½«1molCH4ºÍÒ»¶¨Á¿µÄÑõÆøÔÚÃܱÕÈÝÆ÷Öеãȼ£¬³ä·Ö·´Ó¦ºó£¬¼×ÍéºÍÑõÆø¾ùÎÞÊ£Ó࣬ÇÒ²úÎï¾ùÎªÆøÌå
£¨101kPa£¬120¡æ£©£¬Æä×ÜÖÊÁ¿Îª72g£¬ÏÂÁÐÓйØÐðÊö²»ÕýÈ·µÄÊÇ
[     ]
A£®Èô½«²úÎïͨ¹ý¼îʯ»Ò£¬Ôò²»ÄÜÈ«±»ÎüÊÕ£¬Èôͨ¹ýŨÁòËᣬҲ²»Äܱ»ÍêÈ«ÎüÊÕ
B£®²úÎïµÄƽ¾ùĦ¶ûÖÊÁ¿Îª24g/mol
C£®Èô½«²úÎïͨ¹ýŨÁòËá³ä·ÖÎüÊÕºó»Ö¸´ÖÁ£¨101kPa£¬120¡æ£©£¬Ôòѹǿ±äΪԭÀ´µÄ2/3
D£®·´Ó¦ÖÐÏûºÄµÄÑõÆøÎª56g
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖªÔÚ25¡æ¡¢1.013¡Á105PaÏ£¬1molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö285kJµÄÈÈÁ¿£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Éú³ÉÎïÄÜÁ¿×ܺÍ
СÓÚ
СÓÚ
£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©·´Ó¦ÎïÄÜÁ¿×ܺÍ
£¨2£©Èô2molÇâÆøÍêȫȼÉÕÉú³ÉË®ÕôÆø£¬Ôò·Å³öµÄÈÈÁ¿
£¼
£¼
£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©570kJ
£¨3£©ÏÖÓÐ250C¡¢1.013¡Á105PaϵÄH2ºÍCH4µÄ»ìºÏÆøÌå0.5mol ÍêȫȼÉÕÉú³ÉÒ»¶¨ÖÊÁ¿µÄCO2ÆøÌåºÍ10.8gH2O£¨l£©£¬·Å³ö203kJµÄÈÈÁ¿£¬Ôò1molCH4ÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍH2O£¨l£©·Å³öµÄÈÈÁ¿Îª
890
890
kJ£®
£¨4£©ÃÀ¹ú°¢²¨ÂÞÓîÖæ·É´¬ÉÏʹÓÃÁËÒ»ÖÖÐÂÐÍ×°Öã¬Æä¹¹ÔìÈçͼËùʾ£ºA¡¢BÁ½¸öµç¼«¾ùÓɶà¿×µÄ̼¿é×é³É£®
¸Ãµç³ØµÄÕý¼«Ê½Îª£º
O2+2H2O+4e-=4OH-
O2+2H2O+4e-=4OH-
£®Èô¸Ãµç³Ø¹¤×÷ʱÔö¼ÓÁË1mol H2O£¬µç·ÖÐ×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿Îª
2mol
2mol
£®
£¨5£©Èç¹û½«ÉÏÊö×°ÖÃÖÐͨÈëµÄH2¸Ä³ÉCH4ÆøÌ壬Ҳ¿ÉÒÔ×é³ÉÒ»¸öÔ­µç³Ø×°Öã¬µç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ£ºCH4+2O2+2KOH=K2CO3+3H20£¬Ôò¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª£º
CH4+10OH--8e-=CO32-+7H2O
CH4+10OH--8e-=CO32-+7H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¢ñ£®2007Äêŵ±´¶û»¯Ñ§½±ÊÚÓèÉÆÓÚ×ö¡°±íÃæÎÄÕ¡±µÄµÂ¹ú¿ÆÑ§¼Ò¸ñ¹þµÂ?°£ÌضûÒ»£¬ÒÔ±íÕÃËûÔÚ±íÃæ»¯Ñ§Ñо¿ÁìÓò×÷³öµÄ¿ªÍØÐÔ¹±Ï×£®ËûµÄ³É¾ÍÖ®Ò»ÊÇ֤ʵÁËÆøÌåÔÚ¹ÌÌå´ß»¯¼Á±íÃæ½øÐеķ´Ó¦¹ý³Ì£¬¿ª´´Á˱íÃæ»¯Ñ§µÄ·½·¨ÂÛ£®
£¨1£©Ä³Ð£»¯Ñ§Ñо¿ÐÔѧϰС×éµÄͬѧÔÚ¼¼ÊõÈËÔ±µÄÖ¸µ¼Ï£¬°´ÏÂÁÐÁ÷³Ì̽¾¿²»Í¬´ß»¯¼Á¶ÔNH3»¹Ô­NO·´Ó¦µÄ´ß»¯ÐÔÄÜ£®
¾«Ó¢¼Ò½ÌÍø
Èô¿ØÖÆÆäËûʵÑéÌõ¼þ¾ùÏàͬ£¬ÔÚ´ß»¯·´Ó¦Æ÷ÖÐ×°ÔØ²»Í¬µÄ´ß»¯¼Á£¬½«¾­´ß»¯·´Ó¦ºóµÄ»ìºÏÆøÌåͨ¹ýµÎÓзÓ̪µÄÏ¡ÁòËáÈÜÒº£¨ÈÜÒºµÄÌå»ý¡¢Å¨¶È¾ùÏàͬ£©£®Îª±È½Ï²»Í¬´ß»¯¼ÁµÄ´ß»¯ÐÔÄÜ£¬ÐèÒª²âÁ¿²¢¼Ç¼µÄÊý¾ÝÊÇ
 
£®
£¨2£©Îª±ÜÃâÆû³µÎ²ÆøÖеÄÓк¦ÆøÌå¶Ô´óÆøµÄÎÛȾ£¬¸øÆû³µ°²×°Î²Æø¾»»¯×°Ö㮾»»¯×°ÖÃÀï×°Óк¬PdµÈ¹ý¶ÉÔªËØµÄ´ß»¯¼Á£¬ÆøÌåÔÚ´ß»¯¼Á±íÃæÎü¸½Óë½âÎü×÷ÓõĻúÀíÈçÏÂͼËùʾ£®
¾«Ó¢¼Ò½ÌÍø
ÔÚÆû³µµÄÅÅÆø¹ÜÉϰ²×°¡°´ß»¯×ª»¯Æ÷¡±£¨Óò¬¡¢îٺϽð×÷´ß»¯¼Á£©£¬ËüµÄ×÷ÓÃÊÇʹCO¡¢NO·´Ó¦Éú³É¿É²ÎÓë´óÆøÉú̬»·¾³Ñ­»·µÄÎÞ¶¾ÆøÌ壬²¢´ÙʹÌþÀà³ä·ÖȼÉÕ£®
¢Ùд³öÉÏÊö±ä»¯ÖеÄ×Ü»¯Ñ§·´Ó¦·½³Ìʽ£º
 
£®
¢ÚÓÃCH4´ß»¯»¹Ô­NOxÒ²¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ£®ÀýÈ磺
CH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©£»¡÷H1=-574kJ?mol-1
CH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©£»¡÷H2
Èô1molCH4»¹Ô­NO2ÖÁN2£¬Õû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª867kJ£¬Ôò¡÷H2=
 
£®
£¨3£©¢ÙÓÐÈËÈÏΪ£º¸ÃÑо¿¿ÉÒÔʹ°±µÄºÏ³É·´Ó¦£¬ÔÚÌú´ß»¯¼Á±íÃæ½øÐÐʱµÄЧÂÊ´ó´óÌá¸ß£¬´Ó¶øÊ¹Ô­ÁϵÄת»¯ÂÊ´ó´óÌá¸ß£®ÇëÄãÓ¦Óû¯Ñ§»ù±¾ÀíÂÛ¶Ô´Ë¹Ûµã½øÐÐÆÀ¼Û£º
 
£®
¢ÚÇâÆøÓëµªÆøÔÚ¹ÌÌå´ß»¯¼Á±íÃæºÏ³É°±ÆøµÄ·´Ó¦¹ý³Ì¿É±íʾÈçÏÂͼËùʾ£¬ÕýÈ·µÄ˳ÐòΪ
 

¾«Ó¢¼Ò½ÌÍø
A£®¢Ù¢Ú¢Û¢Ü¢ÝB£®¢Ú¢Ù¢Û¢Ü¢ÝC£®¢Ü¢Ý¢Û¢Ú¢ÙD£®¢Ý¢Ü¢Û¢Ú¢Ù
¢ò£®A¡¢B¡¢C¡¢D¡¢EÎåÖÖÎïÖʺ¬ÓÐÍ¬Ò»ÔªËØX£¬ÔÚÒ»¶¨Ìõ¼þÏÂËüÃÇÓÐÈçÏÂת»¯¹ØÏµ£¬ÈôA¡¢B¡¢C¡¢DÎªÆøÌåÇÒÆäÖÐÓÐÒ»ÖÖΪÓÐÉ«ÆøÌ壬EΪҺÌ壬»Ø´ðÏÂÁÐÎÊÌ⣮
¾«Ó¢¼Ò½ÌÍø
£¨1£©BÖÐÖ»º¬Á½ÖÖÔªËØ£¬ÇÒÁ½ÖÖÔªËØµÄÐòÊýÏà²î6£¬µ±BºÍEÔÚ¿ÕÆøÖÐÏàÓöʱÄÜÐγɴóÁ¿µÄ°×ÑÌ£¬ÔòDµÄ·Ö×ÓʽΪ
 

£¨2£©µ±Ð¡ÐļÓÈÈBÓëE»¯ºÏʱÉú³ÉµÄÎïÖÊFʱ£¬¿ÉµÃµ½Ò»ÖÖÓëCO2¾ßÓÐÏàͬԭ×ÓÊýÄ¿ºÍµç×ÓÊýÄ¿µÄÆøÌåGºÍÁíÒ»ÖÖÎÞÉ«ÎÞζµÄÒºÌåH£¬Ð´³öÏàÓ¦µÄ»¯Ñ§·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìɽ¶«Ê¡Î«·»ÊÐÈýÏØÊиßÈýÉÏѧÆÚÆÚÖÐÁªºÏ¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö£©Ñо¿NOx¡¢SO2¡¢COµÈ´óÆøÎÛÈ¾ÆøÌåµÄ´¦Àí·½·¨¾ßÓÐÖØÒªÒâÒå¡£
£¨1£©´¦Àíº¬CO¡¢SO2Ñ̵ÀÆøÎÛȾµÄÒ»ÖÖ·½·¨Êǽ«ÆäÔÚ´ß»¯¼Á×÷ÓÃÏÂת»¯Îªµ¥ÖÊS¡£
ÒÑÖª£º¢ÙCO£¨g£©+ 1/2O2(g)£½CO2£¨g£©  ¡÷H=£­283.0KJ¡¤mol-1  
¢ÚS£¨s£©+ O2(g)£½SO2£¨g£©    ¡÷H=£­296.0KJ¡¤mol-1
´Ë·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                                          ¡£
£¨2£©µªÑõ»¯ÎïÊÇÔì³É¹â»¯Ñ§ÑÌÎíºÍ³ôÑõ²ãËðºÄµÄÖ÷񻮿Ìå¡£ÒÑÖª£º
¢Ù CO£¨g£©+NO2(g)£½NO(g)+CO2(g)  ¡÷H=£­aKJ¡¤mol-1 (a£¾0)
¢Ú 2CO£¨g£©+2NO (g)£½N2(g)+2CO2(g)  ¡÷H=£­bKJ¡¤mol-1 (b£¾0)
ÈôÓñê×¼×´¿öÏ 3.36LCO»¹Ô­NO2ÖÁN2£¨COÍêÈ«·´Ó¦£©µÄÕû¸ö¹ý³ÌÖÐ×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿Îª             mol£¬·Å³öµÄÈÈÁ¿Îª              KJ£¨Óú¬ÓÐaºÍbµÄ´úÊýʽ±íʾ£©¡£
£¨3£©ÓÃCH4´ß»¯»¹Ô­NOxÒ²¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺
¢ÙCH4£¨g£©+4NO2(g)=4NO(g)+CO2(g)+2H2O(g)   ¡÷H1£½£­574KJ¡¤mol-1 ¢Ù
¢ÚCH4£¨g£©+4NO (g)=2N2(g)+CO2(g)+2H2O(g)   ¡÷H2£½£¿¢Ú
Èô1molCH4»¹Ô­NO2ÖÁN2Õû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª867KJ£¬Ôò¡÷H2=                   ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄêºþ±±Ê¡»Æ¸ÔÊÐӢɽһÖиßÒ»£¨Ï£©ÆÚÖл¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

ÒÑÖªÔÚ25¡æ¡¢1.013×105PaÏ£¬1molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö285kJµÄÈÈÁ¿£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Éú³ÉÎïÄÜÁ¿×ܺÍ______£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©·´Ó¦ÎïÄÜÁ¿×ܺÍ
£¨2£©Èô2molÇâÆøÍêȫȼÉÕÉú³ÉË®ÕôÆø£¬Ôò·Å³öµÄÈÈÁ¿______£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©570kJ
£¨3£©ÏÖÓÐ25C¡¢1.013×105PaϵÄH2ºÍCH4µÄ»ìºÏÆøÌå0.5mol ÍêȫȼÉÕÉú³ÉÒ»¶¨ÖÊÁ¿µÄCO2ÆøÌåºÍ10.8gH2O£¨l£©£¬·Å³ö203kJµÄÈÈÁ¿£¬Ôò1molCH4ÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍH2O£¨l£©·Å³öµÄÈÈÁ¿Îª______kJ£®
£¨4£©ÃÀ¹ú°¢²¨ÂÞÓîÖæ·É´¬ÉÏʹÓÃÁËÒ»ÖÖÐÂÐÍ×°Öã¬Æä¹¹ÔìÈçͼËùʾ£ºA¡¢BÁ½¸öµç¼«¾ùÓɶà¿×µÄ̼¿é×é³É£®
¸Ãµç³ØµÄÕý¼«Ê½Îª£º______£®Èô¸Ãµç³Ø¹¤×÷ʱÔö¼ÓÁË1mol H2O£¬µç·ÖÐ×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿Îª______£®
£¨5£©Èç¹û½«ÉÏÊö×°ÖÃÖÐͨÈëµÄH2¸Ä³ÉCH4ÆøÌ壬Ҳ¿ÉÒÔ×é³ÉÒ»¸öÔ­µç³Ø×°Öã¬µç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ£ºCH4+2O2+2KOH=K2CO3+3H2£¬Ôò¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª£º______£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸