ÌâÄ¿ÄÚÈÝ

ÔÚÊ¢ÓÐÉÙÁ¿ÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬¼ÓÈëһС¿éÐÂÇеġ¢²Á¸É±í̼̾Ó͵ĽðÊôÄÆ£¬Ñ¸ËÙÓÃÅäÓе¼¹ÜµÄµ¥¿×ÈûÈûסÊԹܿڣ¬ÓÃһСÊÔ¹ÜÊÕ¼¯²¢Ñé´¿ÆøÌåºó£¬µãȼ£¬²¢°Ñ¸ÉÔïµÄСÉÕ±­ÕÖÔÚ»ðÑæÉÏ£¬Æ¬¿Ì£¬Ñ¸ËÙµ¹×ªÉÕ±­£¬ÏòÉÕ±­ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®¡£

¹Û²ìÏÖÏó£¬Íê³ÉÏÂ±í¡£

ÒÒ´¼ÓëÄÆ·´Ó¦µÄÏÖÏó

ÆøÌåȼÉÕµÄÏÖÏó

¼ìÑé²úÎï

                   

                   

                   

ÆøÌåȼÉÕʱ»ðÑæ³Ê         £¬

СÉÕ±­ÄÚ±Ú        £¬

³ÎÇåʯ»ÒË®        ¡£

·´Ó¦ÖÐÖ»Éú³ÉÁË

       

д³öÒÒ´¼ÓëÄÆ·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                      

 

¡¾´ð°¸¡¿

£¨10·Ö£©

ÒÒ´¼ÓëÄÆ·´Ó¦µÄÏÖÏó

ÆøÌåȼÉÕµÄÏÖÏó

¼ìÑé²úÎï

¢Ù  Na³Áµ½ÊԹܵײ¿

¢Ú ·´Ó¦ÓÐÆøÅݲúÉú

¢Û     ·´Ó¦Ê±·Å³öÈÈÁ¿

£¨3µã´ð³öÁ½µã¼´¿É£¬Ã¿µã2·Ö£©

À¶É«£¨1·Ö£©

ÓÐË®Ö飨»òÓÐË®Îí£©£¨1·Ö£©

²»±ä»ë×Ç£¨1·Ö£©             

H2   £¨1·Ö£©

2CH3CH2OH + 2Na ¡ú 2CH3CH2ONa + H2¡ü£¨2·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Ϊ¶¨ÐÔ̽¾¿ÒÒ´¼µÄ·Ö×ӽṹ£¬Ä³»¯Ñ§¿ÎÍâС×éÉè¼ÆÁËÈçÏÂʵÑé·½°¸£ºÔÚÊ¢ÓÐÉÙÁ¿ÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬¼ÓÈëÒ»Á£²Á¸ÉúÓ͵ĽðÊôÄÆ£¬ÔÚÊԹܿÚѸËÙÈûÉÏÅäÓÐÒ½ÓÃ×¢ÉäÕëÍ·µÄµ¥¿×Èû£¬µãȼ·Å³öµÄÆøÌ壬²¢°ÑÒ»¸ÉÔïµÄСÉÕ±­ÕÖÔÚ»ðÑæÉÏ£¨Èçͼ£©£¬ÔÚÉÕ±­±ÚÉϳöÏÖÒºµÎºó£¬Ñ¸ËÙµ¹×ªÉÕ±­£¬ÏòÉÕ±­ÖмÓÈëÉÙÁ¿µÄ³ÎÇåʯ»ÒË®£¬¹Û²ìÓÐÎÞ»ì×Ç£®
£¨1£©ÒÔÉÏʵÑéÉè¼Æ´æÔÚÖØ´ó°²È«Òþ»¼£¬ÇëÄã°ïËûÃÇÖ¸³öÀ´
µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È
µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È
£®
£¨2£©È·ÈÏËùÓÃÒÒ´¼ÎªÎÞË®ÒÒ´¼µÄ·½·¨ÊÇ
ÏòÒÒ´¼ÖмÓÈëÎÞË®ÁòËáÍ­²»±äΪÀ¶É«
ÏòÒÒ´¼ÖмÓÈëÎÞË®ÁòËáÍ­²»±äΪÀ¶É«
£®
£¨3£©ÈôÏòÉÕ±­ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®ºó·¢ÏÖÓлì×Ç£¬ÄÇôȼÉÕ²úÉúCO2µÄÎïÖÊ×î¿ÉÄÜÊÇ
ÒÒ´¼£¨ÕôÆø£©
ÒÒ´¼£¨ÕôÆø£©
£®
£¨4£©ÈôÏòÉÕ±­ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®ºóδ·¢ÏÖ»ì×Ç£¬Ôò¿ÉÍÆ¶ÏÒÒ´¼·Ö×ӽṹÖк¬ÓÐ
²»Í¬ÓÚÌþ·Ö×ÓÀïµÄÇâÔ­×Ó´æÔÚ£¨»ò»îÆÃÇâÔ­×Ó¡¢ôÇ»ùµÈ£©
²»Í¬ÓÚÌþ·Ö×ÓÀïµÄÇâÔ­×Ó´æÔÚ£¨»ò»îÆÃÇâÔ­×Ó¡¢ôÇ»ùµÈ£©
£®
´ËÌâ·ÖA¡¢BÌ⣬ÈÎѡһÌâ´ðÌ⣬Á½Ìâ¶¼×ö·ÖÊýÒÔAÌâΪ׼£¬BÌâ½ÏÈÝÒ×£®
AÌ⣺ÔÚÊ¢ÓÐÉÙÁ¿ÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬¼ÓÈëһС¿éÐÂÇеġ¢²Á¸É±í̼̾Ó͵ĽðÊôÄÆ£¬Ñ¸ËÙÓÃÅäÓе¼¹ÜµÄµ¥¿×ÈûÈûסÊԹܿڣ¬ÓÃһСÊÔ¹ÜÊÕ¼¯²¢Ñé´¿ÆøÌåºó£¬µãȼ£¬²¢°Ñ¸ÉÔïµÄСÉÕ±­ÕÖÔÚ»ðÑæÉÏ£¬Æ¬¿Ì£¬Ñ¸ËÙµ¹×ªÉÕ±­£¬ÏòÉÕ±­ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®£®¹Û²ìÏÖÏó£¬Íê³ÉÏÂ±í£®
£¨1£©
ÒÒ´¼ÓëÄÆ·´Ó¦µÄÏÖÏó ÆøÌåȼÉÕµÄÏÖÏó ¼ìÑé²úÎï
¢Ù
Na³Áµ½ÊԹܵײ¿
Na³Áµ½ÊԹܵײ¿

¢Ú
·´Ó¦ÓÐÆøÅݲúÉú
·´Ó¦ÓÐÆøÅݲúÉú

¢Û
·´Ó¦Ê±·Å³öÈÈÁ¿
·´Ó¦Ê±·Å³öÈÈÁ¿
À¶É«
À¶É«

ÓÐË®Ö飨»òÓÐË®Îí£©
ÓÐË®Ö飨»òÓÐË®Îí£©

²»±ä»ë×Ç
²»±ä»ë×Ç
H2
H2
£¨2£©Ð´³öÒÒ´¼ÓëÄÆ·´Ó¦µÄ»¯Ñ§·½³Ìʽ
2CH3CH2OH+2Na¡ú2CH3CH2ONa+H2¡ü
2CH3CH2OH+2Na¡ú2CH3CH2ONa+H2¡ü

BÌ⣺Èçͼ£¬Ä³ÆøÌåX¿ÉÄÜÓÉH2¡¢CO¡¢CH4ÖеÄÒ»ÖÖ×é³É£®½«XÆøÌåȼÉÕ£¬°ÑȼÉÕºóÉú³ÉµÄÆøÌåͨ¹ýA¡¢BÁ½¸öÏ´ÆøÆ¿£®ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÈôAÏ´ÆøÆ¿µÄÖÊÁ¿Ôö¼Ó£¬BÏ´ÆøÆ¿µÄÖÊÁ¿²»±ä£¬ÔòÆøÌåXÊÇ
H2
H2
£®
£¨2£©ÈôAÏ´ÆøÆ¿µÄÖÊÁ¿²»±ä£¬BÏ´ÆøÆ¿µÄÖÊÁ¿Ôö¼Ó£¬ÔòÆøÌåXÊÇ
CO
CO
£®
£¨3£©ÈôA¡¢BÁ½¸öÏ´ÆøÆ¿µÄÖÊÁ¿¶¼Ôö¼Ó£¬ÔòÆøÌåX¿ÉÄÜÊÇ
CH4£»H2¡¢CO£»CH4¡¢H2£»CH4¡¢CO¡¢CH4¡¢H2¡¢CO
CH4£»H2¡¢CO£»CH4¡¢H2£»CH4¡¢CO¡¢CH4¡¢H2¡¢CO
£®
Ϊ¶¨ÐÔ̽¾¿ÒÒ´¼µÄÐÔÖÊ£¬Ä³»¯Ñ§¿ÎÍâС×éͨ¹ý²éÔÄ×ÊÁϲ¢Éè¼ÆÁËʵÑé·½°¸½øÐÐ̽¾¿£®
·½°¸¢ñ£ºÔÚÊ¢ÓÐÉÙÁ¿ÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬¼ÓÈëÒ»Á£³ýȥúÓ͵ĽðÊôÄÆ£¬ÔÚÊԹܿÚѸËÙÈûÉÏÅäÓмâ×ìµ¼¹ÜµÄµ¥¿×Èû£¬µãȼ·Å³öµÄÆøÌ壬²¢°ÑÒ»¸ÉÔïµÄСÉÕ±­ÕÖÔÚ»ðÑæÉÏ£¬ÔÚÉÕ±­±ÚÉϳöÏÖÒºµÎºó£¬Ñ¸ËÙµ¹×ªÉÕ±­£¬ÏòÉÕ±­ÖмÓÈëÉÙÁ¿µÄ³ÎÇåʯ»ÒË®£¬¹Û²ìÓÐÎÞ»ì×DzúÉú£®
£¨1£©Çëд³öÒÒ´¼ÓëÄÆ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü
2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü
£®
£¨2£©ÒÔÉÏʵÑéÉè¼ÆÒòȱÉÙ±ØÒªµÄ²½Öè¶ø´æÔÚ°²È«Òþ»¼£¬ÇëÄãÖ¸³öËùȱÉÙ±ØÒªµÄ²½ÖèÊÇ
µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È
µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È
£®
£¨3£©ÈôÏòÉÕ±­ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®ºó·¢ÏÖÓлì×Ç£¬ÔòȼÉÕ²úÉúCO2µÄÎïÖÊ×î¿ÉÄÜÊÇ
ÒÒ´¼£¨ÕôÆø£©
ÒÒ´¼£¨ÕôÆø£©
£¨Ð´Ãû³Æ£©£®
·½°¸¢ò£º£¨1£©È¡Ò»¸ùÍ­Ë¿£¬°ÑÆäÖÐÒ»¶ËÈÆ³ÉÂÝÐý×´£¨Ôö´ó½Ó´¥Ãæ»ý£©£®µãȼһյ¾Æ¾«µÆ£¬°ÑÈÆ³ÉÂÝÐý×´Ò»¶ËµÄÍ­Ë¿ÒÆµ½¾Æ¾«µÆÍâÑæÉÏׯÉÕ£¨ÈçÓÒͼ£©£¬¹Û²ìµ½µÄʵÑéÏÖÏó£º
Í­Ë¿ÓÉ×ϺìÉ«±äºÚ
Í­Ë¿ÓÉ×ϺìÉ«±äºÚ
£®
 £¨2£©°ÑÂÝÐý×´Í­Ë¿Íù¾Æ¾«µÆÄÚÑæÒÆ¶¯£¬¹Û²ìµ½µÄʵÑéÏÖÏó£º
Í­Ë¿ÓɺÚÉ«±ä³É×ϺìÉ«
Í­Ë¿ÓɺÚÉ«±ä³É×ϺìÉ«
£¬Óû¯Ñ§·½³Ìʽ±íʾ¸ÃÏÖÏó²úÉúµÄÔ­Àí£º
CH3CH2OH+CuO
¡÷
CH3CHO+Cu+H2O
CH3CH2OH+CuO
¡÷
CH3CHO+Cu+H2O
£®

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

¾«Ó¢¼Ò½ÌÍø