ÌâÄ¿ÄÚÈÝ
Ïà¶Ô·Ö×ÓÖÊÁ¿ÓÉСµ½´óÅÅÁеÄX¡¢Y¡¢ZÈýÖÖÆøÌåµ¥ÖÊ£¬×é³ÉÕâÈýÖÖµ¥ÖʵÄÔªËØ·Ö±ðλÓÚ²»Í¬µÄ¶ÌÖÜÆÚ¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬XÓëY»¯ºÏÉú³ÉM£»XÓëZ»¯ºÏÉú³ÉN£»MÄÜÓëN»¯ºÏÉú³ÉA¡£ÊµÑéÊÒ¿ÉÓÃÈçͼËùʾµÄ·¢Éú×°ÖÃÖÆÈ¡X¡¢ZºÍM(¼Ð³Ö×°ÖÃÒÑÂÔ)¡£
![]()
£¨1£©XÆøÌåΪ £¬ÖÆÈ¡MµÄ·¢Éú×°ÖÃÊÇ(ÌîдÐòºÅ)________¡£
£¨2£©AÖк¬ÓеĻ¯Ñ§¼üÀàÐÍÓÐ______________¡£
£¨3£©ÒÑÖªZÄÜÓëMÔÚ³£ÎÂÏ·´Ó¦Éú³ÉY£¬Í¬Ê±Óа×Ñ̲úÉú¡£·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________________¡£
ÓÃÏÂͼËùʾװÖýøÐÐZÓëMÔÚ³£ÎÂÏ·´Ó¦µÄʵÑ飬²¢ÊÕ¼¯Y¡£
![]()
£¨4£©Èô´ÓAÖÐÒݳöµÄÆøÌ庬ÓÐZ£¬ÔòͨÈ뷴ӦװÖÃAÖеÄZºÍMµÄÎïÖʵÄÁ¿Ö®±ÈÓ¦Âú×ã
________¡£
£¨5£©Èô´ÓAÖÐÒݳöµÄÆøÌåÎÞÂÛº¬ÓÐZ»òM£¬¾Ï´ÆøÆ¿Bºó£¬¾ùÄܱ»ÎüÊÕ£¬ÔòÏ´ÆøÆ¿BÖÐÊÔ¼ÁÓëZ¡¢M·´Ó¦µÄÀë×Ó·½³Ìʽ·Ö±ðÊÇ__________________£»__________________¡£
£¨1£©H2£¨2·Ö£© ÒÒ£¨2·Ö£©
£¨2£©Àë×Ó¼ü¡¢¼«ÐÔ¹²¼Û¼ü£¨2·Ö£©
£¨3£©3Cl2£«8NH3===6NH4Cl£«N2£¨2·Ö£©
£¨4£©>3/8£¨2·Ö£©
£¨5£©Cl2£«2Fe2£«===2Cl££«2Fe3£«£¨2·Ö£©
Fe2£«£«2NH3¡¤H2O===Fe(OH)2¡ý£«2NH4£«£¨2·Ö£©
Çë²Î¿¼ÌâÖÐͼ±í£¬ÒÑÖªE1£½134 kJ·mol£1¡¢E2£½368 kJ·mol£1£¬¸ù¾ÝÒªÇ󻨴ðÎÊÌ⣺
¡¡![]()
(1)ͼ¢ñÊÇ1 mol NO2(g)ºÍ1 mol CO(g)·´Ó¦Éú³ÉCO2ºÍNO¹ý³ÌÖеÄÄÜÁ¿±ä»¯Ê¾Òâͼ£¬ÈôÔÚ·´Ó¦ÌåϵÖмÓÈë´ß»¯¼Á£¬·´Ó¦ËÙÂÊÔö´ó£¬E1µÄ±ä»¯ÊÇ________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£¬ÏÂͬ)£¬¦¤HµÄ±ä»¯ÊÇ________¡£Çëд³öNO2ºÍCO·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º________________________________¡£
(2)¼×´¼ÖÊ×Ó½»»»Ä¤È¼ÁÏµç³ØÖн«¼×´¼ÕôÆø×ª»¯ÎªÇâÆøµÄÁ½ÖÖ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
¢ÙCH3OH(g)£«H2O(g)===CO2(g)£«3H2(g)
¦¤H£½49.0 kJ·mol£1
¢ÚCH3OH(g)£«
O2(g)===CO2(g)£«2H2(g)
¦¤H£½£192.9 kJ·mol£1
ÓÖÖª¢ÛH2O(g)===H2O(l)¡¡¦¤H£½£44 kJ·mol£1£¬Ôò¼×´¼ÕôÆøÈ¼ÉÕΪҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________________________________________________¡£
(3)ϱíÊDz¿·Ö»¯Ñ§¼üµÄ¼üÄܲÎÊý£º
| »¯Ñ§¼ü | P—P | P—O | O===O | P===O |
| ¼üÄÜ/kJ·mol£1 | a | b | c | x |
ÒÑÖª°×Á×µÄȼÉÕÈÈΪd kJ·mol£1£¬°×Á×¼°ÆäÍêȫȼÉյIJúÎïµÄ½á¹¹Èçͼ¢òËùʾ£¬Ôò±íÖÐx£½________ kJ·mol£1(Óú¬a¡¢b¡¢c¡¢dµÄ´ú±íÊýʽ±íʾ)¡£