Question

ΪÁ˺ÏÀíÀûÓû¯Ñ§ÄÜ£¬È·±£°²È«Éú²ú£¬»¯¹¤Éè¼ÆÐèÒª³ä·Ö¿¼ÂÇ»¯Ñ§·´Ó¦µÄìʱ䣬²¢²ÉÈ¡ÏàÓ¦´ëÊ©£®»¯Ñ§·´Ó¦µÄìʱäͨ³£ÓÃʵÑé½øÐвⶨ£¬Ò²¿É½øÐÐÀíÂÛÍÆË㣮
£¨1£©ÊµÑé²âµÃ£¬5gҺ̬¼×´¼£¨CH₃OH£©ÔÚÑõÆøÖгä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱÊͷųö113.5kJµÄÈÈÁ¿£¬ÊÔд³ö¼×´¼È¼ÉÕµÄÈÈ»¯Ñ§·½³Ìʽ

2CH₃0H£¨l£©+30₂£¨g£©=2C0₂£¨g£©+4H₂0£¨l£©¡÷H=-1452.8kJ?mol^-1

£®
£¨2£©ÓÉÆø̬»ù̬ԭ×ÓÐγÉ1mol»¯Ñ§¼üÊͷŵÄ×îµÍÄÜÁ¿½Ð¼üÄÜ£®´Ó»¯Ñ§¼üµÄ½Ç¶È·ÖÎö£¬»¯Ñ§·´Ó¦µÄ¹ý³Ì¾ÍÊÇ·´Ó¦ÎïµÄ»¯Ñ§¼üµÄÆÆ»µºÍÉú³ÉÎïµÄ»¯Ñ§¼üµÄÐγɹý³Ì£®ÔÚ»¯Ñ§·´Ó¦¹ý³ÌÖУ¬²ð¿ª»¯Ñ§¼üÐèÒªÏûºÄÄÜÁ¿£¬Ðγɻ¯Ñ§¼üÓÖ»áÊÍ·ÅÄÜÁ¿£®

»¯Ñ§¼ü

H-H

N-H

N¡ÔN

¼üÄÜ/kJ?mol-1

436

391

946

ÒÑÖª·´Ó¦N₂£¨g£©+3H₂ £¨g£©?2NH₃£¨g£©¡÷H=a kJ?mol^-1£®ÊÔ¸ù¾Ý±íÖÐËùÁмüÄÜÊý¾Ý¼ÆËãaΪ

-92

£®
£¨3£©ÒÀ¾Ý¸Ç˹¶¨ÂÉ¿ÉÒÔ¶ÔijЩÄÑÒÔͨ¹ýʵÑéÖ±½Ó²â¶¨µÄ»¯Ñ§·´Ó¦µÄìʱä½øÐÐÍÆË㣮
ÒÑÖª£ºC£¨s£¬Ê¯Ä«£©+0₂£¨g£©=C0₂£¨g£©£»¡÷H₁=-393.5kJ?mol^-1
2H₂£¨g£©+0₂£¨g£©=2H₂0£¨l£©£»¡÷H₂=-571.6kJ?mol^-1
2C₂H₂£¨g£©+50₂£¨g£©=4C0₂£¨g£©+2H₂0£¨l£©£»¡÷H₃=-2599kJ?mol^-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¼ÆËã298Kʱ·´Ó¦2C£¨s£¬Ê¯Ä«£©+H₂£¨g£©=C₂H₂£¨g£©µÄìʱ䣺¡÷H=

+226.7kJ?mol^-1

£®
£¨4£©ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙFe₂O₃£¨s£©+3CO£¨s£©=2Fe£¨s£©+3CO₂£¨g£©£»¡÷H=-25kJ/mol
¢Ú3Fe₂O₃£¨s£©+CO£¨s£©=2Fe₃O₄£¨s£©+CO₂£¨g£©£»¡÷H=-47kJ/mol
¢ÛFe₃O₄£¨s£©+CO£¨s£©=4FeO£¨s£©+CO₂£¨g£©£»¡÷H=+19kJ/mol
д³öFeO£¨s£©±»CO»¹Ô­³ÉFeºÍCO₂µÄÈÈ»¯Ñ§·½³Ìʽ

FeO£¨s£©+CO£¨s£©=Fe£¨s£©+C0₂£¨g£©¡÷H=-11kJ/mol

£®

Answer 1

·ÖÎö£º£¨1£©¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨¿ÉÖª£¬»¯Ñ§¼ÆÁ¿ÊýÓë·´Ó¦ÈȳÉÕý±È£¬²¢×¢Òâ±êÃ÷ÎïÖʵľۼ¯×´Ì¬À´½â´ð£»
£¨2£©ÒÀ¾Ý·´Ó¦¹ý³ÌÖз´Ó¦ìʱä=·´Ó¦Îï¼üÄÜ×ܺÍ-Éú³ÉÎï¼üÄÜ×ܺͼÆË㣻
£¨3£©ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂɼÆËãµÃµ½ËùÐèÈÈ»¯Ñ§·½³Ìʽ£»
£¨4£©ÒÀ¾Ý¸Ç˹¶¨ÂɽáºÏÈÈ»¯Ñ§·½³Ìʽ¼ÆËãµÃµ½£®

½â´ð£º½â£º£¨1£©5gCH₃OHÔÚÑõÆøÖÐȼÉÕÉú³ÉCO₂ºÍҺ̬ˮ£¬·Å³ö113.5kJÈÈÁ¿£¬64g¼´1molCH₃OHÔÚÑõÆøÖÐȼÉÕÉú³ÉCO₂ºÍҺ̬ˮ£¬·Å³ö1452.8kJÈÈÁ¿£¬
ÔòÈÈ»¯Ñ§·½³ÌʽΪ£º2CH₃OH£¨g£©+3O₂£¨g£©¨T2CO₂£¨g£©+4H₂O£¨l£©¡÷H=-1452.8KJ/mol£¬
¹Ê´ð°¸Îª£º2CH₃OH£¨g£©+3O₂£¨g£©¨T2CO₂£¨g£©+4H₂O£¨l£©¡÷H=-1452.8KJ/mol£»
£¨2£©N₂£¨g£©+3H₂ £¨g£©?2NH₃£¨g£©¡÷H=a kJ?mol^-1£¬·´Ó¦ìʱä=·´Ó¦Îï¼üÄÜ×ܺÍ-Éú³ÉÎï¼üÄÜ×ܺͣ¬a=946KJ+3¡Á436KJ-6¡Á391KJ=-92KJ/mol£¬
¹Ê´ð°¸Îª£º-92£»
£¨3£©¢ÙC£¨s£¬Ê¯Ä«£©+0₂£¨g£©=C0₂£¨g£©£»¡÷H₁=-393.5kJ?mol^-1
¢Ú2H₂£¨g£©+0₂£¨g£©=2H₂0£¨l£©£»¡÷H₂=-571.6kJ?mol^-1
¢Û2C₂H₂£¨g£©+50₂£¨g£©=4C0₂£¨g£©+2H₂0£¨l£©£»¡÷H₃=-2599kJ?mol^-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ù¡Á4+¢Ú-¢Û¼ÆËã298Kʱ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º4C£¨s£¬Ê¯Ä«£©+2H₂£¨g£©=2C₂H₂£¨g£©¡÷H=+453.4KJ/mol£»
¼´ÈÈ»¯Ñ§·½³ÌʽΪ£º2C£¨s£¬Ê¯Ä«£©+H₂£¨g£©=C₂H₂£¨g£©¡÷H=+226.7KJ/mol£¬
¹Ê´ð°¸Îª£º+226.7KJ/mol£»
£¨4£©¢ÙFe₂O₃£¨s£©+3CO£¨g£©=2Fe£¨s£©+3CO₂£¨g£©£»¡÷H=-25kJ/mol
¢Ú3Fe₂O₃£¨s£©+CO£¨g£©=2Fe₃O₄£¨s£©+CO₂£¨g£©£»¡÷H=-47kJ/mol
¢ÛFe₃O₄£¨s£©+CO£¨g£©=3FeO£¨s£©+CO₂£¨g£©£»¡÷H=+19kJ/mol
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù¡Á3-£¨¢Û¡Á2+¢Ú£©µÃµ½£º
6CO£¨g£©+6FeO£¨s£©=6Fe£¨g£©+6CO₂£¨g£©¡÷H=-66KJ/mol£»
µÃµ½ÈÈ»¯Ñ§·½³ÌʽΪ£ºCO£¨g£©+FeO£¨s£©=Fe£¨g£©+CO₂£¨g£©¡÷H=-11KJ/mol£¬
¹Ê´ð°¸Îª£ºCO£¨g£©+FeO£¨s£©=Fe£¨g£©+CO₂£¨g£©¡÷H=-11KJ/mol£®

µãÆÀ£º±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨ºÍ¸Ç˹¶¨ÂɵļÆËãÓ¦Óã¬ÕÆÎÕ»ù´¡Êǹؼü£¬ÌâÄ¿ÄѶÈÖеȣ®