ÌâÄ¿ÄÚÈÝ

(5·Ö)µªÊÇÒ»ÖÖµØÇòÉϺ¬Á¿·á¸»µÄÔªËØ£¬µª¼°Æ仯ºÏÎïµÄÑо¿ÔÚÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒªÒâÒå¡£
£¨1£©ÏÂͼÊÇ1 mol NO2ºÍ1 mol CO·´Ó¦Éú³ÉCO2ºÍNO¹ý³ÌÖÐÄÜÁ¿±ä»¯Ê¾Òâͼ£¬Ð´³öNO2ºÍCO·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                            

(2) ×Ô1913Ä깤ҵºÏ³É°±Í¶²úÒÔÀ´£¬ºÏ³É°±¹¤Òµ²»¶Ï·¢Õ¹£¬°±ÓÖ¿ÉÒÔ½øÒ»²½ÖƱ¸ÏõËᣬÔÚ¹¤ÒµÉϿɽøÐÐÁ¬ÐøÉú²ú¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖª£º¢ÙN2(g)+O2(g)=2NO(g)        ¡÷H1= +180.5kJ/mol
¢ÚN2(g)+3H2(g)2NH3(g)    ¡÷H2=£­92.4kJ/mol
¢Û2H2(g)+O2(g)=2H2O(g)       ¡÷H3=£­483.6kJ/mol
д³ö°±Æø¾­´ß»¯Ñõ»¯Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøµÄÈÈ»¯Ñ§·½³Ìʽ£º
                                   
NO2(g)+CO(g)=CO2(g)+NO(g)   ¡÷H=" ¡ª234" mol¡¤L-1 £¨2·Ö£©
4NH3(g) +5 O2(g) ="4NO(g)" + 6H2O(g)      ¡÷H=" ¡ª905.0" mol¡¤L-1£¨3·Ö£©
£¨1£©ÓÉͼÏñ¿ÉÖª£¬¡÷H= ¡ª|E1¡ªE2|=" ¡ª234" mol¡¤L-1
£¨2£©ÒÀ¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª£¬¢Ù¡Á2£­¢Ú¡Á2 + 3¡Á3¼´¿ÉÇóµÃ´ð°¸¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø