ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂÁÐÓйصÄÅжϻò±íʾ·½·¨ÕýÈ·µÄÊÇ

A£®Ò»¶¨Ìõ¼þÏ£¬½«0.5 mol N2£¨g£©ºÍ1.5 mol H2£¨g£©ÖÃÓÚÃܱյÄÈÝÆ÷Öгä·Ö·´Ó¦Éú³ÉNH3£¨g£©£¬·ÅÈÈ19.3 kJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©£«3H2£¨g£©2NH3£¨g£© ¦¤H£¼£­38.6 kJ¡¤mol£­1

B£®ÖкÍÈȵIJⶨʵÑéÖÐÐèÒªÓõÄÖ÷Òª²£Á§ÒÇÆ÷£ºÁ½¸ö´óСÏàͬµÄÉÕ±­¡¢Á½¸ö´óСÏàͬµÄÁ¿Í²¡¢Î¶ȼơ¢»·Ðβ£Á§½Á°è°ô£¬Ò²¿ÉÒÔÓñ£Î±­´úÌæÉÕ±­×öÓйØÖкÍÈȲⶨµÄʵÑé¡£

C£®ÔڲⶨÖкÍÈȵÄʵÑéÖУ¬ÖÁÉÙÐèÒª²â¶¨²¢¼Ç¼µÄζÈÊÇ3´Î

D£®CO£¨g£©µÄȼÉÕÈÈÊÇ283.0 kJ¡¤mol£­1£¬ÔòCO2·Ö½âµÄÈÈ»¯Ñ§·½³ÌʽΪ£º2CO2£¨g£©===2CO£¨g£©£«O2£¨g£© ¦¤H£½£«283.0 kJ¡¤mol£­1

¡¾´ð°¸¡¿A

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£ºA£®N2£¨g£©+3H2£¨g£©2NH3£¨g£©£¬¸Ã·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬Ôò0.5 mol N2£¨g£©ºÍ1.5 mol H2£¨g£©³¹µ×·´Ó¦Íê·ÅÈÈÓ¦´óÓÚ19.3 kJ£¬ËùÒÔÈÈ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©£«3H2£¨g£©2NH3£¨g£© ¦¤H£¼£­38.6 kJ¡¤mol£­1£¬¹ÊAÕýÈ·£»B£®ÖкÍÈȵIJⶨʵÑéÖÐÑ¡ÔñµÄÁ½¸öÉÕ±­´óС²»Ò»Ñù£¬½«Ð¡ÉÕ±­·ÅÔÚ´óÉÕ±­À֮¼äÌîÈëÅÝÄ­»òÕßËéֽƬ£¬¹ÊB´íÎó£»C£®Öкͷ´Ó¦Ö®Ç°£¬Òª·Ö±ð¼Ç¼ÑÎËáºÍÇâÑõ»¯ÄƵÄζȣ¬Öкͷ´Ó¦¿ªÊ¼ºó£¬ÒªÓû·Ðβ£Á§½Á°èÈÜÒº£¬Í¬Ê±¼Ç¼Ï»ìºÏÈÜÒºµÄ×î¸ßζȣ¬¼´×îÖÕζȣ¬ËùÒÔ¸ÃʵÑéÖÁÉÙÐèÒª²â¶¨²¢¼Ç¼Î¶ȵĴÎÊýÊÇ3´Î£¬Îª¼õСÎó²î£¬Öظ´²Ù×÷Á½Èý´Î£¬¹ÊC´íÎó£»D£®CO£¨g£©µÄȼÉÕÈÈÊÇ283.0 kJ¡¤mol£­1£¬ËùÒÔ2CO£¨g£©+O2£¨g£©=2CO2£¨g£© ¦¤H=£­2¡Á283.0 kJmol-1£¬Ôò2CO2£¨g£©=2CO£¨g£©+O2£¨g£©·´Ó¦µÄ¦¤H=+2¡Á283.0 kJmol-1£¬¹ÊD´íÎó¡£¹ÊÑ¡A¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø