ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾ£¬±íʾ·´Ó¦2SO2£¨g£©+O2 
´ß»¯¼Á
¼ÓÈÈ
 2SO3£¨g£©£¬¡÷H£¼0µÄÕý·´Ó¦ËÙÂÊËæÊ±¼äµÄ±ä»¯Çé¿ö£¬ÊÔ¸ù¾Ý´ËÇúÏßÅжÏÏÂÁÐ˵·¨¿ÉÄÜÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®t1ʱÔö¼ÓÁËSO2ºÍO2µÄŨ¶È£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯
B£®t1½µµÍÁËζȣ¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯
C£®t1ʱ¼õСÁËѹǿ£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒÆ¶¯
D£®t1ʱ¼õСÁËSO2µÄŨ¶È£¬Ôö¼ÓÁËSO3µÄŨ¶È£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒÆ¶¯
¾«Ó¢¼Ò½ÌÍø
A¡¢t1ʱÔö¼ÓÁËSO2ºÍO2µÄŨ¶È£¬Õý·´Ó¦ËÙÂÊÓ¦´óÓÚԭƽºâËÙÂÊ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯£¬¹ÊA´íÎó£»
B¡¢t1½µµÍÁËζȣ¬Æ½ºâÏòÕý·´Ó¦Òƶ¯£¬ÖØÐÂÆ½ºâʱµÄËÙÂÊСÓÚԭƽºâËÙÂÊ£¬¹ÊB´íÎó£»
C¡¢t1ʱ¼õСÁËѹǿ£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒÆ¶¯£¬ÖØÐÂÆ½ºâʱµÄËÙÂÊСÓÚԭƽºâËÙÂÊ£¬¹ÊC´íÎó£»
D¡¢t1ʱ¼õСÁËSO2µÄŨ¶È£¬Ôö¼ÓÁËSO3µÄŨ¶È£¬Æ½ºâÏòÄæ·´Ó¦Òƶ¯£¬ÐÂÆ½ºâʱ·´Ó¦ÎïŨ¶È±È¿ÉÒÔÔ­À´»¹´ó£¬Æ½ºâʱÕý·´Ó¦ËÙÂÊ¿ÉÒÔ±ÈԭƽºâËÙÂʴ󣬹ÊDÕýÈ·£®
¹ÊÑ¡D£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
A¡¢B¡¢C¡¢D¡¢EÊÇÖÐѧ»¯Ñ§³£¼ûµÄ5ÖÖ»¯ºÏÎÆäÖÐA¡¢BÊÇÑõ»¯Îµ¥ÖÊX¡¢YÊÇÉú»îÖг£¼ûµÄ½ðÊô£¬Ïà¹ØÎïÖʼäµÄת»¯¹ØÏµÈçͼËùʾ£¨²¿·Ö·´Ó¦ÎïÓë²úÎïÒÑÂÔÈ¥£©£»

£¨1£©ÈôÊÔ¼Á1ÓëÊÔ¼Á2²»ÊôÓÚͬÀàÎïÖÊ£¬ÔòXÓëÊÔ¼Á1·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
2Al+2H2O+2OH-=2AlO2-+3H2¡ü
2Al+2H2O+2OH-=2AlO2-+3H2¡ü
£®
£¨2£©ÈôÊÔ¼Á1ºÍÊÔ¼Á2Ïàͬ£¬ÇÒEÈÜÒº¼ÓÈÈÕô¸É²¢×ÆÉÕºó¿ÉµÃµ½A£¬ÔòAµÄ»¯Ñ§Ê½ÊÇ
Fe2O3
Fe2O3
£®
¢Ù¼ìÑéÎïÖÊDµÄÈÜÒºÖнðÊôÀë×ÓµÄʵÑé²Ù×÷ÊÇ
È¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¼¸µÎKSCNÈÜÒº£¬ÈÜÒº±äºìÉ«£¬ÔòÖ¤Ã÷Ô­ÈÜÒºÖк¬ÓÐFe3+
È¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¼¸µÎKSCNÈÜÒº£¬ÈÜÒº±äºìÉ«£¬ÔòÖ¤Ã÷Ô­ÈÜÒºÖк¬ÓÐFe3+
£®
¢Ú½«ÎïÖÊCÈÜÓÚË®£¬ÆäÈÜÒº³Ê
ËáÐÔ
ËáÐÔ
£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£¬Ô­ÒòÓÃÀë×Ó·½³ÌÀ´±íʾΪ
Al3++3H2OAl£¨OH£©3+3H+
Al3++3H2OAl£¨OH£©3+3H+
£®
£¨3£©ÈôEÈÜÒº¸ô¾ø¿ÕÆø¼ÓÈÈ¡¢Õô¸Éºó£¬¿ÉµÃµ½¸ÃÈÜÒºµÄÈÜÖÊ£¬¹¤ÒµÉÏÓÃE¡¢Ï¡ÁòËáºÍNaNO2ΪԭÁÏÀ´ÖƱ¸¸ßЧ¾»Ë®¼ÁY£¨OH£©SO4£¬·´Ó¦ÖÐÓÐNOÉú³É£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2FeSO4+2NaNO2+H2SO4=2Fe£¨OH£©SO4+Na2SO4+2NO¡ü
2FeSO4+2NaNO2+H2SO4=2Fe£¨OH£©SO4+Na2SO4+2NO¡ü
£®
ÏÂÁÐÈý¸ö»¯Ñ§·´Ó¦µÄƽºâ³£Êý£¨K1¡¢K2¡¢K3£©ÓëζȵĹØÏµ·Ö±ðÈçϱíËùʾ£º
»¯Ñ§·´Ó¦ ƽºâ³£Êý ζÈ
973K 1173K
¢ÙFe£¨s£©+CO2£¨g£©?FeO£¨s£©+CO£¨g£© K1 1.47 2.15
¢ÚFe£¨s£©+H2O£¨g£©?FeO£¨s£©+H2£¨g£© K2 2.38 1.67
¢ÛCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£© K3 £¿ £¿
Çë»Ø´ð£º
£¨1£©·´Ó¦¢ÙÊÇ
ÎüÈÈ
ÎüÈÈ
£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©·´Ó¦£®
£¨2£©Ð´³ö·´Ó¦¢ÛµÄƽºâ³£ÊýK3µÄ±í´ïʽ
C(CO2)£®C(H2)
C(CO)£®C(H2O)
C(CO2)£®C(H2)
C(CO)£®C(H2O)
£®
£¨3£©¸ù¾Ý·´Ó¦¢ÙÓë¢Ú¿ÉÍÆµ¼³öK1¡¢K2ÓëK3Ö®¼äµÄ¹ØÏµ£¬ÔòK3=
K2
K1
K2
K1
£¨ÓÃK1¡¢K2±íʾ£©£®¢ÛÖеġ÷H
£¼
£¼
0£¨Ìîд¡°£¾¡±»ò¡°£¼¡±»ò¡°=¡±£©
£¨4£©ÒªÊ¹·´Ó¦¢ÛÔÚÒ»¶¨Ìõ¼þϽ¨Á¢µÄƽºâÏòÄæ·´Ó¦·½ÏòÒÆ¶¯£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ
CE
CE
£¨Ìîд×ÖĸÐòºÅ£©£®
A£®ËõС·´Ó¦ÈÝÆ÷µÄÈÝ»ý          B£®À©´ó·´Ó¦ÈÝÆ÷µÄÈÝ»ý
C£®Éý¸ßζȠ                   D£®Ê¹ÓúÏÊʵĴ߻¯¼Á
E£®Éè·¨¼õСƽºâÌåϵÖеÄCOµÄŨ¶È  F£®ºãѹµÄÌõ¼þÏÂͨÈëHeÆø
£¨5£©Èô·´Ó¦¢ÛµÄÄæ·´Ó¦ËÙÂÊÓëʱ¼äµÄ¹ØÏµÈçͼËùʾ£º

¢Ù¿É¼û·´Ó¦ÔÚt1¡¢t3¡¢t7ʱ¶¼´ïµ½ÁËÆ½ºâ£¬¶øt2¡¢t8ʱ¶¼¸Ä±äÁËÒ»ÖÖÌõ¼þ£¬ÊÔÅжϸıäµÄÊÇʲôÌõ¼þ£º
t2ʱ
Ôö´óÉú³ÉÎïŨ¶È
Ôö´óÉú³ÉÎïŨ¶È
»ò
Éý¸ßζÈ
Éý¸ßζÈ
£»t8ʱ
Ôö´óѹǿ
Ôö´óѹǿ
 »ò
ʹÓô߻¯¼Á
ʹÓô߻¯¼Á
£®
¢ÚÈôt4ʱ½µÑ¹£¬t6ʱÔö´ó·´Ó¦ÎïµÄŨ¶È£¬ÇëÔÚͼÖл­³öt4¡«t6Ê±Äæ·´Ó¦ËÙÂÊʱ¼äµÄ¹ØÏµÏߣ®
£¨2012?̫ԭһ죩A¡«JÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄ¼¸ÖÖÎïÖÊ£¬ËüÃÇÖ®¼äµÄת»¯¹ØÏµÈçͼËùʾ£¨²¿·Ö·´Ó¦Ìõ¼þÒÑÂÔÈ¥£©£®ÒÑÖª³£ÎÂÏÂAΪ¹ÌÌåµ¥ÖÊ£¬BΪµ­»ÆÉ«·ÛÄ©£¬C¡¢F¡¢IÎªÆøÌ¬µ¥ÖÊ£¬EÔÚ³£ÎÂÏÂΪҺÌ壬J¿ÉÓÃ×÷ɱ¾úÏû¶¾¼Á£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©BµÄµç×ÓʽÊÇ
£®
£¨2£©JµÄË®ÈÜÒºÏÔ
¼îÐÔ
¼îÐÔ
£¨Ìî¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±»ò¡°ÖÐÐÔ¡±£©£¬ÆäÔ­ÒòÊÇ
ClO-+H2O?HClO+OH-
ClO-+H2O?HClO+OH-
£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
£¨3£©ÒÑÖªa g FÔÚ×ãÁ¿CÖÐȼÉÕ£¬»Ö¸´ÖÁÊÒÎÂʱ·Å³öµÄÈÈÁ¿ÊÇb kJ£¬Çëд³öÏàÓ¦µÄÈÈ»¯Ñ§·½³Ìʽ£º
2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-
4b
a
kJ/mol
2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-
4b
a
kJ/mol
£®
£¨4£©ÈôÔÚ1L 0.2mol?L-1 AlCl3ÈÜÒºÖÐÖðµÎ¼ÓÈëµÈÌå»ýµÄ0.7mol?L-1 DÈÜÒº£¬Ôò´Ë¹ý³ÌÖн«¹Û²ìµ½£º
ÏÈÓа×É«³ÁµíÉú³É£¬ºó³Áµí²¿·ÖÈܽâ
ÏÈÓа×É«³ÁµíÉú³É£¬ºó³Áµí²¿·ÖÈܽâ
£®
£¨5£©³£ÎÂÏ£¬ÒÔPtΪµç¼«µç½âµÎ¼ÓÓÐÉÙÁ¿·Ó̪µÄHµÄ±¥ºÍÈÜÒº£¬ÔòÔÚ
Òõ
Òõ
£¨Ìî¡°Òõ¡±»ò¡°Ñô¡±£©¼«¸½½üÈÜÒºÓÉÎÞÉ«±äΪºìÉ«£¬ÈôÔڴ˼«ÊÕ¼¯µ½±ê×¼×´¿öÏÂ2.24LÆøÌ壬Ôò´ËʱÈÜÒºµÄpHÊÇ
13
13
£¨¼ÙÉèÈÜÒºµÄÌå»ýΪ2LÇÒ²»¿¼Âǵç½âºóÌå»ýµÄ±ä»¯£©£®
£¨6£©ÊµÑéÊÒ¿ÉÓò»º¬EµÄÒÒ´¼´¦ÀíÉÙÁ¿²ÐÁôµÄA£¬Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2Na+2C2H5OH¡ú2C2H5ONa+H2¡ü
2Na+2C2H5OH¡ú2C2H5ONa+H2¡ü
£®

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

¾«Ó¢¼Ò½ÌÍø