£¨8·Ö£©ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ¡£Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2Éú²úȼÁϼ״¼¡£Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)+3H2(g)CH3OH(g)+H2O(g)£¬Èçͼ±íʾ¸Ã·´Ó¦½øÐйý³ÌÖÐÄÜÁ¿(µ¥Î»ÎªkJ¡¤mol£­1)µÄ±ä»¯¡£

£¨1£©¹ØÓڸ÷´Ó¦µÄÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ          ¡£(Ìî×Öĸ)
A£®¡÷H>0£¬¡÷S>0        B£®¡÷H>0£¬¡÷S<0
C£®¡÷H<0£¬¡÷S<0        D£®¡÷H<0£¬¡÷S>0
£¨2£©¸Ã·´Ó¦Æ½ºâ³£ÊýKµÄ±í´ïʽΪ                     ¡£
£¨3£©Î¶ȽµµÍ£¬Æ½ºâ³£ÊýK            (Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±)¡£
£¨4£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³äÈë1molCO2ºÍ3molH2£¬²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱÎʱ仯ÈçÏÂͼËùʾ¡£´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÓÃÇâÆøŨ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv(H2)Ϊ         mol¡¤L£­1¡¤min£­1¡£

£¨5£©ÏÂÁдëÊ©ÖÐÄÜʹ£¨4£©ÖÐn(CH3OH)/n(CO2)Ôö´óµÄÓР          ¡£(Ìî×Öĸ)
A£®Éý¸ßζȠ    
B£®¼ÓÈë´ß»¯¼Á   
C£®½«H2O(g)´ÓÌåϵÖзÖÀë
D£®ÔÙ³äÈë1molCO2ºÍ3molH2   
E£®³äÈëHe(g)£¬Ê¹Ìåϵ×ÜѹǿÔö´ó

£¨8·Ö£©£¨1£©C £¨2·Ö£©£¨2£©£¨1·Ö£©£¨3£©Ôö´ó£¨1·Ö£©£¨4£© 0.225 £¨2·Ö£©£¨5£©CD£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ù¾Ý·´Ó¦µÄ·½³Ìʽ¿ÉÖª£¬Õý·´Ó¦ÊÇÌå»ý¼õСµÄ£¬¼´¡÷S£¼0£»¸ù¾ÝͼÏñ¿ÉÖª£¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬ËùÒÔÊÇ·ÅÈÈ·´Ó¦£¬¡÷H£¼0£¬´ð°¸Ñ¡C¡£
£¨2£©»¯Ñ§Æ½ºâ³£ÊýÊÇÔÚÒ»¶¨Ìõ¼þÏ£¬µ±¿ÉÄæ·´Ó¦´ïµ½Æ½ºâ״̬ʱ£¬Éú³ÉÎïŨ¶ÈµÄÃÝÖ®»ýºÍ·´Ó¦ÎïŨ¶ÈµÄÃÝÖ®»ýµÄ±ÈÖµ£¬ËùÒԸ÷´Ó¦µÄƽºâ³£Êý±í´ïʽK£½¡£
£¨3£©Õý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬½µµÍζȣ¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬KÖµÔö´ó¡£
£¨4£©¸ù¾ÝͼÏñ¿ÉÖª£¬Æ½ºâʱÉú³É0.75mol¼×´¼£¬ËùÒÔÏûºÄÇâÆøµÄÎïÖʵÄÁ¿ÊÇ2.25mol£¬ÔòÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊÊÇ¡£
£¨5£©Éý¸ßζȣ¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬n(CH3OH)/n(CO2)¼õС£»¼ÓÈë´ß»¯¼Á£¬Æ½ºâ²»Òƶ¯£¬n(CH3OH)/n(CO2)²»±ä£»½«H2O(g)´ÓÌåϵÖзÖÀ룬ƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬n(CH3OH)/n(CO2)Ôö´ó£»ÔÙ³äÈë1molCO2ºÍ3molH2£¬Ï൱ÓÚÊǼÓѹ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬n(CH3OH)/n(CO2)Ôö´ó£»
³äÈëHe(g)£¬Ê¹Ìåϵ×ÜѹǿÔö´ó£¬µ«Æ½ºâ²»Òƶ¯£¬n(CH3OH)/n(CO2)²»±ä£¬´ð°¸Ñ¡CD¡£
¿¼µã£º¿¼²é·´Ó¦ÈÈ¡¢ìØÖµ¡¢Æ½ºâ³£Êý¡¢·´Ó¦ËÙÂÊÒÔ¼°Íâ½çÌõ¼þ¶Ôƽºâ״̬µÄÓ°Ïì
µãÆÀ£º¸ÃÌâÊǸ߿¼Öеij£¼ûÌâÐÍ£¬ÄѶȲ»´ó£¬ËùÒÔ»ù´¡ÐÔÊÔÌâµÄ¿¼²é£¬ÊÔÌâÄÑÒ×ÊÊÖУ¬ÊôÓÚÖеÈÄѶȵÄÊÔÌâ¡£¸ÃÌâÔ­ÀíÅàÑøѧÉú¹æ·¶´ðÌâ¡¢ÉóÌâÄÜÁ¦£¬ÒÔ¼°ÑϽ÷µÄÂß¼­Ë¼Î¬ÄÜÁ¦£¬Ò²ÓÐÀûÓÚÌá¸ßѧÉúµÄѧϰЧÂÊ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2008?°²ÇìÄ£Ä⣩¢ñ£®ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2Éú²úȼÁϼ״¼£®Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©£¬ÈçÓÒÉÏͼ±íʾ¸Ã·´Ó¦½øÐйý³ÌÖÐÄÜÁ¿£¨µ¥Î»ÎªkJ?mol-1£©µÄ±ä»¯£®
£¨1£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺T1¡æʱ£¬ÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçÓÒÏÂͼËùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÓÃÇâÆøŨ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv£¨H2£©=
0.225
0.225
mol?L-1?min-1£®
£¨2£©¸Ã·´Ó¦Æ½ºâ³£ÊýKµÄ±í´ïʽΪ
5.3
5.3
£®
£¨3£©Î¶ȱäΪT2¡æ£¨T1£¾T2£©£¬Æ½ºâ³£ÊýK
Ôö´ó
Ôö´ó
£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©£®
£¨4£©²»ÄÜÅжϸ÷´Ó¦ÊÇ·ñ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇ
D
D
£®
A£®ÈÝÆ÷ÖÐѹǿ²»±ä        B£®»ìºÏÆøÌåÖР c£¨CO2£©²»±ä
C£®vÕý£¨H2£©=vÄ棨H2O£©    D£®c£¨CO2£©=c£¨CO£©
£¨5£©ÏÂÁдëÊ©ÖÐÄÜʹn£¨CH3OH£©/n£¨CO2£©Ôö´óµÄÓÐ
CD
CD
£®
A£®Éý¸ßζȣ»      B£®¼ÓÈë´ß»¯¼Á£»    C£®½«H2O£¨g£©´ÓÌåϵÖзÖÀ룻
D£®ÔÙ³äÈë1molCO2ºÍ3molH2£»   E£®³äÈëHe£¨g£©£¬Ê¹Ìåϵ×ÜѹǿÔö´ó£®
¢ò£®ÔÚζÈt¡æÏ£¬Ä³NaOHµÄÏ¡ÈÜÒºÖÐc£¨H+£©=10-amol/L£¬c£¨OH-£©=10-bmol/L£¬ÒÑÖªa+b=12
¸ÃζÈÏÂË®µÄÀë×Ó»ýKw=
1¡Á10-12
1¡Á10-12
£»t
´óÓÚ
´óÓÚ
25¡æ£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£®Ïò¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈëpH=cµÄÑÎËᣨt¡æ£©£¬²âµÃ»ìºÏÈÜÒºµÄ²¿·ÖpHÈç±íËùʾ£®
ÐòºÅ NaOHÈÜÒºµÄÌå»ý/mL ÑÎËáµÄÌå»ý/mL ÈÜÒºµÄpH
¢Ù 20.00 0.00 8
¢Ú 20.00 20.00 6
¼ÙÉèÈÜÒº»ìºÏÇ°ºóµÄÌå»ý±ä»¯ºöÂÔ²»¼Æ£¬ÔòcΪ
4
4
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2Éú²úȼÁϼ״¼£®Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©£¬Í¼1±íʾ¸Ã·´Ó¦½øÐйý³ÌÖÐÄÜÁ¿£¨µ¥Î»ÎªkJ?mol-1£©µÄ±ä»¯£®

£¨1£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H=-£¨n-m£©kJ?mol?1
CO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H=-£¨n-m£©kJ?mol?1
£®
£¨2£©¹ØÓڸ÷´Ó¦µÄÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ
C
C
£®
A£®¡÷H£¾0£¬¡÷S£¾0                 B£®¡÷H£¾0£¬¡÷S£¼0
C£®¡÷H£¼0£¬¡÷S£¼0                 D£®¡÷H£¼0£¬¡÷S£¾0
£¨3£©¸Ã·´Ó¦µÄƽºâ³£ÊýKµÄ±í´ïʽΪ£º
c(CH3OH)?c(H2O)
c(CO2)?c3(H2)
c(CH3OH)?c(H2O)
c(CO2)?c3(H2)
£®
£¨4£©Î¶ȽµµÍ£¬Æ½ºâ³£ÊýK
Ôö´ó
Ôö´ó
£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©£®
£¨5£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1molCO2ºÍ3molH2£¬²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯Èçͼ2Ëùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÓÃÇâÆøŨ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv £¨H2£©=
0.225mol/£¨L£®min£©
0.225mol/£¨L£®min£©
£®
£¨6£©ÏÂÁдëÊ©ÖÐÄÜʹn£¨CH3OH£©/n£¨CO2£©Ôö´óµÄÓÐ
C
C
£®
A£®Éý¸ßζȠ                     B£®¼ÓÈë´ß»¯¼Á
C£®½«H2O£¨g£©´ÓÌåϵÖзÖÀë          D£®³äÈëHe£¨g£©£¬Ê¹Ìåϵ×ÜѹǿÔö´ó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2Éú²úȼÁϼ״¼£®Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©£¬Í¼1±íʾ¸Ã·´Ó¦½øÐйý³ÌÖÐÄÜÁ¿µÄ±ä»¯£®

£¨1£©¹ØÓڸ÷´Ó¦µÄÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ
C
C
£®
A£®¡÷H£¾0£¬¡÷S£¾0£» B£®¡÷H£¾0£¬¡÷S£¼0£»
C£®¡÷H£¼0£¬¡÷S£¼0£» D£®¡÷H£¼0£¬¡÷S£¾0£®
£¨2£©¸ÃͼÖÐÄÜÁ¿µÄ×î¸ßµãºÍ×îµÍµãÖ®¼äµÄ²îÖµ´ú±í
Äæ·´Ó¦µÄ»î»¯ÄÜ
Äæ·´Ó¦µÄ»î»¯ÄÜ

£¨3£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2
ºÍ3mol H2£¬²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯Èçͼ2Ëùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâÓÃÇâÆøŨ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv£¨H2£©
0.225
0.225
mol?L-1?min-1£®
£¨4£©ë£¨N2H4£©ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ£®ÒÑÖªÔÚ101kPaʱ£¬32.0gN2H4ÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉµªÆø£¬·Å³öÈÈÁ¿624kJ£¨25¡æʱ£©£¬N2H4ÍêȫȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º
N2H4£¨l£©+O2£¨g£©=N2£¨g£©+2H2O£¨l£©¡÷H=-624KJ/mol
N2H4£¨l£©+O2£¨g£©=N2£¨g£©+2H2O£¨l£©¡÷H=-624KJ/mol
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2012?¿ª·â¶þÄ££©ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®
£¨1£©Ä¿Ç°£¬Óó¬ÁÙ½çCO2£¨Æä״̬½éÓÚÆø̬ºÍҺ̬֮¼ä£©´úÌæ·úÀû°º×÷Àä¼ÁÒѳÉΪһÖÖÇ÷ÊÆ£¬ÕâÒ»×ö·¨¶Ô»·¾³µÄ»ý¼«ÒâÒåÔÚÓÚ
±£»¤³ôÑõ²ã
±£»¤³ôÑõ²ã
£®
£¨2£©½«CO2ת»¯³ÉÓлúÎï¿ÉÓÐЧʵÏÖ̼ѭ»·£®CO2ת»¯³ÉÓлúÎïµÄÀý×Ӻܶ࣬È磺
a.6CO2+6H2O
¹âºÏ
×÷ÓÃ
C6H12O6+6O2      b£®CO2+3H2
´ß»¯¼Á
¡÷
CH3OH+H2O
c£®CO2+CH4
´ß»¯¼Á
¡÷
CH3COOH          d.2CO2+6H2
´ß»¯¼Á
¡÷
CH2=CH2+4H2O
ÒÔÉÏ·´Ó¦ÖУ¬×î½ÚÄܵÄÊÇ
a
a
£¬Ô­×ÓÀûÓÃÂÊ×î¸ßµÄÊÇ
c
c
£®
£¨3£©ÈôÓÐ4.4kg CO2Óë×ãÁ¿H2Ç¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉÆø̬µÄË®ºÍ¼×´¼£¬¿É·Å³ö4947kJµÄÈÈÁ¿£¬ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.47kJ/mol
CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.47kJ/mol
£®
£¨4£©ÎªÌ½¾¿ÓÃCO2À´Éú²úȼÁϼ״¼µÄ·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷£¬³äÈë1mol CO2ºÍ3molH2£¬½øÐз´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâv£¨H2£©=
0.225 mol?L-1?min-1
0.225 mol?L-1?min-1
£»CO2µÄת»¯ÂÊ=
75%
75%
£»¸ÃζÈϵÄƽºâ³£ÊýÊýÖµ=
5.33
5.33
£®ÄÜʹƽºâÌåϵÖÐn£¨CH3OH£©/n£¨CO2£©Ôö´óµÄ´ëÊ©ÓÐ
½«H2O£¨g£©´ÓÌåϵÖзÖÀë
½«H2O£¨g£©´ÓÌåϵÖзÖÀë
 £¨ÈÎдһÌõ£©£®
£¨5£©CO2ÔÚ×ÔÈ»½çÑ­»·Ê±¿ÉÓëCaCO3·´Ó¦£¬CaCO3ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÆäKsp=2.8¡Á10-9£®CaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ¿ÉÐγÉCaCO3³Áµí£¬ÏÖ½«µÈÌå»ýµÄCaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ£¬ÈôNa2CO3ÈÜÒºµÄŨ¶ÈΪ4¡Á10-4mol/L£¬ÔòÉú³É³ÁµíËùÐèCaCl2ÈÜÒºµÄ×îСŨ¶ÈΪ
2.8¡Á10-5mol/L
2.8¡Á10-5mol/L
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¢ñ£ºÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2Éú²úȼÁϼ״¼£®Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©£¬¡÷H£¼0
£¨1£©¸Ã·´Ó¦Æ½ºâ³£ÊýKµÄ±í´ïʽΪ
 
£®Î¶ȽµµÍ£¬Æ½ºâ³£ÊýK
 
£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©£®
£¨2£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯Èçͼ1Ëùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÓÃÇâÆøŨ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv£¨H2£©Îª
 
£®
£¨3£©ÏÂÁдëÊ©ÖÐÄÜʹn£¨CH3OH£©/n£¨CO2£©Ôö´óµÄÓÐ
 
£®
A£®Éý¸ßζȣ»      B£®¼ÓÈë´ß»¯¼Á£»C£®½«H2O£¨g£©´ÓÌåϵÖзÖÀ룻D£®ÔÙ³äÈë1mol CO2ºÍ3mol H2£»
E£®³äÈëHe£¨g£©£¬Ê¹Ìåϵ×ÜѹǿÔö´ó£®
£¨4£©Èçͼ2Ëùʾ£¬Ôڼס¢ÒÒÁ½ÈÝÆ÷Öзֱð³äÈëÎïÖʵÄÁ¿Ö®±ÈΪ1£º3 µÄCO2ºÍH2£¬Ê¹¼×¡¢ÒÒÁ½ÈÝÆ÷³õʼÈÝ»ýÏàµÈ£®ÔÚÏàͬζÈÏ·¢Éú·´Ó¦£¬²¢Î¬³Ö·´Ó¦¹ý³ÌÖÐζȲ»±ä£®¼×ºÍÒÒÏà±È£¬×ª»¯³Ì¶È¸ü´óµÄÊÇ
 
£¬ÇÒÖªÒÒÈÝÆ÷ÖÐCO2µÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏóÈçͼ3Ëùʾ£¬ÇëÔÚͼ3Öл­³ö¼×ÈÝÆ÷ÖÐCO2µÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏó£®²¢Çë˵Ã÷ÒÔCO2ΪԭÁÏÉú²úȼÁϼ״¼µÄÓŵãÊÇ
 
£¨Ð´³öÒ»Ìõ¼´¿É£©£®
¾«Ó¢¼Ò½ÌÍø
¢ò£ºÒ»¶¨³£ÎÂÏ£¬FeSµÄKSP=2.5¡Á10-18£¬H2S±¥ºÍÈÜÒºÔÚ¸ÃζÈÏ£¬[H+]Óë[S2-]´æÔÚ×ÅÒÔϹØϵ£º[H+]2?[S2-]=1.0¡Á10-21£®ÔÚ¸ÃζÈÏ£¬½«ÊÊÁ¿FeSͶÈëH2S±¥ºÍÈÜÒºÖУ¬ÓûʹÈÜÒºÖÐ[Fe2+]´ïµ½1mol/L£¬Ó¦µ÷½ÚÈÜÒºµÄpHΪ
 
£¨ÓöÔÊýÐÎʽ±íʾ£©£®£¨Ð´³ö¼ÆËã¹ý³Ì£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸