£¨2011?¹ã¶«£©ÀûÓùâÄܺ͹â´ß»¯¼Á£¬¿É½«CO2ºÍH2O£¨g£©×ª»¯ÎªCH4ºÍO2£®×ÏÍâ¹âÕÕÉäʱ£¬ÔÚ²»Í¬´ß»¯¼Á£¨I£¬II£¬III£©×÷ÓÃÏ£¬CH4²úÁ¿Ëæ¹âÕÕʱ¼äµÄ±ä»¯ÈçͼËùʾ£®
£¨1£©ÔÚ0-30СʱÄÚ£¬CH4µÄƽ¾ùÉú³ÉËÙÂÊV¢ñ¡¢V¢òºÍV¢ó´Ó´óµ½Ð¡µÄ˳ÐòΪ
V¢ó£¾V¢ò£¾V¢ñ
V¢ó£¾V¢ò£¾V¢ñ
£»
·´Ó¦¿ªÊ¼ºóµÄ12СʱÄÚ£¬ÔÚµÚ
¢ò
¢ò
ÖÖ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬ÊÕ¼¯µÄCH4×î¶à£®
£¨2£©½«ËùµÃCH4ÓëH2O£¨g£©Í¨Èë¾Û½¹Ì«ÑôÄÜ·´Ó¦Æ÷£¬·¢Éú·´Ó¦£ºCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©£¬¸Ã·´Ó¦µÄ¡÷H=+206kJ?mol-1
¢ÙÔÚ´ðÌ⿨µÄ×ø±êͼÖУ¬»­³ö·´Ó¦¹ý³ÌÖÐÌåϵµÄÄÜÁ¿±ä»¯Í¼£¨½øÐбØÒªµÄ±ê×¢£©
¢Ú½«µÈÎïÖʵÄÁ¿µÄCH4ºÍH2O£¨g£©³äÈë1LºãÈÝÃܱÕÈÝÆ÷£¬Ä³Î¶ÈÏ·´Ó¦´ïµ½Æ½ºâ£¬Æ½ºâ³£ÊýK=27£¬´Ëʱ²âµÃCOµÄÎïÖʵÄÁ¿Îª0.10mol£¬ÇóCH4µÄƽºâת»¯ÂÊ£¨¼ÆËã½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£©
£¨3£©ÒÑÖª£ºCH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨g£©¡÷H=-802kJ?mol-1
д³öÓÉCO2Éú³ÉCOµÄÈÈ»¯Ñ§·½³Ìʽ
CO2£¨g£©+3H2O£¨g£©¨T2O2£¨g£©+CO£¨g£©+3H2£¨g£©¡÷H=+1008 kJ?mol-1
CO2£¨g£©+3H2O£¨g£©¨T2O2£¨g£©+CO£¨g£©+3H2£¨g£©¡÷H=+1008 kJ?mol-1
£®
·ÖÎö£º£¨1£©Ïàͬʱ¼ä¼×ÍéµÄÎïÖʵÄÁ¿µÄ±ä»¯Á¿Ô½´ó£¬±íÃ÷ƽ¾ùËÙÂÊÔ½´ó£¬Ïàͬʱ¼ä¼×ÍéµÄÎïÖʵÄÁ¿µÄ±ä»¯Á¿Ô½Ð¡£¬Æ½¾ù·´Ó¦ËÙÂÊԽС£®ÓÉͼ2¿ÉÖª·´Ó¦¿ªÊ¼ºóµÄ12СʱÄÚ£¬ÔÚµÚ¢òÖÖ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬ÊÕ¼¯µÄCH4×î¶à£»
£¨2£©¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©£¬¸Ã·´Ó¦µÄ¡÷H=+206kJ?mol-1£¬·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬·´Ó¦ÎïÄÜÁ¿µÍÓÚÉú³ÉÎïÄÜÁ¿£»ÒÀ¾ÝÄÜÁ¿±ä»¯»­³öͼÏó£»
¢ÚÒÀ¾ÝºÏ³ÉƽºâÈý¶ÎʽÁÐʽ¼ÆËãµÃµ½×ª»¯ÂÊ£»
£¨3£©ÒÀ¾Ý¸Ç˹¶¨ÂɺÍÈÈ»¯Ñ§·½³Ìʽ¼ÆËãµÃµ½£»
½â´ð£º½â£º£¨1£©ÓÉͼ2¿ÉÖª£¬ÔÚ0¡«30hÄÚ£¬¼×ÍéµÄÎïÖʵÄÁ¿±ä»¯Á¿Îª¡÷n£¨¢ñ£©£¼¡÷n£¨¢ò£©£¼¡÷n£¨¢ó£©£¬¹ÊÔÚ0¡«30hÄÚ£¬CH4µÄƽ¾ùÉú³ÉËÙÂÊv£¨¢ó£©£¾v£¨¢ò£©£¾v£¨¢ñ£©£»
ÓÉͼ2¿ÉÖª·´Ó¦¿ªÊ¼ºóµÄ12СʱÄÚ£¬ÔÚµÚ¢òÖÖ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬ÊÕ¼¯µÄCH4×î¶à£»
¹Ê´ð°¸Îª£ºV¢ó£¾V¢ò£¾V¢ñ£»¢ò£®
£¨2£©¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©£¬¸Ã·´Ó¦µÄ¡÷H=+206kJ?mol-1£¬·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬·´Ó¦¹ý³ÌÖÐÌåϵµÄÄÜÁ¿±ä»¯Í¼Îª£º£»
¹Ê´ð°¸Îª£º
¢Ú½«µÈÎïÖʵÄÁ¿µÄCH4ºÍH2O£¨g£©³äÈë1LºãÈÝÃܱÕÈÝÆ÷£¬Ä³Î¶ÈÏ·´Ó¦´ïµ½Æ½ºâ£¬Æ½ºâ³£ÊýK=27£¬´Ëʱ²âµÃCOµÄÎïÖʵÄÁ¿Îª0.10mol£¬¸ù¾ÝƽºâµÄÈý²½¼ÆËã¿ÉÇóCH4µÄƽºâת»¯ÂÊ£º
               CH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©
ÆðʼÁ¿£¨mol£©   x        x         0      0
±ä»¯Á¿£¨mol£©   0.10     0.10     0.10   0.30
ƽºâÁ¿£¨mol£©  x-0.10  x-0.10     0.10   0.30
K=
c(CO)c3(H2)
c(CH4)c(H2O)
=
0.10¡Á0.303
(x-0.10)2
=27
¼ÆËãµÃµ½x=0.11mol
¼×ÍéµÄת»¯ÂÊ=
0.10mol
0£¬11mol
¡Á100%=91%
¹Ê´ð°¸Îª£º91%
£¨3£©¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©£¬¡÷H=+206kJ?mol-1
¢ÚCH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨g£©¡÷H=-802kJ?mol-1
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù-¢ÚµÃµ½£ºCO2£¨g£©+3H2O£¨g£©¨T2O2£¨g£©+CO£¨g£©+3H2£¨g£©¡÷H=+1008 kJ?mol-1
¹Ê´ð°¸Îª£ºCO2£¨g£©+3H2O£¨g£©¨T2O2£¨g£©+CO£¨g£©+3H2£¨g£©¡÷H=+1008 kJ?mol-1
µãÆÀ£º±¾Ì⿼²éÁËͼÏó·ÖÎöºÍ»­Í¼ÏóµÄ·½·¨£¬Æ½ºâ¼ÆËãÓ¦Óã¬ÈÈ»¯Ñ§·½³ÌʽµÄÊéдԭÔòºÍ¸Ç˹¶¨ÂɵļÆËãÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2011?¹ã¶«Ä£Ä⣩Ϊ¼õСºÍÏû³ýCO2¶Ô»·¾³µÄÓ°Ï죬һ·½ÃæÊÀ½ç¸÷¹ú¶¼ÔÚÏÞÖÆÆäÅÅ·ÅÁ¿£¬ÁíÒ»·½Ãæ¿Æѧ¼Ò¼ÓÇ¿Á˶ÔCO2´´ÐÂÀûÓõÄÑо¿£®×î½üÓпÆѧ¼ÒÌá³ö¡°ÂÌÉ«×ÔÓÉ¡±¹¹Ï룺°Ñ¿ÕÆø´µÈë̼Ëá¼ØÈÜÒº£¬È»ºóÔÙ°ÑCO2´ÓÈÜÒºÖÐÌáÈ¡³öÀ´£¬¾­»¯Ñ§·´Ó¦ºóʹ֮±äΪ¿ÉÔÙÉúȼÁϼ״¼£®¡°ÂÌÉ«×ÔÓÉ¡±¹¹Ïë¼¼ÊõÁ÷³ÌÈçÏ£º

£¨1£©ÔںϳÉËþÖУ¬ÈôÓÐ4.4kg CO2Óë×ãÁ¿H2Ç¡ºÃÍêÈ«·´Ó¦£¬¿É·Å³ö4947kJµÄÈÈÁ¿£¬ÊÔд³öºÏ³ÉËþÖз¢Éú·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ
CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.47kJ/mol
CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.47kJ/mol
£®
£¨2£©ÒÔ¼×´¼ÎªÈ¼ÁÏÖÆ×÷ÐÂÐÍȼÁϵç³Ø£¬µç³ØµÄÕý¼«Í¨ÈëO2£¬¸º¼«Í¨Èë¼×´¼£¬ÔÚËáÐÔÈÜÒºÖм״¼ÓëÑõ×÷ÓÃÉú³ÉË®ºÍ¶þÑõ»¯Ì¼£®¸Ãµç³Ø¸º¼«·¢ÉúµÄ·´Ó¦ÊÇ£º
CH3OH+H2O-6e-¨TCO2+6H+
ÔòÕý¼«·¢ÉúµÄ·´Ó¦ÊÇ
O2+4H++4e-=2H2O
O2+4H++4e-=2H2O
£»·Åµçʱ£¬H+ÒÆÏòµç³ØµÄ
Õý
Õý
£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©¼«£®
£¨3£©³£Î³£Ñ¹Ï£¬±¥ºÍCO2Ë®ÈÜÒºµÄpH=5.6£¬c£¨H2CO3£©=1.5¡Á10-5 mol?L-1£®ÈôºöÂÔË®µÄµçÀë¼°H2CO3µÄµÚ¶þ¼¶µçÀ룬ÔòH2CO3?HCO3-+H+µÄƽºâ³£ÊýK=
4.2¡Á10-7mol?L-1
4.2¡Á10-7mol?L-1
£®£¨ÒÑÖª£º10-5.6=2.5¡Á10-6£©
£¨4£©³£ÎÂÏ£¬0.1mol?L-1NaHCO3ÈÜÒºµÄpH´óÓÚ8£¬ÔòÈÜÒºÖÐc£¨H2CO3£©
£¾
£¾
c£¨CO32-£© £¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£¬Ô­ÒòÊÇ
ÈÜÒº´æÔÚƽºâHCO3-?CO32-+H+¡¢HCO3-+H2O?H2CO3+OH-£¬¶øHCO3-µÄË®½â³Ì¶È´óÓÚµçÀë³Ì¶È
ÈÜÒº´æÔÚƽºâHCO3-?CO32-+H+¡¢HCO3-+H2O?H2CO3+OH-£¬¶øHCO3-µÄË®½â³Ì¶È´óÓÚµçÀë³Ì¶È
£¨ÓÃÀë×Ó·½³ÌʽºÍ±ØÒªµÄÎÄ×Ö˵Ã÷£©£®
£¨5£©Ð¡ÀîͬѧÄâÓóÁµí·¨²â¶¨¿ÕÆøÖÐCO2µÄÌå»ý·ÖÊý£¬Ëû²éµÃCaCO3¡¢BaCO3µÄÈܶȻý£¨Ksp£©·Ö±ðΪ4.96¡Á10-9¡¢2.58¡Á10-9£®Ð¡ÀîÓ¦¸ÃÑ¡ÓõÄÊÔ¼ÁÊÇ
Ba£¨OH£©2£¨»òNaOHÈÜÒººÍBaCl2ÈÜÒº£©
Ba£¨OH£©2£¨»òNaOHÈÜÒººÍBaCl2ÈÜÒº£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(8·Ö)(2011¡¤¹ã¶«Ä£Äâ)ÒÑÖªÒ»¸ö̼ԭ×ÓÉÏͬʱÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÈçÏÂת»¯£º

Çë¸ù¾ÝÏÂͼ»Ø´ð£º

(1)EÖк¬ÓеĹÙÄÜÍŵÄÃû³ÆÊÇ________£¬¢ÛµÄ·´Ó¦ÀàÐÍÊÇ________£¬C¸úÐÂÖƵÄÇâÑõ»¯Í­ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________________¡£

(2)ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäȼÉÕ²úÎïÖÐn(CO2)¡Ãn(H2O)£½2¡Ã1¡£FÊǵçÄÔоƬÖеĸ߷Ö×Ó¹â×è¼ÁµÄÖ÷ÒªÔ­ÁÏ£¬FµÄÌصãÊÇ£º¢ÙÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈÈ¡´úÎïÖ»ÓÐÁ½ÖÖ¡£FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________¡£

(3)»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ËüÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦¡£G¿ÉÄÜÓÐ________Öֽṹ£¬Ð´³öÆäÖÐÈÎÒâÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_____________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ì¸ß¿¼»¯Ñ§Ò»ÂÖ¸´Ï°¡¶ÌþµÄÑÜÉúÎרÌâ×ۺϲâÊÔ£¨Ëս̰棩 ÌâÐÍ£ºÌî¿ÕÌâ

(8·Ö)(2011¡¤¹ã¶«Ä£Äâ)ÒÑÖªÒ»¸ö̼ԭ×ÓÉÏͬʱÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÈçÏÂת»¯£º

Çë¸ù¾ÝÏÂͼ»Ø´ð£º

(1)EÖк¬ÓеĹÙÄÜÍŵÄÃû³ÆÊÇ________£¬¢ÛµÄ·´Ó¦ÀàÐÍÊÇ________£¬C¸úÐÂÖƵÄÇâÑõ»¯Í­ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________________¡£
(2)ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäȼÉÕ²úÎïÖÐn(CO2) ¡Ãn(H2O)£½2 ¡Ã1¡£FÊǵçÄÔоƬÖеĸ߷Ö×Ó¹â×è¼ÁµÄÖ÷ÒªÔ­ÁÏ£¬FµÄÌصãÊÇ£º¢ÙÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈÈ¡´úÎïÖ»ÓÐÁ½ÖÖ¡£FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________¡£
(3)»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ËüÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦¡£G¿ÉÄÜÓÐ________Öֽṹ£¬Ð´³öÆäÖÐÈÎÒâÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê¸ß¿¼»¯Ñ§Ò»ÂÖ¸´Ï°¡¶ÌþµÄÑÜÉúÎרÌâ×ۺϲâÊÔ£¨Ëս̰棩 ÌâÐÍ£ºÌî¿ÕÌâ

(8·Ö)(2011¡¤¹ã¶«Ä£Äâ)ÒÑÖªÒ»¸ö̼ԭ×ÓÉÏͬʱÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÈçÏÂת»¯£º

Çë¸ù¾ÝÏÂͼ»Ø´ð£º

(1)EÖк¬ÓеĹÙÄÜÍŵÄÃû³ÆÊÇ________£¬¢ÛµÄ·´Ó¦ÀàÐÍÊÇ________£¬C¸úÐÂÖƵÄÇâÑõ»¯Í­ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________________¡£

(2)ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäȼÉÕ²úÎïÖÐn(CO2) ¡Ãn(H2O)£½2 ¡Ã1¡£FÊǵçÄÔоƬÖеĸ߷Ö×Ó¹â×è¼ÁµÄÖ÷ÒªÔ­ÁÏ£¬FµÄÌصãÊÇ£º¢ÙÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈÈ¡´úÎïÖ»ÓÐÁ½ÖÖ¡£FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________¡£

(3)»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ËüÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦¡£G¿ÉÄÜÓÐ________Öֽṹ£¬Ð´³öÆäÖÐÈÎÒâÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_____________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸