ÌâÄ¿ÄÚÈÝ

¶ÔÓÚƽºâÌåϵ2SO2£¨g£©+O2£¨g£©2SO3£¨g£© ¡÷H<0¡£ÏÂÁнáÂÛÖÐÕýÈ·µÄÊÇ£¨   £©
A£®ÈôζȲ»±ä£¬½«ÈÝÆ÷µÄÌå»ýÔö´óÒ»±¶£¬´ËʱµÄSO2Ũ¶È±äΪԭÀ´µÄ0.5±¶
B£®ÈôƽºâʱSO2¡¢O2µÄת»¯ÂÊÏàµÈ£¬ËµÃ÷·´Ó¦¿ªÊ¼Ê±£¬Á½ÕßµÄÎïÖʵÄÁ¿Õâ±ÈΪ2£º1
C£®Èô´ÓƽºâÌåϵÖзÖÀë³öSO2£¬ÔòÓÐÀûÓÚÌá¸ßSO2µÄת»¯ÂʺͼӿìÕý·´Ó¦ËÙÂÊ
D£®Æ½ºâ״̬ʱSO2¡¢O2¡¢SO3µÄÎïÖʵÄÁ¿Ö®±ÈÒ»¶¨Îª2£º1£º2
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨10·Ö£©£¨1£©ÒÑÖª¿ÉÄæ·´Ó¦£ºM(g)£«N(g)P(g)£«Q(g) ¦¤H£¾0£¬

ÈôÒªÔö´óQµÄŨ¶È£¬ÔÚÆäËüÌõ¼þ²»±äµÄÇé¿öÏ¿ÉÒÔ²ÉÈ¡µÄ´ëʩΪ    £¨ÌîÐòºÅ£©¡£
A£®¼ÓÈëÒ»¶¨Á¿M     B£®½µµÍ·´Ó¦Î¶Ƞ    C£®Éý¸ß·´Ó¦Î¶Ƞ   
D£®ËõСÈÝÆ÷Ìå»ý     E£®¼ÓÈë´ß»¯¼Á       F£®·ÖÀë³öÒ»¶¨Á¿P
£¨2£©¸Ç˹¶¨ÂÉÔÚÉú²úºÍ¿ÆѧÑо¿ÖÐÓкÜÖØÒªµÄÒâÒå¡£ÓÐЩ·´Ó¦µÄ·´Ó¦ÈÈËäÈ»ÎÞ·¨Ö±½Ó²âµÃ£¬µ«¿Éͨ¹ý¼ä½ÓµÄ·½·¨²â¶¨¡£ÏÖ¸ù¾ÝÏÂÁÐ3¸öÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
Fe2O3(s)+3CO(g)=2Fe(s)+3CO2(g)       ¡÷H£½£­24.8 kJ¡¤mol£­1
3Fe2O3(s)+ CO(g)==2Fe3O4(s)+ CO2(g)   ¡÷H£½£­47.2 kJ¡¤mol£­1
Fe3O4(s)+CO(g)==3FeO(s)+CO2(g)      ¡÷H£½£«640.5 kJ¡¤mol£­1
д³öCOÆøÌ廹ԭFeO¹ÌÌåµÃµ½Fe ¹ÌÌåºÍCO2ÆøÌåµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
_________________                                                           ¡£
£¨3£©ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1 mol CO2ºÍ3 mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º CO2(g)+3H2(g)CH3OH(g)+H2O(g)  ¦¤H=£­49.0 kJ/mol£¬ ²â
µÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçÓÒͼËùʾ¡£
¢Ù´Ó3 minµ½10 min£¬v(H2)=     mol/(L¡¤min)¡£
¢ÚÄÜ˵Ã÷ÉÏÊö·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ         £¨Ñ¡Ìî±àºÅ£©¡£
A£®·´Ó¦ÖÐCO2ÓëCH3OHµÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ1©U1£¨¼´Í¼Öн»²æµã£©
B£®»ìºÏÆøÌåµÄÃܶȲ»Ëæʱ¼äµÄ±ä»¯¶ø±ä»¯
C£®µ¥Î»Ê±¼äÄÚÿÏûºÄ3 mol H2£¬Í¬Ê±Éú³É1 mol H2O
D£®CO2µÄÌå»ý·ÖÊýÔÚ»ìºÏÆøÌåÖб£³Ö²»±ä
¢ÛÏÂÁдëÊ©ÖÐÄÜʹn (CH3OH)/n (CO2)Ôö´óµÄÊÇ        £¨Ñ¡Ìî±àºÅ£©¡£
A£®Éý¸ßζȠ                     B£®ºãκãÈݳäÈëHe(g)
C£®½«H2O(g)´ÓÌåϵÖзÖÀë          D£®ºãκãÈÝÔÙ³äÈë1 mol CO2ºÍ3 mol H2

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

¾«Ó¢¼Ò½ÌÍø