ÌâÄ¿ÄÚÈÝ

£¨1£©Ò»¶¨Ìõ¼þÏ£¬NOÓëNO2´æÔÚÏÂÁз´Ó¦£ºNO£¨g£©+NO2£¨g£©?N2O3£¨g£©£¬Æäƽºâ³£Êý±í´ïʽΪK=
 
£®
£¨2£©¶þ¼×ÃÑÊÇÒ»ÖÖÖØÒªµÄÇå½àȼÁÏ£¬¿ÉÒÔͨ¹ýCH3OH·Ö×Ó¼äÍÑË®ÖÆµÃ£º2CH3OH£¨g£©?CH3OCH3£¨g£©+H2O£¨g£©¡÷H=-23.5kJ?mol-1£®ÔÚT1¡æÊ±£¬ºãÈÝÃܱÕÈÝÆ÷Öн¨Á¢ÉÏÊöƽºâ£¬ÌåϵÖи÷×é·ÖŨ¶ÈËæÊ±¼ä±ä»¯ÈçͼËùʾ£®
¢ÙÔÚT1¡æÊ±£¬·´Ó¦µÄƽºâ³£ÊýΪ
 
£»CH3OHµÄת»¯ÂÊΪ
 
£®
¢ÚÏàͬÌõ¼þÏ£¬Èô¸Ä±äÆðʼŨ¶È£¬Ä³Ê±¿Ì¸÷×é·ÖŨ¶ÈÒÀ´ÎΪc£¨CH3OH£©=0.4mol?L-1¡¢c£¨H2O£©=0.6mol?L-1¡¢c£¨CH3OCH3£©=1.2mol?L-1£¬´ËʱÕý¡¢Äæ·´Ó¦ËÙÂʵĴóС£ºv£¨Õý£©
 
v£¨Ä棩£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
£¨3£©²ÝËá¶þ¼×õ¥Ë®½â²úÎï²ÝËᣨH2C2O4£©Îª¶þÔªÖÐÇ¿Ëá
²ÝËáÇâ¼ØÈÜÒºÖдæÔÚÈçÏÂÆ½ºâ£ºH2O?H++OH-¡¢HC2O4-?H++C2O42-ºÍ
 
£®
£¨4£©ÒÔ¼×´¼ÎªÔ­ÁÏ£¬Ê¹ÓÃËáÐÔµç½âÖʹ¹³ÉȼÁÏµç³Ø£¬¸ÃȼÁÏµç³ØµÄ¸º¼«·´Ó¦Ê½Îª
 
£»ÈôÒÔ¼×Íé´úÌæ¸ÃȼÁÏµç³ØÖеļ״¼£¬ÏòÍâ½çÌṩÏàµÈµçÁ¿£¬Ôòÿ´ú Ìæ32g¼×´¼£¬ËùÐè±ê×¼×´¿öϵļ×ÍéµÄÌå»ýΪ
 
L£®
¿¼µã£ºÎïÖʵÄÁ¿»òŨ¶ÈËæÊ±¼äµÄ±ä»¯ÇúÏß,»¯Ñ§µçÔ´ÐÂÐÍµç³Ø
רÌ⣺»¯Ñ§·´Ó¦ÖеÄÄÜÁ¿±ä»¯,µç»¯Ñ§×¨Ìâ
·ÖÎö£º£¨1£©»¯Ñ§Æ½ºâ³£Êý£¬ÊÇÖ¸ÔÚÒ»¶¨Î¶ÈÏ£¬¿ÉÄæ·´Ó¦´ïµ½Æ½ºâʱ¸÷Éú³ÉÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ý³ýÒÔ¸÷·´Ó¦ÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ýËùµÃµÄ±ÈÖµ£»
£¨2£©¢ÙÓÉͼ¿ÉÖª£¬ÔÚt1minʱµ½´ïƽºâ£¬Æ½ºâʱCH3OH¡¢CH3OCH3¡¢H2OŨ¶È·Ö±ðΪ0.4mol/L¡¢1mol/L¡¢0.8mol/L£¬´úÈëÆ½ºâ³£Êý±í´ïʽ¼ÆË㣻¸ù¾Ýת»¯ÂÊ=
ת»¯µÄÁ¿
ÆðʼµÄ×ÜÁ¿
¡Á100%½øÐмÆË㣻
¢Ú¼ÆËã·´Ó¦Öи÷ÎïÖʵÄŨ¶ÈÉÌQc£¬ÓëÆ½ºâ³£Êý±È½Ï£¬ÅжϷ´Ó¦½øÐеķ½Ïò£¬½ø¶øÅжÏÕý¡¢Äæ·´Ó¦ËÙÂʹØÏµ£»
£¨3£©¢ÙH2C2O4Ϊ¶þÔªÖÐÇ¿ËᣬHC2O4-ÔÚÈÜÒºÖмȷ¢ÉúµçÀëÓÖ·¢ÉúË®½â£»
£¨4£©×Ü·´Ó¦Ê½Îª2CH3OH+3O2£¨g£©=2CO2£¨g£©+4H2O£¬Õý¼«ÊÇ»¹Ô­·´Ó¦£¬ÑõÆøÔÚÕý¼«·Åµç£¬ËáÐÔÌõ¼þÏ»ñµÃµç×ÓÉú³ÉË®£¬×Ü·´Ó¦Ê½¼õÈ¥Õý¼«·´Ó¦Ê½¿ÉµÃ¸º¼«µç¼«·´Ó¦Ê½£»¸ù¾Ý×ªÒÆµç×ÓÊýÄ¿ÏàµÈ¼ÆËã¼×ÍéµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝV=nVm¼ÆË㣮
½â´ð£º ½â£º£¨1£©»¯Ñ§Æ½ºâ³£Êý£¬ÊÇÖ¸ÔÚÒ»¶¨Î¶ÈÏ£¬¿ÉÄæ·´Ó¦´ïµ½Æ½ºâʱ¸÷Éú³ÉÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ý³ýÒÔ¸÷·´Ó¦ÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ýËùµÃµÄ±ÈÖµ£¬NO£¨g£©+NO2£¨g£©?N2O3£¨g£©£¬Æäƽºâ³£Êý±í´ïʽΪK=
c(N2O3)
c(NO)?c(NO2)
£¬¹Ê´ð°¸Îª£º=
c(N2O3)
c(NO)?c(NO2)
£»
£¨2£©¢ÙÓÉͼ¿ÉÖª£¬ÔÚt1minʱµ½´ïƽºâ£¬Æ½ºâʱCH3OH¡¢CH3OCH3¡¢H2OŨ¶È·Ö±ðΪ0.4mol/L¡¢1mol/L¡¢0.8mol/L£¬¹Êƽºâ³£Êýk=
c(N2O3)
c(NO)?c(NO2)
=
1¡Á0.8
0.42
=5£»
CH3OHµÄת»¯ÂÊ=
ת»¯µÄÁ¿
ÆðʼµÄ×ÜÁ¿
¡Á100%=
2.0-0.4
2.0
¡Á100%=80%£¬¹Ê´ð°¸Îª£º5£»80%£»
¢Ú´ËʱµÄŨ¶ÈÉÌQc=
0.6¡Á1.2
0.42
=4.5£¬Ð¡ÓÚÆ½ºâ³£Êý5£¬¹Ê·´Ó¦ÏòÕý·´Ó¦½øÐУ¬v £¨Õý£©£¾v £¨Ä棩£¬¹Ê´ð°¸Îª£º£¾£»
£¨3£©H2C2O4Ϊ¶þÔªÖÐÇ¿ËᣬHC2O4-ÔÚÈÜÒºÖз¢ÉúµçÀëÓëË®½â£¬»¹´æÔÚÆ½ºâ£ºHC2O4-+H2O?H2C2O4+OH-£¬¹Ê´ð°¸Îª£ºHC2O4-+H2O?H2C2O4+OH-£»
£¨4£©×Ü·´Ó¦Ê½Îª2CH3OH+3O2£¨g£©=2CO2£¨g£©+4H2O£¬Õý¼«ÊÇ»¹Ô­·´Ó¦£¬ÑõÆøÔÚÕý¼«·Åµç£¬ËáÐÔÌõ¼þÏ»ñµÃµç×ÓÉú³ÉË®£¬Õý¼«µç¼«·´Ó¦Ê½Îª£º3O2+12H++12e-=6H2O£¬×Ü·´Ó¦Ê½¼õÈ¥Õý¼«·´Ó¦Ê½¿ÉµÃ¸º¼«µç¼«·´Ó¦Ê½Îª£º2CH3OH+2H2O-12e-=2CO2+12H+£¬¼´CH3OH+H2O-6e-=CO2+6H+£¬32g¼×´¼µÄÎïÖʵÄÁ¿Îª
32g
32g/mol
=1mol£¬ÍêȫȼÉÕ×ªÒÆµç×ÓÊýΪ1mol¡Á6=6mol£¬¹ÊÐèÒª¼×ÍéµÄÎïÖʵÄÁ¿Îª
6mol
8
=0.75mol£¬ÐèÒª¼×ÍéµÄÌå»ýΪ0.75mol¡Á22.4L/mol=16.8L£¬
¹Ê´ð°¸Îª£ºCH3OH+H2O-6e-=CO2+6H+£»16.8£®
µãÆÀ£º±¾Ì⿼²é»¯Ñ§Æ½ºâ³£ÊýµÄÓйؼÆËãÓëÓ¦Óᢻ¯Ñ§Æ½ºâͼÏóµÈ£¬ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕ»¯Ñ§Æ½ºâ³£ÊýµÄÓ¦Óãº1¡¢ÓÃÀ´ÅжϷ´Ó¦½øÐеij̶ȣ¬2¡¢ÅжϷ´Ó¦½øÐеķ½Ïò£¬3¡¢ÓÃÀ´¼ÆËãÎïÖʵÄת»¯ÂÊ£¬4¡¢×¢ÒâÀûÓÃÕý¡¢¸ºµç¼«·´Ó¦Ê½Ö®ºÍµÈÓÚ×Ü·´Ó¦Ê½½øÐÐÊéд£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø