由于温室效应和资源短缺等问题，如何降低大气中的CO2含量并加以开发利用，引起了各国的普遍重视．目前工
\u968f\u7740\u4eba\u7c7b\u5bf9\u6e29\u5ba4\u6548\u5e94\u548c\u8d44\u6e90\u77ed\u7f3a\u7b49\u95ee\u9898\u7684\u91cd\u89c6\uff0c\u964d\u4f4e\u5927\u6c14\u4e2dCO2\u7684\u542b\u91cf\u53ca\u6709\u6548\u5730\u5f00\u53d1\u5229\u7528CO2\uff0c\u5f15\u8d77\u4e86\u5404\u56fd\u7684\u666e\u904d\uff081\uff09\u53cd\u5e94\u524dCO2\u7684\u6d53\u5ea6\u4e3a1.0mol/L\uff0cH2\u7684\u6d53\u5ea6\u4e3a3.0mol/L\uff0c\u5e73\u8861\u540e\uff0c\u751f\u6210\u7269CH3OH\uff08g\uff09\u548cH2O\uff08g\uff09\u7684\u6d53\u5ea6\u4e3a0.75mol/L\uff0c\u7ed3\u5408\u5176\u53cd\u5e94\u7684\u5316\u5b66\u8ba1\u91cf\u6570\u53ef\u77e5\uff0cCO2\u7684\u53cd\u5e94\u6d53\u5ea6\u4e3a0.75mol/L\uff0cH2\u7684\u53cd\u5e94\u6d53\u5ea6\u4e3a3\u00d70.75mol/L=2.25mol/L\uff0c\u6240\u4ee5\u5176\u5e73\u8861\u6d53\u5ea6\u5206\u522b\u4e3a\uff1a0.25mol/L\u548c0.75mol/L\uff1b\uff082\uff09H2\u7684\u53cd\u5e94\u6d53\u5ea6\u4e3a3\u00d70.75mol/L=2.25mol/L\uff0c\u6240\u4ee5\uff0c\u53cd\u5e94\u901f\u7387\u4e3a\uff1av\uff08H2\uff09=2.25mol?L?110min=0.225mol/\uff08L?min\uff09\uff1b\u53cd\u5e94\u524dCO2\u7684\u6d53\u5ea6\u4e3a1.0mol/L\uff0cH2\u7684\u6d53\u5ea6\u4e3a3.0mol/L\uff0c\u4e8c\u6c27\u5316\u78b3\u4e0e\u6c22\u6c14\u7684\u6d53\u5ea6\u4e4b\u6bd4\u7b49\u4e8e\u5176\u5316\u5b66\u8ba1\u91cf\u6570\u4e4b\u6bd4\uff0c\u53cd\u5e94\u901f\u7387\u4e4b\u6bd4\u7b49\u4e8e\u5316\u5b66\u8ba1\u91cf\u6570\u4e4b\u6bd4\uff0c\u6240\u4ee5\uff0c\u4e8c\u6c27\u5316\u78b3\u548c\u6c22\u6c14\u7684\u8f6c\u5316\u7387\u76f8\u7b49\uff0c\u6545\u7b54\u6848\u4e3a\uff1a0.225\uff1b\u7b49\u4e8e\uff1b\uff083\uff09A\u3001\u751f\u6210CH3OH\uff08g\uff09\u548c\u6d88\u8017CO2\u90fd\u662f\u53cd\u5e94\u5411\u6b63\u5411\u8fdb\u884c\uff0c\u4e0d\u80fd\u8868\u73b0\u6b63\u9006\u53cd\u5e94\u901f\u7387\u76f8\u7b49\uff0c\u6545A\u9519\u8bef\uff1bB\u3001\u751f\u6210n mol H2O\uff08g\uff09\uff0c\u540c\u65f6\u751f\u62103n mol H2\uff0c\u6b63\u9006\u53cd\u5e94\u901f\u7387\u76f8\u7b49\uff0c\u6545B\u6b63\u786e\uff1bC\u3001\u53cd\u5e94\u524d\u540e\u6c14\u4f53\u4f53\u79ef\u6709\u6240\u53d8\u5316\uff0c\u6240\u4ee5\u538b\u5f3a\u4e0d\u53d8\u65f6\u53cd\u5e94\u8fbe\u5230\u5e73\u8861\u72b6\u6001\uff0c\u6545C\u6b63\u786e\uff1bD\u3001\u6052\u5bb9\u5bb9\u5668\uff0c\u53cd\u5e94\u7269\u548c\u751f\u6210\u7269\u90fd\u662f\u6c14\u4f53\uff0c\u6c14\u4f53\u8d28\u91cf\u4e0d\u53d8\uff0c\u6545\u5bc6\u5ea6\u6c38\u8fdc\u4e0d\u53d8\uff0c\u6545D\u9519\u8bef\uff0e\u6545\u9009\uff1aBC\uff0e\uff084\uff09\u53cd\u5e94\u662f\u6b63\u53cd\u5e94\u6c14\u4f53\u4f53\u79ef\u51cf\u5c0f\u7684\u653e\u70ed\u53cd\u5e94\uff0c\u4ece\u56fe\u8c61\u5206\u6790\uff0c\u53cd\u5e94\u6b63\u5411\u79fb\u52a8\u4e14\u901f\u7387\u6bd4\u539f\u6765\u589e\u5927\uff0cA\u3001\u5347\u6e29\u65f6\u5e73\u8861\u9006\u5411\u79fb\u52a8\uff0c\u6545A\u9519\u8bef\uff1bB\u3001\u589e\u5927\u53cd\u5e94\u7269\u6d53\u5ea6\uff0c\u5e73\u8861\u6b63\u5411\u79fb\u52a8\uff0c\u53cd\u5e94\u901f\u7387\u52a0\u5feb\uff0c\u6545B\u6b63\u786e\uff1bC\u3001\u4f53\u79ef\u4e0d\u53d8\u5145\u5165\u6c29\u6c14\uff0c\u5bf9\u53cd\u5e94\u901f\u7387\u548c\u5e73\u8861\u90fd\u65e0\u5f71\u54cd\uff0c\u6545C\u9519\u8bef\uff1bD\u3001\u51cf\u5c0f\u538b\u5f3a\uff0c\u53cd\u5e94\u901f\u7387\u51cf\u6162\uff0c\u6545D\u9519\u8bef\uff0c\u6545\u9009\uff1aB\uff0e\uff085\uff09\u71c3\u6599\u7535\u6c60\u4e2d\u7532\u9187\u5728\u8d1f\u6781\u53d1\u751f\u6c27\u5316\u53cd\u5e94\uff0c\u6c27\u6c14\u5728\u6b63\u6781\u53d1\u751f\u8fd8\u539f\u53cd\u5e94\uff0c\u6ce8\u610f\u7535\u89e3\u8d28\u4e3a\u7194\u878dK2CO3\uff0c\u6240\u4ee5\u6b63\u6781\u53cd\u5e94\u5f0f\u4e3a\uff1a32O2+6e-+3CO2=3CO32-\uff0c\u8d1f\u6781\u53cd\u5e94\u5f0f\u4e3a\uff1aCH3OH-6e-+3CO32-=3CO2+2H2O\uff0c\u6545\u7b54\u6848\u4e3a\uff1a32O2+6e-+3CO2=3CO32-\uff1bCH3OH-6e-+3CO32-=3CO2+2H2O\uff0e
\uff081\uff09\u2460\u5229\u7528\u4e09\u6bb5\u5f0f\u89e3\u9898\u6cd5\u8ba1\u7b97 CO2\uff08g\uff09+3H2\uff08g\uff09 CH3OH\uff08g\uff09+H2O\uff08g\uff09\uff0c\u5f00\u59cb\uff08mol/L\uff09\uff1a1 3 0 0\u53d8\u5316\uff08mol/L\uff09\uff1a0.75 2.25 0.75 0.75 \u5e73\u8861\uff08mol/L\uff09\uff1a0.25 0.75 0.75 0.75\u4ece\u53cd\u5e94\u5f00\u59cb\u5230\u5e73\u8861\uff0c\u6c22\u6c14\u7684\u5e73\u5747\u53cd\u5e94\u901f\u7387v\uff08H2\uff09=2.25mol/L10min=0.225 mol?L-1?min-1\uff0c\u6c22\u6c14\u7684\u8f6c\u5316\u7387=2.253=75%\uff0c\u6545\u7b54\u6848\u4e3a\uff1a0.225\uff1b75%\uff1b\u2461\u8981\u4f7fn\uff08CH3OH\uff09/n\uff08CO2\uff09\u589e\u5927\uff0c\u5e94\u4f7f\u5e73\u8861\u5411\u6b63\u53cd\u5e94\u65b9\u5411\u79fb\u52a8\uff0cA\uff0e\u56e0\u6b63\u53cd\u5e94\u653e\u70ed\uff0c\u5347\u9ad8\u6e29\u5ea6\u5e73\u8861\u5411\u9006\u53cd\u5e94\u65b9\u5411\u79fb\u52a8\uff0c\u5219n\uff08CH3OH\uff09/n\uff08CO2\uff09\u51cf\u5c0f\uff0c\u6545A\u9519\u8bef\uff1bB\uff0e\u5145\u5165He\uff08g\uff09\uff0c\u4f7f\u4f53\u7cfb\u538b\u5f3a\u589e\u5927\uff0c\u4f46\u5bf9\u53cd\u5e94\u7269\u8d28\u6765\u8bf4\uff0c\u6d53\u5ea6\u6ca1\u6709\u53d8\u5316\uff0c\u5e73\u8861\u4e0d\u79fb\u52a8\uff0cn\uff08CH3OH\uff09/n\uff08CO2\uff09\u4e0d\u53d8\uff0c\u6545B\u9519\u8bef\uff1bC\uff0e\u5c06H2O\uff08g\uff09\u4ece\u4f53\u7cfb\u4e2d\u5206\u79bb\uff0c\u5e73\u8861\u5411\u6b63\u53cd\u5e94\u65b9\u6cd5\u79fb\u52a8\uff0cn\uff08CH3OH\uff09/n\uff08CO2\uff09\u589e\u5927\uff0c\u6545C\u6b63\u786e\uff1bD\uff0e\u518d\u5145\u51651mol CO2\u548c3mol H2\uff0c\u7b49\u6548\u4e8e\u5728\u539f\u6765\u57fa\u7840\u4e0a\u7f29\u5c0f\u4f53\u79ef\u4e00\u534a\uff0c\u538b\u5f3a\u589e\u5927\uff0c\u5e73\u8861\u5411\u6b63\u53cd\u5e94\u65b9\u5411\u79fb\u52a8\uff0c\u5219n\uff08CH3OH\uff09/n\uff08CO2\uff09\u589e\u5927\uff0c\u6545D\u6b63\u786e\uff0e\u6545\u7b54\u6848\u4e3a\uff1aCD\uff1b\uff082\uff09\u5728\u6e29\u5ea6\u4e0d\u53d8\u7684\u60c5\u51b5\u4e0b\uff0c\u82e5\u5728\u589e\u5927\u5bb9\u5668\u4f53\u79ef\u7684\u540c\u65f6\u4e0d\u65ad\u5145\u5165\u6c22\u6c14\uff0c\u4f7fH2\u7684\u6d53\u5ea6\u4fdd\u6301\u4e0d\u53d8\uff0c\u4f53\u79ef\u53d8\u5927\u65f6CH3OH\u3001CO\u7684\u6d53\u5ea6\u6210\u6210\u6bd4\u4f8b\u51cf\u5c0f\uff0c\u6240\u4ee5c(CH3OH)c(CO)c2(H2)\u7684\u503c\u4e0d\u53d8\u7b49\u4e8eK\u7684\u503c\uff0c\u5373\u5e73\u8861\u4e0d\u79fb\u52a8\uff0c\u6545\u9009C\uff0c\u6545\u7b54\u6848\u4e3a\uff1aC\uff1bK=c(CH3OH)c(CO)c2(H2)\uff0cCH3OH\u3001CO\u6210\u6bd4\u4f8b\u51cf\u5c0f\uff0cc\uff08H2\uff09\u4e0d\u53d8\uff0c\u5219K\u4e0d\u53d8\uff0e
（1）从图象分析，反应物能量高于生成物，反应放热，说明反应物键能和小于生成物，故答案为：小于；（2）C原子四键，H只有1个化学键，O形成2个化学键，结合乙醇的结构可知甲醇的结构式为：；两非金属元素的原子间形成的是共价键，
故答案为：；共价键；
（3）负极是电子流出的极，甲醇在负极失电子发生氧化反应，故答案为：甲醇；氧化；
（4）欲使反应CO2（g）+3H2 （g）?CH3OH （g）+H2O（g）速率加快，可以升温、加压、加入适当的催化剂等等，故答案为：升温、加压、加入催化剂；
（5）A、CO2 （g）的浓度不再发生变化说明反应达到平衡状态，故A正确；
B、消耗1mol CO2的同时生成3mol H2说明正逆反应速率相等，反应达到了平衡状态，故B正确；
C、温度不再变化，说明正逆反应速率相等，反应达到了平衡状态，故C正确；
D、反应前后气体体积不同，氧气不变说明各组分浓度不变，反应达到了平衡状态，故D正确；
故选ABCD．
绛旓細锛6鍒嗭級锛1锛塁H 3 OH(l)+3/2O 2 (g)=CO 2 (g)+2H 2 O(l) 鈻矵=-725.76kJ路mol -1 锛2锛夆憼0.225 mol/(L路min)銆傗憽B C 锛堟瘡绌2鍒嗭級 鐣
绛旓細锛1锛塁 (1鍒) 锛2锛 (2鍒)锛3锛夊澶 (1鍒)锛4锛0.225 mol路L -1 路min -1 (2鍒) 锛5锛塁 D (2鍒) 鐣
绛旓細骞宠 宸︾Щ,骞宠 甯告暟鍑忓皬,D瀵;鐢变簬鏄亽瀹瑰鍣,鍏呭叆涓庡弽搴斾綋绯绘棤鍏崇殑姘︽皵,瀵归熺巼鏃犲奖鍝,E閿欍(2)缁撳悎鍥剧ず,850 鈩冩椂鍚勭墿璐ㄧ殑骞宠 娴撳害涓篊O涓0.08 mol/L銆丠 2 O涓0.18 mol/L,鐢变簬瀹瑰櫒浣撶Н涓10 L,鏁匟 2 涓0.12 mol/L銆丆O 2 涓0.12 mol/L,姹傚緱骞宠 甯告暟涓1銆傜浉鍚屾俯搴︿笅,鑻ヨ浣垮弽搴斿紑濮...
绛旓細锛1锛夌敱鍥捐薄鍙煡锛屽弽搴旂墿鐨勮兘閲忛珮锛岀敓鎴愮墿鐨勮兘閲忎綆锛岃璐圭敤涓烘斁鐑弽搴旓紝鏁呪柍H锛0锛涘弽搴斿悗姘斾綋浣撶Н鍑忓皬锛屸柍S锛0锛涙晠绛旀涓猴細C锛涳紙2锛夎鍙嶅簲骞宠 甯告暟K鐨勮〃杈惧紡涓 c(C H 3 OH)?c( H 2 O) c(C O 2 )? c 2 ( H 2 ) 2 锛屾晠绛旀涓猴細 c(C ...
绛旓細锛1锛夌敱鍥1鍙煡锛屽弽搴旂墿鐨勬昏兘閲忛珮浜庣敓鎴愮墿鐨勬昏兘閲忥紝璇ュ弽搴斾负鏀剧儹鍙嶅簲锛屽嵆鈻矵锛0锛岀敱鏂圭▼寮忓彲鐭ワ紝璇ュ弽搴旀鍙嶅簲鏄皵浣撶殑鐗╄川鐨勯噺鍑忓皬鐨勫弽搴旓紝鏁呮鍙嶅簲涓虹喌鍑忚繃绋嬶紝鍗斥柍S锛0锛涙晠閫塁锛涳紙2锛夊彲閫嗗弽搴擟O 2 锛坓锛+3H 2 锛坓锛?CH 3 OH锛坓锛+H 2 O锛坓锛夌殑鍖栧骞宠 甯告暟k= c(C H 3 ...
绛旓細浜屾哀鍖栫⒊鐢ㄤ簬鐏伀涓昏鏄噰鐢ㄥ皢鍙噧鐗╀笌姘ф皵闅旂鐨勫姙娉曡揪鍒扮洰鐨勶紙2锛夆憼v(CO 2 )= =0.075mol?L -1 ?min -1 锛岀敱鏂圭▼寮忕殑绯绘暟姣斿彲鐭(H 2 )=3v(CO 2 )=0.225mol?L -1 ?min -1 鈶′緷鎹墍缁欐暟鎹紝骞宠 甯告暟K= 5.33( mol?L -1 ) -2 (3)鐢卞寲鍚堜环鐨勫崌闄嶅彲鐭ワ紝CO 2 涓殑纰冲厓绱...
绛旓細锛1锛夆憼鏀剧儹锛1鍒嗭級 鈶″惁锛1鍒嗭級鍌寲鍓備笉鑳芥敼鍙樺寲瀛﹀弽搴斿鎬佸拰缁堟佺殑鑳介噺锛2鍒嗭級 锛2锛夆憼0.5 锛2鍒嗭級鈶″崌楂樻俯搴︼紙1鍒嗭級澧炲ぇ鍘嬪己锛1鍒嗭級鈶D锛2鍒嗭級锛3锛塁O 2 (g)+4H 2 (g) CH 4 (g)+2H 2 O (l) 鈻矵=-252.9kJ路mol -1 (2鍒) 锛1锛夋牴鎹浘鍍忓彲鐭ワ紝鍙嶅簲鐗╃殑...
绛旓細(1)C銆 (2)銆(3)澧炲ぇ (4)0.225 銆(5)CD 鐣
绛旓細=nB3tBmol?锛圠?min锛-1锛岀敱鏂圭▼寮忓彲鐭锛圚2锛=2v锛圕H3OH锛=2nB3tBmol?锛圠?min锛-1锛屾晠绛旀涓猴細2nB3tBmol?锛圠?min锛-1锛涳紙3锛夊湪鍏朵粬鏉′欢涓嶅彉鐨勬儏鍐典笅锛屽澶勪簬E鐐圭殑浣撶郴浣撶Н鍘嬬缉鍒板師鏉ョ殑12锛屼綋绯荤殑鍘嬪己澧炲ぇ锛屾閫嗗弽搴旈熺巼澧炲ぇ锛屽钩琛″悜姝ｅ弽搴旀柟鍚戠Щ鍔紝鐢查唶鐨勭墿璐ㄧ殑閲忓鍔锛岀敱浜娓╁害涓嶅彉锛...
绛旓細锛1锛夆憼璇ュ伐涓氭柟娉曞埗鍙栫敳閱囧師鐞嗕负浜屾哀鍖栫⒊涓庢阿姘斿弽搴旂敓鎴愮敳閱囧拰姘达紝鍙嶅簲鏂圭▼寮忎负CO 2 锛坓锛+3H 2 锛坓锛 CH 3 OH锛坓锛+H 2 O锛坓锛夛紝鏁呯瓟妗堜负锛欳O 2 锛坓锛+3H 2 锛坓锛 CH 3 OH锛坓锛+H 2 O锛坓锛夛紱鈶℃牴鎹笁娈靛紡璁＄畻 CO 2 锛坓锛+3H 2 锛坓锛 CH 3 OH锛坓锛+H 2 O锛...
