ÌâÄ¿ÄÚÈÝ

£¨6·Ö£©ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ¡£Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2Éú²úȼÁϼ״¼¡£
£¨1£©ÔÚ250C¡¢101KPaʱ£¬3.2¿Ë¼×´¼£¨CH30H£©ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ72.576kJ£¬ÔòÄܱíʾ¼×´¼È¼ÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ___________________¡£
£¨2£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飬ÔÚÌå»ýΪ2 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë2mol CO2ºÍ6mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g) ¡÷H£½¡ª49kJ/mol£¬²âµÃCO2ºÍH2O (g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£

¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊv(H2)£½__________mol/(L¡¤min)¡£
¢ÚÏÂÁдëÊ©ÖÐÄÜʹn(H2O)£¯n(CO2)Ôö´óµÄÊÇ________¡£
A£®Éý¸ßζȠ                      B£®ÔÙ³äÈë3mol H2
C£®½«CH3OH (g)´ÓÌåϵÖзÖÀë        D£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó

£¨6·Ö£©£¨1£©CH3OH(l)+3/2O2(g)=CO2(g)+2H2O(l) ¡÷H=-725.76kJ¡¤mol-1
£¨2£©¢Ù0.225 mol/(L¡¤min)¡£¢ÚB C £¨Ã¿¿Õ2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨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
£®
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£®
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
£®
£¨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
£®
¢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£©

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

¾«Ó¢¼Ò½ÌÍø