ÌâÄ¿ÄÚÈÝ

(9·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Ê±£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÖкÍÈÈ»¯Ñ§·½³Ìʽ______________________________________.
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾­²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈÈ»¯Ñ§·½³Ìʽ___________________£®
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l)     ¦¤H1£½£­870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g)              ¡¡       ¦¤H2£½£­393.5 kJ/mol
¢ÛH2(g)£«O2(g)===H2O(l )                     ¦¤H3£½£­285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ_______________________£®
(9·Ö)   (1)H2SO4(aq)£«NaOH(aq)===Na2SO4(aq)£«H2O(l)¡¡  ¦¤H£½£­57.3 kJ/mol
(2)C4H10(g)£«O2(g)===4CO2(g)£«5H2O(l)   ¦¤H£½£­16Q kJ/mol
(3)2C(s)£«2H2(g)£«O2(g)===CH3COOH(l)    ¦¤H£½£­488.3 kJ/mol
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(8·Ö)µÂ¹úÈ˹þ²®ÔÚ1913ÄêʵÏÖÁ˺ϳɰ±µÄ¹¤Òµ»¯Éú²ú£¬·´Ó¦Ô­Àí£º
N2(g)£«3H2(g)2NH3(g)£»ÒÑÖª298 Kʱ£¬
¦¤H£½£­92.4 kJ¡¤mol£­1£¬¦¤S£½£­198.2 J¡¤mol£­1¡¤K£­1£¬ÊԻشðÏÂÁÐÎÊÌ⣺
(1)¼ÆËã˵Ã÷298 KϺϳɰ±·´Ó¦ÄÜ·ñ×Ô·¢½øÐУ¿________(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)£»ÔÚ298 Kʱ£¬½«10 mol N2ºÍ30 mol H2·ÅÈëºÏ³ÉËþÖУ¬ÎªÊ²Ã´·Å³öµÄÈÈÁ¿Ð¡ÓÚ924 kJ£¿________¡£

(2)ÈçͼÔÚÒ»¶¨Ìõ¼þÏ£¬½«1 mol N2ºÍ3 mol H2»ìºÏÓÚÒ»¸ö10 LµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦´ïµ½Aƽºâʱ£¬»ìºÏÆøÌåÖа±Õ¼25%£¬ÊԻشðÏÂÁÐÎÊÌ⣺
¢ÙN2µÄת»¯ÂÊΪ________£»
¢ÚÔڴﵽ״̬Aʱ£¬Æ½ºâ³£ÊýKA£½________(´úÈëÊýÖµµÄ±í´ïʽ£¬²»ÒªÇóµÃ¾ßÌåÊýÖµ)£¬µ±Î¶ÈÓÉT1±ä»¯µ½T2ʱ£¬KA________KB(Ìî¡°£½¡±¡¢¡°<¡±»ò¡°>¡±)¡£
¢ÛÔڴﵽ״̬Bʱ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)
a£®Í¨Èëë²ÆøʹѹǿÔö´ó£¬»¯Ñ§Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯
b£®N2µÄÕý·´Ó¦ËÙÂÊÊÇH2µÄÄæ·´Ó¦ËÙÂʵÄ1/3±¶
c£®½µµÍζȣ¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±äС
d£®Ôö¼ÓN2µÄÎïÖʵÄÁ¿£¬H2µÄת»¯ÂʽµµÍ
(3)ÈôÔÚºãΡ¢ºãѹÌõ¼þϺϳɰ±·´Ó¦´ïµ½Æ½ºâºó£¬ÔÙÏòƽºâÌåϵÖÐͨÈëë²Æø£¬Æ½ºâ________Òƶ¯(Ìî¡°Ïò×󡱡°ÏòÓÒ¡±»ò¡°²»¡±)¡£
(4)ÔÚ1998ÄêÏ£À°ÑÇÀï˹¶àµÂ´óѧµÄMarnellosºÍStoukides²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É(ÄÜ´«µÝH£«)£¬ÊµÏÖÁ˸ßθßѹϸßת»¯Âʵĵ绯ѧºÏ³É°±£¬ÆäʵÑé×°ÖÃÈçͼ£º

ÔòÒõ¼«µÄµç¼«·´Ó¦Ê½Îª____________________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø