ÇâÆøÊÇÒ»ÖÖÇå½àÄÜÔ´£¬ÇâÆøµÄÖÆÈ¡Óë´¢´æÊÇÇâÄÜÔ´ÀûÓÃÁìÓòµÄÑо¿Èȵ㣮
£¨1£©Ö±½ÓÈȷֽⷨÖÆÇ⣮ijζÈÏ£¬H
2O£¨g£©?H
2£¨g£©+
O
2£¨g£©£®¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK=
£®
£¨2£©ÈÈ»¯Ñ§Ñ»·ÖÆÇ⣮ÖƱ¸H
2µÄ·´Ó¦²½ÖèÈçÏ£º
¢ÙBr
2£¨g£©+CaO£¨s£©¨TCaBr
2£¨s£©+
O
2£¨g£©¡÷H=-73kJ?mol
-1¢Ú3FeBr
2£¨s£©+4H
2O£¨g£©¨TFe
3O
4£¨s£©+6HBr£¨g£©+H
2£¨g£©¡÷H=+384kJ?mol
-1¢ÛCaBr
2£¨s£©+H
2O £¨g£©¨TCaO£¨s£©+2HBr£¨g£©¡÷H=+212kJ?mol
-1¢ÜFe
3O
4£¨s£©+8HBr£¨g£©¨TBr
2£¨g£©+3FeBr
2£¨s£©+4H
2O£¨g£©¡÷H=-274kJ?mol
-1Ôò H
2O£¨g£©?H
2£¨g£©+
O
2£¨g£©¡÷H=
+249
+249
kJ?mol
-1£®
£¨3£©¹âµç»¯Ñ§·Ö½âÖÆÇ⣬ÆäÔÀíÈçͼ£®îÑËáïȹâµç¼«µÄµç¼«·´Ó¦Îª4OH
--4e
-¨TO
2+2H
2O£¬Ôò²¬µç¼«µÄµç¼«·´Ó¦Îª
2H2O+2e-¨TH2¡ü+2OH-£¨»ò2H++2e-¨TH2¡ü£©
2H2O+2e-¨TH2¡ü+2OH-£¨»ò2H++2e-¨TH2¡ü£©
£®
£¨4£©Ë®ÃºÆø·¨ÖÆÇ⣮
CO£¨g£©+H
2O£¨g£©?CO
2£¨g£©+H
2£¨g£©¡÷H£¼0£¬ÔÚ850¡æʱ£¬K=1£®
¢ÙÈôÉý¸ßζȵ½950¡æʱ£¬´ïµ½Æ½ºâʱK
£¼
£¼
1£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
¢Ú850¡æʱ£¬ÈôÏòÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖÐͬʱ³äÈë1.0mol CO¡¢3.0mol H
2O¡¢1.0mol CO
2 ºÍx mol H
2£¬ÈôҪʹÉÏÊö·´Ó¦¿ªÊ¼Ê±ÏòÕý·´Ó¦·½Ïò½øÐУ¬ÔòxÓ¦Âú×ãµÄÌõ¼þÊÇ
x£¼3
x£¼3
£®
£¨5£©¼×ÍéÖÆÇ⣮½«1.0mol CH
4ºÍ2.0mol H
2O £¨g£©Í¨ÈëÈÝ»ýΪ100LµÄ·´Ó¦ÊÒ£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCH
4£¨g£©+H
2O£¨g£©¨TCO£¨g£©+3H
2£¨g£©£®²âµÃ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬CH
4µÄƽºâת»¯ÂÊΪ50%£¬ÔòÓÃH
2±íʾ¸Ã·´Ó¦µÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.003 mol?L-1?min-1
0.003 mol?L-1?min-1
£®
£¨6£©LiBH
4¾ßÓзdz£¸ßµÄ´¢ÇâÄÜÁ¦£¬·Ö½âʱÉú³ÉÇ⻯﮺ÍÁ½ÖַǽðÊôµ¥ÖÊ£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2LiBH4¨T2LiH+2B+3H2¡ü
2LiBH4¨T2LiH+2B+3H2¡ü
£®