问题 填空题

实施以减少能源浪费和降低废气排放为基本内容的节能减排政策,是应对全球气候问题、建设资源节约型、环境友好型社会的必然选择。化工行业的发展必须符合国家节能减排的总体要求。试运用所学知识,解决下列问题:

⑴已知某反应的平衡表达式为:

它所对应的化学反应方程式为:

⑵利用水煤气合成二甲醚的三步反应如下:

①2H2(g)+CO(g)CH3OH(g);ΔH=-90.8kJ·mol

②2CH3OH(g)CH3OCH3(g)+H2O(g);ΔH=-23.5kJ·mol

③CO(g)+H2O(g)CO2(g)+H2(g);ΔH=-41.3kJ·mol

总反应:3H2(g)+3CO(g)CH3OCH3(g)+CO2(g)的ΔH=__________

(3)煤化工通常通过研究不同温度下平衡常数以解决各种实际问题。已知等体积的一氧化碳和水蒸气进入反应器时,会发生如下反应:CO(g)+H2O(g)H2(g)+CO2(g),该反应平衡常数随温度的变化如下表所示:

温度/℃400500800
平衡常数K9.9491
 

该反应的正反应方向是反应(填“吸热”或“放热”),若在500℃时进行,设起始时CO和H2O的起始浓度均为0.020mol/L,在该条件下,CO的平衡转化率为:。

(4)从氨催化氧化可以制硝酸,此过程中涉及氮氧化物,如NO、NO2、N2O4等。

对反应:N2O4(g)2NO2(g)△H>0,在温度为T1、T2时,平衡体系中NO2的体积分数随压强变化曲线如图所示。下列说法正确的是。

A.A、C两点的反应速率:A>C

B.A、C两点气体的颜色:A深,C浅

C.B、C两点的气体的平均相对分子质量:B<C

D.由状态B到状态A,可以用加热的方法

E.A、C两点的化学平衡常数:A>C

(5)CO2(g)+3H2(g)CH3OH(g)+H2O(g)△H=-49.0kJ·mol-1。现在温度、容积相同的3个密闭容器中,按不同方式投入反应物,保持恒温、恒容,测得反应达到平衡时的有关数据如下。下列说法正确的是

答案

⑴C(s)+H2O(g)CO(g)+H2(g)(2分)

⑵-246.4kJ/mol(2分)⑶放热(1分),75%(2分)⑷D(2分)(5)BDE(2分)

⑴由平衡常数表达式可知:生成物及其状态为CO(g)和H2(g),反应物及其状态为H2O(g),再由质量守恒得反应物还有C(s),方程式为C(s)+H2O(g)CO(g)+H2(g),答案:C(s)+H2O(g)CO(g)+H2(g);⑵①×2+②+③得总方程式,△H=-90.8kJ·mol-1×2-23.5kJ·mol-1-41.3kJ·mol-1=-246.4kJ·mol-1,答案:-246.4kJ·mol-1;⑶升高温度,平衡常数减小,反应进行程度越小,平衡向逆反应移动,升高温度平衡向吸热方向移动,故正反应为放热反应;

设CO的浓度变化量为c,三段式法用c表示出平衡时各组分个浓度,

CO(g)+H2O(g)H2(g)+CO2(g),

起始(mol·L-1):0.020.02000

转化(mol·L-1):cccc

平衡(mol·L-1):0.02-c0.02-ccc

代入500℃时反应平衡常数有k===9,解得c=0.015,

CO的最大所以转化率为××100%=75%,

⑷A.由图象可知,A、C两点都在等温线上,C的压强大,则A、C两点的反应速率:A<C,A错误;

B.由图象可知,A、C两点都在等温线上,C的压强大,与A相比C点平衡向逆反应进行,向逆反应进行是减小由于压强增大导致浓度增大趋势,但到达平衡仍比原平衡浓度大,平衡时NO2浓度比A的浓度高,NO2为红棕色气体,则A、C两点气体的颜色:A浅,C深,B错误;C.B、C两点二氧化氮的体积分数相同,则混合气体的平均相对分子质量相同,C错误;

D.在相同压强下,升高温度,平衡向逆反应方向移动,则二氧化氮的体积分数增大,所以由状态B到状态A,可以用加热的方法,故D正确;E由图象可知,A、C两点都在等温线上,平衡常数只受温度影响,温度相同平衡常数相同,所以化学平衡常数KA=KC,E错误;

⑸实验1和实验2是全等等效平衡,实验3相当于实验2放大两倍后将体积压缩,A、CH3OH的平衡浓度2C1<C3,A错误;B、实验1从正向建立平衡和实验2从逆向建立平衡,x+y=49.0,B正确;C、该反应是体积缩小的反应,加压后正向移动,2P2>P3,C错误,又P1=P2,2P1>P3,E正确;D、a1+a2=1,a3<a2,(a1+a3)<1,D正确,F错误,选BDE。

阅读理解

Long long ago, an elephant and a monkey lived in the same forest. They were good friends, but both of them were very proud. The elephant thought of himself strong, and the monkey felt himself quick.

One day they went to ask the old bird, “Can you tell us which one of us is more important?” The old bird didn't give them the answer at once.  However, he asked them to get some bananas on the other side of the river and bring them to him.

So the elephant and the monkey went to the river, but the water ran so fast that the monkey was afraid.

“Get on my back, Monkey,” said the elephant, “I shall take you there. I'm big and strong, and I can swim across the river.”

Soon they got to the other side. The elephant tried to reach the bananas, but they were too high. “Wait a minute, please. I can climb,” said the monkey. He quickly ran up the tree and gave the bananas to the elephant.

Then they came back happily to the old bird and showed him the bananas. “Now, you see, the elephant is strong,” said the old bird with a smile, “but he couldn't get the bananas; the monkey is quick, but he couldn't, either. Only you two could do the work well when you helped each other.” 

小题1:Both the elephant and the monkey are ______ in the story.

A. quick                        B. proud                   C. strong

小题2:According to the story, ______ is more important.

A. the monkey               B. the elephant                C. neither

小题3:Picture ______ shows the right position of the banana tree.

A.                        B.                         C.

小题4:They should ______ to get the bananas.

A. first swim across the river and then come back

B. first climb up the tree and then swim across the river

C. first swim across the river and then climb up the tree

小题5:The best title for the story is “______”.

A. Teamwork is important       B. An elephant and a monkey      C. Strong or quick

填空题