(1)¼×´¼ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤²úÆ·£¬¿ÉÀûÓü״¼´ß»¯ÍÑÇâÖƱ¸¼×È©¡£¼×È©ÓëÆø̬¼×´¼×ª»¯µÄÄÜÁ¿¹ØϵÈçͼËùʾ¡£

·´Ó¦¹ý³ÌÖеÄÄÜÁ¿¹Øϵ
¢Ù¼×´¼´ß»¯ÍÑÇâת»¯Îª¼×È©µÄ·´Ó¦ÊÇ________(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)·´Ó¦¡£
¢Ú¹ý³Ì¢ñÓë¹ý³Ì¢òµÄ·´Ó¦ÈÈÊÇ·ñÏàͬ£¿____________Ô­ÒòÊÇ____________ ______________________________¡£
¢Ûд³ö¼×´¼´ß»¯ÍÑÇâת»¯Îª¼×È©µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ______________ _____________________¡£
(2)ÒÑÖª£º¢ÙCH3OH(g)£«H2O(g)=CO2(g)£«3H2(g)¡¡¦¤H£½£«49.0 kJ¡¤mol£­1
¢ÚCH3OH(g)£« O2(g)=CO2(g)£«2H2(g)¡¡¦¤H£½£­192.9 kJ¡¤mol£­1
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________¡£

A£®CH3OHת±ä³ÉH2µÄ¹ý³ÌÒ»¶¨ÒªÎüÊÕÄÜÁ¿
B£®¢Ù·´Ó¦ÖУ¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿
C£®¸ù¾Ý¢ÚÍÆÖª·´Ó¦£ºCH3OH(l)£«O2(g)=CO2(g)£«2H2(g)µÄ¦¤H£¾£­192.9 kJ¡¤mol£­1
D£®·´Ó¦¢ÚµÄÄÜÁ¿±ä»¯ÈçͼËùʾ

(1)¢ÙÎüÈÈ¡¡¢ÚÏàͬ¡¡Ò»¸ö»¯Ñ§·´Ó¦µÄ·´Ó¦ÈȽöÓ뷴Ӧʼ̬ºÍÖÕ̬Óйأ¬¶øÓë·´Ó¦µÄ;¾¶Î޹ء¡¢ÛCH3OH(g) HCHO(g)£«H2(g)¡¡¦¤H£½£«(E2£­E1)kJ¡¤mol£­1¡¡(2)C

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖª·´Ó¦¢ÙFe(s)+CO2(g)FeO(s)+CO(g)¡¡¦¤H=akJ¡¤mol-1,ƽºâ³£ÊýΪK;
·´Ó¦¢ÚCO(g)+1/2O2(g)CO2(g)¡¡¦¤H=bkJ¡¤mol-1;
·´Ó¦¢ÛFe2O3(s)+3CO(g)2Fe(s)+3CO2(g)¡¡¦¤H=ckJ¡¤mol-1¡£²âµÃÔÚ²»Í¬Î¶ÈÏÂ,KÖµÈçÏÂ:

ζÈ/¡æ
500
700
900
K
1.00
1.47
2.40
 
(1)Èô500 ¡æʱ½øÐз´Ó¦¢Ù,CO2µÄÆðʼŨ¶ÈΪ2 mol¡¤L-1,COµÄƽºâŨ¶ÈΪ¡¡¡¡¡¡¡¡¡£ 
(2)·´Ó¦¢ÙΪ¡¡¡¡¡¡¡¡(Ñ¡Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)·´Ó¦¡£ 
(3)700 ¡æʱ·´Ó¦¢Ù´ïµ½Æ½ºâ״̬,Ҫʹ¸ÃƽºâÏòÓÒÒƶ¯,ÆäËûÌõ¼þ²»±äʱ,¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÓС¡¡¡¡¡¡¡(ÌîÐòºÅ)¡£ 
A.ËõС·´Ó¦Æ÷Ìå»ý  B.ͨÈëCO2
C.ζÈÉý¸ßµ½900 ¡æ   D.ʹÓúÏÊʵĴ߻¯¼Á
E.Ôö¼ÓFeµÄÁ¿
(4)ÏÂÁÐͼÏñ·ûºÏ·´Ó¦¢ÙµÄÊÇ¡¡¡¡¡¡¡¡(ÌîÐòºÅ)(ͼÖÐvΪËÙÂÊ,¦ÕΪ»ìºÏÎïÖÐCOº¬Á¿,TΪζÈÇÒT1>T2)¡£ 

(5)ÓÉ·´Ó¦¢ÙºÍ¢Ú¿ÉÇóµÃ,·´Ó¦2Fe(s)+O2(g)2FeO(s)µÄ¦¤H=¡¡¡¡¡¡¡¡¡£ 
(6)ÇëÔËÓøÇ˹¶¨ÂÉд³öFe(¹ÌÌå)±»O2(ÆøÌå)Ñõ»¯µÃµ½Fe2O3(¹ÌÌå)µÄÈÈ»¯Ñ§·½³Ìʽ:¡¡                                             ¡£

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

ÀûÓõªÆø¡¢ÇâÆøÔÚÒ»¶¨Ìõ¼þÏÂÉú³É°±ÆøÕâÒ»¿ÉÄæ·´Ó¦À´ºÏ³É°±£¬ÊÇÒ»¸öÖØÒªµÄ»¯¹¤·´Ó¦¡£³£ÓÃÀ´Éú²úÒº°±ºÍ°±Ë®¡£
Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©Èçͼ±íʾºÏ³É°±Ê±Éú³É1molÉú³ÉÎïʱµÄÄÜÁ¿±ä»¯£¬EµÄµ¥Î»ÎªkJ¡£Çëд³öºÏ³É°±µÄÈÈ»¯Ñ§·½³Ìʽ____________________¡£

£¨ÈÈÁ¿ÓÃE1¡¢E2»òE3±íʾ£©¡£¸ÃͼÖеÄʵÏßÓëÐéÏß²¿·ÖÊÇʲô·´Ó¦Ìõ¼þ·¢ÉúÁ˱仯£¿
£¨2£©ÔÚÒ»¶¨Î¶ÈÏ£¬Èô½«4a mol H2ºÍ2amol N2·ÅÈëVLµÄÃܱÕÈÝÆ÷ÖУ¬5·ÖÖÓºó²âµÃN2µÄת»¯ÂÊΪ50%£¬Ôò¸Ã¶Îʱ¼äÓÃH2±íʾµÄ·´Ó¦ËÙÂÊΪ__________Ħ¶û/(Éý?Ãë)¡£Èô´ËʱÔÙÏò¸ÃÈÝÆ÷ÖÐͶÈëa mol H2¡¢amol N2ºÍ2amol NH3£¬ÅжÏƽºâÒƶ¯µÄ·½ÏòÊÇ_____£¨Ìî¡°ÕýÏòÒƶ¯¡±¡°ÄæÏòÒƶ¯¡±»ò¡°²»Òƶ¯¡±£©
£¨3£©Òº°±ºÍË®ÀàËÆ£¬Ò²ÄܵçÀ룺2NH3NH4++ NH2£­£¬Ä³Î¶Èʱ£¬ÆäÀë×Ó»ýK=2¡Ál0-30¡£¸ÃζÈÏ£º¢Ù½«ÉÙÁ¿NH4Cl¹ÌÌå¼ÓÈëÒº°±ÖУ¬K____________2¡Á10-30£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£»¢Ú½«ÉÙÁ¿½ðÊôÄÆͶÈëÒº°±ÖУ¬ÍêÈ«·´Ó¦ºóËùµÃÈÜÒºÖи÷΢Á£µÄŨ¶È´óС¹ØϵΪ£º_______
£¨4£©¹¤³§Éú²úµÄ°±Ë®×÷·ÊÁÏʱÐèҪϡÊÍ¡£ÓÃˮϡÊÍ0£®1mol/LÏ¡°±Ë®Ê±£¬ÈÜÒºÖÐËæ×ÅË®Á¿µÄÔö¼Ó¶ø¼õÉÙµÄÊÇ

A£®c(NH4+)/c(NH3?H2O)  B£®c(NH3?H2O)/c(OH-
C£®c(H+)/c(NH4+)  D£®c(OH-)/c(H+

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

ÄÉÃ×¼¶Cu2OÓÉÓÚ¾ßÓÐÓÅÁ¼µÄ´ß»¯ÐÔÄܶøÊܵ½¹Ø×¢£¬Ï±íΪÖÆÈ¡Cu2OµÄÈýÖÖ·½·¨£º

·½·¨¢ñ
ÓÃÌ¿·ÛÔÚ¸ßÎÂÌõ¼þÏ»¹Ô­CuO
·½·¨¢ò
µç½â·¨£¬·´Ó¦Îª2Cu + H2O  Cu2O + H2¡ü¡£
·½·¨¢ó
ÓÃ루N2H4£©»¹Ô­ÐÂÖÆCu(OH)2
 
£¨1£©¹¤ÒµÉϳ£Ó÷½·¨¢òºÍ·½·¨¢óÖÆÈ¡Cu2O¶øºÜÉÙÓ÷½·¨¢ñ£¬ÆäÔ­ÒòÊÇ·´Ó¦Ìõ¼þ²»Ò׿ØÖÆ£¬Èô¿Øβ»µ±Ò×Éú³É          ¶øʹCu2O²úÂʽµµÍ¡£
£¨2£©ÒÑÖª£º2Cu(s)£«1/2O2(g)=Cu2O(s)   ¡÷H = -akJ¡¤mol-1
C(s)£«1/2O2(g)=CO(g)      ¡÷H = -bkJ¡¤mol-1
Cu(s)£«1/2O2(g)=CuO(s)    ¡÷H = -ckJ¡¤mol-1
Ôò·½·¨¢ñ·¢ÉúµÄ·´Ó¦£º2CuO(s)£«C(s)= Cu2O(s)£«CO(g)£»¡÷H =      kJ¡¤mol-1¡£
£¨3£©·½·¨¢ò²ÉÓÃÀë×Ó½»»»Ä¤¿ØÖƵç½âÒºÖÐOH£­µÄŨ¶È¶øÖƱ¸ÄÉÃ×Cu2O£¬×°ÖÃÈçͼËùʾ,¸Ãµç³ØµÄÑô¼«Éú³ÉCu2O·´Ó¦Ê½Îª                                        ¡£

£¨4£©·½·¨¢óΪ¼ÓÈÈÌõ¼þÏÂÓÃҺ̬루N2H4£©»¹Ô­ÐÂÖÆCu(OH)2À´ÖƱ¸ÄÉÃ×¼¶Cu2O£¬Í¬Ê±·Å³öN2¡£¸ÃÖÆ·¨µÄ»¯Ñ§·½³ÌʽΪ             ¡£
£¨5£©ÔÚÏàͬµÄÃܱÕÈÝÆ÷ÖУ¬ÓÃÒÔÉÏÁ½ÖÖ·½·¨ÖƵõÄCu2O·Ö±ð½øÐд߻¯·Ö½âË®µÄʵÑ飺
   ¡÷H >0
Ë®ÕôÆøµÄŨ¶È£¨mol/L£©Ëæʱ¼ät(min)±ä»¯ÈçϱíËùʾ¡£
ÐòºÅ
ζÈ
0
10
20
30
40
50
¢Ù
T1
0.050
0.0492
0.0486
0.0482
0.0480
0.0480
¢Ú
T1
0.050
0.0488
0.0484
0.0480
0.0480
0.0480
¢Û
T2
0.10
0.094
0.090
0.090
0.090
0.090
 
ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ         £¨Ìî×Öĸ´úºÅ£©¡£
A£®ÊµÑéµÄζÈ:T2<T1
B£®ÊµÑé¢ÙÇ°20 minµÄƽ¾ù·´Ó¦ËÙÂÊ v(O2)=7¡Á10-5 mol¡¤L-1 min-1  
C£®ÊµÑé¢Ú±ÈʵÑé¢ÙËùÓõĴ߻¯¼Á´ß»¯Ð§Âʸß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

µªÔªËصĻ¯ºÏÎïÖÖÀà·±¶à£¬ÐÔÖÊÒ²¸÷²»Ïàͬ¡£
£¨1£©NO2ÓнÏÇ¿µÄÑõ»¯ÐÔ£¬Äܽ«SO2Ñõ»¯Éú³ÉSO3£¬±¾Éí±»»¹Ô­ÎªNO£¬ÒÑÖªÏÂÁÐÁ½·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯ÈçͼËùʾ£º

ÔòNO2Ñõ»¯SO2µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________________¡£
£¨2£©ÔÚ2LÃܱÕÈÝÆ÷ÖзÅÈë1mol°±Æø£¬ÔÚÒ»¶¨Î¶ȽøÐÐÈçÏ·´Ó¦£º
2NH3(g)N2£¨g£©+3H2£¨g£©£¬·´Ó¦Ê±¼ä(t)ÓëÈÝÆ÷ÄÚÆøÌå×Üѹǿ(p)µÄÊý¾Ý¼ûϱí

ʱ¼ät/min
0
1
2
3
4
5
×Üѹǿp
100 kPa
5
5.6
6.4
6.8
7
7
 
Ôòƽºâʱ°±ÆøµÄת»¯ÂÊΪ___________¡£
£¨3£©ë£¨N2H4£©ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÔÚ¿ÕÆøÖÐÍêȫȼÉÕÉú³ÉµªÆø£¬µ±·´Ó¦×ªÒÆ0.2molµç×Óʱ£¬Éú³ÉÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ______________¡£Áª°±ÈÜÓÚË®¿ÉÒÔ·¢ÉúÓ백ˮÀàËƵĵçÀ룬ÊÔд³öÁª°±ÔÚË®ÈÜÒºÖеĵçÀë·½³Ìʽ£º
__________________£¨Ð´Ò»²½¼´¿É£©¡£
£¨4£©NH4+ÔÚÈÜÒºÖÐÄÜ·¢ÉúË®½â·´Ó¦¡£ÔÚ25¡æʱ£¬0.1mol/LÂÈ»¯ï§ÈÜÒºÓÉË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ1¡Á10-5 mol/L£¬ÔòÔÚ¸ÃζÈÏ´ËÈÜÒºÖа±Ë®µÄµçÀëƽºâ³£ÊýKb£¨NH3¡¤H2O£©=__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¼×Íé×÷ΪһÖÖÐÂÄÜÔ´ÔÚ»¯Ñ§ÁìÓòÓ¦Óù㷺£¬Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨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×´¿öÏ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Õæ¿Õ̼ÈÈ»¹Ô­¡ªÂÈ»¯·¨¿ÉʵÏÖÓÉÂÁÍÁ¿óÖƱ¸½ðÊôÂÁ,ÆäÏà¹Ø·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏÂ:
Al2O3(s)+AlCl3(g)+3C(s)3AlCl(g)+3CO(g)¡¡¦¤H="a" kJ¡¤mol-1
3AlCl(g)2Al(l)+AlCl3(g)¡¡¦¤H="b" kJ¡¤mol-1
·´Ó¦Al2O3(s) +3C(s)2Al(l)+3CO(g)µÄ¦¤H=¡¡¡¡¡¡¡¡ kJ¡¤mol-1(Óú¬a¡¢bµÄ´úÊýʽ±íʾ)¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸