¾«Ó¢¼Ò½ÌÍø¢ñ¹¤ÒµÉϳ£ÀûÓô×ËáºÍÒÒ´¼ºÏ³ÉÓлúÈܼÁÒÒËáÒÒõ¥£ºCH3COOH£¨l£©+C2H5OH£¨l£©
ŨH2SO4
¡÷
CH3COOC2H5£¨l£©+H2O£¨l£©¡÷H=-8.62kJ?mol-1
ÒÑÖªCH3COOH¡¢C2H5OHºÍCH3COOC2H5µÄ·ÐµãÒÀ´ÎΪ118¡æ¡¢78¡æºÍ77¡æ£®ÔÚÆäËûÌõ¼þÏàͬʱ£¬Ä³Ñо¿Ð¡×é½øÐÐÁ˶à´ÎʵÑ飬ʵÑé½á¹ûÈçͼËùʾ£®£¨1£©¸ÃÑо¿Ð¡×éµÄʵÑéÄ¿µÄÊÇ
 
£®
£¨2£©60¡æÏ·´Ó¦40minÓë70¡æÏ·´Ó¦20minÏà±È£¬Ç°ÕßµÄƽ¾ù·´Ó¦ËÙÂÊ
 
ºóÕߣ¨ÌСÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°´óÓÚ¡±£©£®
£¨3£©ÈçͼËùʾ£¬·´Ó¦Ê±¼äΪ40min¡¢Î¶ȳ¬¹ý80¡æʱ£¬ÒÒËáÒÒõ¥²úÂÊϽµµÄÔ­Òò¿ÉÄÜÊÇ
 
£¨Ð´³öÁ½Ìõ£©£®
¢ò£®Ãº»¯¹¤Öг£ÐèÑо¿²»Í¬Î¶ÈÏÂƽºâ³£Êý¡¢Í¶Áϱȼ°ÈÈÖµµÈÎÊÌ⣮
ÒÑÖª£ºCO£¨g£©+H2O£¨g£©??H2£¨g£©+CO2£¨g£©Æ½ºâ³£ÊýËæζȵı仯ÈçÏÂ±í£º
ζÈ/¡æ 400 500 800
ƽºâ³£ÊýK 9.94 9 1
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚ800¡æ·¢ÉúÉÏÊö·´Ó¦£¬ÒÔ±íÖеÄÎïÖʵÄÁ¿Í¶ÈëºãÈÝ·´Ó¦Æ÷£¬ÆäÖÐÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÓÐ
 
£¨Ñ¡Ìî¡°A¡¢B¡¢C¡¢D¡¢E¡±£©£®
n£¨CO£© n£¨H2O£© n£¨H2£© n£¨CO2£©
A 1 5 2 3
B 2 2 1 1
C 3 3 0 0
D 0.5 2 1 1
E 3 1 2 1
£¨2£©ÒÑÖªÔÚÒ»¶¨Î¶ÈÏ£¬C£¨s£©+CO2£¨g£©??2CO£¨g£©Æ½ºâ³£ÊýΪK£»
C£¨s£©+H2O£¨g£©??CO£¨g£©+H2£¨g£©¡¡Æ½ºâ³£ÊýΪK1£»
CO£¨g£©+H2O£¨g£©??H2£¨g£©+CO2£¨g£©¡¡Æ½ºâ³£ÊýΪK2£»
ÔòK¡¢K1¡¢K2Ö®¼äµÄ¹ØϵÊÇ
 
£®
£¨3£©ÔÚV LÃܱÕÈÝÆ÷ÖÐͨÈë10mol COºÍ10molË®ÕôÆø£¬ÔÚT¡æ´ïµ½Æ½ºâ£¬È»ºó¼±ËÙͨ¹ý¼îʯ»Ò£¬½«ËùµÃ»ìºÏÆøÌåȼÉÕ£¬²âµÃ·Å³öµÄÈÈÁ¿Îª2 842kJ£¨ÒÑÖªCOȼÉÕÈÈΪ283kJ?mol-1£¬H2ȼÉÕÈÈΪ286kJ?mol-1£©£¬ÔòT¡æƽºâ³£ÊýK=
 
£®
·ÖÎö£º¢ñ£®£¨1£©±¾Ìâ̽¾¿·´Ó¦Î¶ȡ¢·´Ó¦Ê±¼ä¶ÔÒÒËáÒÒõ¥²úÂʵÄÓ°Ï죻
£¨2£©¸ù¾Ý²úÂÊÓëʱ¼äµÄ±ÈÖµÅжϣ»
£¨3£©ÒÒ´¼ºÍÒÒËá¾ßÓлӷ¢ÐÔ£¬ÒÒËáÒÒõ¥µÄË®½â·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬¸ù¾ÝζȶԻ¯Ñ§Æ½ºâµÄÓ°Ïì·ÖÎö£»
¢ò£¨1£©·´Ó¦ÔÚͬһÈÝÆ÷ÄÚ½øÐУ¬Ìå»ýÏàͬ£¬·½³ÌʽÖи÷ÎïÖʵĻ¯Ñ§¼ÆÁ¿Êý¶¼ÊÇ1£¬ËùÒÔÔÚ¼ÆËãÖоù¿ÉÓÃÎïÖʵÄÁ¿ÊýÖµ´úÌæŨ¶ÈÊýÖµ£¬Çó³ö´ËʱµÄŨ¶ÈÉÌQc£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬Å¨¶ÈÉÌСÓÚƽºâ³£Êý£»
£¨2£©ÒÀ¾Ý»¯Ñ§·´Ó¦µÄƽºâ³£Êý±í´ïʽ¼ÆËã·ÖÎö£»
£¨3£©CO¡¢H2µÄÎïÖʵÄÁ¿¹²Îª10mol£¬¸ù¾ÝȼÉշųöµÄÈÈÁ¿Çó³öCO¡¢H2¸÷×ÔµÄÎïÖʵÄÁ¿£¬ÀûÓÃÈý¶Îʽ·¨Çó³öƽºâʱ¸÷×é·ÖµÄÎïÖʵÄÁ¿£¬´úÈëƽºâ³£Êý¼ÆË㣮
½â´ð£º½â£º¢ñ£® £¨1£©¸ù¾ÝͼÏóÖª£¬¸ÃʵÑéµÄʵÑéÄ¿µÄÊÇ£ºÌ½¾¿·´Ó¦Î¶ȡ¢·´Ó¦Ê±¼ä¶ÔÒÒËáÒÒõ¥²úÂʵÄÓ°Ï죬
¹Ê´ð°¸Îª£ºÌ½¾¿·´Ó¦Î¶ȡ¢·´Ó¦Ê±¼ä¶ÔÒÒËáÒÒõ¥²úÂʵÄÓ°Ï죻
£¨2£©¸ù¾ÝÒÒËáÒÒõ¥µÄ²úÂʺÍʱ¼äµÄ±ÈÖµÖª£¬60¡æÏ·´Ó¦40minÓë70¡æÏ·´Ó¦20minÏà±È£¬Ç°ÕßµÄƽ¾ù·´Ó¦ËÙÂÊСÓÚºóÕߣ¬
¹Ê´ð°¸Îª£ºÐ¡ÓÚ£»
£¨3£©ÒÒ´¼ºÍÒÒËá¾ßÓлӷ¢ÐÔ£¬ÒÒËáÒÒõ¥µÄË®½â·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâ״̬ʱ£¬ÔÙÉý¸ßζȣ¬´Ù½øÒÒËáÒÒõ¥Ê±¼ä£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬Î¶ȹý¸ß£¬ÒÒ´¼ºÍÒÒËá´óÁ¿»Ó·¢Ê¹·´Ó¦ÎïÀûÓÃÂÊϽµ£¬
¹Ê´ð°¸Îª£º·´Ó¦¿ÉÄÜÒÑ´ïƽºâ״̬£¬Î¶ÈÉý¸ßƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»Î¶ȹý¸ß£¬ÒÒ´¼ºÍÒÒËá´óÁ¿»Ó·¢Ê¹·´Ó¦ÎïÀûÓÃÂÊϽµ£»
¢ò£®£¨1£©·´Ó¦ÔÚͬһÈÝÆ÷ÄÚ½øÐУ¬Ìå»ýÏàͬ£¬·½³ÌʽÖи÷ÎïÖʵĻ¯Ñ§¼ÆÁ¿Êý¶¼ÊÇ1£¬ËùÒÔÔÚ¼ÆËãÖоù¿ÉÓÃÎïÖʵÄÁ¿ÊýÖµ´úÌæŨ¶ÈÊýÖµ£¬800¡æʱ·´Ó¦Æ½ºâ³£ÊýΪ1£®
A¡¢Qc=
c(H2)?c(CO2)
c(CO)?c(H2O)
=
3¡Á2
1¡Á5
=
6
5
£¬´óÓÚ1£¬·´Ó¦ÏòÄæ·´Ó¦½øÐУ¬¹ÊA´íÎó£»
B¡¢Qc=
c(H2)?c(CO2)
c(CO)?c(H2O)
=
1¡Á1
2¡Á2
=
1
4
£¬Ð¡ÓÚ1£¬·´Ó¦ÏòÕý·´Ó¦½øÐУ¬¹ÊBÕýÈ·£»
C¡¢¿ªÊ¼Ö»ÓÐCO¡¢H2£¬·´Ó¦ÏòÕý·´Ó¦½øÐУ¬¹ÊCÕýÈ·£»
D¡¢Qc=
c(H2)?c(CO2)
c(CO)?c(H2O)
=
1¡Á1
0.5¡Á2
=1£¬´¦ÓÚƽºâ״̬£¬¹ÊD´íÎó£»
E¡¢Qc=
c(H2)?c(CO2)
c(CO)?c(H2O)
=
2¡Á1
3¡Á1
=
2
3
£¬Ð¡ÓÚ1£¬·´Ó¦ÏòÕý·´Ó¦½øÐУ¬¹ÊEÕýÈ·£»
¹Ê´ð°¸Îª£ºBCE£»
£¨2£©·´Ó¦¢ÙC£¨s£©+CO2£¨g£©?2CO£¨g£©Æ½ºâ³£ÊýK=
c2(CO)
c(CO2)
£»
·´Ó¦¢ÚC£¨s£©+H2O£¨g£©?CO£¨g£©+H2£¨g£©Æ½ºâ³£ÊýK1=
c(H2)?c(CO)
c(H2O)
£»
·´Ó¦¢ÛCO£¨g£©+H2O£¨g£©?H2£¨g£©+CO2£¨g£© Æ½ºâ³£ÊýK2=
c(H2)?c(CO2)
c(CO)?c(H2O)
£»
ËùÒÔ£º
K1
K2
=
c2(CO)
c(CO2)
=K£¬
¹Ê´ð°¸Îª£ºK=
K1
K2
£»
£¨3£©ÓÉ·½³ÌʽCO£¨g£©+H2O£¨g£©?H2£¨g£©+CO2£¨g£©¿ÉÖª£¬ÓÐ1molCO·´Ó¦ÔòÉú³É1molH2£¬¿ªÊ¼Í¨Èë10molCO£¬ËùÒÔƽºâʱ£¬CO¡¢H2µÄÎïÖʵÄÁ¿¹²Îª10mol£®Ôòƽ¾ùȼÉÕÈÈΪ
2842kJ
10mol
=284.2kJ/mol£¬
ÀûÓÃÊ®×Ö½»²æ·¨¼ÆËãCO¡¢H2µÄÎïÖʵÄÁ¿Ö®±È£¬
¾«Ó¢¼Ò½ÌÍø
¼´CO¡¢H2µÄÎïÖʵÄÁ¿Ö®±ÈΪ1.8kJ/mol£º1.2kJ/mol=3£º2£¬
ËùÒÔn£¨CO£©=10mol¡Á
3
5
=6mol£¬n£¨H2£©=10mol-6mol=4mol£¬
ÀûÓÃÈý¶Îʽ·¨Çó³öƽºâʱ¸÷×é·ÖµÄÎïÖʵÄÁ¿£¬
       CO£¨g£©+H2O£¨g£©?H2£¨g£©+CO2£¨g£©£¬
Æðʼ£º10mol  10mol      0        0   
ת»¯£º4mol   4 mol     4mol     4mol
ƽºâ£º6mol   6mol      4mol     4mol
ËùÒÔ³£Êýƽºâ³£ÊýΪ k=
c(H2)?c(CO2)
c(CO)?c(H2O)
=
4¡Á4
6¡Á6
=
4
9
£¬
¹Ê´ð°¸Îª£º
4
9
£®
µãÆÀ£º±¾Ì⿼²éÁË̽¾¿Î¶ȡ¢Ñ¹Ç¿¶Ô»¯Ñ§Æ½ºâµÄÓ°Ïì¡¢»¯Ñ§Æ½ºâ³£ÊýµÄ¸ÅÄî¼°¼ÆË㣬ÄѵãÖеȣ¬ÊÔÌâÉæ¼°µÄ֪ʶµã½Ï¶à£¬ÌâÁ¿½Ï´ó£¬×¢ÒâÕÆÎÕÓ°Ï컯ѧƽºâµÄÒòËØ£¬Ã÷È·»¯Ñ§Æ½ºâ³£ÊýµÄ¸ÅÄî¼°ÇóËã·½·¨£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?¹ãÖݶþÄ££©¹¤ÒµÉϳ£ÀûÓô×ËáºÍÒÒ´¼ºÏ³ÉÓлúÈܼÁÒÒËáÒÒõ¥£º
CH3COOH£¨l£©+C2H5OH£¨l£©
ŨH2SO4¡÷
CH3COOC2H5£¨l£©+H2O£¨l£©¡÷H=-8.62kJ?mol-1
ÒÑÖªCH3COOH¡¢C2H5OHºÍCH3COOC2H5µÄ·ÐµãÒÀ´ÎΪ118¡æ¡¢78¡æºÍ77¡æ£®ÔÚÆäËûÌõ¼þÏàͬʱ£¬Ä³Ñо¿Ð¡×é½øÐÐÁ˶à´ÎʵÑ飬ʵÑé½á¹ûÈçͼËùʾ£®

£¨1£©¸ÃÑо¿Ð¡×éµÄʵÑéÄ¿µÄÊÇ
̽¾¿·´Ó¦Î¶ȡ¢·´Ó¦Ê±¼ä¶ÔÒÒËáÒÒõ¥²úÂʵÄÓ°Ïì
̽¾¿·´Ó¦Î¶ȡ¢·´Ó¦Ê±¼ä¶ÔÒÒËáÒÒõ¥²úÂʵÄÓ°Ïì
£®
£¨2£©60¡æÏ·´Ó¦40minÓë70¡æÏ·´Ó¦20minÏà±È£¬Ç°ÕßµÄƽ¾ù·´Ó¦ËÙÂÊ
СÓÚ
СÓÚ
ºóÕߣ¨ÌСÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°´óÓÚ¡±£©£®
£¨3£©ÈçͼËùʾ£¬·´Ó¦Ê±¼äΪ40min¡¢Î¶ȳ¬¹ý80¡æʱ£¬ÒÒËáÒÒõ¥²úÂÊϽµµÄÔ­Òò¿ÉÄÜÊÇ
·´Ó¦¿ÉÄÜÒÑ´ïƽºâ״̬£¬Î¶ÈÉý¸ßƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»Î¶ȹý¸ß£¬ÒÒ´¼ºÍÒÒËá´óÁ¿»Ó·¢Ê¹·´Ó¦ÎïÀûÓÃÂÊϽµ
·´Ó¦¿ÉÄÜÒÑ´ïƽºâ״̬£¬Î¶ÈÉý¸ßƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»Î¶ȹý¸ß£¬ÒÒ´¼ºÍÒÒËá´óÁ¿»Ó·¢Ê¹·´Ó¦ÎïÀûÓÃÂÊϽµ
£¨Ð´³öÁ½Ìõ£©£®
£¨4£©Ä³Î¶ÈÏ£¬½«0.10mol CH3COOHÈÜÓÚË®Åä³É1LÈÜÒº£®
¢ÙʵÑé²âµÃÒѵçÀëµÄ´×Ëá·Ö×ÓÕ¼Ô­Óд×Ëá·Ö×Ó×ÜÊýµÄ1.3%£¬Ôò¸ÃζÈÏÂCH3COOHµÄµçÀëƽºâ³£ÊýK=
1.7¡Á10-5
1.7¡Á10-5
£®£¨Ë®µÄµçÀëºöÂÔ²»¼Æ£¬´×ËáµçÀë¶Ô´×Ëá·Ö×ÓŨ¶ÈµÄÓ°ÏìºöÂÔ²»¼Æ£©
¢ÚÏò¸ÃÈÜÒºÖÐÔÙ¼ÓÈë
1.7¡Á10-2
1.7¡Á10-2
mol CH3COONa¿ÉʹÈÜÒºµÄpHԼΪ4£®£¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¸£½¨Ê¡ÈýÃ÷Ò»ÖС¢¶þÖиßÈýÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÊµÑéÌâ

¹¤ÒµÉϳ£ÀûÓô×ËáºÍÒÒ´¼ºÏ³ÉÓлúÈܼÁÒÒËáÒÒõ¥£º
CH3COOH(l)£«C2H5OH(l)CH3COOC2H5(l)£«H2O(l) ¦¤H£½£­8.62kJ¡¤mol£­1
ÒÑÖªCH3COOH¡¢C2H5OHºÍCH3COOC2H5µÄ·ÐµãÒÀ´ÎΪ118¡æ¡¢78¡æºÍ77¡æ¡£ÔÚÆäËûÌõ¼þÏàͬʱ£¬Ä³Ñо¿Ð¡×é½øÐÐÁ˶à´ÎʵÑ飬ʵÑé½á¹ûÈçͼËùʾ¡£

£¨1£©¸ÃÑо¿Ð¡×éµÄʵÑéÄ¿µÄÊÇ__________________¡£
£¨2£©60¡æÏ·´Ó¦40minÓë70¡æÏ·´Ó¦20minÏà±È£¬Ç°ÕßµÄƽ¾ù·´Ó¦ËÙÂÊ___________ºóÕߣ¨ÌСÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°´óÓÚ¡±£©¡£
£¨3£©ÈçͼËùʾ£¬·´Ó¦Ê±¼äΪ40min¡¢Î¶ȳ¬¹ý80¡æʱ£¬ÒÒËáÒÒõ¥²úÂÊϽµµÄÔ­Òò¿ÉÄÜÊÇ______£¨Ð´³öÁ½Ìõ£©¡£
£¨4£©Ä³Î¶ÈÏ£¬½«0.10 mol CH3COOHÈÜÓÚË®Åä³É1 LÈÜÒº¡£
¢ÙʵÑé²âµÃÒѵçÀëµÄ´×Ëá·Ö×ÓÕ¼Ô­Óд×Ëá·Ö×Ó×ÜÊýµÄ1.3%£¬Ôò¸ÃζÈÏÂCH3COOHµÄµçÀëƽºâ³£ÊýK=____________________¡££¨Ë®µÄµçÀëºöÂÔ²»¼Æ£¬´×ËáµçÀë¶Ô´×Ëá·Ö×ÓŨ¶ÈµÄÓ°ÏìºöÂÔ²»¼Æ£©
¢ÚÏò¸ÃÈÜÒºÖÐÔÙ¼ÓÈë__________mol CH3COONa¿ÉʹÈÜÒºµÄpHԼΪ4¡££¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì½­Î÷Ê¡¾°µÂÕòÊиßÈýÒ»¼ì»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

¢ñ£®ºãΣ¬ÈÝ»ýΪ1  LºãÈÝÌõ¼þÏ£¬Áò¿ÉÒÔ·¢ÉúÈçÏÂת»¯£¬Æä·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØϵÈçͼ1Ëùʾ(ÒÑÖª£º2SO2(g)£«O2(g)  2SO3(g)    ¦¤H£½£­196.6  kJ¡¤mol£­1)£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÄܱíʾÁòµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º______________________¡£

(2)¦¤H2£½__________kJ¡¤mol£­1¡£

¢ò.¹¤ÒµÉϳ£ÀûÓô×ËáºÍÒÒ´¼ºÏ³ÉÓлúÈܼÁÒÒËáÒÒõ¥£º

CH3COOH(l)£«C2H5OH(l)  CH3COOC2H5(l)£«H2O(l)    ¦¤H£½£­8.62  kJ¡¤mol£­1

ÒÑÖªCH3COOH¡¢C2H5OHºÍCH3COOC2H5µÄ·ÐµãÒÀ´ÎΪ118  ¡æ¡¢78  ¡æºÍ77  ¡æ¡£ÔÚÆäËûÌõ¼þÏàͬʱ£¬Ä³Ñо¿Ð¡×é½øÐÐÁ˶à´ÎʵÑ飬ʵÑé½á¹ûÈçͼËùʾ¡£

(1)¸ÃÑо¿Ð¡×éµÄʵÑéÄ¿µÄÊÇ___________________________________¡£

(2)60  ¡æÏ·´Ó¦40  minÓë70  ¡æÏ·´Ó¦20  minÏà±È£¬Ç°ÕßµÄƽ¾ù·´Ó¦ËÙÂÊ________ºóÕß(ÌСÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°´óÓÚ¡±)¡£

(3)ÈçͼËùʾ£¬·´Ó¦Ê±¼äΪ40  min¡¢Î¶ȳ¬¹ý80  ¡æʱ£¬ÒÒËáÒÒõ¥²úÂÊϽµµÄÔ­Òò¿ÉÄÜÊÇ_________________________________(д³öÁ½Ìõ)¡£

¢ó.ú»¯¹¤Öг£ÐèÑо¿²»Í¬Î¶ÈÏÂƽºâ³£Êý¡¢Í¶Áϱȼ°ÈÈÖµµÈÎÊÌâ¡£

ÒÑÖª£ºCO(g)£«H2O(g)  H2(g)£«CO2(g)ƽºâ³£ÊýËæζȵı仯ÈçÏÂ±í£º

ζÈ/¡æ

400

500

800

ƽºâ³£ÊýK

9.94

9

1

 

ÊԻشðÏÂÁÐÎÊÌ⣺

(1)ÔÚ800  ¡æ·¢ÉúÉÏÊö·´Ó¦£¬ÒÔ±íÖеÄÎïÖʵÄÁ¿Í¶ÈëºãÈÝ·´Ó¦Æ÷£¬ÆäÖÐÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÓÐ________(Ñ¡Ìî¡°A¡¢B¡¢C¡¢D¡¢E¡±)¡£

            n(CO)         n(H2O)          n(H2)          n(CO2)

A          1                 5                    2                3

B          2                 2                    1                1

C          3                 3                    0                0

D          0.5              2                    1                1

E          3                 1                    2                1

(2)ÒÑÖªÔÚÒ»¶¨Î¶ÈÏ£¬C(s)£«CO2(g)  2CO(g)ƽºâ³£ÊýΪK£»

¢ÙC(s)£«H2O(g)  CO(g)£«H2(g)        Æ½ºâ³£ÊýΪK1£»

¢ÚCO(g)£«H2O(g)  H2(g)£«CO2(g)  ƽºâ³£ÊýΪK2£»

ÔòK¡¢K1¡¢K2Ö®¼äµÄ¹ØϵÊÇ______________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄ긣½¨Ê¡¡¢¶þÖиßÈýÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

¹¤ÒµÉϳ£ÀûÓô×ËáºÍÒÒ´¼ºÏ³ÉÓлúÈܼÁÒÒËáÒÒõ¥£º

CH3COOH(l)£«C2H5OH(l)CH3COOC2H5(l)£«H2O(l) ¦¤H£½£­8.62kJ¡¤mol£­1

ÒÑÖªCH3COOH¡¢C2H5OHºÍCH3COOC2H5µÄ·ÐµãÒÀ´ÎΪ118¡æ¡¢78¡æºÍ77¡æ¡£ÔÚÆäËûÌõ¼þÏàͬʱ£¬Ä³Ñо¿Ð¡×é½øÐÐÁ˶à´ÎʵÑ飬ʵÑé½á¹ûÈçͼËùʾ¡£

£¨1£©¸ÃÑо¿Ð¡×éµÄʵÑéÄ¿µÄÊÇ__________________¡£

£¨2£©60¡æÏ·´Ó¦40minÓë70¡æÏ·´Ó¦20minÏà±È£¬Ç°ÕßµÄƽ¾ù·´Ó¦ËÙÂÊ___________ºóÕߣ¨ÌСÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°´óÓÚ¡±£©¡£

£¨3£©ÈçͼËùʾ£¬·´Ó¦Ê±¼äΪ40min¡¢Î¶ȳ¬¹ý80¡æʱ£¬ÒÒËáÒÒõ¥²úÂÊϽµµÄÔ­Òò¿ÉÄÜÊÇ______£¨Ð´³öÁ½Ìõ£©¡£

£¨4£©Ä³Î¶ÈÏ£¬½«0.10 mol CH3COOHÈÜÓÚË®Åä³É1 LÈÜÒº¡£

¢ÙʵÑé²âµÃÒѵçÀëµÄ´×Ëá·Ö×ÓÕ¼Ô­Óд×Ëá·Ö×Ó×ÜÊýµÄ1.3%£¬Ôò¸ÃζÈÏÂCH3COOHµÄµçÀëƽºâ³£ÊýK=____________________¡££¨Ë®µÄµçÀëºöÂÔ²»¼Æ£¬´×ËáµçÀë¶Ô´×Ëá·Ö×ÓŨ¶ÈµÄÓ°ÏìºöÂÔ²»¼Æ£©

¢ÚÏò¸ÃÈÜÒºÖÐÔÙ¼ÓÈë__________mol CH3COONa¿ÉʹÈÜÒºµÄpHԼΪ4¡££¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸