ÌâÄ¿ÄÚÈÝ


ÔÚÌå»ý¿É±äµÄÈÝÆ÷Öз¢Éú·´Ó¦N2+3H2⇌NH3µ±Ôö´óѹǿʹÈÝÆ÷Ìå»ýËõСʱ£¬»¯Ñ§·´Ó¦ËÙÂʼӿ죬ÆäÖ÷ÒªÔ­ÒòÊÇ£¨¡¡¡¡£©

¡¡

A£®

·Ö×ÓÔ˶¯ËÙÂʼӿ죬ʹ·´Ó¦Îï·Ö×Ó¼äµÄÅöײ»ú»áÔö¶à

¡¡

B£®

·´Ó¦Îï·Ö×ÓµÄÄÜÁ¿Ôö¼Ó£¬»î»¯·Ö×Ó°Ù·ÖÊýÔö´ó£¬ÓÐЧÅöײ´ÎÊýÔö¶à

¡¡

C£®

»î»¯·Ö×Ó°Ù·ÖÊýδ±ä£¬µ«µ¥Î»Ìå»ýÄڻ·Ö×ÓÊýÔö¼Ó£¬ÓÐЧÅöײ´ÎÊýÔö¶à

¡¡

D£®

·Ö×Ó¼ä¾àÀë¼õС£¬Ê¹ËùÓеĻ·Ö×Ó¼äµÄÅöײ¶¼³ÉΪÓÐЧÅöײ

¡¡

D£®

·Ö×Ó¼ä¾àÀë¼õС£¬Ê¹ËùÓеĻ·Ö×Ó¼äµÄÅöײ¶¼³ÉΪÓÐЧÅöײ


¿¼µã£º

»î»¯Äܼ°Æä¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ï죮

רÌ⣺

»¯Ñ§·´Ó¦ËÙÂÊרÌ⣮

·ÖÎö£º

Ôö´óѹǿʹÈÝÆ÷Ìå»ýËõСʱ£¬µ¥Î»Ìå»ýÄÚ·´Ó¦ÎïŨ¶ÈÔö´ó£¬ËùÒÔµ¥Î»Ìå»ýÄڻ·Ö×ÓÊýÔö¼Ó£¬ÓÐЧÅöײ´ÎÊýÔö¶à£¬·´Ó¦ËÙÂʼӿ죬µ«Êǻ·Ö×Ó°Ù·ÖÊýδ±ä£®

½â´ð£º

½â£ºA£®ÓÉÓÚζȲ»±ä£¬ËùÒÔ·Ö×ÓÔ˶¯ËÙÂʲ»±ä£¬¹ÊA´íÎó£»

B£®ÓÉÓÚζȲ»±ä£¬ËùÒÔ·´Ó¦Îï·Ö×ÓµÄÄÜÁ¿²»±ä£¬¹ÊB´íÎó£»

C£®´óѹǿʹÈÝÆ÷Ìå»ýËõСʱ£¬µ¥Î»Ìå»ýÄÚ·´Ó¦ÎïŨ¶ÈÔö´ó£¬ËùÒÔµ¥Î»Ìå»ýÄڻ·Ö×ÓÊýÔö¼Ó£¬ÓÐЧÅöײ´ÎÊýÔö¶à£¬·´Ó¦ËÙÂʼӿ죬µ«Êǻ·Ö×Ó°Ù·ÖÊýδ±ä£¬¹ÊCÕýÈ·£»

D£®»î»¯·Ö×Ó¼äµÄÅöײ²»Ò»¶¨ÊÇÓÐЧÅöײ£¬Ö»ÓÐÈ¡ÏòºÏÊʲÅÄÜ·¢Éú·´Ó¦£¬²ÅÊÇÓÐЧÅöײ£¬¹ÊD´íÎó£®

¹ÊÑ¡C£®

µãÆÀ£º

±¾Ì⿼²éÓ°Ïì»î»¯·Ö×ÓµÄÒòËØ£¬ÌâÄ¿ÄѶȲ»´ó£¬×¢ÒâÍâ½çÌõ¼þ¶Ô»î»¯·Ö×ÓµÄÓ°Ï첻ͬ£¬°ÑÎÕÏà¹Ø»ù´¡ÖªÊ¶µÄ»ýÀÛ£®

¡¡

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ĿǰÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÌáÈ¡µÄ¡£º£Ë®ÌáþµÄÖ÷ÒªÁ÷³ÌÈçÏÂ:

Çë»Ø´ðÏÂÁÐÎÊÌâ:

(1)´ÓÀë×Ó·´Ó¦µÄ½Ç¶È˼¿¼,ÔÚº£Ë®ÖмÓÈëʯ»ÒÈéµÄ×÷ÓÃÊÇ¡¡               ,

д³öÔÚ³Áµí³ØÖз´Ó¦µÄÀë×Ó·½³Ìʽ                                    

¡¡                                                               ¡£

(2)ʯ»ÒÈéÊÇÉúʯ»ÒÓëË®ÐγɵϝºÏÎï,´Ó³ä·ÖÀûÓú£Ñó»¯Ñ§×ÊÔ´,Ìá¸ß¾­¼ÃÐ§ÒæµÄ½Ç¶È,Éú²úÉúʯ»ÒµÄÖ÷ÒªÔ­ÁÏÀ´Ô´ÓÚº£ÑóÖеġ¡¡¡¡¡¡¡¡£

(3)²Ù×÷AÊÇ¡¡¡¡¡¡¡¡,²Ù×÷BÊÇ¡¡¡¡¡¡¡¡¡£

(4)¼ÓÈëµÄ×ãÁ¿ÊÔ¼ÁaÊÇ¡¡¡¡¡¡¡¡(Ìѧʽ)¡£

(5)ÎÞË®MgCl2ÔÚÈÛÈÚ״̬ÏÂ,ͨµçºó»á²úÉúMgºÍCl2,¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ

¡¡                                             ¡£

´Ó¿¼Âdzɱ¾ºÍ·ÏÎïÑ­»·ÀûÓõĽǶÈ,¸±²úÎïÂÈÆø¿ÉÒÔÓÃÓÚ¡¡              ¡£

(6)º£Ë®ÌáþµÄ¹ý³Ì,ΪʲôҪ½«º£Ë®ÖеÄÂÈ»¯Ã¾×ª±äΪÇâÑõ»¯Ã¾,ÔÙת±äΪÂÈ»¯Ã¾?

¡¡                                                               ¡£

(7)ÓÐͬѧÈÏΪ:¿ÉÖ±½Ó¼ÓÈÈMg(OH)2µÃµ½MgO,ÔÙµç½âÈÛÈÚMgOÖÆ½ðÊôþ,ÕâÑù¿É¼ò»¯ÊµÑé²½Öè,ÌåÏÖʵÑéµÄ¼òÔ¼ÐÔÔ­Ôò¡£Äã¡¡¡¡¡¡¡¡(ÌͬÒ⡱»ò¡°²»Í¬Ò⡱)¸ÃͬѧµÄÏë·¨,ÀíÓÉÊÇ¡¡                                      ¡£


Æû³µÔÚÏÖ´úÉú»îÖаçÑÝ×ÅÔ½À´Ô½ÖØÒªµÄ½ÇÉ«£¬µ«ÆäÎ²Æø£¨Ì¼Ç⻯ºÏÎï¡¢µªÑõ»¯Îï¼°Ò»Ñõ»¯Ì¼µÈ£©´øÀ´µÄ»·¾³ÎÛȾԽÀ´Ô½Ã÷ÏÔ£¬»ú¶¯³µ·ÏÆøÅÅ·ÅÒѳÉΪ³ÇÊдóÆøÎÛȾµÄÖØÒªÀ´Ô´£®

£¨1£©ÔÚÆû³µÎ²ÆøÏµÍ³Öа²×°´ß»¯×ª»»Æ÷£¬¿ÉÓÐЧ¼õÉÙÎ²ÆøÖеÄCO¡¢NOx ºÍ̼Ç⻯ºÏÎïµÈ·ÏÆø£®

ÒÑÖª£ºN2£¨g£©+O2£¨g£©=2NO£¨g£©¡÷H1=+180kJ/mol

CO£¨g£©+O2£¨g£©=CO2£¨g£©¡÷H2=﹣283kJ/mol

2NO£¨g£©+2CO£¨g£©=2CO2£¨g£©+N2£¨g£©¡÷H3Ôò¡÷H3=¡¡¡¡kJ•mol﹣1£®

£¨2£©Æø¸×ÖÐÉú³ÉNOµÄ·´Ó¦Îª£ºN2£¨g£©+O2£¨g£©═2NO£¨g£©¡÷H£¾0

¢ÙÆû³µÆô¶¯ºó£¬Æø¸×ÄÚζÈÔ½¸ß£¬µ¥Î»Ê±¼äÄÚNOÅÅ·ÅÁ¿Ô½´ó£¬Ô­ÒòÊÇ¡¡¡¡£®

¢Ú1mol¿ÕÆøÖк¬ÓÐ0.8molN2ºÍ0.2molO2£¬1300¡æÊ±ÔÚº¬1mol¿ÕÆøµÄÃܱÕÈÝÆ÷ÄÚ·´Ó¦´ïµ½Æ½ºâ£¬²âµÃNOΪ8¡Á10﹣4mol£®¼ÆËã¸ÃζÈÏÂµÄÆ½ºâ³£ÊýK¡Ö¡¡¡¡£®

£¨3£©Î²ÆøÖеÄCOÖ÷ÒªÀ´×ÔÓÚÆûÓͲ»ÍêȫȼÉÕ£®

¢ÙÓÐÈËÉèÏë°´ÏÂÁз´Ó¦³ýÈ¥CO£º2CO£¨g£©=2C£¨s£©+O2£¨g£©¡÷H=+221kJ•mol﹣1£¬¼òÊö¸ÃÉèÏëÄÜ·ñʵÏÖµÄÒÀ¾Ý£º¡¡¡¡

¢Ú²âÁ¿Æû³µÎ²ÆøµÄŨ¶È³£ÓÃµç»¯Ñ§ÆøÃô´«¸ÐÆ÷£¬ÆäÖÐCO´«¸ÐÆ÷¿ÉÓÃÈçͼ1¼òµ¥±íʾ£¬ÔòÑô¼«·¢ÉúµÄµç¼«·´Ó¦Îª¡¡£®

£¨4£©Æû³µÎ²ÆøÖеÄCO¿Éת»¯Îª¼×Í飬·½³ÌʽΪCO£¨g£©+3H2£¨g£©⇌CH4£¨g£©+H2O£¨g£©£®ÆäËûÌõ¼þÏàͬʱ£¬H2µÄƽºâת»¯ÂÊÔÚ²»Í¬Ñ¹Ç¿ÏÂËæÎ¶ȵı仯Èçͼ2Ëùʾ£®

¢Ù¸Ã·´Ó¦µÄ¡÷H¡¡¡¡0£¨Ìî¡°£¼¡±¡¢¡°=¡±»ò¡°£¾¡±£©£®

¢Úʵ¼ÊÉú²úÖвÉÓÃͼÖÐMµã¶ø²»ÊÇNµã¶ÔÓ¦µÄ·´Ó¦Ìõ¼þ£¬ÔËÓû¯Ñ§·´Ó¦ËÙÂÊºÍÆ½ºâ֪ʶ£¬Í¬Ê±¿¼ÂÇÉú²úʵ¼Ê£¬ËµÃ÷Ñ¡Ôñ¸Ã·´Ó¦Ìõ¼þµÄÀíÓÉ¡¡¡¡£®

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

¾«Ó¢¼Ò½ÌÍø