·¢²¼Ê±¼ä : ÐÇÆÚÁù ÎÄÕÂÕã½¹¤Òµ´óѧÎïÀí»¯Ñ§ÀúÄêÆÚÄ©¿¼ÊÔÌâÓë´ð°¸ÊÔ¸üÐÂÍê±Ï¿ªÊ¼ÔĶÁd0f91d3d580216fc700afd85
?3?3c?1.2475?10 mol?dmr½âÖ® £¨2·Ö£©
Æß¡¢½â:¢Å dP/dt=k2[A*]£¬
d[A*]/dt=k1[A]2£k£1[A][A*]-k2[A*]=0 £¨1·Ö£©
2
[A*]=k1[A]/(k2+k£1[A]) £¨2·Ö£© ËùÒÔ dP/dt=k2 k1[A]2/(k2+k£1[A]) £¨1·Ö£© ¢Æ k1/k£1=[A][A*]/[A]2øÉ £¨2·Ö£© [A*]=k1[A]2/(k1[A])=k1[A]/k£1øÉ £¨1·Ö£© ËùÒÔ dP/dt=k1k2[A]/k£1øÉ £¨1·Ö£© ¢Ç µ±k2< °Ë¡¢½â£ºÒ©Îï·Ö½â°Ù·ÖÊýÓëŨ¶ÈÎ޹أ¬Ôò¸Ã·Ö½â·´Ó¦ÎªÒ»¼¶·´Ó¦ £¨2·Ö£© 30?Cʱ ln1?kt1?y£¬´úÈëÊý¾ÝÇóÖª ln1?k1?52ÖÜ?2?k1?3.43?10?3?ÖÜ?1?1?0.3 £¨2·Ö£© Éè25?Cʱ£¬ËÙÂÊϵÊýΪk2£¬Ôò ?11?????T2T1?k2/ÖÜ?1130.0?103?11?ln?????3.43?10?38.314?298.15276.15? lnEk2??ak1R k2?0.224?ÖÜ?1? £¨3·Ö£© 25?CÏ·ֽâ30%ÐèÒªµÄ¼äΪt? t??111??1ln??lnÖÜ?1.59ÖÜ?k21?y?0.2241?0.3? £¨2·Ö£© ¼´25?C·Ö½â30%Ö»Ðè1.59ÖÜ£¬ÒÑ·ÅÖÃ2Öܺ󣬷ֽⳬ¹ý30%£¬Ò©ÎïʧЧ £¨1·Ö£© x?¾Å¡¢½â£º£¨1£© ת»¯ÂÊ cA0?cAcB?cC0.08?0.22???60¨º0cA00.50 £¨2·Ö£© £¨2£©¶ÔÓÚƽÐÐÒ»¼¶·´Ó¦ ?dc??A ?dt???k1cA?k2cA?(k1?k2)cA? £¨1·Ö£© ln cA0?(k1?k2)tcA £¨2·Ö£© 0.50mol?dm?3ln?(k1?k2)?30min?3 (0.08?0.22)mol?dm k1?k2?0.03min?1 £¨2·Ö£© k1cB0.08mol?dm?3k1?0.36???3k2 ½âÖ®k20.22mol?dm cC £¨3·Ö£© 2007Äê7ÔÂÎïÀí»¯Ñ§£ÉÏÊÔ¾í´ð°¸ ÐÕÃû£º ѧºÅ£º °à¼¶£º ÈονÌʦ£º ×Ü·Ö£º ÌâºÅ ·ÖÊý Ò» ¶þ1 ¶þ2 ¶þ3 ¶þ4 ¶þ5 ¶þ6 Èý¡¢ Ñ¡ÔñÌ⣨ÿÌâ2·Ö£¬20Ì⣬¹²40·Ö£© ×¢Ò⣺ÌâºÅºó±êÓС°A¡±µÄÌâÄ¿ÒªÇóÐÞÎïÀí»¯Ñ§AµÄͬѧ½â´ð£»ÌâºÅºó±êÓС°B¡±µÄÌâÄ¿ÒªÇóÐÞÎïÀí»¯Ñ§BµÄͬѧ½â´ð¡£ Ç뽫´ð°¸ÌîÔÚ±í¸ñÖУº 1 d 2b 3c 11b 12d 14a 18B, c; 19B, a ; 20B, b 4d 14c 5b 15a 6c 16c 7b 17b 8b 18a 9c 19d 10b 20a ËÄ¡¢ ¼ÆËãÌ⣨ÿÌâ10·Ö£¬6Ì⣬¹²60·Ö£© ×¢Ò⣺ÌâºÅºó±êÓС°A¡±µÄÌâÄ¿ÒªÇóÐÞÎïÀí»¯Ñ§AµÄͬѧ½â´ð£»ÌâºÅºó±êÓС°B¡±µÄÌâÄ¿ÒªÇóÐÞÎïÀí»¯Ñ§BµÄͬѧ½â´ð¡£ 1¡¢ 1 mol ѹÁ¦Îª PµÄÒºÌå A£¬ÔÚÆäÕý³£·ÐµãÏ£¬ÏòÕæ¿ÕÈÝÆ÷ÖÐÕô·¢£¬ÖÕ̬±äΪÓëʼ̬ͬÎÂͬѹµÄ1mol ÕôÆø A¡£ÉèÕôÆøΪÀíÏëÆøÌ壬ҺÌåÌå»ý¿ÉÒÔºöÂÔ£¬²¢ÒÑÖªÒºÌå A ÔÚ 67¡æµÄ ¡Ñ-1 ±¥ºÍÕôÆøѹΪ 0.5P£¬Õô·¢ÈÈΪ 34.92 kJ¡¤mol£¬ÇÒÓëζÈÎ޹ء£¼ÆËãÉÏÊö¹ý³Ì W¡¢ Q¡¢¦¤U¡¢¦¤H¡¢¦¤S¡¢¦¤G¡¢¦¤A [½â] ÓÉ¿Ë--¿Ë·½³Ì: ln(P2/P1)£½ ¦¤Hv,m/R ¡Á(1/Tb- 1/T2) ¼´ ln(0.5/1) £½ 34920/8.314¡Á(1/Tb- 1/340) ¡à Tb£½360.2 K [3·Ö] ÏòÕæ¿ÕÕô·¢: W £½ 0£¬¦¤H £½ ¦¤Hvap£½ 34.920 kJ [1·Ö] ¦¤S £½ ¦¤H/Tb£½ 96.95 J/K [1·Ö] ¦¤U £½ ¦¤H - ¦¤(PV) £½ ¦¤H-PgVg£½¦¤H -nRT £½34920-8.314¡Á360.2£½31925 J [2·Ö] Q £½ ¦¤U £½ 31925 J [1·Ö] ¦¤G £½¦¤H-T¦¤S£½34925 -96.95¡Á360.2£½ 0 [1·Ö] ¦¤A £½¦¤U - T¦¤S £½- 2995 J [1·Ö] øÉ 2¡¢298Kʱ£¬1molCO(g)·ÅÔÚ10molO2Öгä·ÖȼÉÕ£¬Çó£¨1£©ÔÚ298KʱµÄ?rHm£»£¨2£©¸Ã·´ øÉøÉ£1 Ó¦ÔÚ398KʱµÄ?rHm¡£ÒÑÖª£¬CO2ºÍCOµÄ?fHm·Ö±ðΪ£393.509kJ¡¤molºÍ ££££ £110.525kJ¡¤mol1£¬CO¡¢CO2ºÍO2µÄCp,m ·Ö±ðÊÇ29.142 J¡¤K1¡¤mol1¡¢37.11 J¡¤K1¡¤mol£1£1£1 ºÍ29.355 J¡¤K¡¤mol¡£ øÉ£1 (´ð°¸)£¨1£©?rHm= £282.97 kJ¡¤mol [4] £1 £¨2£©?rHm£¨398K£©= £283.65 kJ¡¤mol [6] -2 3-2 3 3¡¢1molÀíÏëÆøÌåºãη´¿¹ºãÍâѹ£¬´Ó1.00¡Á10mÅòÕ͵½10.00¡Á10m,¼ÆËã¸ÃÅòÕ͹ý³ÌÖÐÌåϵ·´¿¹µÄÍâѹ¡¢ÌåϵµÄìر䡢»·¾³µÄìر䡢¹ÂÁ¢ÌåϵµÄìر䡣ÒÑÖª»·¾³µÄζÈΪ121.9 K ¡£ ¡Ñ ½â£ºp = nRT»·/V2 = 1 mol¡Á8.314 J¡¤K¡¤mol¡Á121.9 K/(10.00¡Á10m) = 10.13 kPa [2] ?SÌå= nRln(V2/V1) -1-1-2 3-2 3-1 =1 mol¡Á8.314 J¡¤K¡¤mol¡Áln[10.00¡Á10m/(1.00¡Á10m)] =19.14 J¡¤K [3] Q»·= -p?V = -(10.13¡Á103 Pa) (10.00-1.00)¡Á10-2 m3 = -911.7 J [2] -1 ?S»·= Q»·/T»·= -911.7 J/121.9 K= -7.48J¡¤K [2] -1 ?S¹ÂÁ¢=?SÌå+?S»·= 11.66 J¡¤K [1] 4¡¢ÔÚ25¡æ£¬1kg´¿Ë®ÖУ¬Èܽⲻ»Ó·¢ÐÔÈÜÖÊB22.2g£¬¹¹³ÉһϡÈÜÒº¡£BÔÚË®Öв»µçÀ룬´ËÈÜÒºµÄÃܶÈΪ1010 kg¡¤m-3¡£ÒÑÖªBµÄĦ¶ûÖÊÁ¿Îª111.0g¡¤mol-1£¬Ë®µÄ·ÐµãÉý¸ß³£ÊýKbΪ0.52K¡¤mol-1¡¤Kg, ´¿Ë®ÔÚÕý³£·Ðµã100¡æʱĦ¶ûÆû»¯ìÊΪ40.67kJ¡¤mol-1£¬¿ÉÊÓΪһ³£Êý¡£Çó(1) ´ËÏ¡ÈÜÒºµÄ·ÐµãÉý¸ßÖµ¡£(2) ´ËÏ¡ÈÜÒº25¡æʱµÄÉø͸ѹ¡£ ½â£º(1) bB=(22.2/111.0)mol¡¤kg-1= 0.200mol¡¤kg-1 ?Tb=KbbB =0.52¡Á0.200K=0.104K [5] (2) cB=(22.2/111.0)mol/(1022.2g/1010 g¡¤dm-3)=0.1976mol¡¤dm-3 ?= cBRT=(0.1976¡Á1000¡Á8.315¡Á298.2)Pa=4.90¡Á105Pa [5] 5¡¢Çó·´Ó¦ C3H6(g)£«H2(g)=C3H8(g) ÔÚ 298KʱµÄƽºâ³£Êý¡£ÈôÔÁÏÆø×é³ÉΪC3H630%¡¢ ¡Ñ H240%¡¢C3H80.5%¡¢N229.5%(mol%)£¬ÇÒÌåϵѹÁ¦Îªp£¬·´Ó¦ÏòÄĸö·½Ïò½øÐУ¿ÒÑÖª 298K ʱÏÂÁÐÊý¾Ý£º C3H6 (g) C3H8 (g) H2(g) ??fHm-1 (kJ¡¤mol) -1-1-2 3 20.42 £103.85 0 ?-1Sm (kJ¡¤mol) 266.90 269.90 130.5 ?-1 ½â£º?rHm=£103.85£20.42=-124.27 kJ¡¤mol [1·Ö] ?-1-1?rSm=269.9-266.9£130.5=£127.5 J¡¤K¡¤moløÉ [1·Ö] ???-1?S?rGm?Hrmrm=£T=£86.275 kJ¡¤mol [2·Ö] ?¡Ñ¡Ñ?GrmÓÉ=£RTlnK, K=1.33¡Á1015 ??rGm=?rGm+RTlnQ [2·Ö] p =£86.275+8.314¡Á298¡Áln(0.005/0.3¡Á0.4) -1 =£94.15 kJ¡¤mol [3·Ö] [1·Ö] 6A¡¢ÇóT=300K, V=10mʱë²Æø·Ö×ÓµÄƽ¶¯Åä·Öº¯Êýqt¡¢¸÷ƽ¶¯×ÔÓɶȵÄÅä·Öº¯Êýft¼°Æ½¶¯ -23-1 ìØSt¡££¨ë²µÄÏà¶ÔÔ×ÓÁ¿Îª39.95£¬²£¶ú×ÈÂü³£Êýk=1.381¡Á10J?K £¬Plank³£Êýh=6.626 -34 ¡Á10J?s£© ½â£º¼ûÌì½ò´óѧ½Ì²Äϲáp.116¡£ qt [4·Ö] £» ft [2·Ö]; St¡£(p.134 ¹«Ê½9.8.9) [4·Ö] 6B¡¢ÒÑÖªZn-MgÌåϵÏàͼÈçÏÂ,»Ø´ðÎïϵµãA¡¢B¡¢C¡¢D¡¢EËù±íʾµÄÏà̬¡¢ ÏàÊý¼°Ìõ¼þ×ÔÓɶÈÊý£¨Ö¸ºãѹÌõ¼þϵÄ×ÔÓɶÈÊý£©¡£ -63 ?rGm£¼0,·´Ó¦ÏòÓÒ½øÐÐ ½â£ºA, ÈÜÒº øÉ=1 føÉ=2 (2·Ö) B, ÈÜÒº+MgZnøÉ(s) øÉ=2 føÉ=0 (2·Ö) C, ÈÜÒº+MgZnøÉ(s) øÉ=2 føÉ=1 (2·Ö) D, ÈÜÒº+MgZnøÉ(s)+Mg(s) øÉ=3 føÉ=0 (2·Ö) E, Zn(s)+MgZnøÉ(s) øÉ=2 føÉ=1 (2·Ö) 2008Äê2ÔÂÎïÀí»¯Ñ§-ÏÂÊÔ¾í´ð°¸ Ò»¡¢Ñ¡ÔñÌ⣺ ÌâÐò 1 2 3 4 5 6 7 8 9 10 ´ð°¸ 1¡¢c ÌâÐò 11 2¡¢b 3¡¢b 4¡¢b 5¡¢b 6¡¢d 7¡¢d 8¡¢b 9¡¢b 10¡¢c 12 13 14 15 16 17 18 19 20 ´ð°¸ 11¡¢b 12¡¢c 13¡¢a 14¡¢d 15¡¢d 16¡¢c 17¡¢b 18¡¢d 19¡¢a 20¡¢c BÀàÊÔ¾íÑ¡ÔñÌâ´ð°¸ 16¡¢c 17¡¢d 18¡¢a 19¡¢b 20¡¢d ¶þ¡¢¼ÆËãÌâ´ð°¸ 1¡¢µç³ØAg |AgCl | HCl(aq) | Cl2(100kPa) | PtÔÚ25¡æ£¬p?ʱµÄE =1.1372 V,µç¶¯ÊÆζÈϵÊýΪ -4-1 £5.95¡Á10¡¤V¡¤K¡£ £¨1£©Ð´³öµç¼«·´Ó¦ºÍµç³Ø·´Ó¦£» £¨2£©¿ÉÄæͨµç1Fºó£¬ÇóÆäÈÈЧӦQr£» £¨3£©Èô´Ë·´Ó¦ÎªÈÈ»¯Ñ§·´Ó¦£¬²»ÔÚµç³ØÖнøÐУ¬ÔòQΪ¶àÉÙ£¿ 1¡¢½â£º£¨1£©¸º¼«£ºAg(s)+Cl-?AgCl(s) +e- Õý¼«£º1/2Cl2(100 kPa)+e-=== Cl- µç³Ø·´Ó¦£ºAg(s)+ 1/2 Cl2(100 kPa) === AgCl(s) [3·Ö] (2)Qr= T?rSm= zFT (?E?T)P =£17.11 kJ¡¤mol-1 [3·Ö] (3)Èô·´Ó¦²»ÔÚµç³ØÖнøÐУ¬Ôò£º Qp= ?rHm= ?rGm+T?rSm=£zFE + Qr=£127 kJ¡¤mol-1 [4·Ö] Èç ??Srm×îºó¼ÆËã½á¹ûËã´í£¬Èç¼ÆËãÁË¡¢?rHm¡¢?rGmµÈÖµ²¢¼ÆËãÕýÈ·£¬¿É¸øÿ¸ö1·Ö ?2¡¢ÒÑÖª298 Kʱ£¬E(Hg22+|Hg)=0.7959 V£¬E(Hg2+|Hg)=0.851 V£¬Hg2SO4(s)µÄ»î¶È»ýΪ8.20¡Á10-7¡£ÊÔ¼ÆËãE( Hg2SO4|Hg) ¼°E(Hg2+| Hg22+|Pt) ¡£ 2¡¢½â£º£¨1£©Hg2SO4 (s)Èܽâƽºâʽ£ºHg2SO4 (s) === Hg22+ + SO42- Éè¼Æ³ÉÒ»µç³Ø·´Ó¦£¬Ôòµç³Ø·´Ó¦Îª£º ¸º¼«£º2Hg(l) £¡ú Hg22++2e- Õý¼«£ºHg2SO4 (s) +2e- £¡ú 2Hg(l) + SO42- [2·Ö] E= E(Hg2SO4|Hg) £ E(Hg22+|Hg) = (RT/2F)ln KSP [2·Ö] E( Hg2SO4|Hg)= (0.02958 lg8.2¡Á10-7 +0.7959)= 0.6159 V [2·Ö] £¨2£©µç¼«·´Ó¦£º Hg(l) £¡úHg2++2e- £¨1£© 2 Hg(l) £¡úHg22++2e- £¨2£© £¨1£©¡Á2 ££¨2£©µÃ£ºHg22+£¡úHg2++2e- £¨3£© [2·Ö] ????G?G?Grmrmrm ,3=2,1£,2 ???EEE312 £2 F= £4 F+2 F ?????EEEEE31212 =(4£2)/2 =2£ ???????? =(0.851¡Á2£0.7959) V = 0.9061 V