ÌâÄ¿ÄÚÈÝ

£¨14·Ö£©¸ÖÌúÉú²úÖеÄβÆøÒ×Ôì³É»·¾³ÎÛȾ£¬Çå½àÉú²ú¹¤ÒÕ¿ÉÏû¼õÎÛȾԴ²¢³ä·ÖÀûÓÃ×ÊÔ´¡£ÒÑÖª£º

¢Ù3Fe2O2(s)+CO(g)2Fe3O2(s)+CO2(g)   ¡÷H=¡ª47kJ/mol
¢ÚFe3O3(s)+3CO(g) 2Fe(s)+3CO2(g)   ¡÷H=¡ª25kJ/mol
¢ÛFe3O\4(s)+CO(g) 3FeO(s)+CO2(g)   ¡÷H=+19kJ/mol
£¨1£©ÊÔ¼ÆËã·´Ó¦£ºFeO£¨s£©+CO£¨g£©Fe£¨s£©+CO2£¨g£©µÄ¡÷H=      ¡£ÒÑÖª1092¡æ¸Ã·´Ó¦µÄƽºâ³£ÊýΪ0.357£¬Ôò1200¡æʱ¸Ã·´Ó¦µÄƽºâ³£Êý             0.357£¨Ìî¡°>¡±¡°=¡±»ò¡°<¡±£©£¬ÔÚ1LµÄÃܱÕÈÝÆ÷ÖУ¬Í¶Èë7.2gFeOºÍ0.1molCO2¼ÓÈȵ½1092¡æ²¢±£³Ö¸Ãζȣ¬·´Ó¦´ïƽºâºó£¬ÆøÏàÖÐCOÆøÌåËùÕ¼µÄÌå»ý·ÖÊýΪ       ¡£
£¨2£©Á¶¸ÖβÆø¾»»¯ºó£¬¿ÉÖ±½Ó×÷ÈÛÈÚ̼ËáÑÎȼÁϵç³Ø£¨¹¤×÷Ô­ÀíÈçÓÒͼ£©µÄȼÁÏ£¬Ôò¸º¼«µÄµç¼«·´Ó¦Îª                ¡£
£¨3£©×ªÂ¯Á¶¸Ö£¬Î²ÆøÖÐCOÌå»ý·ÖÊý´ï58%¡ª70%£¬Ä³¸Ö³§ÏÈÓÃNaOHÎüÊÕCOÉú³É¼×ËáÄÆ£¬ÔÙÎüÊÕSO2Éú³É±£ÏÕ·Û£¨Na2S2O3£©£¬ÊÔд³ö¼×ËáÄƺÍÇâÑõ»¯ÄÆ»ìºÏÈÜÒºÓëSO2Éú³É±£ÏÕ·ÛͬʱÉú³É¶þÑõ»¯Ì¼µÄ»¯Ñ§·½³Ìʽ                      ¡£
£¨4£©ÔÚ550¡ª650¡æʱ£¬Î²ÆøÑ̳¾ÖеÄFe2O3ÓëCO¼°H2ÆøÌå¿ÉÓÃÓںϳÉÁ¶¸ÖÔ­ÁÏFe3C£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ           ¡£
£¨5£©Ä¿Ç°ÎÒ¹ú´ó¶àÊýÆóÒµÊǽ«COת»»ÎªH2£¬È»ºóÓÃH2ÓëN2·´Ó¦ºÏ³Éµª£¬ÈôÊÕ¼¯µ½3360m2βÆø£¬ÆäÖÐCOÌå»ý·ÖÊýΪ60%£¬ÓÉÓÚÑ­»·²Ù×÷£¬¼Ù¶¨¸÷²½×ª»¯ÂʾùΪ100%£¬ÀíÂÛÉÏ¿É»ñµÃNH3        1¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
»¯Ñ§ÄܵÄת»¯ÔÚÏÖʵÉú»îÖеõ½Á˹㷺µÄÀûÓ᣻شðÒÔÏÂÎÊÌ⣺
£¨¢ñ£©(1)ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×ÍéÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬H2O£¬·Å³ö55 kJµÄÈÈÁ¿£¬Ð´³ö±íʾ¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º                                   ¡£
(2)2Zn£¨s£©+O2£¨g£©=2ZnO£¨s£©  ¦¤H1=" ¡ª702" kJ/mol
2Hg£¨l£©+O2£¨g£©=2HgO£¨s£©  ¦¤H2=" ¡ª182" kJ/mol
ÓÉ´Ë¿ÉÖªZnO£¨s£©+Hg£¨l£©= Zn£¨s£©+HgO£¨s£© ¡÷H3=          ¡£
£¨3£©20ÊÀ¼Í30Äê´ú£¬EyringºÍPelzerÔÚÅöײÀíÂ۵Ļù´¡ÉÏÌá³ö»¯Ñ§·´Ó¦µÄ¹ý¶É̬ÀíÂÛ£º»¯Ñ§·´Ó¦²¢²»ÊÇͨ¹ý¼òµ¥µÄÅöײ¾ÍÄÜÍê³ÉµÄ£¬¶øÊÇÔÚ·´Ó¦Îïµ½Éú³ÉÎïµÄ¹ý³ÌÖо­¹ýÒ»¸ö¸ßÄÜÁ¿¹ý¶É̬¡£ÏÂͼÊÇNO2ºÍCO·´Ó¦Éú³ÉCO2ºÍNO¹ý³ÌÖÐÄÜÁ¿±ä»¯Ê¾Òâͼ£¬Çëд³öNO2ºÍCO·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                                  

£¨¢ò£©ÏÂͼΪÏ໥´®ÁªµÄ¼×ÒÒÁ½¸öµç½â³Ø£º

Çë»Ø´ð£º
£¨1£©¼×³ØÈôΪÓõç½âÔ­Àí¾«Á¶Í­µÄ×°Öã¬Ôò£º
A¼«ÊÇ¡¡¡¡¡¡¼«£¬²ÄÁÏÊÇ          £¬µç¼«·´Ó¦Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬
B¼«ÊÇ¡¡¡¡¡¡¼«£¬²ÄÁÏÊÇ          £¬µç¼«·´Ó¦Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬
µç½âÖÊÈÜҺΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨2£©ÒÒ³ØÖÐÈôµÎÈëÉÙÁ¿·Ó̪ÊÔÒº£¬¿ªÊ¼Ò»¶Îʱ¼äºó£¬Fe¼«¸½½ü³Ê¡¡¡¡¡¡¡¡¡¡É«¡£
£¨3£©Èô¼×²ÛÒõ¼«ÔöÖØ12£®8g£¬ÔòÒÒ²ÛÑô¼«·Å³öÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý
£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß¡£
£¨4£©Í¬Ê±ÈôÒÒ²ÛÊ£ÓàÒºÌåΪ400mL£¬Ôòµç½âºóµÃµ½¼îÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ
__   ¡¡¡¡¡¡  _____¡£
(16·Ö)ÄÜԴΣ»úÊǵ±Ç°È«ÇòÎÊÌ⣬¿ªÔ´½ÚÁ÷ÊÇÓ¦¶ÔÄÜԴΣ»úµÄÖØÒª¾Ù´ë¡£
¢ÅÏÂÁÐ×ö·¨ÓÐÖúÓÚÄÜÔ´¡°¿ªÔ´½ÚÁ÷¡±µÄÊÇ   ¡ø  £¨Ìî×Öĸ£©¡£
a£®´óÁ¦·¢Õ¹Å©´åÕÓÆø£¬½«·ÏÆúµÄ½Õ¸Ñת»¯ÎªÇå½à¸ßЧµÄÄÜÔ´
b£®´óÁ¦¿ª²Éú¡¢Ê¯ÓͺÍÌìÈ»ÆøÒÔÂú×ãÈËÃÇÈÕÒæÔö³¤µÄÄÜÔ´ÐèÇó
c£®¿ª·¢Ì«ÑôÄÜ¡¢Ë®ÄÜ¡¢·çÄÜ¡¢µØÈȵÈÐÂÄÜÔ´¡¢¼õÉÙʹÓÃú¡¢Ê¯Ó͵Ȼ¯Ê¯È¼ÁÏ
d£®¼õÉÙ×ÊÔ´ÏûºÄ£¬Ôö¼Ó×ÊÔ´µÄÖظ´Ê¹Óá¢×ÊÔ´µÄÑ­»·ÔÙÉú
(2)½ð¸ÕʯºÍʯī¾ùΪ̼µÄͬËØÒìÐÎÌ壬ËüÃÇȼÉÕÑõÆø²»×ãʱÉú³ÉÒ»Ñõ»¯Ì¼£¬³ä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼£¬·´Ó¦ÖзųöµÄÈÈÁ¿ÈçÓÒͼËùʾ¡£          
 
£¨a£©ÔÚͨ³£×´¿öÏ£¬½ð¸ÕʯºÍʯīÖÐ____¡ø___£¨Ìî¡°½ð¸Õʯ¡±»ò¡°Ê¯Ä«¡±£©¸üÎȶ¨£¬Ê¯Ä«µÄȼÉÕÈÈΪ____¡ø___ kJ¡¤mol£­1¡£
£¨b£©12 gʯīÔÚÒ»¶¨Á¿¿ÕÆøÖÐȼÉÕ£¬Éú³ÉÆøÌå36g£¬¸Ã¹ý³Ì·Å³öµÄÈÈÁ¿    ¡ø    kJ¡£
(3)ÒÑÖª£ºN2(g)£«O2(g)£½2NO(g)£»¦¤H£½+180.0 kJ¡¤mol£­1¡£
×ÛºÏÉÏÊöÓйØÐÅÏ¢£¬Çëд³öCO³ýNOµÄÈÈ»¯Ñ§·½³Ìʽ          ¡ø            ¡£
ÃÀ¹ú˹̹¸£´óѧÑо¿ÈËÔ±×î½ü·¢Ã÷Ò»ÖÖ¡°Ë®¡±µç³Ø£¬ÕâÖÖµç³ØÄÜÀûÓõ­Ë®Ó뺣ˮ֮¼äº¬ÑÎÁ¿µÄ²î±ð½øÐз¢µç¡£
£¨4£©Ñо¿±íÃ÷£¬µç³ØµÄÕý¼«ÓöþÑõ»¯ÃÌÄÉÃ×°ôΪ²ÄÁÏ¿ÉÌá¸ß·¢µçЧÂÊ£¬ÕâÊÇÀûÓÃÄÉÃײÄÁϾßÓÐ
     ¡ø       ÌØÐÔ£¬ÄÜÓëÄÆÀë×Ó³ä·Ö½Ó´¥¡£
£¨5£©º£Ë®Öеġ°Ë®¡±µç³Ø×Ü·´Ó¦¿É±íʾΪ£º5MnO2£«2Ag£«2NaCl£½Na2Mn5O10£«2AgCl£¬¸Ãµç³Ø¸º¼«·´Ó¦Ê½Îª      ¡ø      £»µ±Éú³É1 mol Na2Mn5O10תÒÆ     ¡ø   molµç×Ó¡£

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

¾«Ó¢¼Ò½ÌÍø