ÌâÄ¿ÄÚÈÝ

¸ù¾ÝϱíÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ£º                                                                                                       

ÐòºÅ

Ñõ»¯¼Á

»¹Ô­¼Á

ÆäËü·´Ó¦Îï

Ñõ»¯²úÎï

»¹Ô­²úÎï

¢Ù

Cl2

FeBr2

FeCl3

¢Ú

KMnO4

H2O2

H2SO4

O2

MnSO4

¢Û

KClO3

HCl£¨Å¨£©

Cl2

Cl2

¢Ü

KMnO4

HCl£¨Å¨£©

Cl2

MnCl2

A£®±íÖеڢÙ×é·´Ó¦µÄÑõ»¯²úÎïÒ»¶¨Ö»ÓÐFeCl3£¨ÊµÎªFe3+£©

B£®Ñõ»¯ÐԱȽÏ: KMnO4£¾Cl2£¾Fe3+£¾Br2£¾Fe2+

C£®»¹Ô­ÐԱȽÏ: H2O2£¾Mn2+£¾Cl-

D£®¢ÜµÄÀë×Ó·½³ÌʽÅäƽºó£¬H+µÄ»¯Ñ§¼ÆÁ¿ÊýΪ16

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦ÓÚ¡°È˹¤¹Ìµª¡±µÄз½·¨Ñо¿¡£
£¨1£©Ä¿Ç°ºÏ³É°±¼¼ÊõÔ­ÀíΪ£ºN2(g) + 3H2(g)2NH3(g)£» ¡÷H=-92.4kJ¡¤mol-1¡£
¢Ù 673K£¬30MPaÏ£¬ÉÏÊöºÏ³É°±·´Ó¦ÖÐn(NH3)ºÍn(H2)Ëæʱ¼ä±ä»¯µÄ¹ØϵÈçͼËùʾ¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ__________¡£
A£®µãaµÄÕý·´Ó¦ËÙÂʱȵãbµÄ´ó   
B£®µãc´¦·´Ó¦´ïµ½Æ½ºâ
C£®µãdºÍµã e´¦µÄn(N2)Ïàͬ
D£®773K£¬30MPaÏ£¬·´Ó¦ÖÁt2ʱ¿Ì´ïµ½Æ½ºâ£¬Ôòn(NH3)±ÈͼÖÐeµãµÄÖµ´ó
¢Ú ÔÚÈÝ»ýΪ2.0 LºãÈݵÄÃܱÕÈÝÆ÷ÖгäÈë0.80 mol N2(g)ºÍ1.60 mol H2(g)£¬·´Ó¦ÔÚ673K¡¢30MPaÏ´ﵽƽºâʱ£¬NH3µÄÌå»ý·ÖÊýΪ20%¡£¸ÃÌõ¼þÏ·´Ó¦N2(g) + 3H2(g)2NH3(g)µÄƽºâ³£ÊýK= ________¡££¨±£ÁôСÊýһ룩KÖµÔ½´ó£¬±íÃ÷·´Ó¦´ïµ½Æ½ºâʱ_________£¨Ìî±êºÅ£©¡£
A£®H2µÄת»¯ÂÊÒ»¶¨Ô½¸ß       
B£®NH3µÄ²úÁ¿Ò»¶¨Ô½´ó
C£®Õý·´Ó¦½øÐеÃÔ½ÍêÈ«       
D£®»¯Ñ§·´Ó¦ËÙÂÊÔ½´ó
£¨2£©1998ÄêÏ£À°ÑÇÀï˹¶àµÂ´óѧµÄÁ½Î»¿Æѧ¼Ò²ÉÓøßÖÊ×Óµ¼µçÐԵĠSCYÌÕ´É£¨ÄÜ´«µÝH+£©£¬ÊµÏÖÁ˸ßγ£Ñ¹Ï¸ßת»¯Âʵĵç½âºÏ³É°±¡£ÆäʵÑé×°ÖÃÈçͼ¡£Òõ¼«µÄµç¼«·´Ó¦Ê½
____________________¡£
£¨3£©¸ù¾Ý×îС°È˹¤¹Ìµª¡±µÄÑо¿±¨µÀ£¬ÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬ N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢ÉúÏÂÁз´Ó¦£ºN2(g) + 3H2O(1)2NH3(g) + O2(g)¡£¡÷H = a kJ¡¤mol-1 ½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬³£Ñ¹Ï´ﵽƽºâʱ²âµÃ²¿·ÖʵÑéÊý¾ÝÈçϱí
¢Ù´ËºÏ³É·´Ó¦µÄa______0£»¦¤S_______0£¬£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©
¢ÚÒÑÖª£ºN2(g) + 3H2(g)2NH3(g)  ¦¤H= £­92 .4kJ¡¤mol-1        
2H2(g) + O2(g) = 2H2O(l) = £­571.6kJ¡¤mol-1
ÔòN2(g) + 3H2O(l) = 2NH3(g) +  O2(g) ¦¤H=_____________kJ¡¤mol-1

 ¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦ÓÚ¡°È˹¤¹Ìµª¡±µÄз½·¨Ñо¿¡£

£¨1£©Ä¿Ç°ºÏ³É°±¼¼ÊõÔ­ÀíΪ£ºN2(g) + 3H2(g)2NH3(g)£»

¡÷H=¡ª92.4kJ¡¤mol¡ª1¡£

¢Ù 673K£¬30MPaÏ£¬ÉÏÊöºÏ³É°±·´Ó¦ÖÐn(NH3)ºÍn(H2)Ëæ

ʱ¼ä±ä»¯µÄ¹ØϵÈçÓÒͼËùʾ¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ        ¡£

A£®µãaµÄÕý·´Ó¦ËÙÂʱȵãbµÄ´ó

B£®µãc´¦·´Ó¦´ïµ½Æ½ºâ

C£®µãdºÍµã e´¦µÄn(N2)Ïàͬ

D£®773K£¬30MPaÏ£¬·´Ó¦ÖÁt2ʱ¿Ì´ïµ½Æ½ºâ£¬Ôòn(NH3)±ÈͼÖÐeµãµÄÖµ´ó

¢Ú ÔÚÈÝ»ýΪ2.0 LºãÈݵÄÃܱÕÈÝÆ÷ÖгäÈë0.80 mol N2(g)ºÍ1.60 mol H2(g)£¬·´Ó¦ÔÚ673K¡¢30MPaÏ´ﵽƽºâʱ£¬NH3µÄÌå»ý·ÖÊýΪ20%¡£¸ÃÌõ¼þÏ·´Ó¦N2(g) + 3H2(g) 2NH3(g)µÄƽºâ³£ÊýK=        ¡£KÖµÔ½´ó£¬±íÃ÷·´Ó¦´ïµ½Æ½ºâʱ       ¡££¨Ìî±êºÅ£©¡£

A£®»¯Ñ§·´Ó¦ËÙÂÊÔ½´ó  B£®NH3µÄ²úÁ¿Ò»¶¨Ô½´ó  C£®Õý·´Ó¦½øÐеÃÔ½ÍêÈ«

£¨2£©1998ÄêÏ£À°ÑÇÀï˹¶àµÂ´óѧµÄÁ½Î»¿Æѧ¼Ò²ÉÓøßÖÊ×Óµ¼µçÐÔ

µÄ SCYÌÕ´É£¨ÄÜ´«µÝH+£©£¬ÊµÏÖÁ˸ßγ£Ñ¹Ï¸ßת»¯Âʵĵç

½âºÏ³É°±¡£ÆäʵÑé×°ÖÃÈçͼ£¬Òõ¼«µÄµç¼«·´Ó¦Ê½       ¡£

£¨3£©¸ù¾Ý×îС°È˹¤¹Ìµª¡±µÄÑо¿±¨µÀ£¬ÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬

N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢ÉúÏÂÁз´Ó¦£º

N2(g)+ 3H2O(1)  2NH3(g)+ O2(g) ¡÷H = a kJ¡¤mol¡ª1

½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬³£Ñ¹Ï´ﵽƽºâʱ²âµÃ²¿·ÖʵÑéÊý¾ÝÈçÏÂ±í£º

T/K

303

313

323

NH3Éú³ÉÁ¿/£¨10-6mol£©

4.8

5.9

6.0

 

¢Ù´ËºÏ³É·´Ó¦µÄa       0£»¦¤S      0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©£»¸Ã·´Ó¦ÊôÓÚ        

A£®Ò»¶¨×Ô·¢  B£®Ò»¶¨²»×Ô·¢   C£®¸ßÎÂ×Ô·¢   D£®µÍÎÂ×Ô·¢

¢ÚÒÑÖª£ºN2(g)+ 3H2(g)2NH3(g) ¦¤H= £­92 .4kJ¡¤mol¡ª1

       2H2(g) + O2(g) = 2H2O(l) = £­571.6kJ¡¤mol¡ª1

ÔòN2(g)+ 3H2O(1) = 2NH3(g) + O2(g)¦¤H=      kJ¡¤mol¡ª1¡£

 

 ¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦ÓÚ¡°È˹¤¹Ìµª¡±µÄз½·¨Ñо¿¡£

£¨1£©Ä¿Ç°ºÏ³É°±¼¼ÊõÔ­ÀíΪ£ºN2(g) + 3H2(g)2NH3(g)£»

¡÷H=¡ª92.4kJ¡¤mol¡ª1¡£

¢Ù 673K£¬30MPaÏ£¬ÉÏÊöºÏ³É°±·´Ó¦ÖÐn(NH3)ºÍn(H2)Ëæ

ʱ¼ä±ä»¯µÄ¹ØϵÈçÓÒͼËùʾ¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ        ¡£

A£®µãaµÄÕý·´Ó¦ËÙÂʱȵãbµÄ´ó

B£®µãc´¦·´Ó¦´ïµ½Æ½ºâ

C£®µãdºÍµã e´¦µÄn(N2)Ïàͬ

D£®773K£¬30MPaÏ£¬·´Ó¦ÖÁt2ʱ¿Ì´ïµ½Æ½ºâ£¬Ôòn(NH3)±ÈͼÖÐeµãµÄÖµ´ó

¢Ú ÔÚÈÝ»ýΪ2.0 LºãÈݵÄÃܱÕÈÝÆ÷ÖгäÈë0.80 mol N2(g)ºÍ1.60 mol H2(g)£¬·´Ó¦ÔÚ673K¡¢30MPaÏ´ﵽƽºâʱ£¬NH3µÄÌå»ý·ÖÊýΪ20%¡£¸ÃÌõ¼þÏ·´Ó¦N2(g) + 3H2(g)  2NH3(g)µÄƽºâ³£ÊýK=        ¡£KÖµÔ½´ó£¬±íÃ÷·´Ó¦´ïµ½Æ½ºâʱ       ¡££¨Ìî±êºÅ£©¡£

A£®»¯Ñ§·´Ó¦ËÙÂÊÔ½´ó   B£®NH3µÄ²úÁ¿Ò»¶¨Ô½´ó  C£®Õý·´Ó¦½øÐеÃÔ½ÍêÈ«

£¨2£©1998ÄêÏ£À°ÑÇÀï˹¶àµÂ´óѧµÄÁ½Î»¿Æѧ¼Ò²ÉÓøßÖÊ×Óµ¼µçÐÔ

µÄ SCYÌÕ´É£¨ÄÜ´«µÝH+£©£¬ÊµÏÖÁ˸ßγ£Ñ¹Ï¸ßת»¯Âʵĵç

½âºÏ³É°±¡£ÆäʵÑé×°ÖÃÈçͼ£¬Òõ¼«µÄµç¼«·´Ó¦Ê½        ¡£

£¨3£©¸ù¾Ý×îС°È˹¤¹Ìµª¡±µÄÑо¿±¨µÀ£¬ÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬

N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢ÉúÏÂÁз´Ó¦£º

N2(g) + 3H2O(1)  2NH3(g) + O2(g)  ¡÷H = a kJ¡¤mol¡ª1

½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬³£Ñ¹Ï´ﵽƽºâʱ²âµÃ²¿·ÖʵÑéÊý¾ÝÈçÏÂ±í£º

T/K

303

313

323

NH3Éú³ÉÁ¿/£¨10-6mol£©

4.8

5.9

6.0

 

¢Ù´ËºÏ³É·´Ó¦µÄa       0£»¦¤S      0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©£»¸Ã·´Ó¦ÊôÓÚ        

A£®Ò»¶¨×Ô·¢   B£®Ò»¶¨²»×Ô·¢   C£®¸ßÎÂ×Ô·¢    D£®µÍÎÂ×Ô·¢

¢ÚÒÑÖª£ºN2(g) + 3H2(g)2NH3(g)  ¦¤H= £­92 .4kJ¡¤mol¡ª1

        2H2(g) + O2(g) = 2H2O(l) = £­571.6kJ¡¤mol¡ª1

ÔòN2(g) + 3H2O(1) = 2NH3(g) + O2(g) ¦¤H=      kJ¡¤mol¡ª1¡£

 

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

¾«Ó¢¼Ò½ÌÍø