ÌâÄ¿ÄÚÈÝ

ËĸöÌå»ýÏàͬµÄÃܱÕÈÝÆ÷ÖÐÔÚÒ»¶¨µÄÌõ¼þÏ·¢Éú·´Ó¦£º2SO2+O22SO3£¬·´Ó¦¿ªÊ¼Ê±£¬·´Ó¦ËÙÂÊÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÕýÈ·µÄÊÇ£¨    £©
ÈÝÆ÷
ζÈ
SO2(mol)
O2(mol)
´ß»¯¼Á
¼×
5000C
10
5
-
ÒÒ
5000C
8
5
V2O5
±û
4500C
8
5
-
¶¡
5000C
8
5
-
A£®ÒÒ>¼×>¶¡>±û        B£®ÒÒ>¼×>±û>¶¡      C£®¼×>ÒÒ=¶¡>±û   D£®ÒÒ>¼×>±û=¶¡
A
ζÈÔ½¸ß·´Ó¦ËÙÂÊÔ½¿ì£»Å¨¶ÈÔ½¸ß·´Ó¦ËÙÂÊÔ½¿ì£»Óд߻¯¼Á·´Ó¦ËÙÂÊÔ½¿ì£»×ÛÉÏËùÊö£¬ÕýÈ·Ñ¡ÏîΪA£»
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨8·Ö£©¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦ÓÚÑо¿³£Î¡¢³£Ñ¹Ï¡°È˹¤¹Ìµª¡±µÄз½·¨¡£ÔøÓÐʵÑ鱨µÀ£ºÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢Éú·´Ó¦£¬Éú³ÉµÄÖ÷Òª²úÎïΪNH3¡£½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬²¿·ÖʵÑéÊý¾Ý¼ûÏÂ±í£¨¹âÕÕ¡¢N2ѹÁ¦1.0¡Á105Pa¡¢·´Ó¦Ê±¼ä3h£©£º
T/K
303
313
323
353
NH3Éú³ÉÁ¿/£¨10-6mol£©
4.8
5.9
6.0
2.0
ÏàÓ¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
N2£¨g£©+3H2O£¨l£©====2NH3£¨g£©+O2£¨g£© ¦¤H=+765.2kJ¡¤mol-1
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÒͼÊÇÉÏÊö·´Ó¦ÔÚÎÞ´ß»¯¼ÁÇé¿öÏ·´Ó¦¹ý³ÌÖÐÌåϵÄÜÁ¿±ä»¯Ê¾Òâͼ£¬ÇëÔÚͼÖл­³öÔÚÓд߻¯¼ÁÇé¿öÏ·´Ó¦¹ý³ÌÖÐÌåϵÄÜÁ¿±ä»¯Ê¾Òâͼ¡£

£¨2£©ÓëÄ¿Ç°¹ã·ºÊ¹ÓõĹ¤ÒµºÏ³É°±·½·¨Ïà±È£¬¸Ã·½·¨Öй̵ª·´Ó¦ËÙÂÊÂý¡£ÇëÌá³ö¿ÉÌá¸ßÆä·´Ó¦ËÙÂÊÇÒÔö´óNH3Éú³ÉÁ¿µÄ½¨Ò飺                                                  ¡£
£¨3£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2£¨g£©+3H2£¨g£©2NH3£¨g£©¡£ÉèÔÚÈÝ»ýΪ2.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.60molN2£¨g£©ºÍ1.60molH2£¨g£©£¬·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâʱ£¬NH3µÄÎïÖʵÄÁ¿·ÖÊý£¨NH2µÄÎïÖʵÄÁ¿Óë·´Ó¦ÌåϵÖÐ×ܵÄÎïÖʵÄÁ¿Ö®±È£©Îª¡£¼ÆËã
¢Ù¸ÃÌõ¼þÏÂN2µÄƽºâת»¯ÂÊÊÇ                    £»
¢Ú¸ÃÌõ¼þÏ·´Ó¦2NH3£¨g£©N2£¨g£©+3H2£¨g£©µÄƽºâ³£ÊýΪ                    ¡£
¢ÛÈôζȲ»±ä£¬¼õСÈÝÆ÷Ìå»ý£¬ÔòƽºâÏò---------------------------Òƶ¯£¬c(NH3)½«--------------------
c(N2)½«----------------------£¨ÌîÔö´ó¡¢¼õС»ò²»±ä£©                 
»¯Ñ§·´Ó¦ËÙÂʵı仯ºÍ»¯Ñ§Æ½ºâÒƶ¯µÄ¹æÂɶ¼¿ÉÒÔͨ¹ýʵÑéµÃ³ö½áÂÛ£¬ÀÕÏÄÌØÁÐÔ­ÀíÒ²ÊÇÓÉʵÑé×ܽá³öÀ´µÄ¡£
£¨1£©½«Í­Æ¬Í¶È뵽ϡÏõËáÖУ¬¿ªÊ¼Ê±²úÉúÆøÅݵÄËÙÂʺÜÂý£¬Ëæºó·´Ó¦ËÙÂÊÔö´ó£¬¸ù¾ÝÄãѧµ½µÄ֪ʶ·ÖÎö£¬·´Ó¦ËÙÂÊÔö´óµÄÔ­ÒòÊÇ________________ __________________¡£µ±½øÐÐÒ»¶Îʱ¼äºó£¬·´Ó¦ËÙÂÊÓÖ¼õÂý£¬Õâ´ÎËÙÂʼõÂýµÄÔ­ÒòÊÇ_______________________________¡£
£¨2£©±ûͪºÍµâÔÚËáÐÔÈÜÒºÖз¢ÉúÏÂÁз´Ó¦£ºCH3COCH3£«I2CH3COCH2I£«H£«£«I£­¡£25¡æʱ£¬¸Ã·´Ó¦µÄËÙÂÊÓÉÏÂÁо­Ñ鹫ʽ¾ö¶¨£º
¦Ô£½k?c£¨CH3COCH3£©?c£¨H£«£©[mol/£¨L?s£©]
¸ù¾Ý¸Ã¾­Ñ鹫ʽ£¬Èç¹ûc£¨I2£©£¬c£¨CH3COCH3£©£¬c£¨H£«£©µÄÆðʼŨ¶ÈÒÀ´ÎΪ0£®01 mol/L£¬0£®1 mol/L£¬0£®01 mol/L£¬·ÖÎö·´Ó¦ËÙÂʵı仯Ç÷ÊÆÊÇ_______________ ________________¡£
£¨3£©ÔÚÒ»Ö§ÊÔ¹ÜÖÐÏȼÓÈë30 mL 0£®01 mol/L FeCl3ÈÜÒº£¬ÔٵμÓ10 mL 0£®01 mol/L KSCNÈÜÒº£¬Ò¡ÔÈ¡£½«ÖƵõÄÈÜҺƽ¾ù·Ö×°ÓÚ4Ö§ÊÔ¹ÜÖУ¬È»ºó·Ö±ð¼ÓÈ룺¢ÙFeCl3¹ÌÌ壻¢ÚKSCN¹ÌÌ壻¢ÛKCl¹ÌÌ壻¢ÜH2O¡£³ä·ÖÕñµ´ºó£¬ÈÜÒºµÄÑÕÉ«»ù±¾²»±äµÄÊÇ________£¨ÌîÐòºÅ£©¡£
£¨4£©ÔÚ50 mLµÄ×¢ÉäÆ÷ÖÐÎüÈë40 mL NO2ºÍN2O4µÄ»ìºÏÆøÌ壬ÔÚÁíһͬÑù¹æ¸ñµÄ×¢ÉäÆ÷ÖÐÎüÈë40 mL Br2£¨g£©£¬¶Ô±ÈÁ½×¢ÉäÆ÷ÖÐÆøÌåµÄÑÕÉ«ÍêÈ«Ò»Ö¡£È»ºó¶¼½«×¢ÉäÆ÷µÄ»îÈûÂýÂýÍÆÖÁ20 mL´¦£¬·¢ÏÖ×¢ÉäÆ÷ÄÚÆøÌåµÄÑÕÉ«±ä»¯µÄÏÖÏóÊÇ___________________________¡£ÍƲâƽºâÒƶ¯ÊÇͨ¹ý¹Û²ì×°ÈëNO2ºÍN2O4µÄ»ìºÏÆøÌåµÄ×¢ÉäÆ÷ÄÚÆøÌåµÄÑÕÉ«±È×°ÈëBr2£¨g£©µÄ×¢ÉäÆ÷ÄÚÆøÌåµÄÑÕÉ«_______¡£
£¨5£©Èç¹ûÒª¼ø±ðÃÜ·âÓÚÓÉÁ½ÇòÁ¬Í¨µÄÃܱÕÈÝÆ÷ÖгäÂúµÄºì×ØÉ«ÆøÌåÊÇNO2ºÍN2O4»ìºÏÆø»¹ÊÇBr2£¨g£©£¬¿ÉÒÔ____________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø