ÔÚÒ»¶¨Ìõ¼þÏ£¬COºÍCH4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º

    2CO£¨g£©£«O2£¨g£© 2CO2£¨g£©£»¡÷H  = £­566kJ¡¤mol£­1

  CH4£¨g£©+ 2O2£¨g£©CO2£¨g£©£« 2H2O£¨l£©£»¡÷H  = £­890kJ¡¤mol£­1

ÓÉ1molCOºÍ3molCH4×é³ÉµÄ»ìºÏÆøÌåÔÚÉÏÊöÌõ¼þÏÂÍêȫȼÉÕʱ£¬ÊͷŵÄÈÈÁ¿Îª£¨   £©

  A£®2912 kJ         B£®2953 kJ          C£®3236 kJ         D£®3867 kJ

B


½âÎö:

ÒÀ¾ÝìʵÄÐÔÖÊ£¬Èô»¯Ñ§·½³ÌʽÖи÷ÎïÖʵÄϵÊý¼Ó±¶£¬Ôò¡÷HµÄÊýÖµÒ²¼Ó±¶£¬¿ÉÒÔ½øÐмӡ¢¼õ¡¢³Ë¡¢³ýËÄÔòÔËË㣺Q £½ 1mol¡Á566kJ¡¤mol£­1/2 £« 3mol¡Á890kJ¡¤mol£­1 £½ 2953 Kj¡£ 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÒ»¶¨Ìõ¼þÏ£¬COºÍCH4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ
2CO£¨g£©+O2£¨g£©¨T2CO2£¨g£©¡÷H=-566kJ?mol-1
CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890kJ?mol-1
ÓÉ1mol CO£¨g£©ºÍ3mol CH4£¨g£©×é³ÉµÄ»ìºÏÆøÌåÔÚÉÏÊöÌõ¼þϳä·ÖȼÉÕ£¬»Ö¸´ÖÁÊÒÎÂÊͷŵÄÈÈÁ¿Îª£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÒ»¶¨Ìõ¼þÏ£¬COºÍCH4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º
2CO£¨g£©+O2£¨g£©=2CO2£¨g£©£»¡÷H=-566kJ?mol-1
CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©£»¡÷H=-890kJ?mol-1
ÓɵÈÎïÖʵÄÁ¿µÄCOºÍCH4µÄ»ìºÏÆøÌå¹²3mol£¬ÔÚÉÏÊöÌõ¼þÏÂÍêȫȼÉÕʱÊͷŵÄÈÈÁ¿Îª£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÒ»¶¨Ìõ¼þÏ£¬COºÍCH4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º
2CO£¨Æø£©+O2£¨Æø£©=2CO2£¨Æø£©+566ǧ½¹
CH4£¨Æø£©+2O2£¨Æø£©=CO2£¨Æø£©+2H2O£¨Òº£©+890ǧ½¹
ÓÉ1ĦCOºÍ3ĦCH4×é³ÉµÄ»ìºÍÆøÔÚÉÏÊöÌõ¼þÏÂÍêȫȼÉÕʱ£¬ÊͷŵÄÈÈÁ¿Îª£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÒ»¶¨Ìõ¼þÏ£¬COºÍCH4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º

2CO£¨g£©+O2(g)====2CO2(g);¦¤H=-566 kJ¡¤mol-1

CH4(g)+2O2(g)====CO2(g)+2H2O(l);¦¤H=-890 kJ¡¤mol-1

ÓÉ1 mol COºÍ3 mol CH4×é³ÉµÄ»ìºÏÆøÔÚÉÏÊöÌõ¼þϳä·ÖȼÉÕ£¬ÊͷŵÄÈÈÁ¿Îª

A.2912 kJ                B.2953 kJ                  C.3236 kJ                D.3867 kJ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÉϺ£ÊÐãÉÐÐÇø¸ßÈýÏÂѧÆÚÈýÄ£¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

¡°½à¾»Ãº¼¼Êõ¡±Ñо¿ÔÚÊÀ½çÉÏÏ൱Æձ飬¿ÆÑÐÈËԱͨ¹ýÏòµØÏÂú²ãÆø»¯Â¯Öн»Ìæ¹ÄÈë¿ÕÆøºÍË®ÕôÆøµÄ·½·¨£¬Á¬Ðø²ú³öÁ˸ßÈÈÖµµÄú̿Æø£¬ÆäÖ÷Òª³É·ÖÊÇCOºÍH2¡£COºÍH2¿É×÷ΪÄÜÔ´ºÍ»¯¹¤Ô­ÁÏ£¬Ó¦ÓÃÊ®·Ö¹ã·º¡£Éú²úú̿ÆøµÄ·´Ó¦Ö®Ò»ÊÇ£ºC(s)+H2O(g)CO(g)+H2(g)−131.4 kJ¡£

£¨1£©ÔÚÈÝ»ýΪ3LµÄÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬5minºóÈÝÆ÷ÄÚÆøÌåµÄÃܶÈÔö´óÁË0.12g£¯L£¬ÓÃH2O±íʾ0¡«5miinµÄƽ¾ù·´Ó¦ËÙÂÊΪ_________________________¡£

£¨2£©ÄÜ˵Ã÷¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ________£¨Ñ¡Ìî±àºÅ£©¡£

a£®vÕý (C)= vÄæ(H2O)    b£®ÈÝÆ÷ÖÐCOµÄÌå»ý·ÖÊý±£³Ö²»±ä

c£®c(H2)=c(CO)        d£®Ì¿µÄÖÊÁ¿±£³Ö²»±ä

£¨3£©ÈôÉÏÊö·´Ó¦ÔÚt0ʱ¿Ì´ïµ½Æ½ºâ(Èçͼ)£¬ÔÚt1ʱ¿Ì¸Ä±äijһÌõ¼þ£¬ÇëÔÚÓÒͼÖмÌÐø»­³öt1ʱ¿ÌÖ®ºóÕý·´Ó¦ËÙÂÊËæʱ¼äµÄ±ä»¯£º

¢ÙËõСÈÝÆ÷Ìå»ý£¬t2ʱµ½´ïƽºâ(ÓÃʵÏß±íʾ)£»

¢Út3ʱƽºâ³£ÊýKÖµ±ä´ó£¬t4µ½´ïƽºâ(ÓÃÐéÏß±íʾ)¡£

£¨4£©ÔÚÒ»¶¨Ìõ¼þÏÂÓÃCOºÍH2¾­ÈçÏÂÁ½²½·´Ó¦ÖƵü×Ëá¼×õ¥£º

¢ÙCO(g) + 2H2(g)CH3OH(g)    ¢ÚCO(g) + CH3OH(g)HCOOCH3(g)

¢Ù·´Ó¦¢ÙÖÐCOµÄƽºâת»¯ÂÊ(¦Á)Óëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ¡£ÔÚ²»¸Ä±ä·´Ó¦ÎïÓÃÁ¿µÄÇé¿öÏ£¬ÎªÌá¸ßCOµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ                                 ¡£

¢ÚÒÑÖª·´Ó¦¢ÙÖÐCOµÄת»¯ÂÊΪ80%£¬·´Ó¦¢ÚÖÐÁ½ÖÖ·´Ó¦ÎïµÄת»¯ÂʾùΪ85%£¬Ôò5.04kgCO×î¶à¿ÉÖƵü×Ëá¼×õ¥         kg¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸