ÌâÄ¿ÄÚÈÝ
°±ÔÚ¹úÃñ¾¼ÃÖÐÕ¼ÓÐÖØÒªµÄµØ룬Çë²ÎÓëÏÂÁÐ̽¾¿£®
£¨1£©Éú²úÇâÆø£º½«Ë®ÕôÆøͨ¹ýºìÈȵÄÌ¿¼´²úÉúˮúÆø£®
C£¨s£©+H2O£¨g£©?H2£¨g£©+CO£¨g£©¡÷H=+131.3kJ£¬¡÷S=+133.7J/K
¸Ã·´Ó¦ÔÚµÍÎÂÏÂÄÜ·ñ×Ô·¢ £¨ÌÄÜ»ò·ñ£©£®
£¨2£©ÒÑÖªÔÚ400¡æʱ£¬N2 £¨g£©+3H2£¨g£©?2NH3£¨g£©µÄK=0.5£¬
¢Ù2NH3£¨g£©?N2 £¨g£©+3H2£¨g£©µÄK= £¨ÌîÊýÖµ£©£®
¢Ú400¡æʱ£¬ÔÚ0.5LµÄ·´Ó¦ÈÝÆ÷ÖнøÐкϳɰ±·´Ó¦£¬Ò»¶Îʱ¼äºó£¬²âµÃN2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿·Ö±ðΪ2mol¡¢1mol¡¢2mol£¬Ôò´Ëʱ·´Ó¦V£¨N2£©Õý V£¨N2£©Ä棨Ì£¾¡¢£¼¡¢=¡¢²»ÄÜÈ·¶¨£©
£¨3£©ÔÚÈý¸öÏàͬÈÝÆ÷Öи÷³äÈë1molN2ºÍ3molH2£¬ÔÚijһ²»Í¬Ìõ¼þÏ·´Ó¦²¢´ïµ½Æ½ºâ£¬°±µÄÌå»ý·ÖÊýËæʱ¼ä±ä»¯ÇúÏßÈçͼ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ £¨ÌîÐòºÅ£©£®
A£®Í¼¢ñ¿ÉÄÜÊDz»Í¬Ñ¹Ç¿¶Ô·´Ó¦µÄÓ°Ï죬ÇÒP2£¾P1
B£®Í¼¢ò¿ÉÄÜÊDz»Í¬Ñ¹Ç¿¶Ô·´Ó¦µÄÓ°Ï죬ÇÒP1£¾P2
C£®Í¼¢ó¿ÉÄÜÊDz»Í¬Î¶ȶԷ´Ó¦µÄÓ°Ï죬ÇÒT1£¾T2
D£®Í¼¢ò¿ÉÄÜÊÇͬÎÂͬѹÏ£¬´ß»¯¼ÁÐÔÄÜ£¬1£¾2£®
£¨1£©Éú²úÇâÆø£º½«Ë®ÕôÆøͨ¹ýºìÈȵÄÌ¿¼´²úÉúˮúÆø£®
C£¨s£©+H2O£¨g£©?H2£¨g£©+CO£¨g£©¡÷H=+131.3kJ£¬¡÷S=+133.7J/K
¸Ã·´Ó¦ÔÚµÍÎÂÏÂÄÜ·ñ×Ô·¢
£¨2£©ÒÑÖªÔÚ400¡æʱ£¬N2 £¨g£©+3H2£¨g£©?2NH3£¨g£©µÄK=0.5£¬
¢Ù2NH3£¨g£©?N2 £¨g£©+3H2£¨g£©µÄK=
¢Ú400¡æʱ£¬ÔÚ0.5LµÄ·´Ó¦ÈÝÆ÷ÖнøÐкϳɰ±·´Ó¦£¬Ò»¶Îʱ¼äºó£¬²âµÃN2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿·Ö±ðΪ2mol¡¢1mol¡¢2mol£¬Ôò´Ëʱ·´Ó¦V£¨N2£©Õý
£¨3£©ÔÚÈý¸öÏàͬÈÝÆ÷Öи÷³äÈë1molN2ºÍ3molH2£¬ÔÚijһ²»Í¬Ìõ¼þÏ·´Ó¦²¢´ïµ½Æ½ºâ£¬°±µÄÌå»ý·ÖÊýËæʱ¼ä±ä»¯ÇúÏßÈçͼ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®Í¼¢ñ¿ÉÄÜÊDz»Í¬Ñ¹Ç¿¶Ô·´Ó¦µÄÓ°Ï죬ÇÒP2£¾P1
B£®Í¼¢ò¿ÉÄÜÊDz»Í¬Ñ¹Ç¿¶Ô·´Ó¦µÄÓ°Ï죬ÇÒP1£¾P2
C£®Í¼¢ó¿ÉÄÜÊDz»Í¬Î¶ȶԷ´Ó¦µÄÓ°Ï죬ÇÒT1£¾T2
D£®Í¼¢ò¿ÉÄÜÊÇͬÎÂͬѹÏ£¬´ß»¯¼ÁÐÔÄÜ£¬1£¾2£®
·ÖÎö£º£¨1£©¸ù¾Ý¡÷G=¡÷H-T¡÷SÅжϣ¬¡÷G£¼0×Ô·¢½øÐУ¬¡÷G£¾0·Ç×Ô·¢£»
£¨2£©¢Ùͬһ¿ÉÄæ·´Ó¦£¬Õý¡¢Äæ·´Ó¦µÄƽºâ³£Êý»¥Îªµ¹Êý£»
¢Ú¼ÆËãŨ¶ÈÉÌQc£¬Óëƽºâ³£Êý±È½ÏÅжϷ´Ó¦½øÐз½Ïò£¬½ø¶øÅжÏÕý¡¢ÄæËÙÂʹØϵ£»
£¨3£©A£®Ôö´óѹǿƽºâÓÒÒÆ£»
B£®Ñ¹Ç¿²»Í¬£¬Æ½ºâ״̬²»Í¬£»
C£®Éý¸ßζÈƽºâÄæÏòÒƶ¯£»
D£®´ß»¯¼Á²»Ó°ÏìƽºâÒƶ¯£®
£¨2£©¢Ùͬһ¿ÉÄæ·´Ó¦£¬Õý¡¢Äæ·´Ó¦µÄƽºâ³£Êý»¥Îªµ¹Êý£»
¢Ú¼ÆËãŨ¶ÈÉÌQc£¬Óëƽºâ³£Êý±È½ÏÅжϷ´Ó¦½øÐз½Ïò£¬½ø¶øÅжÏÕý¡¢ÄæËÙÂʹØϵ£»
£¨3£©A£®Ôö´óѹǿƽºâÓÒÒÆ£»
B£®Ñ¹Ç¿²»Í¬£¬Æ½ºâ״̬²»Í¬£»
C£®Éý¸ßζÈƽºâÄæÏòÒƶ¯£»
D£®´ß»¯¼Á²»Ó°ÏìƽºâÒƶ¯£®
½â´ð£º½â£º£¨1£©C£¨s£©+H2O£¨g£©?H2£¨g£©+CO£¨g£©¡÷H=+131.3kJ£¬¡÷S=+133.7J/K£¬¡÷G=¡÷H-T¡÷S£¬¡÷G£¼0×Ô·¢½øÐУ¬·´Ó¦ÖÐìʱ䡢ìر䶼Ôö´ó£¬¸ßÎÂÏ¿ÉÄÜ¡÷G£¼0£¬·´Ó¦×Ô·¢½øÐУ¬¹Ê´ð°¸Îª£º·ñ£»
£¨2£©¢Ù·´Ó¦2NH3£¨g£©?N2 £¨g£©+3H2£¨g£©ºÍ·´Ó¦N2 £¨g£©+3H2£¨g£©?2NH3£¨g£©ÊÇ»¥Îª¿ÉÄæ·´Ó¦£¬¹Ê2NH3£¨g£©?N2 £¨g£©+3H2£¨g£©µÄKƽºâ³£Êý=
=2£¬
¹Ê´ð°¸Îª£º2£»
¢ÚÒ»¶Îʱ¼äºó£¬µ±N2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿·Ö±ðΪ4mol/L¡¢2mol/L¡¢4mol/Lʱ£¬Qc=
=0.5£¬ËùÒÔ¸Ã״̬ÊÇƽºâ״̬£¬ÕýÄæ·´Ó¦ËÙÂÊÏàµÈ£¬
¹Ê´ð°¸Îª£º=£»
£¨3£©A£®Ôö´óѹǿƽºâÓÒÒÆ£¬°±ÆøµÄº¬Á¿Ó¦Ôö´ó£¬Í¼ÏóÓëʵ¼Ê²»·û£¬¹ÊA´íÎó£»
B£®Ñ¹Ç¿²»Í¬£¬Æ½ºâ״̬²»Í¬£¬²»¿ÉÄÜÔÚͬһƽºâ״̬£¬Í¼ÏóÓëʵ¼Ê²»·û£¬¹ÊB´íÎó£»
C£®Éý¸ßζÈƽºâÄæÏòÒƶ¯£¬°±ÆøµÄº¬Á¿¼õС£¬Í¼ÏóÓëʵ¼Ê²»·û£¬¹ÊC´íÎó£»
D£®´ß»¯¼Á²»Ó°ÏìƽºâÒƶ¯£¬ÓÉͼÏó¿ÉÖª1µ½´ïƽºâʱ¼ä¹ý¶Ì£¬¹Ê´ß»¯¼ÁÐÔÄÜ1£¾2£¬¹ÊDÕýÈ·£®
¹Ê´ð°¸Îª£ºD£®
£¨2£©¢Ù·´Ó¦2NH3£¨g£©?N2 £¨g£©+3H2£¨g£©ºÍ·´Ó¦N2 £¨g£©+3H2£¨g£©?2NH3£¨g£©ÊÇ»¥Îª¿ÉÄæ·´Ó¦£¬¹Ê2NH3£¨g£©?N2 £¨g£©+3H2£¨g£©µÄKƽºâ³£Êý=
1 |
0.5 |
¹Ê´ð°¸Îª£º2£»
¢ÚÒ»¶Îʱ¼äºó£¬µ±N2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿·Ö±ðΪ4mol/L¡¢2mol/L¡¢4mol/Lʱ£¬Qc=
42 |
4¡Á23 |
42 |
4¡Á22 |
¹Ê´ð°¸Îª£º=£»
£¨3£©A£®Ôö´óѹǿƽºâÓÒÒÆ£¬°±ÆøµÄº¬Á¿Ó¦Ôö´ó£¬Í¼ÏóÓëʵ¼Ê²»·û£¬¹ÊA´íÎó£»
B£®Ñ¹Ç¿²»Í¬£¬Æ½ºâ״̬²»Í¬£¬²»¿ÉÄÜÔÚͬһƽºâ״̬£¬Í¼ÏóÓëʵ¼Ê²»·û£¬¹ÊB´íÎó£»
C£®Éý¸ßζÈƽºâÄæÏòÒƶ¯£¬°±ÆøµÄº¬Á¿¼õС£¬Í¼ÏóÓëʵ¼Ê²»·û£¬¹ÊC´íÎó£»
D£®´ß»¯¼Á²»Ó°ÏìƽºâÒƶ¯£¬ÓÉͼÏó¿ÉÖª1µ½´ïƽºâʱ¼ä¹ý¶Ì£¬¹Ê´ß»¯¼ÁÐÔÄÜ1£¾2£¬¹ÊDÕýÈ·£®
¹Ê´ð°¸Îª£ºD£®
µãÆÀ£º±¾Ì⿼²é»¯Ñ§Æ½ºâͼÏóÓëÓ°ÏìÒòËØ¡¢»¯Ñ§Æ½ºâ³£ÊýµÈ£¬ÄѶÈÖеȣ¬£¨2£©ÖÐ×¢Òâ¸ù¾Ýƽºâ³£Êý±í´ïʽ½øÐÐÀí½â£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿