£¨1£©ÒÑÖªH2£¨g£©+1/2O2£¨g £©=H2O£¨g£©£¬·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯Èçͼ1£ºÇë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ùͼ1ÖÐa£¬b·Ö±ð´ú±íʲôÒâÒ壿a£®
»î»¯ÄÜ
»î»¯ÄÜ
£»b£®
ìʱä
ìʱä
£®
¢Ú¸Ã·´Ó¦ÊÇ
·ÅÈÈ
·ÅÈÈ
·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©£¬¡÷H
£¼0
£¼0
£¨Ìî¡°£¼0¡±»ò¡°£¾0¡±£©£®
£¨2£©ÏÖÓпÉÄæ·´Ó¦A£¨Æø£©+B£¨Æø£©?3C£¨Æø£©¡÷H£¼0£¬Í¼2Öмס¢ÒÒ¡¢±û·Ö±ð±íʾÔÚ²»Í¬µÄÌõ¼þÏ£¬Éú³ÉÎïCÔÚ·´Ó¦»ìºÏÎïÖеİٷֺ¬Á¿£¨C%£©ºÍ·´Ó¦Ê±¼äµÄ¹Øϵ£º
¢ÙÈô¼×ͼÖÐÁ½ÌõÇúÏß·Ö±ð±íʾÓд߻¯¼ÁºÍÎÞ´ß»¯¼ÁʱµÄÇé¿ö£¬Ôò_
a
a
ÇúÏßÊDZíʾ
Óд߻¯¼ÁʱµÄÇé¿ö£®
¢ÚÈôÒÒͼÖеÄÁ½ÌõÇúÏß±íʾ100¡æ¡¢200¡æʱµÄÇé¿ö£¬Ôò
b
b
ÇúÏßÊDZíʾ100¡æµÄÇé¿ö£®
¢ÛÈô±ûͼÖÐÁ½ÌõÇúÏß·Ö±ð±íʾ²»Í¬Ñ¹Ç¿ÏµÄÇé¿ö£¬Ôò
b
b
ÇúÏßÊDZíʾѹǿ½Ï´óµÄÇé¿ö£®
·ÖÎö£º£¨1£©¢ÙÒÀ¾Ý»¯Ñ§·´Ó¦µÄ·´Ó¦ÈÈ¿ÉÒÔ¸ù¾Ý·´Ó¦µÄʵÖÊ£¬¶ÏÁÑ»¯Ñ§¼üÎüÊÕÄÜÁ¿£¬Éú³É»¯Ñ§¼ü·Å³öÄÜÁ¿£¬¶þÕߵIJîÖµÊÇ·´Ó¦µÄìʱ䣻¢ÚÒÀ¾Ý·´Ó¦ÎïºÍÉú³ÉÎïµÄÄÜÁ¿¸ßµÍºÍ·´Ó¦µÄÄÜÁ¿Êغã·ÖÎöÅжÏ
£¨2£©¢Ù´ß»¯¼Á¸Ä±ä·´Ó¦Àú³Ì£¬Ìá¸ß·´Ó¦ËÙÂÊ£¬Ëõ¶Ìµ½´ïƽºâµÄʱ¼ä£»
¢ÚζÈÉý¸ß£¬Ìá¸ß·´Ó¦ËÙÂÊ£¬Ëõ¶Ìµ½´ïƽºâµÄʱ¼ä£»
¢ÛѹǿÉý¸ß£¬Ìá¸ß·´Ó¦ËÙÂÊ£¬Ëõ¶Ìµ½´ïƽºâµÄʱ¼ä£®
½â´ð£º½â£º£¨1£©¢ÙͼÏóÖпÉÒÔ·ÖÎöÅжϣ¬aΪ²ð»¯Ñ§¼üÎüÊÕµÄÄÜÁ¿£¬¼´·´Ó¦µÄ»î»¯ÄÜ£»bΪ·´Ó¦Éú³É»¯Ñ§¼ü·Å³öµÄÈÈÁ¿¼õÈ¥·´Ó¦¶ÏÁÑ»¯Ñ§¼üÎüÊÕµÄÈÈÁ¿£¬¼´·´Ó¦µÄìʱ䣬
¹Ê´ð°¸Îª£º»î»¯ÄÜ£»ìʱ䣻
¢Ú·´Ó¦ÎïµÄÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄÄÜÁ¿£¬ÒÀ¾Ý»¯Ñ§·´Ó¦µÄÄÜÁ¿Êغ㣬·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬ìʱäСÓÚ0£¬¹Ê´ð°¸Îª£º·ÅÈÈ£»£¼0£»
£¨2£©¢Ù´ß»¯¼Á¸Ä±ä·´Ó¦Àú³Ì£¬Ìá¸ß·´Ó¦ËÙÂÊ£¬Ëõ¶Ìµ½´ïƽºâµÄʱ¼ä£¬aÇúÏßµ½´ïƽºâʱ¼ä¶Ì£¬bÇúÏßµ½´ïƽºâʱ¼ä³¤£¬ËùÒÔaÇúÏß±íʾʹÓô߻¯¼Á£¬bÇúÏßδʹÓô߻¯¼Á£¬
¹Ê´ð°¸Îª£ºa£»     
¢ÚζÈÉý¸ß£¬Ìá¸ß·´Ó¦ËÙÂÊ£¬Ëõ¶Ìµ½´ïƽºâµÄʱ¼ä£¬aÇúÏßµ½´ïƽºâʱ¼ä¶Ì£¬bÇúÏßµ½´ïƽºâʱ¼ä³¤£¬ËùÒÔaÇúÏß±íʾ200¡æ£¬bÇúÏß±íʾ100¡æ£¬¹Ê´ð°¸Îª£ºb£»  
¢ÛѹǿÉý¸ß£¬Ìá¸ß·´Ó¦ËÙÂÊ£¬Ëõ¶Ìµ½´ïƽºâµÄʱ¼ä£¬aÇúÏßµ½´ïƽºâʱ¼ä³¤£¬bÇúÏßµ½´ïƽºâʱ¼ä¶Ì£¬ËùÒÔaÇúÏß±íʾѹǿС£¬bÇúÏß±íʾѹǿ´ó£¬¹Ê´ð°¸Îª£ºb£®
µãÆÀ£º±¾ÌâÒÔͼÏóΪÔØÌ忼²éÍâ½çÌõ¼þ¶Ô·´Ó¦ËÙÂÊ¡¢Æ½ºâµÄÓ°Ï죬ÒÔ¼°Ñ§Éú¶ÁͼÌáÈ¡ÐÅÏ¢µÄÄÜÁ¦£¬ÄѶȲ»´ó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿Æѧ¼ÒÀûÓÃÌ«ÑôÄÜ·Ö½âË®Éú³ÉµÄÇâÆøÔÚ´ß»¯¼Á×÷ÓÃÏÂÓë¶þÑõ»¯Ì¼·´Ó¦Éú³É¼×´¼£®ÒÑÖªH2£¨g£©¡¢CO£¨g£©ºÍCH3OH£¨l£©µÄȼÉÕÈÈ¡÷H·Ö±ðΪ-285.8kJ?mol-1¡¢-283.0kJ?mol-1ºÍ-726.5kJ?mol-1£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃÌ«ÑôÄÜ·Ö½â10molË®ÏûºÄµÄÄÜÁ¿ÊÇ
2858
2858
kJ£»
£¨2£©¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ
CH3OH£¨l£©+O2£¨g£©=CO£¨g£©+2 H2O£¨l£©¡÷H=-443.5kJ?mol-1
CH3OH£¨l£©+O2£¨g£©=CO£¨g£©+2 H2O£¨l£©¡÷H=-443.5kJ?mol-1
£»
£¨3£©ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬ÓÉCO2ºÍH2ºÏ³É¼×´¼£¬ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¿¼²ìζȶԷ´Ó¦µÄÓ°Ï죬ʵÑé½á¹ûÈçͼËùʾ£¨×¢£ºT1¡¢T2¾ù´óÓÚ300¡æ£©£»ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
¢Û¢Ü
¢Û¢Ü
£¨ÌîÐòºÅ£©
¢ÙζÈΪT1ʱ£¬´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬Éú³É¼×´¼µÄƽ¾ùËÙÂÊΪv£¨CH3OH£©=
nA
tA
 
nA
tA
mol?L-1?min-1
¢Ú¸Ã·´Ó¦ÔÚT1ʱµÄƽºâ³£Êý±ÈT2ʱµÄС
¢Û¸Ã·´Ó¦Îª·ÅÈÈ·´Ó¦
¢Ü´¦ÓÚAµãµÄ·´Ó¦Ìåϵ´ÓT1±äµ½T2£¬´ïµ½Æ½ºâʱÔö´ó
£¨4£©ÔÚT1ζÈʱ£¬½«1mol CO2ºÍ3mol H2³äÈëÒ»ÃܱպãÈÝÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦´ïµ½Æ½ºâºó£¬ÈôCO2ת»¯ÂÊΪ¦Á£¬ÔòÈÝÆ÷ÄÚµÄѹǿÓëÆðʼѹǿ֮±ÈΪ
1-
a
2
1-
a
2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêºÓÄÏÊ¡ÓÀ³ÇʵÑé¸ßÖи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©£¨1£©ÒÑÖªH2£¨g£©È¼ÉÕÈÈΪ-285£®8kJ¡¤mol-1ÓÃÌ«ÑôÄÜ·Ö½â10molË®ÏûºÄµÄÄÜÁ¿ÊÇ_____________kJ£»
£¨2£©ÒÑÖª0.4 molҺ̬ëÂ(N2 H4)Óë×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.65 kJµÄÈÈÁ¿¡£Ð´³ö¸Ã¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________________________________________¡£
ÓÖÒÑÖª H2O£¨l£©=== H2O£¨g£©¦¤H£½£«44kJ¡¤mol£­1Ôò16gҺ̬ëÂÓë×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍҺ̬ˮ·Å³öµÄÈÈÁ¿ÊÇ_____________ kJ¡£
ÉÏÊö·´Ó¦Ó¦ÓÃÓÚ»ð¼ýÍƽø¼Á£¬³ýÊÍ·Å´óÁ¿µÄÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜÍ»³öµÄÓŵãÊÇ____________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìºÓÄÏÊ¡¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©£¨1£©ÒÑÖªH2£¨g£©È¼ÉÕÈÈΪ-285£®8kJ¡¤mol-1ÓÃÌ«ÑôÄÜ·Ö½â10molË®ÏûºÄµÄÄÜÁ¿ÊÇ_____________kJ£»

£¨2£©ÒÑÖª0.4 molҺ̬ëÂ(N2 H4)Óë×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.65 kJµÄÈÈÁ¿¡£Ð´³ö¸Ã¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________________________________________¡£

ÓÖÒÑÖª H2O£¨l£©=== H2O£¨g£©¦¤H£½£«44kJ¡¤mol£­1Ôò16gҺ̬ëÂÓë×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍҺ̬ˮ·Å³öµÄÈÈÁ¿ÊÇ_____________ kJ¡£

ÉÏÊö·´Ó¦Ó¦ÓÃÓÚ»ð¼ýÍƽø¼Á£¬³ýÊÍ·Å´óÁ¿µÄÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜÍ»³öµÄÓŵãÊÇ____________________________________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêɽÎ÷Ê¡ÐÃÖÝÊÐԭƽһÖи߶þ£¨ÉÏ£©ÆÚÖл¯Ñ§ÊÔ¾í£¨ÆÕͨ°à£©£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

£¨1£©ÒÑÖªH2£¨g£©+1/2O2£¨g £©=H2O£¨g£©£¬·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯Èçͼ1£ºÇë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ùͼ1ÖÐa£¬b·Ö±ð´ú±íʲôÒâÒ壿a£®______£»b£®______£®
¢Ú¸Ã·´Ó¦ÊÇ______·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©£¬¡÷H______£¨Ìî¡°£¼0¡±»ò¡°£¾0¡±£©£®
£¨2£©ÏÖÓпÉÄæ·´Ó¦A£¨Æø£©+B£¨Æø£©?3C£¨Æø£©¡÷H£¼0£¬Í¼2Öмס¢ÒÒ¡¢±û·Ö±ð±íʾÔÚ²»Í¬µÄÌõ¼þÏ£¬Éú³ÉÎïCÔÚ·´Ó¦»ìºÏÎïÖеİٷֺ¬Á¿£¨C%£©ºÍ·´Ó¦Ê±¼äµÄ¹Øϵ£º
¢ÙÈô¼×ͼÖÐÁ½ÌõÇúÏß·Ö±ð±íʾÓд߻¯¼ÁºÍÎÞ´ß»¯¼ÁʱµÄÇé¿ö£¬Ôò_______ÇúÏßÊDZíʾ
Óд߻¯¼ÁʱµÄÇé¿ö£®
¢ÚÈôÒÒͼÖеÄÁ½ÌõÇúÏß±íʾ100¡æ¡¢200¡æʱµÄÇé¿ö£¬Ôò______ÇúÏßÊDZíʾ100¡æµÄÇé¿ö£®
¢ÛÈô±ûͼÖÐÁ½ÌõÇúÏß·Ö±ð±íʾ²»Í¬Ñ¹Ç¿ÏµÄÇé¿ö£¬Ôò______ÇúÏßÊDZíʾѹǿ½Ï´óµÄÇé¿ö£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸