1¡¢Ìî¿ÕÌâ ÒÑÖªÏÂÁз´Ó¦µÄìʱä
£¨1£©CH3COOH£¨l£©+2O2£¨g£©¨T2CO2£¨g£©+2H2O£¨l£©¡÷H1=-870.3kJ/mol
£¨2£©C£¨s£©+O2£¨g£©¨TCO2£¨g£©¡÷H2=-393.5kJ/mol
£¨3£©2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H3=-285.8kJ/mol
ÊÔ¼ÆËã·´Ó¦?2C£¨s£©+2H2£¨g£©+O2£¨g£©¨TCH3COOH£¨l£©µÄìʱä¡÷H=______£®
²Î¿¼´ð°¸£ºÒÑÖª£º£¨1£©CH3COOH£¨l£©+2O2£¨g£©¨T2CO2£¨g
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
2¡¢Ñ¡ÔñÌâ ÔÚ³£Î³£Ñ¹Ï£¬ÒÑÖª£º4Fe£¨s£©+3O2£¨g£©=2Fe2O3£¨s£©¡÷H1
4Al£¨s£©+3O2£¨g£©=2Al2O3£¨s£©¡÷H2
2Al£¨s£©+Fe2O3£¨s£©=Al2O3£¨s£©+2Fe£¨s£©¡÷H3
Ôò¡÷H3Óë¡÷H1ºÍ¡÷H2¼äµÄ¹ØÏµÕýÈ·µÄÊÇ£¨ ? £©
A£®¡÷H3=
£¨¡÷H1+¡÷H2£©
B£®¡÷H3=¡÷H2-¡÷H1
C£®¡÷H3=2£¨¡÷H2+¡÷H1£©
D£®¡÷H3=
£¨¡÷H2-¡÷H1£©
²Î¿¼´ð°¸£ºD
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣº¼òµ¥
3¡¢¼ò´ðÌâ ÄÜÔ´µÄ¿ª·¢¡¢ÀûÓÃÓëÈËÀàÉç»áµÄ¿É³ÖÐø·¢Õ¹Ï¢Ï¢Ïà¹Ø£¬ÔõÑù³ä·ÖÀûÓúÃÄÜÔ´ÊǰÚÔÚÈËÀàÃæÇ°µÄÖØ´ó¿ÎÌ⣮
I£®ÒÑÖª£ºFe2O3£¨s£©+3C£¨Ê¯Ä«£©=2Fe£¨s£©+3CO£¨g£©¡÷H=akJ?mol-1
CO£¨g£©+1/2O2£¨g£©=CO2£¨g£©¡÷H=bkJ?mol-1
C£¨Ê¯Ä«£©+O2£¨g£©=CO2£¨g£©¡÷H=ckJ?mol-1
Ôò·´Ó¦£º4Fe£¨s£©+3O2£¨g£©=2Fe2O3£¨s£©µÄìʱä¡÷H=______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£®CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©
ÈôÒÔKOHÈÜҺΪµç½âÖÊÈÜÒº£¬ÒÀ¾ÝËùÑ¡·´Ó¦¿ÉÒÔÉè¼Æ³ÉÒ»¸öÔµç³Ø£¬Çëд³ö¸ÃÔµç³ØµÄµç¼«·´Ó¦£®
¸º¼«£º______£¬
Õý¼«£º______£®
£¨2£©¶þÑõ»¯ÂÈ£¨ClO2£©ÊÇÒ»ÖÖ¸ßЧ°²È«µÄ×ÔÀ´Ë®Ïû¶¾¼Á£®ClO2ÊÇÒ»ÖÖ»ÆÂÌÉ«ÆøÌ壬Ò×ÈÜÓÚË®£®ÊµÑéÊÒÒÔNH4Cl¡¢ÑÎËá¡¢NaClO2ΪÔÁÏÖÆ±¸ClO2Á÷³ÌÈçÏ£º

ÒÑÖª£ºµç½â¹ý³ÌÖз¢ÉúµÄ·´Ó¦Îª£º
NH4Cl+2HCl
ͨµç
.
.
NCl3+3H2¡ü£»¼ÙÉèNCl3ÖеªÔªËØÎª+3¼Û£®
¢Ùд³öµç½âʱÒõ¼«µÄµç¼«·´Ó¦Ê½______£®
¢ÚÔÚÑô¼«ÉϷŵçµÄÎïÖÊ£¨»òÀë×Ó£©ÊÇ______£®
¢Û³ýÈ¥ClO2ÖеÄNH3¿ÉÑ¡ÓõÄÊÔ¼ÁÊÇ______£¨ÌîÐòºÅ£©
A£®Éúʯ»ÒB£®¼îʯ»ÒC£®Å¨H2SO4D£®Ë®
¢ÜÔÚÉú²ú¹ý³ÌÖУ¬Ã¿Éú³É1molClO2£¬ÐèÏûºÄ______molNCl3£®
²Î¿¼´ð°¸£º¢ñ¡¢¢ÙFe2O3£¨s£©+3C£¨Ê¯Ä«£©=2Fe£¨s£©+3CO£¨
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
4¡¢Ñ¡ÔñÌâ ÓÃCH4´ß»¯»¹ÔNOx¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺
¢ÙCH4(g)£«4NO2(g)===4NO(g)£«CO2(g)£«2H2O(g)??¦¤H£½£574?kJ¡¤mol£1
¢ÚCH4(g)£«4NO(g)===2N2(g)£«CO2(g)£«2H2O(g)???¦¤H£½£1160?kJ¡¤mol£1
ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ [???? ]
A.?ÓÉ·´Ó¦¢Ù¡¢¢Ú¿ÉÍÆÖª£ºCH4(g)£«2NO2(g)==N2(g)£«CO2(g)£«2H2O(g) ¦¤H£½£867?kJ?mol£1
B. µÈÎïÖʵÄÁ¿µÄ¼×Íé·Ö±ð²ÎÓë·´Ó¦¢Ù¡¢¢Ú£¬Ôò·´Ó¦×ªÒƵĵç×ÓÊýÏàµÈ
C.?ÉÏÊö·´Ó¦·½³ÌʽÖеĻ¯Ñ§¼ÆÁ¿Êý¼È¿ÉÒÔ±íʾÎïÖʵÄÁ¿£¬ÓÖ¿ÉÒÔ±íʾ·Ö×ÓÊý
D.?ÈôÓñê×¼×´¿öÏÂ4.48?L?CH4»¹ÔNO2ÖÁN2£¬Õû¸ö¹ý³ÌÖÐ×ªÒÆµÄµç×Ó×ÜÊýΪ1.6NA
²Î¿¼´ð°¸£ºC
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
5¡¢Ìî¿ÕÌâ 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=-393.5 kJ/mol
CO(g) +l/2O2(g)=CO2(g) ¡÷H= -283.0 kJ/mol
H2(g) +l/2O2(g)=H2O(g) ¡÷H= -242.0 kJ/mol
Çë»Ø´ð£º
(1)¸ù¾ÝÒÔÉÏÐÅÏ¢£¬Ð´³öC(s)ÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º________________________¡£
(2)±È½Ï·´Ó¦ÈÈÊý¾Ý¿ÉÖª£¬1 mol CO(g)ºÍ1 mol H2(g)ÍêȫȼÉշųöµÄÈÈÁ¿Ö®ºÍ±È1 mol C(s)ÍêȫȼÉշųöµÄ ÈÈÁ¿¶à¡£¼×ͬѧ¾Ý´ËÈÏΪ¡°Ãº×ª»¯ÎªË®ÃºÆø¿ÉÒÔʹúȼÉշųö¸ü¶àµÄÈÈÁ¿¡±£»ÒÒͬѧ¸ù¾Ý¸Ç˹¶¨ÂÉ×ö³öÈçͼËùʾµÄÑ»·Í¼£¬²¢¾Ý´ËÈÏΪ¡°Ãº×ª»¯ÎªË®ÃºÆøÔÙȼÉշųöµÄÈÈÁ¿Óëúֱ½ÓȼÉշųöµÄÈÈÁ¿ÏàµÈ¡±¡£

Çë·ÖÎö£º¼×¡¢ÒÒÁ½Í¬Ñ§¹ÛµãÕýÈ·µÄÊÇ_____£¨Ìî ¡°¼×¡±»ò¡°ÒÒ¡±£©£»ÅжϵÄÀíÓÉÊÇ_____________¡£½«Ãº×ª»¯ÎªË®ÃºÆø×÷ΪȼÁϺÍúֱ½ÓȼÉÕÏà±ÈÓкܶàÓŵ㣬ÇëÁÐ¾ÙÆäÖеÄÁ½¸öÓŵã____________¡£
²Î¿¼´ð°¸£º(1)C(s) +H2O(g)=CO(g) +H2(g) ¡÷
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
΢ÐÅËÑË÷¹Ø×¢"91¿¼ÊÔÍø"¹«ÖÚºÅ,Áì30Ôª,»ñÈ¡¹«ÎñÔ±ÊÂÒµ±à½Ìʦ¿¼ÊÔ×ÊÁÏ40G