ѧϰ Ï°Ìâ¼°´ð°¸ ÁªÏµ¿Í·þ

·¢²¼Ê±¼ä : ÐÇÆÚ¶þ ÎÄÕÂѧϰ Ï°Ìâ¼°´ð°¸¸üÐÂÍê±Ï¿ªÊ¼ÔĶÁ117b606a561252d380eb6e96

26£®¶¨Î¡¢¶¨Ñ¹¡¢W=0 27£®¦ÌB(l)=¦ÌB(g)£» ¼õС£¬ ¼õСµÃÂý

*

28£®¦ÌB=¦ÌB(T,p)+RTlnxB 2 9£®¦Ì1=¦Ì2¡£

Èý¡¢1¡¢B£¬2¡¢A£¬3¡¢C£¬4¡¢B£¬5¡¢B£¬6 ¡¢C£¬ 7 ¡¢A£¬ 8¡¢C£¬ 9¡¢A£¬10¡¢C£¬ 11¡¢B£¬ 12¡¢B£¬13¡¢D£¬14 ¡¢B£¬ 15¡¢A £¬16¡¢B £¬17¡¢A£¬ 18¡¢B £¬19¡¢A £¬20 ¡¢C£¬21 ¡¢C £¬ 22 ¡¢C£¬ 23¡¢C£¬ 24£®B£¬25£®C £¬26. D£¬27£®B£¬28¡¢B£¬29¡¢D£¬30£®A£¬31£®A£¬32£®B£¬ 33£®D£¬34£®D£¬35£®D£¬36£®B£¬37£®B£¬38£®A¡£

/

Ïàƽºâ

Ò»¡¢ÅжÏÌ⣺

1¡¢ÏàÊÇָϵͳ´¦ÓÚƽºâʱ,ϵͳÖÐÎïÀíÐÔÖʼ°»¯Ñ§ÐÔÖʶ¼¾ùÔȵIJ¿·Ö¡£( ) 2¡¢ÒÀ¾ÝÏàÂÉ,´¿ÒºÌåÔÚÒ»¶¨Î¶ÈÏÂ,ÕôÆøѹӦ¸ÃÊǶ¨Öµ¡£( ) 3¡¢ÒÀ¾ÝÏàÂÉ,ºã·ÐκÏÎïµÄ·Ðµã²»ËæÍâѹµÄ¸Ä±ä¶ø¸Ä±ä¡£( ) 4¡¢Ë«×é·ÖÏàͼÖкã·Ð»ìºÏÎïµÄ×é³ÉËæÍâѹÁ¦µÄ²»Í¬¶ø²»Í¬¡£( )

5¡¢²»¿ÉÄÜÓüòµ¥¾«ÁóµÄ·½·¨½«¶þ×é·Öºã·Ð»ìºÏÎï·ÖÀëΪÁ½¸ö´¿×é·Ö¡£( ) 6¡¢¶þ×é·ÖµÄÀíÏëҺ̬»ìºÏÎïµÄÕôÆø×ÜѹÁ¦½éÓÚ¶þ´¿×é·ÖµÄÕûÆëѹ֮¼ä¡££¨ £©

7. ÔÚÒ»¸ö¸ø¶¨µÄÌåϵÖУ¬ÎïÖÖÊý¿ÉÒÔÒò·ÖÎöÎÊÌâµÄ½Ç¶È²»Í¬¶ø²»Í¬£¬µ«¶ÀÁ¢×é·ÖÊýÊÇÒ»¸öÈ·¶¨µÄÊý¡£( )

8£®×ÔÓɶȾÍÊÇ¿ÉÒÔ¶ÀÁ¢±ä»¯µÄ±äÁ¿¡£( )

/

9£®I2£¨s£©= I2£¨g£©Æ½ºâ¹²´æ£¬ÒòS = 2£¬ R = 1£¬ R= 0ËùÒÔC = 1¡£( ) 10£®µ¥×é·ÖÌåϵµÄÏàͼÖÐÁ½ÏàƽºâÏ߶¼¿ÉÒÔÓÿËÀ­±´Áú·½³Ì¶¨Á¿ÃèÊö¡£( ) 11£®ÔÚÏàͼÖÐ×Ü¿ÉÒÔÀûÓøܸ˹æÔò¼ÆËãÁ½ÏàƽºâʱÁ½ÏàµÄÏà¶ÔµÄÁ¿¡£( ) 12£®¶ÔÓÚ¶þÔª»¥ÈÜҺϵ£¬Í¨¹ý¾«Áó·½·¨×Ü¿ÉÒԵõ½Á½¸ö´¿×é·Ö¡£( ) 13£®²¿·Ö»¥ÈÜ˫Һϵ×ÜÒÔÏ໥¹²éîµÄÁ½Ïàƽºâ¹²´æ¡£( ) 14£®ºã·ÐÎïµÄ×é³É²»±ä¡£( )

15£®ÏàͼÖеĵ㶼ÊÇ´ú±íÌåϵ״̬µÄµã¡£( ) 16£®Èý×é·ÖÌåϵ×î¶àͬʱ´æÔÚ4¸öÏà¡£( )

17£®ÍêÈ«»¥ÈÜ˫ҺϵT~xͼÖУ¬ÈÜÒºµÄ·ÐµãÓë´¿×é·ÖµÄ·ÐµãµÄÒâÒåÊÇÒ»ÑùµÄ¡£( ) 18£®¾Ý¶þԪҺϵµÄp~xͼ£¬¿ÉÒÔ׼ȷµÄÅжϸÃÌåϵµÄÒºÏàÊÇ·ñÊÇÀíÏëÒºÌå»ìºÏÎï¡£( ) 19£®¶þԪҺϵÖÐÈôA×é·Ö¶ÔÀ­ÎÚ¶û¶¨ÂɲúÉúÕýÆ«²î£¬ÄÇôB×é·Ö±Ø¶¨¶ÔRaoult¶¨Âɲú Éú¸ºÆ«²î¡£( )

20£®A¡¢BÁ½ÒºÌåÍêÈ«²»»¥ÈÜ£¬ÄÇôµ±ÓÐB´æÔÚʱ£¬AµÄÕôÆøѹÓëÌåϵÖÐAµÄĦ¶û·ÖÊý ³ÉÕý±È¡£( )

21

¶þ¡¢Ìî¿ÕÌ⣺

1¡¢¶ÔÈý×é·ÖÏàͼ, ×î¶àÏàÊýΪ £»×î´óµÄ×ÔÓɶÈÊýΪ ,ËüÃÇ·Ö±ðÊÇ µÈ Ç¿¶È±äÁ¿¡£

2 ¡¢ÔÚ³é¿ÕµÄÈÝÆ÷ÖзÅÈëNH4HCO3(s)£¬·¢Éú·´Ó¦NH4HCO3(s) ===== NH3 (g) + CO2 (g) + H2O (g) ÇҴﵽƽºâ£¬ÔòÕâ¸öϵͳµÄ×é·ÖÊý(¶ÀÁ¢)=______£»×ÔÓɶÈÊý=_______¡£

3¡¢AlCl3 ÈÜÓÚË®ºóË®½â²¢ÓÐAl(OH)3³ÁµíÉú³É.´ËϵͳµÄ×é·ÖÊýΪ £¬×ÔÓɶÈÊýΪ ¡£ 4¡¢½«CaCO3(s)¡¢CaO(s)ºÍCO2(g)ÒÔÈÎÒâ±ÈÀý»ìºÏ£¬·ÅÈëÒ»ÃܱÕÈÝÆ÷ÖУ¬Ò»¶¨Î¶ÈϽ¨Á¢»¯Ñ§Æ½ºâ£¬ÔòϵͳµÄ×é·ÖÊýC=______£»ÏàÊý? =_______£»Ìõ¼þ×ÔÓɶÈÊý??=________¡£ 5¡¢½«Ag2O(s)·ÅÔÚÒ»¸ö³é¿ÕµÄÈÝÆ÷ÖУ¬Ê¹Ö®·Ö½âµÃµ½Ag(s)ºÍO2(g)²¢´ïµ½Æ½ºâ£¬Ôò´ËʱϵͳµÄ×é·ÖÊý£¨¶ÀÁ¢£©=______£»×ÔÓɶÈÊý=________¡£ 6¡¢CH4(g)ÓëH2O(g)·´Ó¦£¬²¿·Öת»¯ÎªCO(g)ºÍCO2(g)¼°H2(g)£¬²¢´ïƽºâ£¬ÔòϵͳµÄS =______£»R =______£»R¡ä=______£»C =______£»? =______£»f =______¡£

7¡¢½«Ò»¶¨Á¿NaCl(s)ÈÜÓÚË®ÖÐÐγɲ»±¥ºÍÈÜÒº£¬¼ÙÉèNaClÍêÈ«µçÀ룬H2O(l) ¿É½¨Á¢µçÀëƽºâ£¬Àë×ÓÎÞË®ºÏ·´Ó¦£¬ÔòϵͳµÄS =______£»R =______£»R¡ä=______£»C =______£»? =______£»f =______¡£

8 ¡¢ÒÑÖªNaHCO3(s)ÈȷֽⷴӦΪ2NaHCO3 == Na2CO3(s) + CO2(g) + H2O(g)

½ñ½«NaHCO3(s)£¬Na2CO3(s)£¬CO2ºÍH2O(g)°´ÈÎÒâ±ÈÀý»ìºÏ£¬·ÅÈëÒ»¸öÃܱÕÈÝÆ÷ÖУ¬µ±·´Ó¦½¨Á¢Æ½ºâʱϵͳµÄR¡ä=______£»C =______£»? =______£»f =______¡£

9¡¢ 450¡æʱ£¬½«Óë»ìºÏ£¬ÓÉÓڵķֽ⣬×îÖյõ½ºÍƽºâ¹²´æµÄϵͳ£¬ÔòϵͳµÄ×é·ÖÊýC =______£»ÏàÊý? =______£»Ìõ¼þ×ÔÓɶÈÊýf¡¯ =______¡£

10¡¢Na2CO3(s)Óë H2O(l) ¿ÉÉú³ÉNa2CO3? H2O (s)¡¢ Na2CO3? 7H2O (s) ¡¢ Na2CO3? 10H2O (s)£¬Ôò30¡æʱ£¬ÓëNa2CO3 Ë®ÈÜÒº¡¢±ùƽºâ¹²´æµÄË®ºÏÎï×î¶àÓÐ ÖÖ¡£ 11¡¢ÏÂÁл¯Ñ§·´Ó¦£¬Í¬Ê±¹²´æ²¢´ïµ½Æ½ºâ£¨Î¶ÈÔÚ900~1200K£©£º

CaCO3(s)==CaO(s)+CO2(g)£» CO2(g)+H2(g)==CO(g)+H2O(g)

H2O(g)+CO(g)+CaO(s)==CaCO3(s)+H2(g)¡£

¸ÃϵͳµÄ×ÔÓɶÈ=______¡£

12¡¢CaCO3(s)¡¢BaCO3(s)¡¢CaO(s)¡¢BaO(s)ºÍCO2(g)¹¹³ÉÒ»¸ö¶àÏàƽºâϵͳ£¬Õâ¸öϵͳµÄ×é·ÖÊý£¨¶ÀÁ¢£©=______£»×ÔÓɶÈÊý=________¡£

13¡¢ FeCl3Óë H2O ¿ÉÉú³ÉËÄÖÖË®ºÏÎïFeCl3 ?£¶H2O (s)¡¢ £²FeCl3 ?£³H2O (s) ¡¢ £²FeCl3 ?

£µH2O (s)¡¢ FeCl3 ?£²H2O (s) £¬Õâ¸öϵͳµÄ×é·ÖÊý£¨¶ÀÁ¢£©=______£»ÔÚ¶¨Ñ¹ÏÂ×î¶àÓÐ Ïàƽºâ¹²´æÖÖ¡£

14¡¢Ò»¸ö´ïµ½Æ½ºâµÄϵͳÖÐÓÐÈÎÒâÁ¿ZnO(s)¡¢Zn(s)¡¢CO(g)¡¢CO2(g)¡¢C(s)ÎåÖÖÎïÖÊ£¬ÔòÕâ¸öϵͳµÄ×é·ÖÊý£¨¶ÀpÒ»¶¨

t

Á¢£©=______£»×ÔÓɶÈÊý=________¡£

15¡¢ÀíÏëҺ̬»ìºÏÎﶨÎÂp-xB(yB)Ïàͼ×îÏÔÖøµÄÌØÕ÷ÊÇÒºÏàÏßΪ ¡£

16¡¢Ò»°ãÓлúÎï¿ÉÒÔÓÃË®ÕôÆøÕôÁó·¨Ìá´¿£¬µ±ÓлúÎïµÄ ºÍ Ô½´óʱ£¬Ìá´¿Ò»¶¨ÖÊÁ¿ÓлúÎïÐèÒªµÄË®ÕôÆøÔ½ÉÙ£¬È¼ÁÏÔ½½ÚÊ¡¡£

17¡¢ÓÒͼΪÁ½×é·Ö¾ßÓÐ×î¸ß·ÐµãµÄÏàͼ£¬ÈôÓÐ×é³ÉΪx0

x0

µÄÈÜÒº£¬¾­¹ý¾«Áó£¬ÔòÔÚËþ¶¥µÃµ½ £¬Ëþµ×A xB B

22

µÃµ½ ¡£

18¡¢ÍêÈ«»¥ÈܵÄA£¬B¶þ×é·ÖÈÜÒº£¬ÔÚxB=0.6´¦£¬Æ½ºâÕôÆøѹÓÐ×î¸ßÖµ£¬ÄÇô×é³ÉxB=0.4µÄÈÜÒºÔÚÆø-ƽºâʱ£¬yB(g)£¬xB(1)£¬xB(×Ü)µÄ´óС˳ÐòΪ__________¡£½«xB=0.4µÄÈÜÒº½øÐо«Áó, Ëþ¶¥½«µÃµ½______________¡£

19¡¢¶¨Ñ¹ÏÂA£¬B¿ÉÐγɾßÓÐ×îµÍºã·ÐµãµÄϵͳ£¬×îµÍºã·Ðµã×é³ÉΪxB=0.475¡£ Èô½øÁÏ×é³ÉΪxB=0.800µÄϵͳ£¬ÔÚ¾ßÓÐ×ã¹»Ëþ°åÊýµÄ¾«ÁóËþÖо«Áó£¬Ëþ¶¥½«µÃµ½______________¡£Ëþµ×½«µÃµ½______________¡£

20¡¢35¡æʱ£¬´¿CH3COCH3µÄ±¥ºÍÕôÆøѹÁ¦Îª43.06kPa¡£ CH3COCH3ÓëCHCl3×é³ÉÈÜÒº£¬µ±CHCl3µÄĦ¶û·ÖÊýΪ0.30ʱ£¬ÈÜÒºÉÏCH3COCH3µÄÕôÆøѹÁ¦Îª26.77kPa£¬Ôò¸ÃÈÜÒº¶ÔCH3COCH3Ϊ Æ«²î¡££¨Ñ¡ÔñÕý¡¢¸º£©

21¡¢ÓÒͼÊÇË®µÄÏàͼ¡£ (1)oaÊÇ¡ª¡ª¡ªÆ½ºâÇúÏߣ»

C obÊÇ¡ª¡ª¡ªÆ½ºâÇúÏߣ»

ocÊÇ______ƽºâÇúÏߣ» (2)oµã³ÆΪ________µã£»

oaÏßÖÐÖ¹ÓÚaµã£¬aµã³ÆΪ_______µã£»

(3)´¦ÓÚdµãµÄË®³ÆΪ______Ë®£¬ËüÓëË®ÕôÆø´ïµ½µÄƽºâ³ÆΪ

_______ƽºâ¡£

22¡¢ÒÒÏ©ëæ(A)-Ë®(B)-ÒÒÃÑ(C)ÔÚζÈʱµÄÈý×é·ÖҺ̬²¿·Ö»¥

A B ÈÜϵͳÏàͼÈçͼËùʾ£¬Ôò¸ÃÏàͼÖÐÓÐ ¸ö¶þÏàÇø

23. ÔÚ25¡æʱ£¬A£¬B£¬CÈýÖÖÎïÖÊ(²»·¢Éú»¯Ñ§·´Ó¦)ËùÐγɵÄÈÜÒºÓë¹ÌÏàAºÍB¡¢C×é³ÉµÄÆøÏà³Êƽºâ£¬ÔòÌåϵµÄ×ÔÓɶÈf = ¡£ A£¬B£¬CÈýÎïÖÊ×é³ÉµÄÌåϵÄÜƽºâ¹²´æµÄ×î´óÏàÊýÊÇ ¡£

24£®ÔÚʯ»ÒÒ¤ÖУ¬·Ö½â·´Ó¦CaCO3(s)=CaO(s)+CO2 (g)ÒÑƽºâ£¬Ôò¸ÃÌåϵµÄ×é·ÖÊýC= £¬ÏàÊýP = £¬×ÔÓɶÈÊýf = ¡£

25£®CaCO3(s),CaO(s),BaCO3(s),BaO(s)ºÍCO2(g)´ïµ½Æ½ºâʱ£¬´ËÌåϵµÄÏàÊýÊÇ £¬ ×é·ÖÊýÊÇ £¬×ÔÓɶÈÊýÊÇ ¡£

26£®´¿ÒºÌåÔÚÆäÕý³£·ÐµãÏ·ÐÌÚ±äΪÆøÌ壬ÏÂÊö¸÷Á¿ÖÐÔö¼ÓµÄÁ¿ÊÇ £¬¼õÉÙµÄÁ¿ÊÇ £¬²»±äµÄÁ¿ÊÇ ¡£

A£®ÕôÆøѹ B£®Ä¦¶ûÆø»¯ÈÈ C£®Ä¦¶ûìØ D£®Ä¦¶ûìÊ E£®¼ª²¼Ë¹×ÔÓÉÄÜ F£®ÎÂ¶È G£®Íâѹ H£®Ä¦¶ûÌå»ý I£®º¥Ä·»ô×È×ÔÓÉÄÜ

-1

27£®ÒÑ֪ˮµÄƽ¾ùÆø»¯ÈÈΪ40.67kJ¡¤mol,ÈôѹÁ¦¹øÔÊÐíµÄ×î¸ßζÈΪ423K£¬´ËʱѹÁ¦¹øÄÚµÄѹÁ¦Îª kPa. 28£®ÔÚһ͸Ã÷µÄÕæ¿ÕÈÝÆ÷ÖÐ×°Èë×ã¹»Á¿µÄ´¿ÒºÌ壬Èô¶ÔÆä²»¶Ï¼ÓÈÈ£¬¿É¼ûµ½ ÏÖÏó£¬Èô½«ÈÝÆ÷²»¶ÏÀäÈ´£¬Óֿɼûµ½ ÏÖÏó¡£

++£­2£­

29£®Ò»¸öº¬ÓÐK£¬Na,NO3,SO4,ËÄÖÖÀë×ӵIJ»±¥ºÍË®ÈÜÒº£¬Æä×é·ÖÊýΪ ¡£ 30£®ÒÑ֪ζÈTʱ£¬ÒºÌåAµÄÕôÆøѹΪ13330Pa£¬ÒºÌåBµÄÕôÆøѹΪ6665Pa£¬ÉèAÓëB¹¹³ÉÀíÏëÒºÌå»ìºÏÎÔòµ±AÔÚÈÜÒºÖеÄĦ¶û·ÖÊý0.5ʱ£¬ÆäÔÚÆøÏàÖеÄĦ¶û·ÖÊýΪ ¡£

31£®¶¨ÎÂÏÂË®£¬±½¼×Ëᣬ±½Æ½ºâÌåϵÖпÉÒÔ¹²´æµÄ×î´óÏàÊýΪ ¡£ 32£®¶þԪҺϵÏàͼÖУ¬¹²·ÐµãµÄ×ÔÓɶÈÊýf = ¡£

Èý¡¢Ñ¡ÔñÌ⣺

1 Ò»¸öË®ÈÜÒº¹²ÓÐSÖÖÈÜÖÊ£¬Ï໥֮¼äÎÞ»¯Ñ§·´Ó¦¡£ÈôʹÓÃÖ»ÔÊÐíË®³öÈëµÄ°ë͸Ĥ½«´ËÈÜÒº

23

Óë´¿Ë®·Ö¿ª£¬µ±´ïµ½Éø͸ƽºâʱ£¬Ë®ÃæÉϵÄÍâѹÊÇpW£¬ÈÜÒºÃæÉϵÄÍâѹÊÇps£¬Ôò¸ÃϵͳµÄ×ÔÓɶÈÊýΪ£º( )

(A)f = S (B) f = S + 1 (C) f = S + 2 (D) f = S + 3 2 NH4HS(s)ºÍÈÎÒâÁ¿µÄNH3(g)¼°H2 S(g)´ïƽºâʱÓУº( )¡£ (A)C=2£¬f=2£¬f =2 (B) C=1£¬f=2£¬f =1 (C) C=1£¬f=3£¬f =2 (D) C=1£¬f=2£¬f =3

3 ÈôA(l)ÓëB(l)¿ÉÐγÉÀíÏëҺ̬»ìºÏÎï,ζÈTʱ,´¿A¼°´¿BµÄ±¥ºÍÕôÆøѹp£Â£¾pA,Ôòµ±»ìºÏÎïµÄ×é³ÉΪ0£¼x£Â£¼1ʱ,ÔòÔÚÆäÕôÆøѹ-×é³ÉͼÉÏ¿É¿´³öÕôÆø×ÜѹpÓëpA£¬p£ÂµÄÏà¶Ô´óСΪ:( )

(A) p£¾p£Â

*

*

**

*

(B)p£¼pA

*

(C)pA£¼p£¼p£Â

**

4 ¶ÔÓÚºã·Ð»ìºÏÎÏÂÁÐ˵·¨ÖдíÎóµÄÊÇ£º( )¡£

(A) ²»¾ßÓÐÈ·¶¨×é³É (B)ƽºâʱÆøÏà×é³ÉºÍÒºÏà×é³ÉÏàͬ£» (C)Æä·ÐµãËæÍâѹµÄ±ä»¯¶ø±ä»¯ (D)Ó뻯ºÏÎïÒ»Ñù¾ßÓÐÈ·¶¨×é³É

5 ÒÑÖªÁò¿ÉÒÔÓе¥Ð±Áò£¬Ð±·½Áò£¬ÒºÌ¬ÁòºÍÆø̬ÁòËÄÖÖ´æÔÚ״̬¡£ÁòµÄÕâËÄÖÖ״̬____Îȶ¨¹²´æ¡£

(A) Äܹ» (B) ²»Äܹ» (C) ²»Ò»¶¨

6 ¶þ×é·ÖºÏ½ð´¦Óڵ͹²ÈÛζÈʱϵͳµÄÌõ¼þ×ÔÓɶÈÊýΪ£º( )¡£

(A)0 (B) 1 (C) 2 (D) 3 7¡¢ A(l)ÓëB(l)¿ÉÐγÉÀíÏëҺ̬»ìºÏÎÈôÔÚÒ»¶¨Î¶ÈÏ£¬´¿A¡¢´¿BµÄ±¥ºÍÕôÆøѹp*A£¾

p*B£¬ÔòÔڸöþ×é·ÖµÄÕôÆøѹ×é³ÉͼÉϵÄÆø¡¢ÒºÁ½ÏàƽºâÇø£¬³ÊƽºâµÄÆø¡¢ÒºÁ½ÏàµÄ×é³É±Ø

ÓÐ:( )

(A)y£Â£¾x£Â (B)y£Â£¼x£Â (C)y£Â£½x£Â

8¡¢ ÔÚ101 325PaµÄѹÁ¦Ï£¬I2ÔÚҺ̬ˮºÍCCl4Öдﵽ·ÖÅäƽºâ£¨ÎÞ¹Ì̬µâ´æÔÚ£©Ôò¸ÃϵͳµÄÌõ¼þ×ÔÓɶÈÊýΪ£¨ £©

(A) 1 (B) 2 (C) 0 (D) 3

9 ¡¢ÁòËáÓëË®¿ÉÐγÉH2SO4?H2O(s)£¬H2SO4?2H2O(s)£¬H2SO4?4H2O(s)ÈýÖÖË®ºÏÎÎÊÔÚ101 325PaµÄѹÁ¦Ï£¬ÄÜÓëÁòËáË®ÈÜÒº¼°±ùƽºâ¹²´æµÄÁòËáË®ºÏÎï×î¶à¿ÉÓжàÉÙÖÖ£¿( ) (A) 3ÖÖ£»(B) 2ÖÖ£»(C) 1ÖÖ£»(D) ²»¿ÉÄÜÓÐÁòËáË®ºÏÎïÓë֮ƽºâ¹²´æ¡£

10 ¡¢½«¹ÌÌåNH4HCO3(s) ·ÅÈëÕæ¿ÕÈÝÆ÷ÖУ¬ºãε½400 K£¬NH4HCO3 °´ÏÂʽ·Ö½â²¢´ïµ½Æ½ºâ£º NH4HCO3(s) === NH3(g) + H2O(g) + CO2(g) ϵͳµÄ×é·ÖÊýCºÍ×ÔÓɶÈÊýfΪ£º( ) (A) (C)

£Ã£½2£¬f £½2£» (B) £Ã£½2£¬f £½2£» £Ã£½2£¬f £½0£» (D) £Ã£½3£¬f £½2¡£

24