(1)еġ¶»·¾³¿ÕÆøÖÊÁ¿±ê×¼¡·(GB 30952012)½«ÓÚ2016Äê1ÔÂ1ÈÕÔÚÎÒ¹úÈ«Ãæʵʩ¡£¾Ý´Ë,»·¾³¿ÕÆøÖÊÁ¿Ö¸Êý(AQI)ÈÕ±¨ºÍʵʱ±¨¸æ°üÀ¨ÁËSO2¡¢NO2¡¢CO¡¢O3¡¢PM10¡¢PM2.5µÈÖ¸±ê,Ϊ¹«ÖÚÌṩ½¡¿µÖ¸Òý,Òýµ¼µ±µØ¾ÓÃñºÏÀí°²ÅųöÐкÍÉú»î¡£
¢ÙÆû³µÅųöµÄβÆøÖк¬ÓÐCOºÍNOµÈÆøÌå,Óû¯Ñ§·½³Ìʽ½âÊͲúÉúNOµÄÔ­Òò                                ¡£
¢ÚÆû³µÅÅÆø¹ÜÄÚ°²×°µÄ´ß»¯×ª»¯Æ÷,¿ÉʹÆû³µÎ²ÆøÖеÄÖ÷ÒªÎÛȾÎïת»¯ÎªÎÞ¶¾µÄ´óÆøÑ­»·ÎïÖÊ¡£ÒÑÖª:
N2(g)+O2(g)="2NO(g)" ¦¤H="+180.5" kJ/mol
2C(s)+O2(g)="2CO(g)" ¦¤H="-221.0" kJ/mol
C(s)+O2(g)=CO2(g) ¦¤H="-393.5" kJ/mol
Ôò·´Ó¦2NO(g)+2CO(g)=N2(g)+2CO2(g)µÄ¦¤H=   kJ/mol¡£
(2)Ö±½ÓÅŷŵªÑõ»¯Îï»áÐγÉËáÓê¡¢Îíö²,´ß»¯»¹Ô­·¨ºÍÑõ»¯ÎüÊÕ·¨Êdz£ÓõĴ¦Àí·½·¨¡£ÀûÓÃNH3ºÍCH4µÈÆøÌå³ýÈ¥ÑÌÆøÖеĵªÑõ»¯Îï¡£ÒÑÖª:CH4(g)+2O2(g)=CO2(g)+2H2O(l) ¦¤H1="a" kJ/mol;Óû¼ÆËã·´Ó¦CH4(g)+4NO(g)=CO2(g)+2H2O(l)+2N2(g)µÄìʱ䦤H2Ôò»¹ÐèÒª²éѯij·´Ó¦µÄìʱ䦤H3,µ±·´Ó¦Öи÷ÎïÖʵĻ¯Ñ§¼ÆÁ¿ÊýÖ®±ÈΪ×î¼òÕûÊý±Èʱ,¦¤H3="b" kJ/mol,¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                 ,¾Ý´Ë¼ÆËã³ö¦¤H2=   kJ/mol(Óú¬a¡¢bµÄʽ×Ó±íʾ)¡£
(3)ϱíÁгöÁ˹¤ÒµÉÏÎüÊÕSO2µÄÈýÖÖ·½·¨¡£

·½·¨¢ñ
Óð±Ë®½«SO2ת»¯(NH4)2SO3,ÔÙÑõ»¯³É(NH4)2SO4
·½·¨¢ò
ÓÃÉúÎïÖÊÈȽâÆø(Ö÷Òª³É·ÖCO¡¢CH4¡¢H2)½«SO2ÔÚ¸ßÎÂÏ»¹Ô­³Éµ¥ÖÊÁò
·½·¨¢ó
ÓÃNa2SO3ÈÜÒºÎüÊÕSO2,ÔÙ¾­µç½âת»¯ÎªH2SO4
·½·¨¢òÖ÷Òª·¢ÉúÁËÏÂÁз´Ó¦:
2CO(g)+SO2(g)=S(g)+2CO2(g) ¦¤H="+8.0" kJ/mol
2H2(g)+SO2(g)=S(g)+2H2O(g)¦¤H="+90.4" kJ/mol
2CO(g)+O2(g)=2CO2(g) ¦¤H="-566.0" kJ/mol
2H2(g)+O2(g)=2H2O(g) ¦¤H="-483.6" kJ/mol
ÔòS(g)ÓëO2(g)·´Ó¦Éú³ÉSO2(g)µÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ                     ¡£
(4)ºÏ³É°±ÓõÄÇâÆø¿ÉÒÔ¼×ÍéΪԭÁÏÖƵá£Óйػ¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯ÈçͼËùʾ,ÔòCH4(g)ÓëH2O(g)·´Ó¦Éú³ÉCO(g)ºÍH2(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ                          ¡£ 

(1)¢ÙN2+O22NO ¢Ú-746.5
(2)N2(g)+O2(g)="2NO(g)" ¦¤H3="b" kJ/mol a-2b
(3)S(g)+O2(g)=SO2(g) ¦¤H="-574.0" kJ/mol
(4)CH4(g)+H2O(g)=CO(g)+3H2(g)¦¤H="+161.1" kJ/mol

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿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

¶þ¼×ÃÑ£¨CH3OCH3£©ÊÇÒ»ÖÖÖØÒªµÄ¾«Ï¸»¯¹¤²úÆ·£¬±»ÈÏΪÊǶþʮһÊÀ¼Í×îÓÐDZÁ¦µÄȼÁÏ[ ÒÑÖª£ºCH3OCH3(g)+3O2(g)£½2CO2(g)+3H2O£¨1£© ¡÷H£½£­1455kJ/mol ]¡£Í¬Ê±ËüÒ²¿ÉÒÔ×÷ΪÖÆÀä¼Á¶øÌæ´ú·úÂÈ´úÌþ¡£¹¤ÒµÉÏÖƱ¸¶þ¼×ÃѵÄÖ÷Òª·½·¨¾­ÀúÁËÈý¸ö½×¶Î£º
¢Ù¼×´¼ÒºÌåÔÚŨÁòËá×÷ÓÃÏ»ò¼×´¼ÆøÌåÔÚ´ß»¯×÷ÓÃÏÂÖ±½ÓÍÑË®Öƶþ¼×ÃÑ£» 2CH3OH CH3OCH3£«H2O
¢ÚºÏ³ÉÆøCOÓëH2Ö±½ÓºÏ³É¶þ¼×ÃÑ£º 3H2(g)£«3CO(g)CH3OCH3(g)£«CO2(g)  ¡÷H£½£­247kJ/mol
¢ÛÌìÈ»ÆøÓëË®ÕôÆø·´Ó¦ÖƱ¸¶þ¼×ÃÑ¡£ÒÔCH4ºÍH2OΪԭÁÏÖƱ¸¶þ¼×ÃѺͼ״¼¹¤ÒµÁ÷³ÌÈçÏ£º

£¨1£©Ð´³öCO(g)¡¢H2(g)¡¢O2(g)·´Ó¦Éú³ÉCO2(g)ºÍH2O£¨1£©µÄÈÈ»¯Ñ§·½³Ìʽ£¨½á¹û±£ÁôһλСÊý£©                                                
£¨2£©ÔÚ·´Ó¦ÊÒ2ÖУ¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦3H2(g)£«3CO(g)CH3OCH3(g)£«CO2(g)ÔÚÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬ÒªÌá¸ßCOµÄת»¯ÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ     
A£®µÍθßѹ   B£®¼Ó´ß»¯¼Á    C£®Ôö¼ÓCOŨ¶È   D£®·ÖÀë³ö¶þ¼×ÃÑ
£¨3£©ÔÚ·´Ó¦ÊÒ3ÖУ¬ÔÚÒ»¶¨Î¶ȺÍѹǿÌõ¼þÏ·¢ÉúÁË·´Ó¦£º3H2(g)£«CO2(g) CH3OH(g)£«H2O (g) ¡÷H£¼0·´Ó¦´ïµ½Æ½ºâʱ£¬¸Ä±äζȣ¨T£©ºÍѹǿ£¨P£©£¬·´Ó¦»ìºÏÎïCH3OH¡°ÎïÖʵÄÁ¿·ÖÊý¡±±ä»¯Çé¿öÈçͼËùʾ£¬¹ØÓÚζȣ¨T£©ºÍѹǿ£¨P£©µÄ¹ØϵÅжÏÕýÈ·µÄÊÇ   £¨ÌîÐòºÅ£©

A£®P3£¾P2   T3£¾T2       B£®P2£¾P4   T4£¾T2
C£®P1£¾P3   T1£¾T3       D£®P1£¾P4   T2£¾T3
£¨4£©·´Ó¦ÊÒ1Öз¢Éú·´Ó¦£ºCH4(g)£«H2O(g)CO(g)£«3H2(g) ¡÷H£¾0д³öƽºâ³£ÊýµÄ±í´ïʽ£º                          
Èç¹ûζȽµµÍ£¬¸Ã·´Ó¦µÄƽºâ³£Êý             £¨Ìî¡°²»±ä¡±¡¢¡°±ä´ó¡±¡¢¡°±äС¡±£©

£¨5£©ÈçͼΪÂÌÉ«µçÔ´¡°¶þ¼×ÃÑȼÁϵç³Ø¡±µÄ¹¤×÷Ô­ÀíʾÒâͼÔòaµç¼«µÄ·´Ó¦Ê½Îª£º________________

£¨6£©ÏÂÁÐÅжÏÖÐÕýÈ·µÄÊÇ_______
A£®ÏòÉÕ±­aÖмÓÈëÉÙÁ¿K3[Fe(CN)6]ÈÜÒº£¬ÓÐÀ¶É«³ÁµíÉú³É
B£®ÉÕ±­bÖз¢Éú·´Ó¦Îª2Zn-4e¡¥ £½2Zn2+
C£®µç×Ó´ÓZn¼«Á÷³ö£¬Á÷ÈëFe¼«£¬¾­ÑÎÇŻص½Zn¼«
D£®ÉÕ±­aÖз¢Éú·´Ó¦O2 + 4H++ 4e¡¥ £½ 2H2O£¬ÈÜÒºpH½µµÍ

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>>

¿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

Ò»¶¨Ìõ¼þÏ£¬ÔÚÌå»ýΪ3 LµÄÃܱÕÈÝÆ÷Öз´Ó¦£ºCO£¨g£©+ 2H2£¨g£©CH3OH£¨g£©´ïµ½»¯Ñ§Æ½ºâ״̬¡£
£¨1£©¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽK=                 £»¸ù¾ÝÏÂͼ£¬Éý¸ßζȣ¬KÖµ½«       £¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£

£¨2£©500¡æʱ£¬´Ó·´Ó¦¿ªÊ¼µ½´ïµ½»¯Ñ§Æ½ºâ£¬ÒÔH2µÄŨ¶È±ä»¯±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊÊÇ  £¨ÓÃnB¡¢tB±íʾ£©¡£
£¨3£©ÅжϸÿÉÄæ·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄ±êÖ¾ÊÇ      £¨Ìî×Öĸ£©¡£
a¡¢CO¡¢H2¡¢CH3OHµÄŨ¶È¾ù²»Ôٱ仯
b¡¢»ìºÏÆøÌåµÄÃܶȲ»Ôٸıä
c¡¢»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٸıä
d¡¢vÉú³É£¨CH3OH£©= vÏûºÄ£¨CO£©
£¨4£©300¡æʱ£¬½«ÈÝÆ÷µÄÈÝ»ýѹËõµ½Ô­À´µÄ1/2£¬ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¶ÔƽºâÌåϵ²úÉúµÄÓ°ÏìÊÇ       £¨Ìî×Öĸ£©¡£
a¡¢c£¨H2£©¼õÉÙ
b¡¢Õý·´Ó¦ËÙÂʼӿ죬Äæ·´Ó¦ËÙÂʼõÂý
c¡¢CH3OH µÄÎïÖʵÄÁ¿Ôö¼Ó
d¡¢ÖØÐÂƽºâʱc£¨H2£©/ c£¨CH3OH£©¼õС
£¨5£©¸ù¾ÝÌâÄ¿ÓйØÐÅÏ¢£¬ÇëÔÚÓÒÏÂ×ø±êͼÖбêʾ³ö¸Ã»¯Ñ§·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯£¨±êÃ÷ÐÅÏ¢£©¡£

£¨6£©ÒÔ¼×´¼¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«¿É¹¹³ÉȼÁϵç³Ø¡£ÒÑÖª¸ÃȼÁϵç³ØµÄ×Ü·´Ó¦Ê½Îª£º2CH3OH +3O2+4OH- = 2CO32- + 6H2O£¬¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ£º2CH3OH¨C12e£­+16OH£­£½ 2CO32£­+ 12H2O £¬ÔòÕý¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª£º                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ºãκãÈÝÌõ¼þÏÂ,Áò¿ÉÒÔ·¢ÉúÈçÏÂת»¯,Æä·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØϵÈçͼ1Ëùʾ¡£ÒÑÖª:2SO2(g)+O2(g)2SO3(g)¡¡
¦¤H="-196.6" kJ/mol¡£
  
Çë»Ø´ðÏÂÁÐÎÊÌâ:
(1)д³öÄܱíʾÁòµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ:                   ¡£
(2)¦¤H2=¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
(3)ºãκãÈÝʱ,1 mol SO2ºÍ2 mol O2³ä·Ö·´Ó¦,·Å³öÈÈÁ¿µÄÊýÖµ±È¨O¦¤H2¨O¡¡¡¡¡¡¡¡(Ìî¡°´ó¡±¡¢¡°Ð¡¡±»ò¡°ÏàµÈ¡±)¡£
(4)½«¢óÖеĻìºÏÆøÌåͨÈë×ãÁ¿µÄNaOHÈÜÒºÖÐÏûºÄNaOHµÄÎïÖʵÄÁ¿Îª¡¡¡¡¡¡¡¡,ÈôÈÜÒºÖз¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦,Ôò¸Ã¹ý³ÌµÄÀë×Ó·½³ÌʽΪ¡¡                                                   ¡£
(5)ºãÈÝÌõ¼þÏÂ,ÏÂÁдëÊ©ÖÐÄÜʹn(SO3)/ n(SO2)Ôö´óµÄÓС¡¡¡¡¡¡¡¡£
a.Éý¸ßζÈ
b.³äÈëHeÆø
c.ÔÙ³äÈë1 mol SO2(g)ºÍ1 mol O2(g)
d.ʹÓô߻¯¼Á
(6)ijSO2(g)ºÍO2 (g)Ìåϵ,ʱ¼ät1´ïµ½Æ½ºâºó,¸Ä±äijһÍâ½çÌõ¼þ,·´Ó¦ËÙÂÊvÓëʱ¼ätµÄ¹ØϵÈçͼ2Ëùʾ,Èô²»¸Ä±äSO2(g)ºÍO2 (g)µÄÁ¿,ÔòͼÖÐt4ʱÒýÆðƽºâÒƶ¯µÄÌõ¼þ¿ÉÄÜÊÇ¡¡¡¡¡¡¡¡;ͼÖбíʾƽºâ»ìºÏÎïÖÐSO3µÄº¬Á¿×î¸ßµÄÒ»¶Îʱ¼äÊÇ¡¡¡¡¡¡¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÄÜÔ´µÄ¿ª·¢ÀûÓÃÓëÈËÀàÉç»áµÄ¿É³ÖÐø·¢Õ¹Ï¢Ï¢Ïà¹Ø¡£
¢ñ.ÒÑÖª£ºFe2O3£¨s£©£«3C£¨s£©=2Fe£¨s£©£«3CO£¨g£©
¦¤H1£½a kJ¡¤mol£­1
CO£¨g£©£«O2£¨g£©=CO2£¨g£©¡¡    ¦¤H2£½b kJ¡¤mol£­1
4Fe£¨s£©£«3O2£¨g£©=2Fe2O3£¨s£©¡¡ ¦¤H3£½c kJ¡¤mol£­1
ÔòCµÄȼÉÕÈȦ¤H£½________kJ¡¤mol£­1¡£
¢ò.£¨1£©ÒÀ¾ÝÔ­µç³ØµÄ¹¹³ÉÔ­Àí£¬ÏÂÁл¯Ñ§·´Ó¦ÔÚÀíÂÛÉÏ¿ÉÒÔÉè¼Æ³ÉÔ­µç³ØµÄÊÇ________£¨ÌîÐòºÅ£©¡£
A£®C£¨s£©£«CO2£¨g£©=2CO£¨g£©
B£®NaOH£¨aq£©£«HCl£¨aq£©=NaCl£¨aq£©£«H2O£¨l£©
C£®2H2O£¨l£©=2H2£¨g£©£«O2£¨g£©
D£®2CO£¨g£©£«O2£¨g£©=2CO2£¨g£©
ÈôÒÔÈÛÈÚµÄK2CO3ÓëCO2Ϊ·´Ó¦µÄ»·¾³£¬ÒÀ¾ÝËùÑ¡·´Ó¦Éè¼Æ³ÉÒ»¸öÔ­µç³Ø£¬Çëд³ö¸ÃÔ­µç³ØµÄ¸º¼«·´Ó¦£º_____________________________________¡£
£¨2£©Ä³ÊµÑéС×éÄ£Ä⹤ҵºÏ³É°±·´Ó¦N2£¨g£©£«3H2£¨g£©2NH3£¨g£©¡¡¦¤H£½£­92.4 kJ¡¤mol£­1£¬¿ªÊ¼ËûÃǽ«N2ºÍH2»ìºÏÆøÌå20 mol£¨Ìå»ý±È1¡Ã1£©³äÈë5 LºÏ³ÉËþÖУ¬·´Ó¦Ç°Ñ¹Ç¿ÎªP0£¬·´Ó¦¹ý³ÌÖÐѹǿÓÃP±íʾ£¬·´Ó¦¹ý³ÌÖÐÓëʱ¼ätµÄ¹ØϵÈçͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù·´Ó¦´ïƽºâµÄ±êÖ¾ÊÇ__________________________£¨Ìî×Öĸ´úºÅ£¬ÏÂͬ£©¡£
A£®Ñ¹Ç¿±£³Ö²»±ä
B£®ÆøÌåÃܶȱ£³Ö²»±ä
C£®NH3µÄÉú³ÉËÙÂÊÊÇN2µÄÉú³ÉËÙÂʵÄ2±¶
¢Ú0¡«2 minÄÚ£¬ÒÔc£¨N2£©±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ________________¡£
¢ÛÓûÌá¸ßN2µÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ_____________________________¡£
A£®ÏòÌåϵÖа´Ìå»ý±È1¡Ã1ÔÙ³äÈëN2ºÍH2
B£®·ÖÀë³öNH3
C£®Éý¸ßζÈ
D£®³äÈ뺤ÆøʹѹǿÔö´ó
E£®¼ÓÈëÒ»¶¨Á¿µÄN2
£¨3£©25¡æʱ£¬BaCO3ºÍBaSO4µÄÈܶȻý³£Êý·Ö±ðÊÇ8¡Á10£­9ºÍ1¡Á10£­10£¬Ä³º¬ÓÐBaCO3³ÁµíµÄÐü×ÇÒºÖУ¬c£¨CO32-£©£½0.2 mol¡¤L£­1£¬Èç¹û¼ÓÈëµÈÌå»ýµÄNa2SO4ÈÜÒº£¬ÈôÒª²úÉúBaSO4³Áµí£¬¼ÓÈëNa2SO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È×îСÊÇ________mol¡¤L£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖª·´Ó¦¢ÙFe(s)+CO2(g)FeO(s)+CO(g)¡¡¦¤H="a" kJ¡¤mol-1,ƽºâ³£ÊýΪK;·´Ó¦¢ÚCO(g)+1/2O2(g)CO2(g)¡¡¦¤H="b" kJ¡¤mol-1;·´Ó¦¢ÛFe2O3(s)+3CO(g)2Fe(s)+3CO2(g)¡¡¦¤H="c" kJ¡¤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>>

ͬ²½Á·Ï°²á´ð°¸