Question

¡¾ÌâÄ¿¡¿ÒÑÖª25¡æ¡¢101kPaʱ£¬Ò»Ð©ÎïÖʵÄȼÉÕÈÈΪ£º

»¯Ñ§Ê½	CO(g)	H2(g)	CH3OH(l)	CH4(g)
¦¤H/(kJ/mol)	-283.0	-285.8	-726.51	-890. 31

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¸ù¾Ý¸Ç˹¶¨ÂÉÍê³ÉÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º

CO(g)+2H2(g)==CH3OH(l) ¦¤H=_________¡£

£¨2£©ÏÖÓÐH2ºÍCH4µÄ»ìºÏÆøÌå112 L£¨±ê×¼×´¿ö£©£¬Ê¹ÆäÍêȫȼÉÕÉú³ÉCO2(g)ºÍH2O(l)£¬¹²·Å³öÈÈÁ¿3 242.5 kJ£¬ÔòÔ­»ìºÏÆøÌåÖÐH2ºÍCH4µÄÎïÖʵÄÁ¿Ö®±ÈÊÇ£º____________¡£

A.1:1 B.1:3 C.1:4 D.2:3

Answer 1

¡¾´ð°¸¡¿-128.09 kJ/molD

¡¾½âÎö¡¿£¨1£©COȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£ºCO£¨g£©+1/2O2£¨g£©¨TCO2£¨g£©¡÷H=-283.0 kJ¡¤mol£­1 ¢Ù£¬H2ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H=-285.8¡Á2 kJ¡¤mol£­1 ¢Ú£¬CH3OHȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£ºCH3OH£¨l£©+3/2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-726.5 kJ¡¤mol£­1 ¢Û£¬½«¢Ù+¢Ú+£¨-¢Û£©¿ÉµÃ£ºCO£¨g£©+2H2£¨g£©¨TCH3OH£¨l£©¡÷H=-128.1KJ¡¤mol£­1

¹Ê´ð°¸Îª£º-128.1KJ¡¤mol£­1£»

(2)H2ºÍCH4µÄ»ìºÏÆøÌå112L£¨±ê×¼×´¿ö£©ÎïÖʵÄÁ¿Îª112L/22.4L¡¤mol£­1=5mol£¬ËùÒÔ»ìºÏÆøÌåµÄƽ¾ùȼÉÕÈÈΪΪ3242.5/5kJ¡¤mol£­1=648.5kJ¡¤mol£­1£¬ÓÉH2£¨g£©+1/2O2£¨g£©=H2O£¨1£©¡÷H=-285.8kJ¡¤mol£­1¿ÉÖª£¬ÇâÆøµÄȼÉÕÈÈΪ285.8kJ¡¤mol£­1£»ÓÉCH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨1£©¡÷H=-890.3kJ¡¤mol£­1¿ÉÖª£¬¼×ÍéµÄȼÉÕÈÈΪ890.3kJ¡¤mol£­1£®

¸ù¾ÝÊ®×Ö½»²æ·¨¼ÆËãÔ­»ìºÏÆøÌåÖÐH2ºÍCH4µÄÎïÖʵÄÁ¿Ö®±È£º

ËùÒÔÔ­»ìºÏÆøÌåÖÐH2ºÍCH4µÄÎïÖʵÄÁ¿Ö®±ÈΪ241.8kJ¡¤mol£­1£º362.7kJ¡¤mol£­1=2£º3£®

¹ÊÑ¡£ºD£®