Question

2£®¼×´¼ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÓֿɳÆΪȼÁÏ£®¹¤ÒµÉÏÀûÓúϳÉÆø£¨Ö÷Òª³É·ÖΪCO¡¢CO₂ºÍH₂£©ÔÚ´ß»¯¼ÁµÄ×÷ÓÃϺϳɼ״¼£¬·¢ÉúµÄÖ÷Òª·´Ó¦ÈçÏ£º
¢ÙCO£¨g£©+2H₂£¨g£©¨TCH₃OH£¨g£©¡÷H
¢ÚCO₂£¨g£©+3H₂£¨g£©¨TCH₃OH£¨g£©+H₂O£¨g£©¡÷H₂=-58kJ•mol^-1
¢ÛCO₂£¨g£©+H₂£¨g£©¨TCO£¨g£©+H₂O£¨g£©¡÷H₃=+41kJ•mol^-1
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖª·´Ó¦¢ÙÖеÄÏà¹ØµÄ»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º

»¯Ñ§¼ü	H¨TH	C¨TO	C¡ÔO	H-O	C-H
F/£¨kJ•mol-1£©	435	343	1076	465	X

Ôòx=413£»
£¨2£©ÈôT¡æʱ½«6molCO₂ºÍ8molH₂³äÈë2LÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ú£¬²âµÃH₂µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯¹ØϵÈçͼÖÐ״̬I£¨Í¼ÖÐʵÏߣ©Ëùʾ£®Í¼ÖÐÊý¾ÝA£¨1£¬6£©´ú±íÔÚ1minʱH₂µÄÎïÖʵÄÁ¿ÊÇ6mol£®
¢ÙT¡æʱ״̬IÌõ¼þÏ£¬0¡«3minÄÚCH₃OHµÄƽ¾ù·´Ó¦ËÙÂÊv=0.28mol/£¨L•min£©£¬Æ½ºâ³£ÊýK=0.5£»
¢ÚÆäËûÌõ¼þ²»±äʱ£¬½ö¸Ä±äijһÌõ¼þºó²âµÃH₂µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼÖÐ״̬¢òËùʾ£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇÔö´óѹǿ£»
¢ÛÆäËûÌõ¼þ²»±ä£¬½ö¸Ä±äζÈʱ£¬²âµÃH₂µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼÖÐ״̬¢óËùʾ£¬Ôò״̬¢ó¶ÔÓ¦µÄζȣ¾£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©T¡æ£»
¢ÜÈô״̬¢òµÄƽºâ³£ÊýΪK₂£¬×´Ì¬¢óµÄƽºâ³£ÊýΪK₃£¬ÔòK₂£¾£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©K₃£»
¢ÝÒ»¶¨Î¶ÈÏÂͬ£¬´Ë·´Ó¦ÔÚºãÈÝÈÝÆ÷ÖнøÐУ¬ÄÜÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâÒÀ¾ÝµÄÊÇac£®
a£®ÈÝÆ÷ÖÐѹǿ²»±ä              b£®¼×´¼ºÍË®ÕôÆûµÄÌå»ý±È±£³Ö²»±ä
c£®v_Õý£¨H₂£©=3v_Ä棨CH₃OH£©      d£®2¸öC¨TO¶ÏÁѵÄͬʱÓÐ6¸öH-H¶ÏÁÑ£®

Answer 1

·ÖÎö £¨1£©ÒÑÖª£º¢ÙCO£¨g£©+2H₂£¨g£©?CH₃OH£¨g£©¡÷H
¢ÚCO₂£¨g£©+3H₂£¨g£©?CH₃OH£¨g£©+H₂O£¨g£©¡÷H=-58kJ/mol
¢ÛCO₂£¨g£©+H₂£¨g£©?CO£¨g£©+H₂O£¨g£©¡÷H=+41kJ/mol
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ú-¢Û=¢Ù£¬¾Ý´Ë¼ÆËã¡÷H£¬ÔÙ½áºÏ¼üÄܼÆËãx£»
£¨2£©¢Ù¸ù¾Ý·´Ó¦ËÙÂʵļÆË㹫ʽ½øÐмÆËã0¡«3minÄÚCH₃OHµÄƽ¾ù·´Ó¦ËÙÂÊ£»¸ù¾ÝͼÏó¿ÉÖª£¬·´Ó¦ÔÚ8min´ïµ½Æ½ºâ£¬´ËʱÇâÆøµÄÎïÖʵÄÁ¿Îª2mol£¬ÁÐÈý¶Îʽ£¬¼ÆË㻯ѧƽºâ³£ÊýK£»
¢Ú¢òÔÚ4min´ïµ½Æ½ºâ£¬ÇÒƽºâʱÇâÆøµÄÎïÖʵÄÁ¿±È¢ñС£¬ËµÃ÷·´Ó¦ËÙÂʼӿ죬ÇÒƽºâÕýÏòÒƶ¯£¬ÔÙ¸ù¾ÝÓ°Ï컯ѧƽºâµÄÒòËØ·ÖÎö¿ÉµÃ£»
¢Û״̬¢ó·´Ó¦ËÙÂʿ죬ƽºâʱÇâÆøµÄÎïÖʵÄÁ¿½Ï¢ñ´ó£¬ËµÃ÷ƽºâÄæÏòÒƶ¯£¬¹ÊζÈÉý¸ßÁË£»
¢Ü¸ù¾ÝζȶԻ¯Ñ§Æ½ºâµÄÒƶ¯¿ÉµÃKµÄ´óС£»
¢Ý¿ÉÄ淴Ӧƽºâʱ£¬ÕýÄæ·´Ó¦ËÙÂÊÏàµÈ£¬¸÷×é·Öº¬Á¿±£³Ö²»±ä£¬¾Ý´Ë·ÖÎö£®

½â´ð ½â£º£¨1£©ÒÑÖª£º¢ÙCO£¨g£©+2H₂£¨g£©?CH₃OH£¨g£©¡÷H
¢ÚCO₂£¨g£©+3H₂£¨g£©?CH₃OH£¨g£©+H₂O£¨g£©¡÷H=-58kJ/mol
¢ÛCO₂£¨g£©+H₂£¨g£©?CO£¨g£©+H₂O£¨g£©¡÷H=+41kJ/mol
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ú-¢Û=¢Ù£¬Ôò¢ÙÖеġ÷H=-58kJ/mol-41kJ/mol=-99kJ/mol£»
´Ëʱ¡÷H=·´Ó¦ÎïµÄ¼üÄܺÍ-Éú³ÉÎïµÄ¼üÄܺÍ=£¨1076kJ•mol^-1+435¡Á2kJ•mol^-1£©-£¨xkJ•mol^-1¡Á3+465kJ•mol^-1+343kJ•mol^-1£©=-99kJ/mol£¬½âµÃx=413£¬
¹Ê´ð°¸Îª£º413£»
£¨2£©¢ÙT¡æʱ״̬IÌõ¼þÏ£¬3minʱH₂µÄÎïÖʵÄÁ¿Îª3mol£»±ä»¯5mol£¬¸ù¾Ý·´Ó¦  CO₂£¨g£©+3H₂£¨g£©?CH₃OH£¨g£©+H₂O£¨g£©¿ÉÖªCH₃OH£¨g£©µÄ±ä»¯ÎïÖʵÄÁ¿Îª$\frac{5}{3}$mol£¬´Ëʱ0¡«3minÄÚCH₃OHµÄƽ¾ù·´Ó¦ËÙÂÊv=$\frac{£¨\frac{\frac{5}{3}mol}{2L}£©}{3min}$¡Ö0.28mol/£¨L•min£©£»
T¡æʱ½«6molCO₂ºÍ8molH₂³äÈë2LÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ú£¬¸ù¾ÝͼÏó¿ÉÖª£¬·´Ó¦ÔÚ8min´ïµ½Æ½ºâ£¬´ËʱÇâÆøµÄÎïÖʵÄÁ¿Îª2mol£¬ÁÐÈý¶Îʽ£º
                          CO₂£¨g£©+3H₂£¨g£©?CH₃OH£¨g£©+H₂O£¨g£©
Æðʼ£¨mol/L£©£º3                4                 0                   0
ת»¯£¨mol/L£©£º1                3                 1                   1
ƽºâ£¨mol/L£©£º2                1                 1                   1
Ôò·´Ó¦¢ÚµÄƽºâ³£ÊýK=$\frac{1¡Á1}{2¡Á{1}^{3}}$=0.5£»
¹Ê´ð°¸Îª£º0.28mol/£¨L•min£©£»0.5£»
¢Ú´ÓͼÏó¿ÉÖª£º¢òÔÚ4min´ïµ½Æ½ºâ£¬ËµÃ÷·´Ó¦ËÙÂʼӿ죬¿ÉΪÉý¸ßζȣ¬¼Ó´ß»¯¼Á£¬Ôö´óѹǿ£¬Æ½ºâʱÇâÆøµÄÎïÖʵÄÁ¿±È¢ñС£¬ËµÃ÷ƽºâÕýÏòÒƶ¯£¬¸Ã·´Ó¦·ÅÈÈ·´Ó¦£¬Éý¸ßζȣ¬Æ½ºâÄæÏòÒƶ¯£¬´ß»¯¼Á²»Ó°Ïìƽºâ£¬¸Ã·´Ó¦ÊǸöÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬Ôö´óѹǿ£¬Æ½ºâÕýÏòÒƶ¯£¬ÔòÓ¦¸ÃÊÇÔö´óѹǿ£»
¹Ê´ð°¸Îª£ºÔö´óѹǿ£»
¢Û¸ù¾ÝͼÏó£¬×´Ì¬¢ó´ïµ½Æ½ºâÄØËùÓÃʱ¼ä¶Ì£¬·´Ó¦ËÙÂʽϿ죬ƽºâʱÇâÆøµÄÎïÖʵÄÁ¿Ò²½Ï´ó£¬ËµÃ÷·´Ó¦ÄæÏòÒƶ¯£¬¹ÊӦζȽÏT¡æ¸ß£»
¹Ê´ð°¸Îª£º£¾£»
 ¢Ü״̬¢òÔö´óѹǿ£¬Î¶Ȳ»±ä£¬ÈÔΪT¡æ£¬×´Ì¬¢óζȱÈT¡æ¸ß£¬¸Ã·´Ó¦ÊÇÕý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Î¶ÈÉý¸ß£¬Æ½ºâÄæÏòÒƶ¯£¬»¯Ñ§Æ½ºâ³£Êý¼õС£¬K₂£¾K₃£»
¹Ê´ð°¸Îª£º£¾£»
 ¢Ý·´Ó¦CO₂£¨g£©+3H₂£¨g£©?CH₃OH£¨g£©+H₂O£¨g£©¡÷H=-58kJ/molÔÚºãÈÝÈÝÆ÷ÖнøÐУ¬
a£®·´Ó¦ÆøÌåÌå»ý¼õС£¬ºãÈÝʱ£¬Ëæ×Å·´Ó¦µÄ½øÐУ¬Ñ¹Ç¿¼õС£¬Æ½ºâºó£¬ÈÝÆ÷ÖÐѹǿ±£³Ö²»±ä£¬¹ÊaÕýÈ·£»
b£®¼×´¼ºÍË®ÕôÆø×÷ΪÉú³ÉÎÌå»ý±ÈʼÖÕδ1£º1±£³Ö²»±ä£¬¹Êb´íÎó£»
c£®v_Õý£¨H₂£©=3v_Õý£¨CH₃OH£©=3v_Ä棨CH₃OH£©£¬ÕýÄæ·´Ó¦ËÙÂÊÏàµÈ£¬·´Ó¦Æ½ºâ£¬¹ÊcÕýÈ·£»       
d£®ÈκÎʱ¿Ì£¬2¸öC=O¶ÏÁѵÄͬʱ¶¼ÓÐ6¸öH-H¶ÏÁÑ£¬¹Êd´íÎó£»
¹Ê´ð°¸Îª£ºac£®

µãÆÀ ±¾Ì⿼²éÁ˸Ç˹¶¨ÂÉ¡¢»¯Ñ§Æ½ºâ³£ÊýµÄ¼ÆË㣬ӰÏ컯ѧƽºâµÄÒòËØÒÔ¼°»¯Ñ§Æ½ºâ״̬µÄÅжϣ¬ÌâÄ¿ÄѶȲ»´ó£¬ÖØÔÚ¿¼²é¶Ô»ù´¡ÖªÊ¶µÄÀí½âÕÆÎÕ£®