浙江高考选考科目化学模拟试题(二)

浙江高考选考科目化学模拟试题(二)
浙江高考选考科目化学模拟试题(二)

浙江省普通高校招生选考科目化学模拟试题(二)

(时间:90分钟满分:100分)

可能用到的相对原子质量:H—1C—12N—14O—16Na—23S—32 Cl—35.5Ca—40Cr—52Fe—56Cu—64Ba—137

选择题部分

一、选择题(本大题共20小题,共50分。第1~10每小题2分,第11~20每小题3分。每个小题列出的四个备选项中只有一个是符合题目要求的,不选、多选、错选均不得分)

1.人类生活离不开化学知识,下列叙述不正确的是()

A.臭氧是饮用水的理想消毒剂之一,因为它杀菌能力强且不影响水质

B.因患“禽流感”而被捕杀的家禽尸体常用生石灰处理

C.医疗上常用体积分数为75%的酒精进行皮肤消毒

D.食品加工、消毒、防腐常使用福尔马林

解析福尔马林有毒,不能用于食品加工,D项错误。

答案 D

2.下列属于酸性氧化物的是()

A.NO

B.MgO

C.H2O2

D.CO2

解析酸性氧化物是指能够与碱反应生成盐和水的氧化物,且反应为非氧化还原反应,氧元素价态为-2价;A.NO属于非成盐氧化物,不符合;B.MgO属于碱性氧化物,不符合;C.H2O2属于过氧化物,不符合;D.CO2属于酸性氧化物,符合;故应选D。

答案 D

3.下列说法不正确的是()

A.二氧化硫可用于漂白纸浆

B.热的纯碱溶液可以去除厨房里的油污

C.钠钾合金常温下是液体,可用于快中子反应堆的热交换剂

D.浓硫酸可使有机物炭化,是由于它的吸水性,实验室常用作干燥剂

解析 A.二氧化硫与有机色素结合生成无色物质,具有漂白性,因此二氧化硫可

用于漂白纸浆,故A正确;B.油污中含—COOC—,热的纯碱溶液促进水解后碱性增强,且碱性利于油污水解,则热的纯碱溶液可以去除油污,故B正确;C.钠钾合金常温下是液体,可用于热量交换,则可用于快中子反应堆的热交换剂,故C 正确;D.浓硫酸使有机物炭化,与其脱水性有关,而浓硫酸具有吸水性可用作干燥剂,故D错误;故选D。

答案 D

4.下列表示正确的是()

A.HClO的结构式:H—O—Cl

B.乙醇分子的球棍模型:

C.18O2-离子结构示意图:

D.胆矾的化学式:CuSO4

解析 A.HClO的结构式:H—O—Cl,正确;B.该模型为比例模型,不是球棍模型,错误;C.核电荷数为+8,核外电子排布为2,8,错误;D.胆矾的化学式为CuSO4·5H2O,故D错误。

答案 A

5.下列说法不正确的是()

A.可用pH试纸测新制氯水的pH

B.容量瓶使用前要检漏

C.蒸发皿可直接用酒精灯加热

D.蒸馏时忘加沸石,需冷却后补加

解析氯水具有漂白性,会使pH试纸颜色变浅或褪色,错误,故应选A。

答案 A

6.下列各组中,互称为同分异构体的是()

A.CH3CH2OH和CH3OH

B.35Cl与37Cl

C.CH3CHO与

D.CH2Cl2与CH3Cl

解析A项两者是同系物关系,B项是同位素关系,C项具有相同的分子式C2H4O,不同的结构,属于同分异构体关系,D项是一氯取代物和二氯取代物不属于同分异构体。

答案 C

7.在恒温恒容的密闭容器中通入1 mol X和2 mol Y,发生下列反应:X(g)+2Y(g)M(g)+2N(g)ΔH=a kJ/mol(a>0),下列说法中正确的是()

A.到达平衡状态时,加入催化剂,平衡不移动,反应速率也不会发生改变

B.反应达到平衡时,X和Y的转化率相等

C.v正(X)=2v逆(N)时,反应到达平衡状态

D.到达平衡状态时,反应吸收a kJ能量

解析 A.加入催化剂对化学平衡移动无影响,但化学反应速率加快,故A错误;

B.假设消耗x mol X,同时消耗2x mol Y,X的转化率为x,Y的转化率为2x/2=x,

因此X和Y的转化率相等,故B正确;C.化学反应速率之比等于化学计量数之比,

应是2v

正(X)=v

(N),故C错误;D.此反应是可逆反应,不能进行到底,达到平

衡时,吸收能量小于a kJ,故D错误。

答案 B

8.下列实验方法合理且离子方程式正确的是()

A.用稀硫酸除去Cu中的Fe2O3杂质:Fe2O3+6H+===2Fe3++3H2O

B.实验室利用稀硝酸和石灰石制CO2: CaCO3+2H+===Ca2++H2O+CO2↑

C.用盐酸制硅胶:Na2SiO3+2H+===2Na++H2SiO3↓

D.检验溶液中的Fe3+:Fe3++3SCN-===Fe(SCN)3↓(红色)

解析 A.用稀硫酸除去Cu中的Fe2O3杂质时,Fe2O3+6H+===2Fe3++3H2O,生成的铁离子能够溶解铜,故A错误;B.硝酸是强酸,碳酸钙难溶于水,实验室利用稀硝酸和石灰石制CO2的离子方程式为CaCO3+2H+===Ca2++H2O+CO2↑,故B 正确;C.硅酸钠易溶于水,应该用离子表示,故C错误;D.Fe(SCN)3不是沉淀,故D错误;故选B。

答案 B

9.短周期元素W、X、Y和Z在周期表中的相对位置如图所示,这四种元素原子的最外层电子数之和为17。下列说法正确的是()

A.原子半径:W<X<Y<Z

B.X与Y可形成化合物Y3X2

C.X和Z的气态氢化物水溶液均显酸性

D.最高价氧化物对应水化物的酸性:Z<W

解析根据四种元素原子的最外层电子数之和为17,以及图中的位置,W、X、Y 和Z分别为C、N、Mg、S。A.原子半径:X<W<Z<Y,错误;B.可形成化合物Mg3N2,正确;C.NH3的水溶液呈碱性,错误;D.最高价氧化物对应水化物的酸性:H2CO3<H2SO4,错误;故应选B。

答案 B

10.下列叙述正确的是()

A.酒越陈越香与酯化反应有关

B.乙烷、丙烷和丁烷都没有同分异构体

C.乙烯和聚氯乙烯都能使溴的四氯化碳溶液褪色

D.甲烷与足量氯气在光照下反应生成物都难溶于水

解析酒中含乙醇,乙醇可被氧化成乙醛,乙醛进一步被氧化成乙酸,乙醇和乙酸在一定条件下反应生成乙酸乙酯,乙酸乙酯具有香味,故酒是陈的香,故A正确;丁烷存在正丁烷、异丁烷两种同分异构体,故B错误;聚氯乙烯不含碳碳双键,不能使溴的四氯化碳溶液褪色,故C错误;甲烷与足量氯气反应除卤代烃外还有氯化氢,氯化氢易溶于水,故D错误。

答案 A

11.下列说法不正确的是()

A.一定条件下,氨基酸之间能发生反应合成更复杂的化合物(多肽),构成蛋白质

B.淀粉和纤维素的组成都可用(C6H10O5)n表示,它们都可转化为乙醇

C.除去油脂中的少量水分,既可以用无水硫酸钠也可以用碱石灰

D.食用植物油在烧碱溶液中水解的主要产物是不饱和高级脂肪酸钠和甘油

解析A项氨基酸之间能发生缩聚反应合成更复杂的化合物(多肽),进而形成蛋白质,正确;B项淀粉和纤维素能水解生成葡萄糖,葡萄糖能生成乙醇,正确;C项碱石灰的碱性能使油脂水解,故不能用于油脂的除水,错误;D项食用植物油是不饱和高级脂肪酸甘油酯,水解产物为不饱和高级脂肪酸钠和甘油,正确。

答案 C

12.H2S废气资源化的原理为:2H2S(g)+O2(g)===S2(s)+2H2O(l)ΔH=-632 kJ·mol-1,如图为H2S燃料电池示意图。下列说法正确的是()

A.电极a为电池的正极

B.电极b上发生的电极反应为:O2+4H++4e-===2H2O

C.电路中每流过4 mol电子,电池会产生632 kJ热能

D.22.4 L H2S参与反应时,2 mol H+经质子交换膜进入正极区

解析 A.H2S燃料电池,通入H2S的电极a为负极,通氧气的电极b为正极,错误;

B.由于存在质子交换膜,因此溶液为酸性,电极b上发生的电极反应为:O2+4H++4e-===2H2O,正确;

C.原电池是化学能转化为电能的装置,因此电路中每流过4 mol电子,电池会产生632 kJ热能,错误;

D.气体体积没有标明状态不能确定物质的量,错误;故应选B。

答案 B

13.相同温度下,关于氢氧化钠和氨水两种溶液的比较,下列说法正确的是()

A.pH 相等的两溶液中:c(Na+)=c(NH+4)

B.分别中和pH 相等、体积相等的两溶液,所需HNO3的物质的量相同

C.相同浓度的两溶液,其导电能力相同

D.相同浓度的两溶液,分别通入HCl 气体反应后呈中性的溶液中(忽略溶液体积变化):c(NH+4)=c(Na+)

解析 A.根据电荷守恒,pH相等的两溶液中c(Na+)=c(NH+4)=c(OH-)-c(H+),正确;B.分别中和pH相等、体积相等的两溶液含有的NaOH的物质的量比NH3·H2O 少得多,因此所需HNO3的物质的量前者少的多,错误;C.相同浓度的两溶液,前者的离子浓度比后者大得多,因此其导电能力不同,错误;D.相同浓度的两溶液,

分别通入HCl气体反应后呈中性的溶液中(忽略溶液体积变化),根据电荷守恒和溶液呈中性的要求,c(Na+)=c1(Cl-);c(NH+4)=c2(Cl-),但是要达到中性,前者通HCl气体要略多,因此c(Na+)>c(NH+4),错误;故应选A。

答案 A

14.十氢萘是具有高储氢密度的氢能载体,经历“十氢萘(C10H18)→四氢萘(C10H12)→萘(C10H8)”的脱氢过程释放氢气。

已知:

Ⅰ.C10H18(l)===C10H12(l)+3H2(g)ΔH1

Ⅱ.C10H12(l)===C10H8(l)+2H2(g)ΔH2

假设某温度下,ΔH1>ΔH2>0。在下图中绘制下列“C10H18→C10H12→C10H8”的“能量~反应过程”示意图正确的是()

解析某温度下,ΔH1>ΔH2>0。说明该反应的正反应均为吸热反应,即反应物的总能量小于生成物的总能量,A 、C错误;由于ΔH1>ΔH2,说明第一个反应吸收的热量比第二个反应多,从B、D两个图像看出,B中ΔH2>ΔH1,不符合题意,D正确。

答案 D

15.下列说法正确的是()

A.HClO溶于水能电离,破坏了H—O键和O—Cl键

B.CaCl2晶体中存在共价键

C.SiO2属于原子晶体,熔化破坏共价键和分子间作用力

D.NH4Cl受热分解既破坏了离子键,也破坏了共价键

解析HClO溶于水能电离,破坏了H—O键没有破坏O—Cl键;CaCl2晶体中不存在共价键;SiO2属于原子晶体不存在分子间作用力;NH4Cl受热分解既破坏了铵根离子和氯离子的离子键,也破坏了氮氢原子之间的共价键。

答案 D

16.设N A为阿伏加德罗常数的值,下列说法不正确的是()

A.标准状况下的22.4 L辛烷完全燃烧,生成二氧化碳分子数为8N A

B.18 g水中含有的电子数为10N A

C.46 g 二氧化氮和46 g四氧化二氮含有的原子数均为3N A

D.在1 L 2 mol·L-1的硝酸镁溶液中含有的硝酸根离子数为4N A

解析 A.标况下,辛烷是液体,气体摩尔体积对其不适用,所以无法计算辛烷的物质的量,导致无法计算辛烷燃烧生成的二氧化碳分子数,故A错误;B.18 g水

的物质的量=18 g

18 g/mol=1 mol,一个水分子含有10个电子,所以含有电子总数是

10N A,故B正确;C.二氧化氮的物质的量=

46 g

46 g/mol=1 mol,四氧化二氮的物质

的量=46 g

92 g/mol=0.5 mol,一个二氧化氮分子中含有3个原子,一个四氧化二氮分子中含有6个原子,所以46 g二氧化氮和46 g四氧化二氮含有的原子数均为3N A,故C正确;D.1 L 2 mol·L-1的硝酸镁的物质的量=2 mol/L×1 L=2 mol,一个硝酸镁的化学式中含有2个硝酸根离子,所以含有硝酸根离子总数4N A,故D正确。答案 A

17.在调节好pH和Fe2+浓度的废水中加H2O2,能氧化降解污染物。现用该方法降解污染物p-CP,控制H2O2、Fe2+和p-CP的初始浓度相同,设计如下对比实验探究温度、pH对该降解反应速率的影响,测得p-CP的浓度随时间变化的关系如下图。

下列结论或推断不正确的是()

A.实验①、②表明适当升高温度,降解反应速率增大

B.根据推断,在原来条件下适当升高温度和调高pH,p-CP的分解速率可能保持不变

C.313 K、pH=3时,在0~100 s内,p-CP的分解平均速率约为1.8×10-2mol·L -1·s-1

D.室温下调节溶液pH=10,可使催化剂失去活性,p-CP几乎不分解

解析A项,研究的是温度变量对反应速率的影响,②曲线温度高反应时间短,降解速率大,正确;B项,根据推断,在原来条件下适当升高温度能使分解速率增大,而调高pH则降低分解速率,则p-CP的分解速率有可能在原基础上保持不变;C项,根据图像可求得313 K、pH=3时,在0~100 s内,p-CP的分解平均速率约为1.8×10-5mol·L-1·s-1;D项,曲线③表明在调节溶液pH=10时,p-CP 的浓度不变,分解速率几乎为零。

答案 C

18.常温时,向10 mL浓度为0.01 mol/L的氨水溶液中逐滴加入0.01 mol/L的CH3COOH溶液,混合液的pH 与加入CH3COOH 的体积V之间的变化曲线如下图所示。下列说法正确的是()

A.a点溶液的pH=12

B.b点时c(CH3COO-)>c(NH+4)

C.c点时水的电离程度最大

D.d点处微粒间存在:2c(H+)+c(CH3COO-)=2c(OH-)+c(NH3·H2O)

解析 A.氨水是弱电解质,部分电离,则0.01 mol·L-1的氨水,其pH<12,故A 错误;B.b点溶液的pH>7,溶液中存在c(OH-)+c(CH3COO-)=c(NH+4)+c(H+),则c(OH-)>c(H+),c(CH3COO-)

答案 C

19.某同学通过以下流程制取NaClO2晶体:

NaClO3(s)Na2SO3(s)――→

浓硫酸

①ClO2――→

NaOH、30% H2O2

②NaClO2溶液――→

结晶

③――→

过滤

洗涤④

NaClO2晶体

已知:ClO2浓度过高时易发生分解,NaClO2高于60 ℃时易分解,NaClO2饱和溶液在低于38 ℃时析出NaClO2·3H2O,高于38 ℃时析出NaClO2。

下列说法不正确的是()

A.步骤①中工业常用H2C2O4代替Na2SO3

B.步骤②发生反应的离子方程式为2ClO2+2OH-+H2O2===2ClO-2+O2+2H2O

C.步骤③的结晶方法为:蒸发浓缩,冷却结晶

D.步骤④中制得的NaClO2晶体中可用略高于38 ℃的热水洗涤2~3遍

解析 A.步骤①中Na2SO3作为还原剂,H2C2O4也是常见的还原剂,工业常用H2C2O4代替Na2SO3,正确;B.步骤②发生反应的离子方程式为2ClO2+2OH-+H2O2===2ClO-2+O2+2H2O,正确;C.由于NaClO2高于60 ℃时易分解,因此结晶方法不可以用蒸发浓缩和冷却结晶的方法,应该用减压蒸发的办法,错误;D.由于NaClO2饱和溶液在低于38 ℃时析出NaClO2·3H2O,高于38 ℃时析出NaClO2,步骤④中制得的NaClO2晶体中可用略高于38 ℃的热水洗涤2~3遍,正确;故应选C。

答案 C

20.某试液中只可能含有K+、NH+4、Fe2+、Al3+、Cl-、CO2-3、SO2-4、NO-3中的若干种离子,离子浓度均为0.1 mol/L,某同学进行如下实验。下列说法正确的是()

A.无法确定沉淀C 的成分

B.无法确定原试液中是否存在K+、Al3+、Cl-、

C.试液中可能含有CO2-3、K+、NH+4

D.试液中一定存在的离子为NH+4、Fe2+、SO2-4、NO-3

解析试液中加入稀硫酸产生气体,说明一定含有硝酸根离子和二价铁离子或是含有碳酸根,滤液中加入氢氧化钠生成气体,说明一定含有铵根离子,产生沉淀C,该沉淀只能是氢氧化铁,因此一定存在Fe2+、NO-3,滤液中通入二氧化碳,得不到沉淀,则一定不含铝离子,即原溶液中一定存在的离子为NH+4、Fe2+、NO-3;根据离子浓度均为0.1 mol·L-1,结合电荷守恒,原溶液中一定存在SO2-4,因此原溶液中一定存在的离子为NH+4、Fe2+、SO2-4、NO-3,无法确定原试液中有K+、Cl -,一定不含Al3+,故选D。

答案 D

非选择题部分

二、非选择题(本大题共7小题,共50分)

21.(6分)已知A为石油裂解的主要产物,相对分子质量为28。B为气体,在标准状况下的密度为1.965 g·L-1。D与F互为同分异构体,F是环状化合物。G的结构

简式为:。在一定条件下,它们有如图所示的转化关系。

请回答:

(1)E中所含官能团名称________。

(2)B与F反应生成G的反应类型________。

(3)C →D 的化学方程式______________________________________________ _______________________________________________________________。

(4)下列说法正确的是________。(填编号)

A.A 与O 2在催化剂作用下反应也可以生成D

B.不可以用金属钠检验C 中是否含有少量水

C.可以用NaHCO 3溶液鉴别C 、D 、E 的水溶液

D.G 可以在NaOH 溶液或稀硫酸中发生水解

解析 本题考查有机物的推断,A 为石油裂解的主要产物,且相对分子质量为28,即A 为CH 2===CH 2,B 为气体,在标准状况下的密度为1.965 g·L -1,则B 的摩尔质量为1.965×22.4 g·mol -1=44 g·mol -1,则B 为CO 2,乙烯和水发生加成反应,生成乙醇,即C 为CH 3CH 2OH ,C →D 发生氧化反应,即D 为乙醛,结构简式为CH 3CHO ,D →E 发生醛基的氧化反应,即E 为CH 3COOH ,D 和F 互为同分异构体,且F 为环状化合物,即F 为,(1)根据上述分析,E 为乙酸,含有的官能团是羧基;(2)B 和F 发生的反应是加聚反应;

(3)根据上述分析,其反应方程式为2CH 3CH 2OH +O 2――→催化剂△

2CH 3CHO +2H 2O ;(4)A.CH 2===CH 2与氧气在一定条件下,生成CH 3CHO ,其反应方程式为2CH 2===CH 2+O 2――→一定条件下2CH 3CHO ,故A 正确;B.C 为乙醇,金属钠能与水反应,也能与乙醇反应,不可以鉴别乙醇中是否含有水,故B 正确;C.乙醇、乙醛不与NaHCO 3反应,乙酸能与NaHCO 3反应生成CO 2,因此不能鉴别乙醇和乙醛,故C 错误;D.根据G 的结构简式,G 中含有酯基,因此在碱或酸中发生水解,

故D 正确。

答案 (1)羧基 (2)加聚反应 (3)2CH 3CH 2OH +O 2――→催化剂△

2CH 3CHO +2H 2O (4)ABD

22.(6分)某研究小组为了探究一种不溶性盐X(仅含四种元素)的组成和性质,设计并完成了如下实验:

请回答:

(1)X 除了含有H 、O 元素外,还有________元素(填元素符号)。

(2)混合气体通入溴水中发生反应③的离子方程式____________________________________________________________________ ___________________________________________________________________。

(3)写出X 的化学式________。

解析 (1)根据黑色固体A 与稀硫酸反应生成蓝色溶液可知,溶液中含有铜离子,X 中还有铜元素;溶液B 通过盐酸酸化的氯化钡产生白色沉淀,因此白色沉淀为硫酸钡,所以X 中还有硫元素;

(2)由于X 中还有硫元素,所以气体中有二氧化硫,二氧化硫会与溴单质发生氧化

还原反应,反应方程式为:SO 2+Br 2+2H 2O===4H ++SO 2-4+2Br -;

(3)X 隔绝空气产生的黑色固体为氧化铜,其物质的量为16.0 g÷80 g/mol =0.2 mol ,

1.80 g 水的物质的量为1.80 g÷18 g/mol =0.1 mol ,由白色沉淀的质量23.3 g ,可知硫酸根的物质的量为23.3 g÷233 g/mol =0.1 mol ,因此X 的化学式为Cu 2(OH)2SO 4。

答案 (1)Cu 、S (2)SO 2+Br 2+2H 2O===4H ++SO 2-4+2Br - (3)Cu 2(OH)2SO 4

23.(4分)某研究性学习小组采用氢气还原法测定碱式碳酸铜[x CuCO 3·y Cu(OH) 2·z H 2O]的组成,反应原理为x CuCO 3·y Cu(OH)2·z H 2O +(x +y )H 2=====△(x +y )Cu +x CO 2+(x +2y +z )H 2O 。装置如下:

请回答:

(1)仪器 D 的名称________, D 中所装药品及目的是________,________;

(2)请设计实验验证硬质玻璃管 A 中的碱式碳酸铜是否完全反应___________________________________________________________________。 解析 (1)仪器 D 的名称为球形干燥管。因为x CuCO 3·y Cu(OH)2·z H 2O +(x +

y )H 2=====△(x +y )Cu +x CO 2+(x +2y +z )H 2O 。为了防止空气中的水蒸气和二氧化碳干扰,所以 D 中所装药品为碱石灰。

(2)因为x CuCO 3·y Cu(OH)2·z H 2O +(x +y )H 2=====△(x +y )Cu +x CO 2+(x +2y +z )H 2O 。

可取硬质玻璃管A 中的物质加入稀盐酸中,若溶液变蓝,则反应还未完全(或取上层清液滴加NaOH 溶液,若产生蓝色沉淀则反应还未完全)。其他答案符合题意也可。

答案 (1)球形干燥管 碱石灰 防止空气中的二氧化碳和水进入C 中 (2)将硬质玻璃管中固体加入稀盐酸中,若溶液变蓝,则反应还未完全(或取上层清液滴加NaOH 溶液,若产生蓝色沉淀则反应还未完全。其他合理答案均可)

24.(4分)向Fe 和Fe 2O 3组成的3.84 g 混合物中加入120 mL 某浓度的盐酸,恰好完全反应,生成672 mL H 2(标准状况下),向反应后的溶液中滴加几滴KSCN 溶液,

溶液无明显变化。(结果保留三位有效数字)

(1)原混合物中Fe的质量为________ g。

(2)所用盐酸的物质的量浓度为________ mol/L。

解析(1)根据题意,与盐酸反应的铁粉的物质的量为0.03 mol,质量为1.68 g,则Fe与Fe2O3反应的总质量为2.16 g,且物质的量之比为1∶1,因此与Fe2O3反应的Fe的物质的量为0.01 mol,所以原混合物中Fe的物质的量为0.04 mol,质量为2.24 g。

(2)反应的溶液为FeCl2溶液,根据守恒,n(HCl)=2n(铁元素)=2×(0.04+0.02) mol =0.12 mol,因此浓度为1.00 mol/L。

答案(1)2.24(2)1.00

25.(10分)氢气因热值高、来源广、产物无污染,常被人们看做一种理想的“绿色能源”,氢气的制取是氢能源利用领域的研究热点。

图1

(1)图1为用电解法制备H2,写出阳极的电极反应式:____________________________。

(2)从可持续发展考虑,太阳能光解水制氢是获取H2的最好途径,但迄今仍然存在诸多问题,如光催化剂大多仅在紫外光区稳定有效,能够在可见光区使用的光催化剂不但催化活性低,而且几乎都存在光腐蚀现象,需使用牺牲剂进行抑制,能量转化效率低等,这些都阻碍了光解水的实际应用,需设计课题进一步研究解决。下列设想符合研究方向的是________。

A.将研究方向专注于紫外光区,无需考虑可见光区

B.研究光腐蚀机理,寻找高稳定性,不产生光腐蚀的制氢材料

C.研制具有特殊结构的新型光催化剂,开发低成本、高性能光催化材料

D.研究新的光解水的催化机制,使利用红外光进行光解水制氢成为可能

(3)以甲烷为原料制取氢气是工业上常用的制氢方法。

已知:CH4(g)+H2O(g)===CO(g)+3H2(g)

ΔH1=-206.2 kJ·mol-1

CH4(g)+CO2(g)===2CO(g)+2H2(g)

ΔH2=-247.4 kJ·mol-1

则CH4(g)与H2O(g)反应生成CO2(g)和H2(g)的热化学方程式为:___________________________________________________________________

___________________________________________________________________。(4)利用废弃的H2S的热分解可生产H2:2H2S(g)2H2(g)+S2(g)。现将0.30 mol H2S(g)通入到压强为p的恒压密闭容器中,在不同温度下测得H2S的平衡转化率如图2所示:

图2

①温度为T4℃时,初始体积为1 L,t分钟后,反应达到平衡,该温度下的平衡常数K=________。

②若保持恒温T℃,将一定量的H2S(g)通入到恒压密闭容器中,反应一段时间后,压强为p的体系恰好达到平衡,试在图3上画出不同起始压强下H2S转化率的趋

势图。

图3

解析(1)根据图示,阳极放电的离子应为水电离出的OH-,放电生成氧气,从而导致溶液中H+浓度增大,H+与CO2-3再反应生成HCO-3。电极反应式为4CO2-3+2H2O-4e-===O2↑+4HCO-3

(2)A.可见光区更有普遍性,更有发展前途,错误;B.根据题意,研究光腐蚀机理,寻找高稳定性,不产生光腐蚀的制氢材料,正确;C.根据题意,研制具有特殊结构的新型光催化剂,开发低成本、高性能光催化材料,正确;D.红外光也是生活中常见光源,研究新的光解水的催化机制,使利用红外光进行光解水制氢成为可能,正确;故应该选BCD。

(3)CH4(g)与H2O(g)反应生成CO2(g)和H2(g)的化学方程式应为CH4(g)+2H2O(g)===CO2(g)+4H2(g),根据盖斯定律,该反应的ΔH=2ΔH1-ΔH2=-165 kJ·mol-1,故该反应的热化学方程式为CH4(g)+2H2O(g)===CO2(g)+4H2(g)ΔH=-165 kJ·mol-1

(4)①根据图示,T4℃达到平衡时,平衡转化率为40%,则平衡时H2S为0.18 mol,H2为0.12 mol,S2为0.06 mol,总物质的量为0.36 mol,因为是恒压容器,刚开始时体积为1 L,则平衡时体积为1.2 L,因此平衡时H2S、H2和S2的浓度分别为0.15 mol·L-1、0.1 mol·L-1和0.05 mol·L-1,则平衡常数K=1/45≈2.2 ×10-2。

②纵坐标为转化率,非平衡转化率,因此在压强小于p时,未达到平衡,随着压强的增大,反应速率增大,转化率增大;压强达到p后,达到平衡,此时随着压

强增大,平衡向左移动,转化率下降。

答案(1)4CO2-3+2H2O-4e-===O2↑+4HCO-3

(2)BCD(3)CH4(g)+2H2O(g)===CO2(g)+4H2(g)ΔH=-165.0 kJ·mol-1

(4)①1/45或2.2 ×10-2

26.(10分)(2019·上海卷节选)(1)从绿色化学、环保的角度选择制取硫酸铜的原料________。

a.Cu

b.CuO

c.CuS

d.Cu(OH)2·CuSO4

(2)写出从硫酸铜溶液中得到硫酸铜晶体的操作:________、________、过滤、洗涤……

(3)如图球形管装置的作用:_________________________________________。

(4)反应后溶液中存在Cu 2+,但浓硫酸有吸水性,不能通过溶液颜色来判断Cu 2+的存在,请写出一种简便的方法验证Cu 2+的存在:_________________________________________________________________。

(5)在实际反应中,由于条件控制不好,容易产生CuS 和Cu 2S 固体。

已知:2CuS ――→△Cu 2S +S 。取a g 固体样品(只含CuS 和Cu 2S),加热后测得Cu 2S

固体质量为b g ,求固体样品中CuS 质量分数________:若Cu 2S 在高温下会分解为Cu ,则测得CuS 的质量分数________(偏大、偏小或不变)。

(6)根据以上信息,说明Cu 与浓硫酸反应需要控制哪些因素:__________________________________________________________________。 解析 (1)a 选项Cu 与浓硫酸反应才能制取硫酸铜,并且产生污染环境的气体二氧化硫,故不选;b 选项CuO 与稀硫酸反应即可生成硫酸铜,且无污染环境的产物;c 选项CuS 不溶于酸,如果要制备硫酸铜,需要先与氧气反应生成氧化铜和二氧化硫,氧化铜再与稀硫酸反应得到硫酸铜,过程中产生二氧化硫污染环境,故不选;d 选项只需要稀硫酸并无有害产物;综上,答案选bd 。

(2)大多数盐的溶解度随温度降低而下降,因此用降温结晶的方法得到更多的晶体。

(3)球形管提供较大的空间,防止形成过大的负压,防倒吸。

(4)由于浓硫酸的吸水性,导致Cu 2+无法与水结合产生蓝色溶液,因此需要加水稀释检验铜离子,需要注意的是原反应液中存在浓硫酸,因此稀释时应注酸入水,即原反应液倒入水中进行稀释,若溶液呈蓝色,则说明铜离子的存在。

(5)列方程组:设CuS 质量为x g ,Cu 2S 质量为y g ,根据题意可列出方程x +y =a ,

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