ÌâÄ¿ÄÚÈÝ

ÏÂÁÐÓйØÈÈ»¯Ñ§·½³ÌʽµÄÐðÊöÕýÈ·µÄÊÇ(    )
A£®ÒÑÖªC(ʯī£¬s)C(½ð¸Õʯ£¬s)£»¡÷H>0£¬Ôò½ð¸Õʯ±ÈʯīÎȶ¨
B£®ÒÑÖªC(s)+O2(g)=CO2(g)£»¡÷H1ºÍC(s)+1/2O2(g)=CO(g)£»¡÷H2£¬Ôò¡÷H1>¡÷H2
C£®101kPaʱ£¬2H2(g)+O2(g)=2H2O(l)£»¡÷H=-5716kJ/mol£¬ÔòÇâÆøµÄȼÉÕÈÈΪ285.8kJ/mol
D£®º¬20.0gNaOHµÄÏ¡ÈÜÒºÓëÏ¡ÑÎËáÍêÈ«ÖкÍʱ·Å³ö28.7kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º NaOH(aq)+HCl(aq)="NaCl(aq)+" H2O(1)£»¡÷H="+57.4" kJ/mol
C

ÊÔÌâ·ÖÎö£ºA¡¢ÓÉ·´Ó¦£ºC(ʯī£¬s)C(½ð¸Õʯ£¬s)£»¡÷H>0Öª£¬Ê¯Ä«¾ßÓеÄÄÜÁ¿µÍ£¬ÎïÖʾßÓеÄÄÜÁ¿Ô½µÍÔ½Îȶ¨£¬Ôòʯī±È½ð¸ÕʯÎȶ¨£¬´íÎó£»B¡¢¡÷H´óС±È½ÏÒª´ø·ûºÅ±È½Ï£¬¶ÔÓÚ·ÅÈÈ·´Ó¦¶øÑÔ£¬·Å³öµÄÈÈÁ¿Ô½¶à£¬¡÷HԽС£¬Ôò¡÷H1£¼¡÷H2£¬´íÎó£»C¡¢101kPaʱ£¬2H2(g)+O2(g)=2H2O(l)£»¡÷H=-5716kJ/mol£¬ÔòÇâÆøµÄȼÉÕÈÈΪ285.8kJ/mol£¬ÕýÈ·£»D¡¢º¬20.0gNaOHµÄÏ¡ÈÜÒºÓëÏ¡ÑÎËáÍêÈ«ÖкÍʱ·Å³ö28.7kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º NaOH(aq)+HCl(aq)="NaCl(aq)+" H2O(1)£»¡÷H=¡ª57.4 kJ/mol£¬´íÎó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨14·Ö£©ÔËÓû¯Ñ§·´Ó¦Ô­Àí·ÖÎö½â´ðÒÔÏÂÎÊÌâ
£¨1£©ÒÑÖª£º ¢ÙCO(g)+2H2(g)  CH3OH(g)              ¡÷Hl= £­91kJ¡¤mol£­l
¢Ú2CH3OH(g) CH3OCH3(g)+H2O(g) ¡÷H2= £­24 kJ¡¤mol£­l                ¢ÛCO(g) +H2O(g)  CO2(g)+H2(g)       ¡÷H3= £­41 kJ¡¤mol£­l
ÇÒÈý¸ö·´Ó¦µÄƽºâ³£ÊýÒÀ´ÎΪK1¡¢K2¡¢K3
    Ôò·´Ó¦ 3CO(g) +3H2(g)       CH3OCH3(g) +CO2(g) ¡÷H=               .
»¯Ñ§Æ½ºâ³£ÊýK=        £¨Óú¬K1¡¢K2¡¢K3µÄ´úÊýʽ±íʾ£©¡£
£¨2£©Ò»¶¨Ìõ¼þÏ£¬Èô½«Ìå»ý±ÈΪ1:2µÄCOºÍH2ÆøÌåͨÈëÌå»ýÒ»¶¨µÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦
         3CO(g) +3H2(g)      CH3OCH3(g) +CO2(g)£¬ÏÂÁÐÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬ÊÇ         ¡£
a£®Ìåϵѹǿ±£³Ö²»±ä    B£®»ìºÏÆøÌåÃܶȱ£³Ö²»±ä
c£® COºÍH2µÄÎïÖʵÄÁ¿±£³Ö²»±ä       d£®COµÄÏûºÄËٶȵÈÓÚCO2µÄÉú³ÉËÙÂÊ
£¨3£©°±ÆøÈÜÓÚË®µÃµ½°±Ë®¡£ÔÚ25¡æÏ£¬½«x mol£®L£­lµÄ°±Ë®Óëy mol£®L£­1µÄÑÎËáµÈÌå»ý»ìºÏ£¬·´Ó¦ºóÈÜÒºÏÔÖÐÐÔ£¬Ôòc(NH4+)____c£¨Cl£­£©£¨Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±£©£»Óú¬xºÍyµÄ´úÊýʽ±íʾ³ö°±Ë®µÄµçÀëƽºâ³£Êý             .
£¨4£©¿Æѧ¼Ò·¢Ã÷ÁËʹNH3Ö±½ÓÓÃÓÚȼÁϵç³ØµÄ·½·¨£¬Æä×°ÖÃÓò¬ºÚ×÷µç¼«¡¢¼ÓÈëµç½âÖÊÈÜÒºÖУ¬Ò»¸öµç¼«Í¨Èë¿ÕÆø£¬ÁíÒ»µç¼«Í¨ÈëNH3¡£Æäµç³Ø·´Ó¦Ê½Îª£º4NH3+3O2 = 2N2+6H2O£¬µç½âÖÊÈÜÒºÓ¦ÏÔ           £¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±¡¢¡°¼îÐÔ¡±£©£¬
д³öÕý¼«µÄµç¼«·´Ó¦·½³Ìʽ                .

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø