¸ß¿¼»¯Ñ§ÖªÊ¶µã×ܽᡶ»¯Ñ§Æ½ºâ³£Êý¡·¸ßƵ¿¼µã¹®¹Ì£¨2017ÄêÄ£Äâ°æ£©(Ê®)

ʱ¼ä:2017-07-18 00:53:01

΢ÐÅËÑË÷¹Ø×¢"91¿¼ÊÔÍø"¹«ÖÚºÅ,Áì30Ôª,»ñÈ¡ÊÂÒµ±à½Ìʦ¹«ÎñÔ±µÈ¿¼ÊÔ×ÊÁÏ40G

1¡¢Ñ¡ÔñÌâ  ·´Ó¦N2O4£¨g£©¨T2NO2£¨g£©¡÷H=+57kJ?mol-1£¬ÔÚζÈΪT1¡¢T2ʱ£¬Æ½ºâÌåϵÖÐNO2µÄÌå»ý·ÖÊýËæÑ¹Ç¿±ä»¯ÇúÏßÈçͼËùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®A¡¢CÁ½µãµÄ·´Ó¦ËÙÂÊ£ºA£¾C
B£®A¡¢BÁ½µãÆøÌåµÄÑÕÉ«£ºAÉBdz
C£®T1£¾T2
D£®A¡¢CÁ½µãÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£ºA£¾C
91¿¼ÊÔÍø


²Î¿¼´ð°¸£ºB


±¾Ìâ½âÎö£º


±¾ÌâÄѶȣºÒ»°ã



2¡¢Ìî¿ÕÌâ  ÎÂÊÒÆøÌå¶þÑõ»¯Ì¼¼õÅŵÄÒ»ÖÖ·½·¨ÊÇ£º´Óȼúµç³§»òÌìÈ»Æøµç³§ÅÅ·ÅÆøÖлØÊÕCO2£¬ÔÙÓëCH4ÈȽ⯲úÉúµÄH2·´Ó¦Éú³ÉCH3OH£®
CO2£¨g£©+3H2£¨g£©

Ò»¶¨Ìõ¼þ
CH3OH£¨g£©+H2O£¨g£©
£¨1£©500¡æÊ±£¬ÔÚÌå»ýΪ1LµÄÈÝ»ý¹Ì¶¨µÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1molCO2ºÍ3molH2£¬²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæÊ±¼ä±ä»¯Èçͼ1Ëùʾ£®

¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊv£¨H2£©=______mol?L-1?min-1£®
¢Ú¸Ã·´Ó¦µÄƽºâ³£ÊýK=______£¨±£ÁôһλСÊý£©£¬Æ½ºâʱH2µÄת»¯ÂÊΪ______£®
¢ÛÒÑ֪ζÈÉý¸ß£¬KÖµ¼õС£®ÏÂÁдëÊ©ÖÐÄÜʹ
n(CH3OH)
n(CO2)
Ôö´óµÄÊÇ______£®
A£®Éý¸ßζÈB£®³äÈëHe£¨g£©£¬Ê¹ÌåϵѹǿÔö´ó
C£®½«H2O£¨g£©´ÓÌåϵÖзÖÀëD£®ÔÙ³äÈë1molCO2ºÍ3molH2
£¨2£©CH3OH×÷ΪÄÚȼ»úȼÁÏ»òͨ¹ýȼÁÏµç³ØÇý¶¯³µÁ¾£®¿É¼õÉÙCO2ÅŷŶà´ï45%£®
¢ÙÒÑÖª£º
CH3OH£¨g£©+H2O£¨g£©¨TCO2£¨g£©+3H2£¨g£©¡÷H1=+49.0kJ?mol-1
2H2£¨g£©+O2£¨g£©¨T2H2O£¨g£©¡÷H=-483.6kJ?mol-1
Ôò·´Ó¦CH3OH£¨g£©+
1
2
O2£¨g£©¨TCO2£¨g£©+2H2£¨g£©µÄ¡÷H=______kJ?mol-1£®
¢Ú¼×´¼ÖÊ×Ó½»»»Ä¤È¼ÁÏµç³ØµÄ½á¹¹Ê¾ÒâͼÈçͼ2£¬ÒÑÖªH+ÒÆÏòͨÈëO2µÄµç¼«£¬¼×´¼½øÈë______¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬Õý¼«·¢ÉúµÄµç¼«·´Ó¦Îª______£®


²Î¿¼´ð°¸£º£¨1£©¢ÙÓÉͼ¿ÉÖª£¬10minµ½´ïƽºâ£¬Æ½ºâʱ¼×´¼µÄŨ¶È±ä»¯Îª0.75mol/L£¬¹Êv£¨CH3OH£©=0.75mol/L10min=0.075mol/£¨L£®min£©£¬ËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬¹Êv£¨H2£©=3v£¨CH3OH£©=0.075mol/£¨L£®min£©¡Á3=0.225mol/£¨L£®mon£©£¬
¹Ê´ð°¸Îª£º0.225£»
¢ÚCO2£¨g£©+3H2£¨g£©Ò»¶¨Ìõ¼þ


±¾Ìâ½âÎö£º


±¾ÌâÄѶȣºÒ»°ã



3¡¢Ìî¿ÕÌâ  PCl5µÄÈȷֽⷴӦÈçÏ£ºPCl5£¨g£©?PCl3£¨g£©+Cl2£¨g£©
£¨1£©¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ£ºK=______£®
£¨2£©ÈôijζÈÏ£¬ÔÚÈÝ»ýΪ1.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.20mol?PCl5£¬´ïµ½Æ½ºâºó£¬²âµÃÈÝÆ÷ÄÚPCl3µÄŨ¶ÈΪ0.15mol/L£¬
¢Ùƽºâʱ£¬PCl5µÄŨ¶ÈΪ£º______
¢Úƽºâʱ£¬Cl2µÄŨ¶ÈΪ£º______
¢Û¸ÃζÈÏÂµÄÆ½ºâ³£ÊýΪ£º______£®
£¨3£©ÏàͬÌõ¼þÏ£¬ÔÚÈÝ»ýΪ1.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.20mol?PCl3ºÍ0.20mol?Cl2£¬´ïµ½Æ½ºâºó£¬Æäƽºâ³£ÊýΪ£º______£®£¨±£ÁôÁ½Î»Ð¡Êý£©?
£¨4£©Ä³Î¶ÈÏ£¬ºÏ³É°±·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-92.4kJ/molÔÚ¸ÃζÈÏ£¬½«1mol?N2ºÍ3mol?H2·ÅÈëÃܱÕÈÝÆ÷ÖУ¬ÔÚÓд߻¯¼Á´æÔÚµÄÌõ¼þϳä·Ö·´Ó¦£¬²âµÃ·´Ó¦·Å³öµÄÈÈÁ¿×ÜÊÇСÓÚ92.4kJ£¬ÆäÔ­ÒòÊÇ£º______£®


²Î¿¼´ð°¸£º£¨1£©¿ÉÄæ·´Ó¦PCl5£¨g £©?PCl3£¨g£©+Cl2£¨g£©µÄƽºâ³£Êýk=c(PCl3)?c(Cl2)c(PCl5)£¬
¹Ê´ð°¸Îª£ºc(PCl3)?c(Cl2)c(PCl5)£»
£¨2£©Æ½ºâºóPCl3µÄŨ¶ÈΪ0.15mol/L£¬Ôò£º
? ? PCl5£¨g£©?PCl3£¨g£©+Cl2£¨g£©
¿ªÊ¼£¨mol/L£©£º0.2? 0? 0
±ä»¯£¨mol/L£©£º0.15? 0.15? 0.15
ƽºâ£¨mol/L£©£º0.05? 0.15?0.15
¢ÙÓÉÉÏÊö¼ÆËã¿ÉÖª£¬Æ½ºâʱ£¬PCl5µÄŨ¶ÈΪ0.05mol/L£¬
¹Ê´ð°¸Îª£º0.05mol/L£»
¢ÚÓÉÉÏÊö¼ÆËã¿ÉÖª£¬Æ½ºâʱ£¬Cl2µÄŨ¶ÈΪ0.15mol/L£¬
¹Ê´ð°¸Îª£º0.15mol/L£»
¢Û¹Ê¸ÃζÈÏÂÆ½ºâ³£Êýk=c(PCl3)?c(Cl2)c(PCl5)=0.15¡Á0.150.05=0.45£¬
¹Ê´ð°¸Îª£º0.45£»
£¨3£©¶ÔÓÚͬһ¿ÉÄæ·´Ó¦£¬ÔÚÏàͬζÈÏ£¬Õý¡¢ÄæÁ½¸ö¹ý³ÌµÄƽºâ³£Êý»¥Îªµ¹Êý£¬¹Ê PCl3£¨g£©+Cl2£¨g£©?PCl5£¨g£©£¬ÔÚÏàͬζÈÏÂµÄÆ½ºâ³£Êýk=10.45=2.22£¬
¹Ê´ð°¸Îª£º2.22£»
£¨4£©ÈÈ»¯Ñ§·½³ÌʽΪN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-92.4kJ/mol±íʾ1molµªÆø£¨g£©Óë3molÇâÆø£¨g£©Éú³É2mol°±Æø£¨g£©·´Ó¦µÄÈÈÁ¿Îª92.4kJ£¬ÓÉÓڸ÷´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬¼ÓÈë1molN2ºÍ3molH2²»¿ÉÄÜÍêÈ«·´Ó¦£¬ËùÒԷųöµÄÈÈÁ¿×ÜÊÇСÓÚ92.4kJ£¬
¹Ê´ð°¸Îª£º¸Ã·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬1molN2ºÍ3molH2²»¿ÉÄÜÍêÈ«·´Ó¦£¬ËùÒԷųöµÄÈÈÁ¿×ÜÊÇСÓÚ92.4kJ£®


±¾Ìâ½âÎö£º


±¾ÌâÄѶȣºÒ»°ã



4¡¢Ñ¡ÔñÌâ  ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©£¬ÆäËûÌõ¼þ²»±ä£¬ÔÚ300¡æºÍ500¡æÊ±£¬ÎïÖʵÄÁ¿n£¨CH3OH£©-·´Ó¦Ê±¼ätµÄ±ä»¯ÇúÏßÈçͼ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¸Ã·´Ó¦µÄ¡÷H£¼O
B£®ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȷ´Ó¦µÄƽºâ³£ÊýÔö´ó
C£®300¡æÊ±£¬O-t1minÄÚCH3OHµÄƽ¾ùÉú³ÉËÙÂÊΪmol?L-1?min?-1
D£®AµãµÄ·´Ó¦Ìåϵ´Ó300¡æÉý¸ßµ½500¡æ£¬´ïµ½Æ½ºâʱ¼õС
91¿¼ÊÔÍø


²Î¿¼´ð°¸£ºAC


±¾Ìâ½âÎö£º


±¾ÌâÄѶȣºÒ»°ã



5¡¢Ìî¿ÕÌâ  80¡æÊ±£¬½«0£®40molµÄN2O4ÆøÌå³äÈë2 LÒѾ­³é¿ÕµÄ¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦£º
N2O42NO2£¬¡÷H£¾0¸ôÒ»¶Îʱ¼ä¶Ô¸ÃÈÝÆ÷ÄÚµÄÎïÖʽøÐзÖÎö£¬µÃµ½ÈçÏÂÊý¾Ý

£¨1£©¼ÆËã20s-40sÄÚÓÃN2O4±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ______________mol¡¤L-1¡¤s-1
£¨2£©¼ÆËãÔÚ80¡æÊ±¸Ã·´Ó¦µÄƽºâ³£ÊýK=_______________£»
£¨3£©ÒªÔö´ó¸Ã·´Ó¦µÄKÖµ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ(ÌîÐòºÅ)__________________£»
A£®Ôö´óN2O4µÄÆðʼŨ¶È
B£®Ïò»ìºÏÆøÌåÖÐͨÈëNO2
C£®Ê¹ÓøßЧ´ß»¯¼Á
D£®Éý¸ßζÈ
£¨4£©ÈçͼÊÇ80¡æÊ±ÈÝÆ÷ÖÐN2O4ÎïÖʵÄÁ¿µÄ±ä»¯ÇúÏߣ¬ÇëÔÚ¸ÃͼÖв¹»­³ö¸Ã·´Ó¦ÔÚ60¡æÊ±N2O4ÎïÖʵÄÁ¿µÄ±ä»¯ÇúÏß¡£


²Î¿¼´ð°¸£º£¨1£©0.0020mol¡¤L-1¡¤s-1
£¨2£©1.8mol¡¤L-1£¨»ò1.8£©
£¨3£©D
£¨4£©¡°ÂÔ¡±


±¾Ìâ½âÎö£º


±¾ÌâÄѶȣºÒ»°ã




Ê×Ò³ ÉÏÒ³ 9 10 ÏÂÒ³ βҳ 10/10/10
΢ÐÅËÑË÷¹Ø×¢"91¿¼ÊÔÍø"¹«ÖÚºÅ,Áì30Ôª,»ñÈ¡¹«ÎñÔ±ÊÂÒµ±à½Ìʦ¿¼ÊÔ×ÊÁÏ40G
¡¾Ê¡ÊÐÏØµØÇøµ¼º½¡¿¡ï¡¾¿¼ÊÔÌâ¿âµ¼º½¡¿
 ¡ï ¸ß¿¼Ê¡¼¶µ¼º½ ¡ï 
È«¹ú A°²»Õ B±±¾© CÖØÇì F¸£½¨ G¹ã¶« ¹ãÎ÷ ¸ÊËà ¹óÖÝ HºÓÄÏ ºÓ±± ºþÄÏ ºþ±± ºÚÁú½­ º£ÄÏ J½­ËÕ ½­Î÷ ¼ªÁÖ LÁÉÄþ NÄÚÃɹŠÄþÏÄ QÇຣ Sɽ¶« ɽÎ÷ ÉÂÎ÷ ËÄ´¨ ÉϺ£ TÌì½ò Xн® Î÷²Ø YÔÆÄÏ ZÕã½­
 ¡ï ¸ß¿¼ÐÅÏ¢»ã×Ü ¡ï 
 ¡ï ¸ß¿¼ÀúÄêÕæÌâ ¡ï 
 ¡ï ¸ß¿¼ÀúÄêÕæÌâ ¡ï 
 ¡ï ¸ß¿¼Ìâ¿â ¡ï 
 ¡ï ¸ß¿¼Ìâ¿â ¡ï 

µçÄÔ°æ  |  ÊÖ»ú°æ  |  ·µ»Ø¶¥²¿