ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ò»¶¨Î¶ÈÏ£¬ÏòÈÝ»ýΪV LµÄºãÈÝÃܱÕÈÝÆ÷ÖÐͨÈë1 mol CH3Cl(g)ºÍ1 mol H2O(g)£¬ÌåϵÄÚͬʱ´æÔÚÏÂÁÐÁ½¸öƽºâ£º

·´Ó¦¢Ù£ºCH3Cl(g)£«H2O(g)CH3OH(g)£«HCl(g) K1

·´Ó¦¢Ú£º2CH3OH(g)(CH3)2O(g)£«H2O(g) K2

·´Ó¦t minºóÌåϵ´ïµ½Æ½ºâ£¬´Ëʱ(CH3)2O(g)µÄÎïÖʵÄÁ¿Îª9¡Á10-3mol£¬CH3Cl(g)µÄƽºâת»¯ÂÊΪ4.8%£¬»Ø´ðÏÂÁÐÎÊÌ⣺

(1)0¡«t minÄÚ£¬CH3Cl(g)µÄ·´Ó¦ËÙÂÊΪ___________¡£

(2)·´Ó¦´ïµ½Æ½ºâʱ£¬CH3OH(g)µÄÎïÖʵÄÁ¿Îª___________¡£

(3)¼ÆËã·´Ó¦¢ÚµÄƽºâ³£ÊýK2£½___________¡£

(4)µ±·´Ó¦µ½´ïƽºâʱ£¬ÔÙÏòÌåϵÄÚͨÈëÒ»¶¨Á¿µÄCH3OH(g)£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________(Ìî×Öĸ)¡£

A.·´Ó¦¢ÙµÄƽºâÄæÏòÒƶ¯£¬·´Ó¦¢ÚµÄƽºâ²»·¢ÉúÒƶ¯

B.ƽºâʱ·´Ó¦¢Ù¡¢·´Ó¦¢ÚµÄ·´Ó¦ËÙÂʶ¼Ôö´ó

C.K1Ôö´ó£¬K2¼õС

¡¾´ð°¸¡¿molL-1min-1 0.03mol 9.61 B

¡¾½âÎö¡¿

Ò»¶¨Î¶ÈÏ£¬ÏòÈÝ»ýΪV LµÄºãÈÝÃܱÕÈÝÆ÷ÖÐͨÈë1 mol CH3Cl(g)ºÍ1 mol H2O(g)£¬ÌåϵÄÚͬʱ´æÔÚÏÂÁÐÁ½¸öƽºâ£º

·´Ó¦¢Ù£ºCH3Cl(g)£«H2O(g)CH3OH(g)£«HCl(g) K1

·´Ó¦¢Ú£º2CH3OH(g)(CH3)2O(g)£«H2O(g) K2

·´Ó¦´ïµ½Æ½ºâʱҪ¿¼Âǵ½CH3OH(g)µÄŨ¶ÈÊÇÁ½¸ö»¯Ñ§·´Ó¦×ۺϺóµÃµ½µÄ½á¹û¡£

£¨1£©·´Ó¦¿ªÊ¼Ê±£¬CH3Cl(g)Ϊ1 mol£¬t minºóCH3Cl(g)µÄƽºâת»¯ÂÊΪ4.8%£¬ËùÒÔCH3Cl(g)µÄÎïÖʵÄÁ¿±ä»¯Á¿Îª1mol4.8%=0.048mol£¬»¯Ñ§·´Ó¦ËÙÂÊ===molL-1min-1£¬¹Ê´ð°¸ÎªmolL-1min-1£»

£¨2£©ÌåϵÖÐÁ½¸ö·´Ó¦Í¬Ê±·¢Éú£¬ÓÉ·´Ó¦¢ÙCH3Cl(g)µÄÎïÖʵÄÁ¿±ä»¯Á¿Îª0.048mol£¬¿ÉÖª¹²Éú³ÉCH3OH(g)µÄÎïÖʵÄÁ¿Îª0.048mol£¬ÓÉ·´Ó¦¢ÚÖÐ(CH3)2O(g)µÄÎïÖʵÄÁ¿Îª9¡Á10-3mol£¬¿ÉÖªÏûºÄµÄCH3OH(g)µÄÎïÖʵÄÁ¿Îª9¡Á10-3mol¡Á2=0.018mol£¬ËùÒÔ·´Ó¦´ïµ½Æ½ºâʱ£¬CH3OH(g)µÄÎïÖʵÄÁ¿Îª0.048mol-0.018mol=0.03mol£¬¹Ê´ð°¸Îª0.03mol£»

£¨3£©·´Ó¦¢Ù´ïµ½Æ½ºâºóûÓвμӷ´Ó¦µÄH2O(g)µÄÎïÖʵÄÁ¿Îª1mol-0.048mol=0.952mol£¬·´Ó¦¢ÚÉú³ÉµÄH2O£¨g£©µÄÎïÖʵÄÁ¿Îª9¡Á10-3mol£¬ËùÒÔƽºâʱÈÝÆ÷ÄÚH2O£¨g£©µÄÎïÖʵÄÁ¿Å¨¶ÈΪ= molL-1£¬(CH3)2O(g)µÄÎïÖʵÄÁ¿Å¨¶ÈΪ=molL-1£¬ÓÉ£¨2£©¿É֪ƽºâʱCH3OH(g)µÄÎïÖʵÄÁ¿Å¨¶ÈΪ= molL-1£¬Æ½ºâ³£ÊýK2£½=9.61£¬¹Ê´ð°¸Îª9.61£»

£¨4£©µ±·´Ó¦µ½´ïƽºâʱ£¬ÔÙÏòÌåϵÄÚͨÈëÒ»¶¨Á¿µÄCH3OH(g)£¬CH3OH(g)ÎïÖʵÄÁ¿Å¨¶ÈÔö´ó£¬·´Ó¦¢ÙµÄƽºâÄæÏòÒƶ¯£¬·´Ó¦¢ÚµÄƽºâÕýÏòÒƶ¯£¬Æ½ºâʱ·´Ó¦¢Ù¡¢·´Ó¦¢ÚµÄ·´Ó¦ËÙÂʶ¼Ôö´ó£¬Î¶Ȳ»±äʱ£¬·´Ó¦µÄƽºâ³£Êý¶¼²»¸Ä±ä£¬¹Ê´ð°¸Ñ¡B¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¹¤Òµ·ÏÆø¡¢Æû³µÎ²ÆøÅŷųöµÄNOx¡¢SO2µÈ£¬ÊÇÐγÉÎíö²µÄÖ÷ÒªÎïÖÊ£¬Æä×ÛºÏÖÎÀíÊǵ±Ç°ÖØÒªµÄÑо¿¿ÎÌâ¡£

£¨1£©ÒÑÖª£º¢ÙCOȼÉÕÈȵġ÷H1=£­283.0kJ¡¤mol-l£¬¢ÚN2(g)+O2(g) 2NO(g) ¡÷H2=+180.5kJ¡¤mol-1£¬Æû³µÎ²ÆøÖеÄNO(g)ºÍCO(g)ÔÚÒ»¶¨Î¶Ⱥʹ߻¯¼ÁÌõ¼þÏ¿ɷ¢ÉúÈçÏ·´Ó¦£º2NO(g)+2CO(g) N2(g)+2CO2(g)£» ¡÷H=___¡£

£¨2£©½«0£®20mol NOºÍ0£®10molCO³äÈëÒ»¸öÈÝ»ýºã¶¨Îª1LµÄÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦¹ý³ÌÖв¿·ÖÎïÖʵÄŨ¶È±ä»¯ÈçÏÂͼËùʾ£®

¢Ù¸Ã·´Ó¦µÚÒ»´Î´ïµ½Æ½ºâʱµÄƽºâ³£ÊýΪ________¡£

¢ÚµÚ12minʱ¸Ä±äµÄÌõ¼þÊÇ________¡£

¢ÛÔÚµÚ24minʱ£¬Èô±£³ÖζȲ»±ä£¬ÔÙÏòÈÝÆ÷ÖгäÈëCOºÍN2¸÷0£®060mol£¬Æ½ºâ½«________Òƶ¯(Ìî¡°ÕýÏò¡±¡¢¡°ÄæÏò¡±»ò¡°²»¡±)£®

(3)SNCR-SCRÍÑÏõ¼¼ÊõÊÇÒ»ÖÖÐÂÐ͵ijýÈ¥ÑÌÆøÖеªÑõ»¯ÎïµÄÍÑÏõ¼¼Êõ£¬Ò»°ã²ÉÓð±Æø»òÄòËØ¡£

¢ÙSNCRÍÑÏõ¼¼ÊõÖУºÔÚ´ß»¯¼Á×÷ÓÃÏÂÓÃNH3×÷»¹Ô­¼Á»¹Ô­NO£¬ÆäÖ÷Òª·´Ó¦Îª£º4NH3(g)+4NO(g)+O2(g)=4N2(g)+6H2O(g)£¬¡÷H<0¡£ÌåϵζÈÖ±½ÓÓ°ÏìSNCR¼¼ÊõµÄÍÑÏõЧÂÊ£¬ÈçͼËùʾ¡£µ±ÌåϵζÈԼΪ925¡æʱ£¬SNCRÍÑÏõЧÂÊ×î¸ß£¬Æä¿ÉÄܵÄÔ­ÒòÊÇ________¡£

¢ÚSCRÍÑÏõ¼¼ÊõÖÐÔòÓÃÄòËØ[CO(NH2)2]×÷»¹Ô­¼Á»¹Ô­NO2µÄ»¯Ñ§·½³ÌʽΪ____________¡£

¡¾ÌâÄ¿¡¿ÄÉÃ×̼Ëá¸Æ¹ã·ºÓ¦ÓÃÓÚÏ𽺡¢ËÜÁÏ¡¢ÔìÖ½¡¢»¯Ñ§½¨²Ä¡¢ÓÍÄ«¡¢Í¿ÁÏ¡¢Ãܷ⽺Ó뽺ճ¼ÁµÈÐÐÒµ¡£ÔÚŨCaCl2ÈÜÒºÖÐͨÈëNH3ºÍCO2£¬¿ÉÒÔÖƵÃÄÉÃ×¼¶Ì¼Ëá¸Æ¡£Ä³Ð£Ñ§ÉúʵÑéС×éÉè¼ÆÏÂͼËùʾװÖã¬ÖÆÈ¡¸Ã²úÆ·¡£DÖÐ×°ÓÐպϡÁòËáµÄÍÑÖ¬ÃÞ£¬Í¼ÖмгÖ×°ÖÃÒÑÂÔÈ¥¡£

¢ñ£®¿ÉÑ¡ÓõÄÒ©Æ·ÓУº

a£®Ê¯»Òʯ£»b£®±¥ºÍÂÈ»¯¸ÆÈÜÒº£»c£®6 mol/LÑÎË᣻d£®ÂÈ»¯ï§£»e£®ÇâÑõ»¯¸Æ

£¨1£©AÖÐÖƱ¸ÆøÌåʱ£¬ËùÐèÒ©Æ·ÊÇ£¨Ñ¡Ìî×ÖĸÐòºÅ£©______________£»

£¨2£©BÖÐÊ¢Óб¥ºÍ̼ËáÇâÄÆÈÜÒº£¬Æä×÷ÓÃÊÇ______________________________£»

£¨3£©Ð´³öÖÆÈ¡°±ÆøµÄ»¯Ñ§·½³Ìʽ__________________________________£»

£¨4£©ÔÚʵÑé¹ý³ÌÖУ¬ÏòCÖÐͨÈëÆøÌåÊÇÓÐÏȺó˳ÐòµÄ£¬Ó¦ÏÈͨÈëÆøÌåµÄ»¯Ñ§Ê½______________£»

£¨5£©¼ìÑéD³ö¿Ú´¦ÊÇ·ñÓа±ÆøÒݳöµÄ·½·¨ÊÇ__________________________£»

£¨6£©Ð´³öÖÆÄÉÃ×¼¶Ì¼Ëá¸ÆµÄ»¯Ñ§·½³Ìʽ______________________________¡£

£¨7£©ÈôʵÑé¹ý³ÌÖÐÓа±ÆøÒݳö£¬Ó¦Ñ¡ÓÃÏÂÁÐ_____________×°ÖûØÊÕ£¨Ìî´úºÅ£©¡£

¢ò£®¾­·ÖÎöÔÚÉÏÊöÂÈ»¯ï§ÑùÆ·Öк¬ÓÐÔÓÖÊ̼ËáÇâÄÆ¡£ÎªÁ˲ⶨÂÈ»¯ï§µÄÖÊÁ¿·ÖÊý£¬¸ÃѧÉúʵÑéС×éÓÖÉè¼ÆÁËÈçÏÂʵÑéÁ÷³Ì£º

ÊԻشð£º

£¨1£©Ëù¼ÓÊÔ¼ÁAµÄ»¯Ñ§Ê½Îª______________________________________£»

£¨2£©B²Ù×÷·½·¨ÊÇ_______________________________________________£»

£¨3£©ÑùÆ·ÖÐÂÈ»¯ï§µÄÖÊÁ¿·ÖÊýΪ___________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø