2O3(s)H4和6CO(g)2Fe2O3(s)=6CO2(g)4Fe(s)2H5相加，得2CO(g)O2(g)=2CO2(g)H3(H42H5)/3，D項(xiàng)錯(cuò)誤。72017江西南昌摸底調(diào)研已知：2C(s)O2(g)=2CO(g)H221.0 kJmol1C(s)O2(g)=CO2(g)H393.5 kJmol12H2(g)O2(g)=2H2O(g)H483.6 kJmol1則制備水煤氣的反應(yīng)C(s)H2O(g)=CO(g)H2(g)的H為()A131.3 kJmol1 B131.3 kJmol1C373.1 kJmol1 D373.1 kJmol1答案A解析根據(jù)蓋斯定律，由可得C(s)H2O(g)=CO(g)H2(g)，則有H(221.0 kJmol1)(483.6 kJmol1)131.3 kJmol1。8已知：C(s)H2O(g)=CO(g)H2(g)Ha kJmol12C(s)O2(g)=2CO(g) H220 kJmol1HH、O=O和OH鍵的鍵能分別為436、496和462 kJmol1，則a為()A332 B118 C350 D130答案D解析根據(jù)蓋斯定律由題給的兩個(gè)熱化學(xué)方程式可得：2H2O(g)=2H2(g)O2(g)H(2a220) kJmol1，則有：4462 kJmol12436 kJmol1496 kJmol1(2a220) kJmol1，解得a130，故選項(xiàng)D正確。92018北京朝陽統(tǒng)考聯(lián)氨(N2H4)是一種應(yīng)用廣泛的化工原料，可用作火箭燃料。已知：N2H4(l)N2O4(l)=N2(g)2H2O(l)H546.45 kJmol1H2O(g)=H2O(l) H44.0 kJmol1則2N2H4(l)N2O4(l)=3N2(g)4H2O(g)的H是()A916.9 kJmol1 B458.45 kJmol1C916.9 kJmol1 D458.45 kJmol1答案A解析根據(jù)蓋斯定律，由24可得2N2H4(l)N2O4(l)=3N2(g)4H2O(g)，則有H(546.45 kJmol1)2(44.0 kJmol1)4916.9 kJmol1。10在1200 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)答案A解析首先找到式中含有S2(g)、式中含有S(g)，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H4H2H1H32。112018蘭州一中高三月考已知：2H2(g)O2(g)=2H2O(l)H571.6 kJmol12CH3OH(l)3O2(g)=2CO2(g)4H2O(l)H1452 kJmol1H(aq)OH(aq)=H2O(l)H57.3 kJmol1下列說法正確的是()AH2(g)的燃燒熱為142.9 kJmol1B同質(zhì)量的H2(g)和CH3OH(l)完全燃燒，H2(g)放出的熱量多C.H2SO4(aq)Ba(OH)2(aq)=BaSO4(s)H2O(l)H57.3 kJmol1D3H2(g)CO2(g)=CH3OH(l)H2O(l)H131.4 kJmol1答案B解析根據(jù)燃燒熱的定義可知，H2(g)的燃燒熱為285.8 kJmol1，A項(xiàng)錯(cuò)誤；2 mol (即4 g) H2(g)完全燃燒放出571.6 kJ的熱量，2 mol(即64 g) CH3OH(l)完全燃燒放出1452 kJ的熱量，故單位質(zhì)量的H2(g)放出的熱量多，B項(xiàng)正確；HOH=H2O和Ba2SO=BaSO4都是放熱的，所以反應(yīng)H<57.3 kJmol1，C項(xiàng)錯(cuò)誤；將題干中的熱化學(xué)方程式依次編號(hào)為，根據(jù)蓋斯定律，由，可得熱化學(xué)方程式：3H2(g)CO2(g)=CH3OH(l)H2O(l)H131.4 kJmol1，D項(xiàng)錯(cuò)誤。二、非選擇題(共23分)12. (12分)現(xiàn)利用如圖裝置進(jìn)行中和熱的測(cè)定，請(qǐng)回答下列問題：(1)該圖中有兩處未畫出，它們是_、_。(2)把溫度為15.0 ，濃度為0.5 molL1的酸溶液和0.55 molL1的堿溶液各50 mL混合(溶液密度均為1 gmL1)，生成溶液的比熱容c4.18 Jg11，輕輕攪動(dòng)，測(cè)得酸堿混合液的溫度變化數(shù)據(jù)如下：反應(yīng)物起始溫度t1/終止溫度t2/中和熱HClNaOH15.018.4H1HClNH3H2O15.018.1H2試計(jì)算上述兩組實(shí)驗(yàn)測(cè)出的中和熱H1_，H2_。兩組實(shí)驗(yàn)結(jié)果差異的原因是_。寫出HClNH3H2O的熱化學(xué)方程式：_。答案(1)環(huán)形玻璃攪拌棒燒杯上方的泡沫塑料蓋(2)56.8 kJmol1 51.8 kJmol1NH3H2O是弱堿，在中和過程中NH3H2O發(fā)生電離，要吸熱，因而總體放熱較少HCl(aq)NH3H2O(aq)=NH4Cl(aq)H2O(l)H51.8 kJmol1解析(2)根據(jù)題目給出的酸、堿的物質(zhì)的量，酸為0.025 mol，堿為0.0275 mol，堿是過量的，應(yīng)根據(jù)酸的量進(jìn)行計(jì)算，即反應(yīng)生成了0.025 mol H2O。根據(jù)實(shí)驗(yàn)數(shù)據(jù)算出生成1 mol H2O所放出的熱量，即得出兩組實(shí)驗(yàn)測(cè)出的中和熱數(shù)值：H14.18(5050)(18.415.0)1030.02556.8 kJmol1，H24.18(5050)(18.115.0)1030.02551.8 kJmol1。132017四川綿陽診斷(11分)汽車尾氣中排放的NOx和CO污染環(huán)境，在汽車尾氣系統(tǒng)中裝置催化轉(zhuǎn)化器，可有效降低NOx和CO的排放。.已知：2CO(g)O2(g)=2CO2(g)H1566.0 kJmol1N2(g)O2(g)=2NO(g)H2180.5 kJmol12NO(g)O2(g)=2NO2(g)H3116.5 kJmol1(1)CO的燃燒熱為_。若1 mol N2(g)、1 mol O2(g)分子中化學(xué)鍵斷裂時(shí)分別需要吸收946 kJ、498 kJ的能量，則1 mol NO(g)分子中化學(xué)鍵斷裂時(shí)需吸收的能量為_。(2)CO將NO2還原為單質(zhì)的熱化學(xué)方程式為_。.利用水煤氣合成二甲醚的總反應(yīng)為：3CO(g)3H2(g)=CH3OCH3(g)CO2(g)H246.4 kJmol1(3)它可以分為兩步，反應(yīng)分別如下：2CO(g)4H2(g)=CH3OCH3(g)H2O(g)H1205.1 kJmol1CO(g)H2O(g)=CO2(g)H2(g)H2_。(4)已知CH3OCH3(g)的燃燒熱為1455 kJmol1，寫出表示其燃燒熱的熱化學(xué)方程式：_。若二甲醚燃燒生成的CO2恰好能被100 mL 0.2 molL1 NaOH溶液吸收生成Na2CO3，則燃燒過程中放出的熱量為_。答案(1)283.0 kJmol1631.75 kJ(2)2NO2(g)4CO(g)=N2(g)4CO2(g)H1196 kJmol1(3)41.3 kJmol1(4)CH3OCH3(g)3O2(g)=2CO2(g)3H2O(l)H11455 kJmol17.275 kJ解析(1)由可知，2 mol CO(g)完全燃燒生成CO2(g)時(shí)放出566.0 kJ熱量，則CO(g)的燃燒熱為283.0 kJmol1。由可知，1 mol N2(g)和1 mol O2(g)反應(yīng)生成2 mol NO(g)吸收180.5 kJ熱量，根據(jù)反應(yīng)熱與鍵能的關(guān)系可得946 kJmol1498 kJmol12E(氮氧鍵)180.5 kJmol1，則有E(氮氧鍵)631.75 kJmol1。(2)CO將NO2還原為單質(zhì)的反應(yīng)為2NO2(g)4CO(g)=N2(g)4CO2(g)，分析題給三個(gè)熱化學(xué)方程式，根據(jù)蓋斯定律，由2可得2NO2(g)4CO(g)=N2(g)4CO2(g)H( 566.0 kJmol1)2(180.5 kJmol1)(116.5 kJmol1)1196 kJmol1。(3)根據(jù)蓋斯定律，由總反應(yīng)減去反應(yīng)可得反應(yīng)，則H2HH1(246.4 kJmol1)(205.1 kJmol1)41.3 kJmol1。(4)n(NaOH)0.1 L0.2 molL10.02 mol，可與0.01 mol CO2反應(yīng)生成Na2CO3，則有Q kJmol10.01 mol7.275 kJ。