ÌâÄ¿ÄÚÈÝ
(´´ÐÂ)¿ÉÄæ·´Ó¦£ºC(s)£«H2O(g)
CO(g)£«H2(g)ÔÚÒ»¶¨Î¶ÈÏ´ﵽƽºâºó£¬²âµÃ»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Îªa£®ÏÖÔö´óѹǿ£¬´ïµ½ÐÂµÄÆ½ºâ״̬ºó²âµÃ»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Îªb£®
(1)¢ÙÈôÔÆ½ºâÊÇÒÔC(s)ºÍH2O(g)ΪÆðʼ·´Ó¦ÎィÁ¢µÄ£¬ÔòaÓëbµÄ¹ØÏµÎª________£»
(2)¢ÚÈôÔÆ½ºâÊÇÒÔCO(g)ºÍH2(g)ΪÆðʼ·´Ó¦ÎィÁ¢µÄ£¬ÔòaÓëbµÄ¹ØÏµÎª________£®
A£®a£¾b
B£®a£¼b
C£®a£½b
D£®ÒÔÉÏÈýÖÖÇé¿ö¶¼ÓпÉÄÜ
(2)ÈôÔÆ½ºâÊÇÒÔCO(g)ºÍH2(g)ΪÆðʼ·´Ó¦ÎィÁ¢µÄ£¬ÇÒa£½b£¬ÔòÆðʼʱCOÓëH2µÄÌå»ý±ÈΪ________£®
½âÎö£º
|
|
£¨11·Ö£©Ñо¿ºÍ¿ª·¢CO2ºÍCOµÄ´´ÐÂÀûÓÃÊÇ»·¾³±£»¤ºÍ×ÊÔ´ÀûÓõÄ˫ӮµÄ¿ÎÌâ¡£
£¨1£©CO¿ÉÓÃÓںϳɼ״¼¡£ÔÚѹǿΪ0.1MpaÌõ¼þÏ£¬ÔÚÌå»ýΪbLµÄÃܱÕÈÝÆ÷ÖгäÈëamolCOºÍ2amolH2£¬ÔÚ´ß»¯¼Á×÷ÓÃϺϳɼ״¼£º
CO£¨g£©+2H2£¨g£©
CH3OH£¨g£©Æ½ºâʱCOµÄת»¯ÂÊÓëζȣ¬Ñ¹Ç¿µÄ¹ØÏµÈçÏÂͼ£º![]()
£¨i£©¸Ã·´Ó¦ÊôÓÚ_____________·´Ó¦£º£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£
£¨ii£©100¡æÊ±£¬¸Ã·´Ó¦µÄƽºâ³£Êý£ºK=_____________£»£¨ÓÃa¡¢bµÄ´úÊýʽ±íʾ£©¡£
ÈôÒ»¸ö¿ÉÄæ·´Ó¦µÄƽºâ³£ÊýKÖµºÜ´ó£¬¶Ô´Ë·´Ó¦µÄ˵·¨ÕýÈ·µÄÊÇ£º_________________ÌîÐòºÅ£©
| A£®¸Ã·´Ó¦Ê¹Óô߻¯¼ÁÒâÒå²»´ó£» |
| B£®¸Ã·´Ó¦·¢Éú½«ÔںܶÌʱ¼äÄÚÍê³É£» |
| C£®¸Ã·´Ó¦´ïµ½Æ½ºâʱÖÁÉÙÓÐÒ»ÖÖ·´Ó¦Îï°Ù·Öº¬Á¿ºÜС£» |
| D£®¸Ã·´Ó¦Ò»¶¨ÊÇ·ÅÈÈ·´Ó¦£» |
£¨iv£©ÔÚijζÈÏ£¬ÏòÒ»ÈÝ»ý²»±äµÄÃܱÕÈÝÆ÷ÖгäÈë2.5molCO£¬7.5molH2£¬·´Ó¦Éú³ÉCH3OH£¨g£©£¬´ïµ½Æ½ºâʱ£¬COת»¯ÂÊΪ90%£¬´ËʱÈÝÆ÷ÄÚѹǿΪ¿ªÊ¼Ê±µÄѹǿ__________±¶¡£
£¨2£©Ä³Î¶ÈÌõ¼þÏ£¬Èô½«CO2£¨g£©ºÍH2£¨g£©ÒÔÌå»ý±È1£º4»ìºÏ£¬ÔÚÊʵ±Ñ¹Ç¿ºÍ´ß»¯¼Á×÷ÓÃÏ¿ÉÖÆµÃ¼×Í飬ÒÑÖª£º
CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890.3KJ/mol
H2£¨g£©+1/2O2£¨g£©= H2O£¨l£© ¡÷H=-285.8KJ/mol
ÔòCO2£¨g£©ºÍH2£¨g£©·´Ó¦Éú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ£º_________________________¡£