£¨16·Ö£©ÎªÁ˺ÏÀíÀûÓû¯Ñ§ÄÜ£¬È·±£°²È«Éú²ú£¬»¯¹¤Éè¼ÆÐèÒª³ä·Ö¿¼ÂÇ»¯Ñ§·´Ó¦µÄìʱ䣬²¢²ÉÈ¡ÏàÓ¦´ëÊ©¡£»¯Ñ§·´Ó¦µÄìʱäͨ³£ÓÃʵÑé½øÐвⶨ£¬Ò²¿É½øÐÐÀíÂÛÍÆËã¡£
¢ÅʵÑé²âµÃ£¬5g¼×´¼ÔÚÑõÆøÖгä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱÊͷųö113.5kJµÄÈÈÁ¿£¬ÊÔд³ö¼×´¼È¼ÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º____________________________________¡£
¢ÆÏÂͼÊÇij±Ê¼Ç±¾µçÄÔÓü״¼È¼Áϵç³ØµÄ½á¹¹Ê¾Òâͼ¡£
·Åµçʱ¼×´¼Ó¦´Ó______´¦Í¨È루Ìî¡°a¡±»ò¡°b¡±£©£¬µç³ØÄÚ²¿H£«Ïò_____£¨Ìî¡°×ó¡±»ò¡°ÓÒ¡±£©Òƶ¯¡£Ð´³öµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½£º_______________________________¡£
¢ÇÓÉÆø̬»ù̬Ô×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
| a
| 945
|
ÒÑÖª·´Ó¦N
2(g)£«3H
2(g)
2NH
3(g) ¡÷
H£½£93 kJ¡¤mol
£1¡£ÊÔ¸ù¾Ý±íÖÐËùÁмüÄÜÊý¾Ý¼ÆËã
aµÄÊýÖµ£º_______________¡£
¢ÈÒÀ¾Ý¸Ç˹¶¨ÂÉ¿ÉÒÔ¶ÔijЩÄÑÒÔͨ¹ýʵÑéÖ±½Ó²â¶¨µÄ»¯Ñ§·´Ó¦µÄìʱä½øÐÐÍÆËã¡£
ÒÑÖª£ºC(s£¬Ê¯Ä«)£«O
2(g)£½CO
2(g) ¡÷
H1£½
£393.5kJ¡¤mol
£12H
2(g)£«O
2(g)£½2H
2O(l) ¡÷
H2£½£571.6kJ¡¤mol
£12C
2H
2(g)£«5O
2(g)£½4CO
2(g)£«2H
2O(l) ¡÷
H3£½
£2599kJ¡¤mol
£1¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¼ÆËã2C(s£¬Ê¯Ä«)£«H
2(g)£½C
2H
2(g)·´Ó¦µÄìʱä¡÷
H£½________¡£