ÒÑÖª£º¢Ù½«Ãº×ª»¯ÎªË®ÃºÆøµÄÖ÷Òª»¯Ñ§·´Ó¦ÎªC(s)£«H2O(g) CO(g)£«H2(g)£»¢ÚC(s)¡¢CO(g)ºÍH2(g)ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º
C(s)£«O2(g)=CO2 (g)£»¦¤H£½£­393.5 kJ¡¤mol£­1
CO(g)£«1/2O2(g)=CO2(g)£»¦¤H£½£­283.0 kJ¡¤mol£­1
H2(g)£«1/2O2(g)=H2O(g)£»¦¤H£½£­242.0 kJ¡¤mol£­1
Çë»Ø´ð£º
(1)¸ù¾ÝÒÔÉÏÐÅÏ¢£¬Ð´³öCOÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º____________________________¡£

(2)ÈçͼÊǸù¾Ý¸Ç˹¶¨ÂÉ×ö³öµÄÑ­»·Í¼£¬¸ù¾ÝͼµÄת»¯¹ØϵºÍÈÈ»¯Ñ§·½³Ìʽ¼ÆË㦤H3£½________kJ/mol¡£
Çë±È½Ï¦¤H1Ó릤H3ÊýÖµÊÇ·ñ¿ÉÒÔ˵Ã÷ÓÃˮúÆø×öȼÁÏÒª±ÈÖ±½Óȼú·Å³öµÄÈÈÁ¿¶à________(ÊÇ»ò·ñ)Ô­ÒòÊÇ___________________________________¡£
(3)ĿǰúµÄÔËÊ仹Ö÷Òª¿¿Ìú·ÔËÊäºÍ¹«Â·ÔËÊ䣬ÄãÄÜÔÚËùѧ֪ʶ»ù´¡ÉÏÌá³ö»º½âÌú·ºÍ¹«Â·ÔËÊäµÄ·½·¨£º____________________________¡£

(1)CO(g)£«H2O(g)=CO2(g)£«H2(g)£¬¦¤H£½£­41.0 kJ/mol
(2)·ñ¡¡¸ù¾Ý¸Ç˹¶¨ÂÉûÓмÆË㦤H2ÎüÊÕµÄÈÈÁ¿
(3)½«Ãº±äΪˮúÆøºó¹ÜµÀÔËÊä

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÔCO2Ϊ̼ԴÖÆÈ¡µÍ̼ÓлúÎï³ÉΪ¹ú¼ÊÑо¿½¹µã£¬ÏÂÃæΪCO2¼ÓÇâÖÆÈ¡µÍ̼´¼µÄÈÈÁ¦Ñ§Êý¾Ý£º
·´Ó¦¢ñ£º CO2(g)+3H2(g)CH3OH(g)+H2O(g)          ?H = ¡ª49£®0  kJ¡¤mol-1
·´Ó¦¢ò£º2CO2(g)+6H2(g)CH3CH2OH(g)+3H2O(g)     ?H = ¡ª173£®6 kJ¡¤mol-1
£¨1£©Ð´³öÓÉCH3OH(g)ºÏ³ÉCH3CH2OH(g)µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º                     
£¨2£©¶Ô·´Ó¦¢ñ£¬ÔÚÒ»¶¨Î¶ÈÏ·´Ó¦´ïµ½Æ½ºâµÄ±êÖ¾ÊÇ     £¨Ñ¡Ìî±àºÅ)
a£®·´Ó¦Îï²»ÔÙת»¯ÎªÉú³ÉÎï b£®Æ½ºâ³£ÊýK²»ÔÙÔö´ó
c£®CO2µÄת»¯Âʲ»ÔÙÔö´ó d£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٸı䡡
£¨3£©ÔÚÃܱÕÈÝÆ÷ÖУ¬·´Ó¦¢ñÔÚÒ»¶¨Ìõ¼þ´ïµ½Æ½ºâºó£¬ÆäËüÌõ¼þºã¶¨£¬ÄÜÌá¸ßCO2ת»¯ÂʵĴëÊ©ÊÇ    £¨Ñ¡Ìî±àºÅ£©

A£®½µµÍζÈB£®²¹³äCO2C£®¼ÓÈë´ß»¯¼ÁD£®ÒÆÈ¥¼×´¼
£¨4£©Ñо¿Ô±ÒÔÉú²úÒÒ´¼ÎªÑо¿¶ÔÏó£¬ÔÚÃܱÕÈÝÆ÷ÖУ¬°´H2ÓëCO2µÄÎïÖʵÄÁ¿Ö®±ÈΪ3:1½øÐÐͶÁÏ£¬ÔÚ5MPaϲâµÃ²»Í¬Î¶ÈÏÂƽºâÌåϵÖи÷ÖÖÎïÖʵÄÌå»ý·ÖÊý£¨y%£©ÈçÏÂͼËùʾ¡£±íʾCH3CH2OH×é·ÖµÄÇúÏßÊÇ     £»Í¼ÖÐÇúÏߢòºÍ¢óµÄ½»µãa¶ÔÓ¦µÄÌå»ý·ÖÊýya=     %£¨¼ÆËã½á¹û±£ÁôÈýλÓÐЧÊý×Ö£©

(5)Ò»ÖÖÒÔ¼×´¼×÷ȼÁϵĵç³ØʾÒâͼÈçͼ¡£Ð´³ö¸Ãµç³Ø·Åµçʱ¸º¼«µÄµç¼«·´Ó¦Ê½£º                     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

̼ºÍµªµÄ»¯ºÏÎïÓëÈËÀàÉú²ú¡¢Éú»îÃÜÇÐÏà¹Ø¡£
£¨1£©ÔÚÒ»ºãΡ¢ºãÈÝÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£º Ni(s)+4CO(g)Ni(CO)4(g)£¬H<0¡£ÀûÓø÷´Ó¦¿ÉÒÔ½«´ÖÄøת»¯Îª´¿¶È´ï99£®9£¥µÄ¸ß´¿Äø¡£¶Ô¸Ã·´Ó¦µÄ˵·¨ÕýÈ·µÄÊÇ     (Ìî×Öĸ±àºÅ)¡£

A£®Ôö¼ÓNiµÄÁ¿¿ÉÌá¸ßCOµÄת»¯ÂÊ£¬NiµÄת»¯ÂʽµµÍ
B£®ËõСÈÝÆ÷ÈÝ»ý£¬Æ½ºâÓÒÒÆ£¬H¼õС
C£®·´Ó¦´ïµ½Æ½ºâºó£¬³äÈëCOÔٴδﵽƽºâʱ£¬COµÄÌå»ý·ÖÊý½µµÍ
D£®µ±4v[Ni(CO)4]=v(CO)ʱ»òÈÝÆ÷ÖлìºÏÆøÌåÃܶȲ»±äʱ£¬¶¼¿É˵Ã÷·´Ó¦ÒѴﻯѧƽºâ״̬
£¨2£©COÓëÄø·´Ó¦»áÔì³ÉÄø´ß»¯¼ÁÖж¾¡£Îª·ÀÖ¹Äø´ß»¯¼ÁÖж¾£¬¹¤ÒµÉϳ£ÓÃSO2½«COÑõ»¯£¬¶þÑõ»¯Áòת»¯Îªµ¥ÖÊÁò¡£
ÒÑÖª£ºC(s)+O2(g)=CO(g)  H=-Q1 kJmol-1
C(s)+ O2(g)=CO2(g)   H=-Q2 kJmol-1
S(s)+O2(g)=SO2(g)    H=-Q3 kJmol-1
ÔòSO2(g)+2CO(g)=S(s)+2CO2(g)  H=          ¡£
£¨3£©½ðÊôÑõ»¯Îï¿É±»Ò»Ñõ»¯Ì¼»¹Ô­Éú³É½ðÊôµ¥ÖʺͶþÑõ»¯Ì¼¡£Í¼28£¨3£©ÊÇËÄÖÖ½ðÊôÑõ»¯ÎCr2O3¡¢SnO2¡¢PbO2¡¢Cu2O)±»Ò»Ñõ»¯Ì¼»¹Ô­Ê±Óëζȣ¨t£©µÄ¹ØϵÇúÏßͼ¡£
700oCʱ£¬ÆäÖÐ×îÄѱ»»¹Ô­µÄ½ðÊôÑõ»¯ÎïÊÇ         (Ìѧʽ)£¬ÓÃÒ»Ñõ»¯Ì¼»¹Ô­¸Ã½ðÊôÑõ»¯Îïʱ£¬Èô·´Ó¦·½³ÌʽϵÊýΪ×î¼òÕûÊý±È£¬¸Ã·´Ó¦µÄƽºâ³£Êý(K)ÊýÖµµÈÓÚ             ¡£

£¨4£©NO2¡¢O2ºÍÈÛÈÚNaNO3¿ÉÖÆ×÷ȼÁϵç³Ø£¬ÆäÔ­ÀíÈçÉÏͼ28£¨4£©Ëùʾ¡£¸Ãµç³ØÔÚʹÓùý³ÌÖÐʯīIµç¼«ÉÏÉú³ÉÑõ»¯ÎïY£¬Æäµç¼«·´Ó¦Ê½Îª              ¡£
Èô¸ÃȼÁϵç³ØʹÓÃÒ»¶Îʱ¼äºó£¬¹²ÊÕ¼¯µ½20mol Y£¬ÔòÀíÂÛÉÏÐèÒªÏûºÄ±ê×¼×´¿öÏÂÑõÆøµÄÌå»ýΪ       L¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÇâÆø¡¢¼×´¼ÊÇÓÅÖʵÄÇå½àȼÁÏ£¬¿ÉÖÆ×÷ȼÁϵç³Ø¡£
£¨1£©ÒÑÖª£º¢Ù=mol
¢Ú=mol 
¢Û=mol
д³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ                      ¡£
£¨2£©Éú²ú¼×´¼µÄÔ­ÁÏCOºÍH2À´Ô´ÓÚÏÂÁз´Ó¦£º

¢ÙÒ»¶¨Ìõ¼þϵÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼa¡£Ôò          (Ìî¡°<¡±¡¢¡°>¡±»ò¡°="£¬ÏÂͬ)£»A¡¢B¡¢CÈýµã´¦¶ÔӦƽºâ³£Êý()µÄ´óС¹ØϵΪ          £»

¢Ú100¡æʱ£¬½«1 mol ºÍ2 mol ͨÈëÈÝ»ýΪ1LµÄ¶¨ÈÝÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£¬ÄÜ˵Ã÷¸Ã·´Ó¦ÒѾ­´ïµ½Æ½ºâ״̬µÄÊÇ            (ÌîÐòºÅ)¡£
a£®ÈÝÆ÷µÄѹǿºã¶¨
b£®µ¥Î»Ê±¼äÄÚÏûºÄ0.1 mol CH4ͬʱÉú³É0.3 molH2
c£®ÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨
d£®
Èç¹û´ïµ½Æ½ºâʱµÄת»¯ÂÊΪ0£®5£¬Ôò100¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK=           ¡£
£¨3£©Ä³ÊµÑéС×éÀûÓÃCO(g)¡¢¡¢KOH(aq)Éè¼Æ³ÉÈçͼbËùʾµÄµç³Ø×°Öã¬Ôò¸Ãµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½Îª            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ËÄ´¨ÓзḻµÄÌìÈ»Æø×ÊÔ´£¬ÒÔÌìÈ»ÆøΪԭÁϺϳÉÄòËصÄÖ÷Òª²½ÖèÈçÏÂͼËùʾ(ͼÖÐijЩת»¯²½Öè¼°Éú³ÉÎïδÁгö)£º

ÇëÌîдÏÂÁпհףº
(1)ÒÑÖª0.5 mol¼×ÍéÓë0.5 molË®ÕôÆøÔÚt ¡æ¡¢p kPaʱ£¬ÍêÈ«·´Ó¦Éú³ÉÒ»Ñõ
»¯Ì¼ºÍÇâÆø(ºÏ³ÉÆø)£¬ÎüÊÕÁËa kJÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º______________________¡£
(2)Ôںϳɰ±µÄʵ¼ÊÉú²ú¹ý³ÌÖУ¬³£²ÉÈ¡µÄ´ëÊ©Ö®Ò»ÊÇ£º½«Éú³ÉµÄ°±´Ó»ìºÏÆøÌåÖм°Ê±·ÖÀë³öÀ´£¬²¢½«·ÖÀë³ö°±ºóµÄµªÆøºÍÇâÆøÑ­»·ÀûÓã¬Í¬Ê±²¹³äµªÆøºÍÇâÆø¡£ÇëÔËÓû¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâµÄ¹Ûµã˵Ã÷²ÉÈ¡¸Ã´ëÊ©µÄÀíÓÉ£º
________________________________________________________________¡£
(3)µ±¼×ÍéºÏ³É°±ÆøµÄת»¯ÂÊΪ75%ʱ£¬ÒÔ5.60¡Á107 L¼×ÍéΪԭÁÏÄܹ»ºÏ³É________L°±Æø¡£(¼ÙÉèÌå»ý¾ùÔÚ±ê×¼×´¿öϲⶨ)
(4)ÒÑÖªÄòËصĽṹ¼òʽΪ£¬Çëд³öÁ½ÖÖº¬ÓÐ̼ÑõË«¼üµÄÄòËصÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º
¢Ù__________________£»  ¢Ú_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¼×Íé×÷ΪһÖÖÐÂÄÜÔ´ÔÚ»¯Ñ§ÁìÓòÓ¦Óù㷺£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¸ß¯ұÌú¹ý³ÌÖУ¬¼×ÍéÔÚ´ß»¯·´Ó¦ÊÒÖвúÉúˮúÆø(COºÍH2)»¹Ô­Ñõ»¯Ìú£¬Óйط´Ó¦Îª£ºCH4(g)£«CO2(g)=2CO(g)£«2H2(g)¡¡¦¤H£½260 kJ¡¤mol£­1
ÒÑÖª£º2CO(g)£«O2(g)=2CO2(g)¡¡¦¤H£½£­566 kJ¡¤mol£­1
ÔòCH4ÓëO2·´Ó¦Éú³ÉCOºÍH2µÄÈÈ»¯Ñ§·½³ÌʽΪ_____________________£»
(2)ÈçÏÂͼËùʾ£¬×°ÖâñΪ¼×ÍéȼÁϵç³Ø(µç½âÖÊÈÜҺΪKOHÈÜÒº)£¬Í¨¹ý×°ÖâòʵÏÖÌú°ôÉ϶ÆÍ­¡£

¢Ùa´¦Ó¦Í¨Èë________(Ìî¡°CH4¡±»ò¡°O2¡±)£¬b´¦µç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½ÊÇ________£»
¢Úµç¶Æ½áÊøºó£¬×°ÖâñÖÐÈÜÒºµÄpH________(Ìîд¡°±ä´ó¡±¡°±äС¡±»ò¡°²»±ä¡±£¬ÏÂͬ)£¬×°ÖâòÖÐCu2£«µÄÎïÖʵÄÁ¿Å¨¶È________£»
¢Ûµç¶Æ½áÊøºó£¬×°ÖâñÈÜÒºÖеÄÒõÀë×Ó³ýÁËOH£­ÒÔÍ⻹º¬ÓÐ________(ºöÂÔË®½â)£»
¢ÜÔڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬×°ÖâòÖÐÒõ¼«ÖÊÁ¿±ä»¯12.8 g£¬Ôò×°ÖâñÖÐÀíÂÛÉÏÏûºÄ¼×Íé________L(±ê×¼×´¿öÏÂ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

NOxÊÇÆû³µÎ²ÆøÖеÄÖ÷ÒªÎÛȾÎïÖ®Ò»¡£
(1)NOxÄÜÐγÉËáÓ꣬д³öNO2ת»¯ÎªHNO3µÄ»¯Ñ§·½³Ìʽ£º__________________________¡£
(2)Æû³µ·¢¶¯»ú¹¤×÷ʱ»áÒý·¢N2ºÍO2·´Ó¦£¬ÆäÄÜÁ¿±ä»¯Ê¾ÒâͼÈçÏ£º

¢Ùд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º_______________________________¡£
¢ÚËæζÈÉý¸ß£¬¸Ã·´Ó¦»¯Ñ§Æ½ºâ³£ÊýµÄ±ä»¯Ç÷ÊÆÊÇ____¡£
(3)ÔÚÆû³µÎ²ÆøϵͳÖÐ×°Öô߻¯×ª»¯Æ÷£¬¿ÉÓÐЧ½µµÍNOxµÄÅÅ·Å¡£
¢Ùµ±Î²ÆøÖпÕÆø²»×ãʱ£¬NOxÔÚ´ß»¯×ª»¯Æ÷Öб»»¹Ô­³ÉN2Åųö¡£Ð´³öNO±»CO»¹Ô­µÄ»¯Ñ§·½³Ìʽ£º______________________________
¢Úµ±Î²ÆøÖпÕÆø¹ýÁ¿Ê±£¬´ß»¯×ª»¯Æ÷ÖеĽðÊôÑõ»¯ÎïÎüÊÕNOxÉú³ÉÑΡ£ÆäÎüÊÕÄÜÁ¦Ë³ÐòÈçÏ£º12MgO£¼20CaO£¼38SrO£¼56BaO¡£Ô­ÒòÊÇ___________________________________________£¬
ÔªËصĽðÊôÐÔÖð½¥ÔöÇ¿£¬½ðÊôÑõ»¯Îï¶ÔNOxµÄÎüÊÕÄÜÁ¦Öð½¥ÔöÇ¿¡£
(4)ͨ¹ýNOx´«¸ÐÆ÷¿É¼à²âNOxµÄº¬Á¿£¬Æ乤×÷Ô­ÀíʾÒâͼÈçÏ£º

¢ÙPtµç¼«ÉÏ·¢ÉúµÄÊÇ________·´Ó¦(Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±)
¢Úд³öNiOµç¼«µÄµç¼«·´Ó¦Ê½£º______________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ºÏ³ÉÂÈÊÇÈËÀàÑо¿µÄÖØÒª¿ÎÌ⣬Ŀǰ¹¤ÒµºÏ³É°±µÄÔ­ÀíΪ£º
N2(g)+3H2(g) 2NH3(g) ¦¤H=-93£®0kJ/mol
£¨1£©Ä³Î¶ÈÏ£¬ÔÚ2 LÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬²âµÃÊý¾ÝÈçÏÂ

           Ê±¼ä/h
ÎïÖʵÄÁ¿/mol
0
1
2
3
4
N2
2£®0
1£®83
1£®7
1£®6
1£®6
H2
6£®0
5£®49
5£®1
4£®8
4£®8
NH3
0
0£®34
0£®6
0£®8
0£®8
 
¢Ù0~2 hÄÚ£¬v£¨N2£©=           ¡£
¢Úƽºâʱ£¬H2µÄת»¯ÂÊΪ____£»¸ÃζÈÏ£¬·´Ó¦2NH3(g)  N2(g)+3H2(g)µÄƽºâ³£ÊýK=  ¡£
¢ÛÈô±£³ÖζȺÍÌå»ý²»±ä£¬ÆðʼͶÈëµÄN2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿·Ö±ðΪa mol¡¢b mol¡¢c mol£¬´ïµ½Æ½ºâºó£¬NH3±ÈµÄŨ¶ÈÓëÉϱíÖÐÏàͬµÄΪ      £¨ÌîÑ¡Ïî×Öĸ£©¡£
A£®a=l¡¢b=3£®c=0        B£®a=4¡¢b=12¡¢c=0
C£®a=0¡¢b=0£®c=4       D£®a=l¡¢b=3¡¢c=2
£¨2£©Áí¾Ý±¨µÀ£¬³£Î¡¢³£Ñ¹Ï£¬N2ÔÚ²ôÓÐÉÙÁ¿Ñõ»¯ÌúµÄ¶þÑõ»¯îÑ´ß»¯¼Á±íÃæÄÜÓëË®·¢Éú·´Ó¦£¬Éú³ÉNH3ºÍO2¡£ÒÑÖª£ºH2µÄȼÉÕÈȦ¤H=-286kJ/mol£¬ÔòÅãÖÆNH3·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ        ¡£
£¨3£©²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É£¨ÄÜ´«µÝH'£©£¬Í¨¹ýµç½â·¨Ò²¿ÉºÏ³É°±£¬Ô­ÀíΪ£º
N2(g)+3H2(g) 2NH3(g)¡£ÔÚµç½â·¨ºÏ³É°±µÄ¹ý³ÌÉ꣬Ӧ½«N2²»¶ÏµØͨÈë  ___¼«£¬¸Ãµç¼«·´Ó¦Ê½Îª                      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©Æû³µ·¢¶¯»ú¹¤×÷ʱ»áÒý·¢N2ºÍO2·´Ó¦£¬ÆäÄÜÁ¿±ä»¯Ê¾ÒâͼÈçÏ£º

д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º________________¡£
£¨2£©¶þ¼×ÃÑ(CH3OCH3)ÊÇÎÞÉ«ÆøÌ壬¿É×÷ΪһÖÖÐÂÐÍÄÜÔ´¡£ÓɺϳÉÆø(×é³ÉΪH2¡¢COºÍÉÙÁ¿µÄCO2)Ö±½ÓÖƱ¸¶þ¼×ÃÑ£¬ÆäÖеÄÖ÷Òª¹ý³Ì°üÀ¨ÒÔÏÂËĸö·´Ó¦£º
¼×´¼ºÏ³É·´Ó¦£º
(¢¡)CO(g)£«2H2(g)=CH3OH(g)¦¤H1£½£­90.1 kJ¡¤mol£­1
(¢¢)CO2(g)£«3H2(g)=CH3OH(g)£«H2O(g)¦¤H2£½£­49.0 kJ¡¤mol£­1
ˮúÆø±ä»»·´Ó¦£º
(¢£)CO(g)£«H2O(g)=CO2(g)£«H2(g)¦¤H3£½£­41.1 kJ¡¤mol£­1
¶þ¼×ÃѺϳɷ´Ó¦£º
(¢¤)2CH3OH(g)=CH3OCH3(g)£«H2O(g)¦¤H4£½£­24.5 kJ¡¤mol£­1
ÓÉH2ºÍCOÖ±½ÓÖƱ¸¶þ¼×ÃÑ(ÁíÒ»²úÎïΪˮÕôÆø)µÄÈÈ»¯Ñ§·½³ÌʽΪ_______________¡£
¸ù¾Ý»¯Ñ§·´Ó¦Ô­Àí£¬·ÖÎöÔö¼Óѹǿ¶ÔÖ±½ÓÖƱ¸¶þ¼×ÃÑ·´Ó¦µÄÓ°Ïì_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸