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¿Õ¼ä¹¹ÐÍ ËùÊôÄÚ£¨Í⣩¹ìÐÍ ËùÊô¸ß£¨µÍ£©×ÔÐý £¨¼Û¼üÀíÂÛ£© £¨¾§Ì峡ÀíÂÛ£© ¸ß×ÔÐý ¸ß×ÔÐý ¸ß×ÔÐý µÍ×ÔÐý ¸ß×ÔÐý µÍ×ÔÐý 4 5 3 1 2 0 sp3d2 sp3d2 sp3 dsp2 sp3d2 dsp2 °ËÃæÌå °ËÃæÌå ËÄÃæÌå ËıßÐÎ °ËÃæÌå ËıßÐÎ Íâ¹ìÐÍ Íâ¹ìÐÍ Íâ¹ìÐÍ ÄÚ¹ìÐÍ Íâ¹ìÐÍ ÄÚ¹ìÐÍ µÚ5Õ »¯Ñ§ÈÈÁ¦Ñ§
5-6£®²âµÃ2.96gÂÈ»¯¹¯ÔÚ407¡æµÄ1LÈÝ»ýµÄÕæ¿ÕϵͳÀïÍêÈ«Õô·¢´ïµ½µÄѹÁ¦Îª60kPa, ÇóÂÈ»¯¹¯ÕôÆøµÄĦ¶ûÖÊÁ¿ºÍ»¯Ñ§Ê½¡£
½â£º M?mRT2.96?8.314?(407?273)??278.9PV60?1
? HgCl2µÄʽÁ¿?271.496
? ·Ö×ÓʽΪ: HgCl25-8£®25¡æÊ±½«ÏàͬѹÁ¦µÄ5.0LµªÆøºÍ15LÑõÆøÑ¹Ëõµ½Ò»¸ö10.0LµÄÕæ¿ÕÈÝÆ÷ÖУ¬²âµÃ×ÜѹΪ150kPa£¬¢ÙÇóÁ½ÖÖÆøÌåµÄ³õʼѹÁ¦£»¢ÚÇó»ìºÏÆøÌåÖеªºÍÑõµÄ·Öѹ£»¢Û½«Î¶ÈÉÏÉýµ½210¡æ£¬ÈÝÆ÷µÄ×Üѹ¡£
½â£º (1) ? P1V1?P2V2 ? P1?P2V2150?10??75kPaV120(2) PN2?XN2?P? PO25?150?37.5kPa20?150?37.5?112.5kPa (3) P1?P2 P?150?(210?273)?243kPa2T1T2298
5-9. 25¡æ£¬1.47MPaϰѰ±ÆøÍ¨ÈëÈÝ»ýΪ1.00L¸ÕÐÔÃܱÕÈÝÆ÷ÖУ¬ÔÚ350¡æÏÂÓô߻¯¼Áʹ²¿·Ö°±·Ö½âΪµªÆøºÍÇâÆø£¬²âµÃ×ÜѹΪ5MPa£¬Ç󰱵ĽâÀë¶ÈºÍ¸÷×é·ÖµÄĦ¶û·ÖÊýºÍ·Öѹ¡£
PV1.47?103?11½â n1???0.593molRT8.314?298 1
3PV5?10?1 n2?2??0.965molRT8.314?(350?273) 22NH3 ==== N2 + 3H2
nƽ 0.593-2x x 3x
0.593 -2x +x +4x = 0.965 x =0. 186mol
5-10. ijÒÒÏ©ºÍ×ãÁ¿µÄÇâÆøµÄ»ìºÏÆøÌåµÄ×ÜѹΪ6930Pa£¬ÔÚ²¬´ß»¯¼ÁÏ·¢ÉúÈçÏ·´Ó¦£ºC2H4(g) + H2(g) = C2H6(g) ,·´Ó¦½áÊøÊ±Î¶ȽµÖÁÔζȺó²âµÃ×ÜѹΪ4530Pa¡£ÇóÔ»ìºÏÆøÌåÖÐÒÒÏ©µÄĦ¶û·ÖÊý¡£ ½â£º C2H4(g) + H2(g) = C2H6(g)
23
??2?0.186?100%?62.7%0.5930.593?2?0.186?0.2290.965XNH3?PNH3?0.229?5?1.14MPaXH2?0.186?0.1930.965PN2?0.193?5?0.965MPaPH2?5?0.578?2.89MPaXN2?1?0.229?0.193?0.578 Pʼ X Y 0 PÖÕ 0 Y-X X
Pʼ= X+Y = 6930Pa PÖÕ= Y-X+X = 4530Pa
? XÒÒÏ©?
X6930?4530??0.346X?Y6930
5-14.·´Ó¦CaC2(s) + 2H2O(l) = Ca(OH)2(s) +C2H2(g) ÔÚ298Kϵıê׼Ħ¶ûÈÈÁ¦Ñ§Äܱ仯Á¿Îª-128.0kJ.mol-1,Çó¸Ã·´Ó¦µÄ±ê׼Ħ¶ûìʱ䡣 ½â£º¡÷H ? =¡÷U ? +P¡÷V = ¡÷U ? +RT¡Æ?B(g)
= -128.0 +8.314¡Á10-3¡Á298¡Á1 = -125.5 kJ.mol-1
5-16. ÒÑÖªAl2O3(s)ºÍMnO2(s)µÄ±ê׼Ħ¶ûÉú³ÉìÊΪ
-1676kJ.mol-1ºÍ-521kJ.mol-1,¼ÆËã1gÂÁÓë×ãÁ¿MnO2(s)·´Ó¦²úÉúµÄÈÈÁ¿¡£ ½â£º 4Al(s) + 3MnO2(s) = 2Al2O3(s) + 3Mn(s)
¡÷fH ? m /kJ¡¤mol-1 0 -521 -1676 0
¡÷rHm? = 2¡Á(-1676) - 3¡Á(-521) = -1789kJ.mol-1
? Q?1?(?1789)??16.56kJ4?275-17. ÒÑÖªCl-(aq)µÄ±ê׼Ħ¶ûÉú³ÉìÊΪ-167.5kJ.mol-1£¬¼ÆËã1molHCl(g)ÈÜÓÚ×ãÁ¿Ë®ÖÐÊͷųö¶àÉÙÈÈ£¿ ½â£º HCl (g) +¡Þ(aq) = H+ (aq) + Cl- (aq)
¡÷fH ? m /kJ¡¤mol-1 -92.307 0 -167.5
¡÷rH ? m= (-167.5) - (-92.307) = -75.2kJ.mol-1
5-21
?rHm??12?fHm?(CO2,g)?10?fHm?(H2O,g)?4?fHm?[C3H5(NO3)3,l)½â´ð£º?12?(?393.509)?10?(?241.818)?4?(?355)kJmol?1
??5720.264kJmol?1??5.72MJmol?15-25. ÒÑÖªN2¡¢NOºÍO2µÄ½âÀëìÊ·Ö±ðΪ941.7kJ.mol-1¡¢631.8 kJ.mol-1ºÍ493.7kJ.mol-1,½öÀûÓÃÕâЩÊý¾ÝÅжÏNOÔÚ³£Î³£Ñ¹ÏÂÄÜ·ñ×Ô·¢·Ö½â¡£ ½â£º 2NO(g) == N2(g) + O2(g)
¡÷rH ? m= 2¡Á631.8 ¨C 941.7 ¨C 493.7 = -171.8kJ.mol-1
¡ß ·´Ó¦Ç°ºóÆøÌå·Ö×ÓÊýÏàͬ£¬¡÷S ?±ä»¯²»´ó¡£
¡à ·´Ó¦µÄ¡÷G ?Ö÷Ҫȡ¾öÓÚ¡÷H ? £¬ ¹Ê¡÷G ? £¼0£¬·´Ó¦×Ô·¢¡£
5-26. ÏÂÁз´Ó¦ÊÇìØÔö»¹ÊÇìØ¼õ·´Ó¦£¿¢ÙÆÏÌÑÌÇȼÉÕ£»¢ÚÒÒȲȼÉÕ£»¢Û̼ËáÇâÄÆ·Ö½â£»¢ÜÌú˿ȼÉÕ£»¢Ý¼×ÍéÓëË®ÕôÆø·´Ó¦Éú³ÉË®ÃºÆø£»¢Þ¼×ÍéÓëÑõÆø·´Ó¦Éú³ÉºÏ³ÉÆø¡£ ½â£º¢Ù C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l)
24
¢Ú 2C2H2(g) + 5O2(g) = 4CO2(g) + 2H2O(l) ¢Û 2NaHCO3(s) = Na2CO3(s) + CO2(g) + H2O(l) ¢Ü 3Fe(s) + 2O2(g) = Fe3O4(s) ¢Ý CH4(g) + H2O(g) = CO (g) + 3H2(g) ¢Þ 2CH4(g) + O2(g) = 2CO(g) + 4H2(g)
¢Ù¡¢¢Û¡¢¢Ý¡¢¢Þ ¡÷S£¾0 ¢Ú¡¢¢Ü ¡÷S£¼0
5-28. µâÎٵƵÆÄÚ·¢ÉúÈçÏ¿ÉÄæ·´Ó¦£º
W(s) + I2(g) = WI2(g) ¡÷fG? m /kJ.mol-1 0 19.327 -8.37 S? m /J.mol-1.K-1 33.5 260.69 251
¢Ù Çó623Kʱ£¬ÉÏʽ·´Ó¦µÄ¡÷rG? m
¢Ú ¹ÀËãWI2(g)ÔÚÎÙË¿ÉÏ·Ö½âËùÐèµÄ×îµÍζȡ£ ½â£º¢Ù ¡÷rG ? 298 = -8.37-19.327 = -27.697kJ.mol-1 ¡÷rS ? 298 = 251-33.5-260.69 = -43.19 J.mol-1.K-1
¡÷rH?298 = ¡÷rG ?298 +298¡÷rS ?298
= -27.697 + 298¡Á10-3¡Á( -43.19) = - 40.568 kJ.mol-1
¡÷rG ?623 ¡Ö ¡÷rH?298 - 623¡÷rS?298
= - 40.568 - 623¡Á (-43.19)¡Á10-3 = -13.66 kJ.mol-1 ¢Ú WI2(g) = W(s) + I2(g)
¡÷rH?298 = 40.568 kJ.mol-1 ¡÷rS ? 298 = 43.19 J.mol-1.K-1
5-29£®ÓÃ͹͸¾µ¾Û¼¯Ì«Ñô¹â¼ÓÈȷֽ⵹ÖÃÔÚÒº¹¯ÉϵÄ×°ÂúÒº¹¯ÊÔ¹ÜÄÚµÄÑõ»¯¹¯£¬Ê¹Æä·Ö½â¡£¹ÀË㣺ʹÑõÆøµÄѹÁ¦´ïµ½±ê̬ѹÁ¦ºÍ1kPaËùÐèµÄ×îµÍζȣ¬²¢¹À¼ÆÎªÊ¹ÑõÆøÑ¹Á¦´ï1kPa £¬ÊԹܵij¤¶ÈÖÁÉÙΪ¶à³¤£¿
½â£º HgO(s) == Hg(s) + ? O2(g)
¡÷fH?m/kJ.mol-1 -90.46 0 0
?-1-1
Sm/J.mol.K
T?40.568?1000?939.3K43.19 71.1 76.02 205.138
¡÷rH?298 = 90.46 kJ.mol-1
¡÷rS?298 = 1/2¡Á205.138 + 76.02 -71.1 = 107.5J.mol-1 K-1
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T?90460?841.5K107.5¢Ú1kPaϵķֽâÎÂ¶È ÊԹܵij¤¶ÈÓ¦´óÓÚ760mm¡£ 5-30
½â´ð£º2NO(g) + 2CO(g) = N2(g) + 2CO2(g)
T?9046011/2107.5?8.314ln()100?714.3K?rH298??2?fHm?(CO2,g)?2?fHm?(NO,g)?2?fHm?(CO,g)?2?(?393.509)?2?90.25?2?(?110.525)kJmol?1??746.648kJmol?1???? ?rS298??Sm(N2,g)?2Sm(CO2,g)?2Sm(NO,g)?2Sm(CO,g)
?191.61?2?213.74?2?197.674?2?210.761??197.78JK?1mol?1?rH298?-476.648?103T?==3775K??rS298?197.785-31£®Ê¯»ÒÒ¤µÄ̼Ëá¸ÆÐè¼ÓÈȵ½¶àÉٶȲÅÄֽܷ⣿ÈôÔÚÒ»¸öÓÃÕæ¿Õ±Ã²»¶Ï³éÕæ¿ÕµÄϵͳÄÚ£¬ÏµÍ³Ä򵀮øÌåѹÁ¦±£³Ö10Pa£¬¼ÓÈȵ½¶àÉٶȲÅÄֽܷ⣿ ½â£º CaCO3(s) == CaO(s) + CO2(g) ¡÷fH?m /kJ.mol-1 -1206.92 -635.09 -393.509 S?m /J.mol-1.K ¨C1 92.9 39.75 213.74 ¡÷rH?298=(-393.509)+(-635.09)¨C(-1206.92)=178.321 kJ.mol-1 ¡÷rS?298 = 213.74 + 39.75 ¨C 92.9 = 160.59 J.mol-1.K ¨C1 ¢Ù±ê̬ϵķֽâζÈ
¢Ú10PaϵķֽâζÈ
26
T?178321?1110.4KT160.591783210.010160.59?8.314ln()100?H?T????S?RlnJ?751.9KµÚ6Õ »¯Ñ§Æ½ºâ
6-1 д³öÏÂÁи÷·´Ó¦µÄ±ê׼ƽºâ³£Êý±í´ïʽºÍʵÑ鯽ºâ³£Êý±í´ïʽ ¢Ù. 2SO2(g) + O2(g) = 2SO3(g)
¢Ú. NH4HCO3(s) = CO2(g)+NH3(g) +H2O(g) ¢Û CaCO3(s) = CaO(s) + CO2(g) ¢Ü CO2(g) = CO2(aq) ½â£º¢Ù
¢Ú K??{PCO/P?}{PNH/P?}{PHO/P?}223K??{PSO3/P?}2{PSO2/P?}2{PO2/P?}KP?{PSO3}2{PSO2}2{PO2}KP?{PCO2}{PNH3}{PH2O}KP?PCO2 ¢Û ¢Ü
K??{PCO2/P?}[CO2]/C?K?PCO2/P??K?[CO2]PCO26-2 (1) K1 = K?2 = (2.2¡Á10-3)2 = 4.84¡Á10-6 (2) K2 = 1/ K? = 1/2.2¡Á10-3 = 4.55¡Á102
6-4 ²âµÃºÏ³É°±·´Ó¦ÔÚ500¡æµÄƽºâŨ¶È·Ö±ðΪ£º[H2]=1.15mol/L,[N2]=0.75mol/L,[NH3]= 0.26mol/L,ÇóK?¡¢KcÒÔ¼°KP£¨·Ö±ðÓÃPaºÍbarÎªÆøÌåµÄѹÁ¦µ¥Î»£©¡£ ½â£º N2(g) + 3H2(g) = 2NH3(g)
[NH3]20.262KC???0.0593 (mol/l)?233[N2][H2]0.75?1.15KP?KC(RT)?n?0.0593?(0.08314?773)?2?1.44?10?5bar?2KP?KC(RT)?n?0.0593?(8314?773)?2?1.44?10?15Pa?2K??KP(P?)??n?1.44?10?5?(1)2?1.44?10?56-5 ÒÑÖª£º¢Ù HCN = H+ + CN- K1 ? = 4.9¡Á10-10 ¢Ú NH3+H2O =NH4++OH- K2 ? = 1.8¡Á10-5 ¢Û H2O = H+ + OH- K3 ? = 1.0¡Á10-14
Çó·´Ó¦ NH3 + HCN = NH4+ +CN- µÄƽºâ³£Êý¡£ ½â£º¢Ù + ¢Ú - ¢ÛµÃ£º NH3 + HCN = NH4+ +CN-
K??K1.K2K3???4.9?10?10?1.8?10?5??0.881.0?10?146-6 ·´Ó¦CO + H2O = CO2 + H2 ÔÚ749KʱµÄƽºâ³£ÊýK ? =2.6 Éè¢Ù·´Ó¦ÆðʼʱCOºÍH2OµÄŨ¶È¾ùΪ1mol/L; ¢Ú ÆðʼʱCOºÍH2OµÄĦ¶û±ÈΪ1£º3£¬ÇóCOµÄƽºâת»¯ÂÊ¡£ÓüÆËã½á¹û˵Ã÷ÀÕÉ³ÌØÁÐÔÀí¡£
27
½â£º¡ß ¡÷n = 0 ¡à Kc= KP = K ?
¢Ù CO + H2O = CO2 + H2
Cƽ 1-x 1-x x x
X2KC??2.6 X?0.62 ??62%2(1?X)¢Ú CO + H2O = CO2 + H2 Cƽ 1-x 3-x x x
X2KC??2.6 X?0.86 ??86%(1?X)(3?X)¼ÆËã½á¹û˵Ã÷£¬Ôö´ó·´Ó¦ÎïµÄŨ¶È£¬Æ½ºâÏòÉú³ÉÎï·½ÏòÒÆ¶¯¡£
6-7 ½«SO3¹ÌÌåÖÃÓÚÒ»·´Ó¦Æ÷ÄÚ£¬¼ÓÈÈʹSO3Æø»¯²¢ÁîÆä·Ö½â£¬²âµÃζÈΪ900K£¬×ÜѹΪP ?ʱ£¬ÆøÌå»ìºÏÎïµÄÃܶÈΪ? = 0.925g/dm3£¬ÇóSO3µÄ½âÀë¶È¡£ ½â£º
M??RTP?0.925?8.314?900?69.21100
SO3 = SO2 + ? O2
nƽ 1-X X X/2 n = 1+X/2
MSO380??69.21 X?0.32XX1?1?226-8 ÒÑÖª·´Ó¦N2O4 = 2NO2 ÔÚ308Kϵıê׼ƽºâ³£ÊýK?Ϊ0.32¡£Çó·´Ó¦ÏµÍ³µÄ×ÜѹÁ¦ÎªP ?ºÍ2P ? £¬N2O4µÄ½âÀë¶È¼°Æä±È¡£ÓüÆËãµÄ½á¹û˵Ã÷ÀÕÉ³ÌØÁÐÔÀí¡£ ½â£º N2O4 == 2NO2
nƽ 1-x 2x n = 1+x
1?x2xPƽ ?P ?P1?x1?x2x?P}2{P/P}4x2?PNO2?1?xK???1?xPN2O4/P?(1?x2)P???P?P1?x?2{½«P = P ?ºÍ 2P ? ´úÈë½âµÃ£º x1= 27.2% x2=19.6% ¼´Ôö´óѹÁ¦£¬Æ½ºâ³¯ËõÐ¡ÆøÌåÌå»ýµÄ·½ÏòÒÆ¶¯¡£
6-9 ΪʹAg2OÔÚ³£ÎÂÏ·ֽ⣬ÑõµÄ·ÖѹӦ½µÎª¶à´ó£¿ ½â£º Ag2O(s) = 2Ag(s) + ? O2(g) ¦¤fGm ? /kJ¡¤mol-1 -11.2 0 0
¦¤rGm ? = 11.2 kJ¡¤mol-1
28
??G??11200lnK????4.52 K??0.011 RT8.314?298?K??(PO2/P?)1/2 PO2?(K?)2?P??0.0112?100?1.21?10?2kPa6-12 ÔÚ693KºÍ723KÏÂÑõ»¯¹¯·Ö½âΪ¹¯ÕôÆøºÍÑõµÄƽºâ×Üѹ·Ö±ðΪ5.16¡Á104ºÍ1.08¡Á105Pa£¬ÇóÔÚ¸ÃζÈÇøÓòÄڷֽⷴӦµÄ±ê׼Ħ¶ûìʺͱê׼Ħ¶ûìØ±ä¡£ ½â£º HgO(s) == Hg(g) + ? O2(g)
Pƽ X X/2
P×Ü1 = X1 +X1/2 = 5.16¡Á104 Pa X1=3.44¡Á104 Pa P×Ü2 = X2 +X2/2 = 1.08¡Á105 Pa X2=7.2¡Á104 Pa
K1?3.44?1043.44?1041/2?{}{}?0.1431.0?1052?1.0?1057.2?1047.2?1041/2?{}{}?0.432551.0?102?1.0?10?K2?K2?H?T2?T1ln??()RT2?T1K10.432?H?723?693ln?() ?H??1.54?105J.mol?10.1438.314723?693¡÷G693? =-RTlnK? =- 8.314¡Á693ln0.143=1.12¡Á104 J.mol-1 ¡÷GT ? = ¡÷H ? - T¡÷S ?
?H???GT1.54?105?1.12?104?S???206J.mol?1.K?1T693??6-14 À×Óêµ¼ÖÂ¿ÕÆøÖеĵªÆøºÍÑõÆø»¯ºÏΪNO¡£¾ÈÈÁ¦Ñ§¼ÆËãµÃÖª£¬ÔÚ2033KºÍ3000Kϸ÷´Ó¦´ïƽºâʱϵͳÖÐNOµÄÌå»ý·ÖÊý·Ö±ðΪ0.8%ºÍ4.5%£¬ÊÔÎʢٸ÷´Ó¦ÊÇÎüÈÈ·´Ó¦»¹ÊÇ·ÅÈÈ·´Ó¦£¿¢Ú¼ÆËã2033KʱµÄƽºâ³£Êý¡£
½â£º ¢Ù ζÈÉý¸ß£¬Æ½ºâÓÒÒÆ£¬ËùÒԸ÷´Ó¦ÊÇÎüÈÈ·´Ó¦
¢Ú N2 + O2 == 2NO
P³õ/kPa 78 21 0 Pƽ /kPa 78-0.4 21-0.4 0.8
6-19 ÒÑÖªÂÈÆøÔÚ±¥ºÍʳÑÎË®ÀïµÄÈܽâ¶ÈСÓÚÔÚ´¿Ë®ÀïµÄÈܽâ¶È¡£ÊÔ´ÓÆ½ºâÒÆ¶¯µÄÔÀí½âÊÍ¡£ ½â£º ¢Ù Cl2(g) === Cl2(aq)
29
{0.8/p?}2K??4.0?10?4??{78?0.4}/p{21?0.4}/p?¢ÚCl2(aq) + H2O(l) ===== Cl-(aq) + HClO(aq) + H+(aq)
¼ÓÈëʳÑΣ¬Ôö´óÁËCl-Ũ¶È£¬´Ùʹƽºâ¢ÚºÍ¢ÙÏò×óÒÆ¶¯£¬¹ÊÂÈÆøÔÚʳÑÎÖеÄÈܽâ¶È½µµÍ¡£
6-20 ʵÑé²âµÃÂÈÆøÈÜÓÚË®ºóÔ¼ÓÐÈý·ÖÖ®Ò»µÄCl2·¢ÉúÆç»¯×ª»¯ÎªÑÎËáºÍ´ÎÂÈËᣬÇó¸Ã·´Ó¦µÄƽºâ³£Êý¡£293KÏÂÂÈÆøÔÚË®ÖеÄÈܽâ¶ÈΪ0.09mol/L. ½â£º ¢Ù Cl2(aq)+H2O(l) ==== Cl-(aq)+HClO(aq)+H+(aq) Cƽ 0.06 0.03 0.03 0.03
0.033K??4.5?10?40.06?6-22 ³¬ÒôËÙ·É»úÔÚ·ÉÐзųöµÄȼÉÕÎ²ÆøÖеÄNO»áͨ¹ýÏÂÁз´Ó¦ÆÆ»µ³ôÆø²ã£ºNO(g)+O3(g)= NO2(g)+O2(g) ¢ÙÒÑÖª298KºÍ100KPaÏÂNO£¬NO2ºÍO3µÄÉú³É×ÔÓÉÄÜ·Ö±ðΪ£º86.7¡¢51.8 ¡¢163.6kJ.mol-1£¬ÇóÉÏÃæ·´Ó¦µÄKPºÍKC ¢Ú¼Ù¶¨·´Ó¦ÔÚ298KÏ·¢Éúǰ£¬¸ß²ã´óÆøÀïµÄNO¡¢O3ºÍO2µÄŨ¶È·Ö±ðΪ2¡Á10-9mol/L, 1¡Á10-9mol/L, 2¡Á10-3mol/L£¬NO2µÄŨ¶ÈΪÁ㣬ÊÔ¼ÆËãO3µÄƽºâŨ¶È¡£ ½â£º¢Ù NO(g) + O3(g) = NO2(g) + O2(g)
¡÷G? = 51.8 ¨C 86.7 ¨C 163.6 = -198.5 kJ.mol-1
lnK??198500?80.12 K??6.24?10348.314?298? ?n?0 ? KP?KC?K?¢Ú NO(g) + O3(g) = NO2(g) + O2(g)
-9
C³õ 2¡Á 1¡Á10-9 0 2¡Á10-3 10
Cƽ 10-9+X X 10-9-X 2¡Á10-3 +1¡Á10-9-X ¡ß KÖµºÜ´ó£¬X¡Ö0 6-23
½â£º CO (g) + 1/2 O2(g) = CO2 (g) ³õʼŨ¶È(mol.L-1) 4.0¡Á10-5 4.0¡Á10-4 ƽºâŨ¶È(mol.L-1) 4.0¡Á10-5-x 4.0¡Á10-4 4.0¡Á10-4 +x
2?10?3?10?9K??6.24?1034 X?3.2?10?38mol/L?910XKc?cCO2cCOc1/2O2 (4.0?10-4 +x)4??1?10 (4.0?10-5-x )( 4.0?10-4 )1/2?x?3.8?10?5mol.L?1Òò¶øCOµÄƽºâŨ¶ÈΪ[CO] = 4.0¡Á10-5-x = 4.0¡Á10-5-3.8¡Á10-5 = 2¡Á10-6mol.L-1 COµÄת»¯ÂÊΪ¦Á = 3.8¡Á10-5/4.0¡Á10-5 = 96%¡£
6-26 ÎÞË®ÈýÂÈ»¯ÂÁÔÚÈÈÁ¦Ñ§±ê׼ѹÁ¦ÏµÄÒÔϸ÷ζÈʱ²â¶¨µÄÃܶÈΪ£º
30
T/¡æ 200 600 800 d/kg.L-1 6.8¡Á10-3 2.65¡Á10-3 1.51¡Á10-3 ¢Ù ÇóÈýÂÈ»¯ÂÁÔÚ200¡æºÍ800¡æÊ±µÄ·Ö×Óʽ¡£ ¢Ú Çó600¡æÏÂµÄÆ½ºâÎïÖÖ¡£ ¢Û Çó600¡æÏ¸÷ÎïÖֵį½ºâ·Öѹ¡£ ¢Ü Çó600¡æµÄKCºÍKP ½â£º¢Ù Md1?1RT16.8?8 P?.314?473100?267.4 Md3RT31.51?8.314?10733?
P?100?134.7¡ß ʽÁ¿AlCl3 = 133.34
¡à 200¡æµÄ·Ö×ÓʽΪAl2Cl6 £¬800¡æÊ±ÎªAlCl3 ¢Ú d2RT2P?2.65?8.314?873
M2?100?192.3ËùÒÔ600¡æÏÂµÄÆ½ºâÎïÖÖΪAlCl3ºÍAl2Cl6¡£
¢Û ÉèAlCl3º¬Á¿ÎªX£¬ Al2Cl6º¬Á¿Îª1-X¡£
Ôò£º133.4X + 266.78(1-X) = 192.3 X= 55.78%
PAlCl3?55.78KPa P Al2Cl6?100?55.78?44.22KPa¢Ü½â£º Al2Cl6 == 2AlCl3
55.782 Kp?44.22?70.36KPa Kc?Kp(RT)??n?70.36?(8.314?873)?1?9.7?10?3mol/L
²¹³äÁ·Ï°Ìâ
1. ÒÑÖªÔÚBr2ÓëNOµÄ»ìºÏÎïÖУ¬¿ÉÄÜ´ïµ½ÏÂÁÐÆ½ºâ£º ¢Ù NO(g) + ? Br2(l) = NOBr(g) ¢Ú Br2(l) = Br2(g)
¢Û NO(g) + ? Br2(g) = NOBr(g)
A:Èç¹ûÔÚÃܱÕÈÝÆ÷ÖУ¬ÓÐÒºÌåBr2´æÔÚ£¬µ±Î¶ÈÒ»¶¨Ê±£¬Ñ¹ËõÈÝÆ÷£¬Æ½ºâÈçºÎÒÆ¶¯£¿B£ºÈç¹ûÔÚÃܱÕÈÝÆ÷ÖУ¬ÎÞÒºÌåBr2´æÔÚ£¬µ±Î¶ÈÒ»¶¨Ê±£¬Ñ¹ËõÈÝÆ÷£¬Æ½ºâÈçºÎÒÆ¶¯£¿£¨ÉèÈÔÎÞÒºäå³öÏÖ£©
½â£ºA£º¢Ù²»Òƶ¯ ¢ÚƽºâÏò×óÒÆ¶¯ ¢Û²»Òƶ¯£¨Î¶ÈÒ»¶¨Ê±£¬ÒºäåµÄÕôÆøÑ¹²»±ä£©¡£
B£º¢Ù¡¢¢Úƽºâ²»´æÔÚ£¬¢ÛƽºâÏòÓÒÒÆ¶¯¡£
31
2. ÒÑÖª·´Ó¦H2(g) = 2H(g) ¡÷H? = 412.5kJ.mol-1£¬ÔÚ3000K¼°P ?ʱ£¬H2ÓÐ9%Àë½â£¬ÎÊÔÚ3600Kʱ£¬H2µÄÀë½âÂÊΪ¶àÉÙ£¿
½â£º H2(g) == 2H(g)
nƽ 1 ¨C 0.09 2¡Á0.09 n×Ü = 1+0.09 (0.18/10.9)2 K1?0.91/1.09?0.032
lnK20.032?4125008.314(3600?30003600?3000)?2.76 K2?0.50 H2(g) == 2H(g)
nƽ 1 ¨C x 2x n×Ü = 1+x
p1-x?2x?ƽ 1?xp 1?xp
K{2x/(1?x)}24x22?(1?x)/(1?x)?1?x2?0.50
32
x?0.33µÚ7Õ »¯Ñ§¶¯Á¦Ñ§
7-2. 970KÏ£¬·´Ó¦2N2O =2N2+O2 ÆðʼʱN2OµÄѹÁ¦Îª2.93¡Á104Pa£¬²âµÃ·´Ó¦ÖÐϵͳµÄ×Üѹ±ä»¯ÈçÏÂËùʾ£º
t/s P(×Ü)/104Pa 300 3.33 900 3.63 2000 3.93 4000 4.14 Çó×î³õ300sÓë×îºó2000sµÄʱ¼ä¼ä¸ôÄ򵀮½Ô¼ËÙÂÊ¡£ ½â£º 2N2O = 2N2 + O2
(3.33?2.93)?104?1??13.3Pa.S?1300(4.14?3.93)?104?2??1.05Pa.S?1200027-3. (1) r?kc(NO)c(O2)
(2) k = 2.5¡Á103 (3) r = 1.4 ¡Á10-2
7-4. N2OÔÚ½ð±íÃæÉÏ·Ö½âµÄʵÑéÊý¾ÝÈçÏ£º
¢ÙÇó·Ö½â·´Ó¦µÄ·´Ó¦¼¶Êý¡£ ¢ÚÇóËÙÂʳ£Êý¡£
¢ÛÇóN2OÏûºÄÒ»°ëʱµÄ·´Ó¦ËÙÂÊ¡£ ¢Ü¸Ã·´Ó¦µÄ°ëË¥ÆÚÓë³õʼŨ¶È³Êʲô¹ØÏµ?
½â£º¢ÙÔÈËÙ·´Ó¦£¬ËÙÂÊÓë·´Ó¦ÎïµÄŨ¶ÈÎ޹أ¬Áã¼¶·´Ó¦¡£
¢Ú
¢Û v = 1.0¡Á10-3 mol¡¤l-1¡¤min-1
dc? ¢Ü dt ? k ? dc ? kdt ¶¨»ý·ÖµÃ£º
½«Ct = ? C¡ã´úÈëµÃ:
C(N2O)/mol.L-1 0.10 0.08 0.06 0.04 0.02 0 t/min 0 20 40 60 80 100 k???0.10?0.08?1.0?10?3mol.l?1.min?120c??ct?ktt1/2c0?2k7-7 ̼-14°ëË¥ÆÚΪ5720a£¬½ñ²âµÃ±±¾©Öܿڵêɽ¶¥¶´ÒÅÖ·³öÍÁµÄ¹Å°ß¹¹Ç»¯Ê¯ÖÐ14C/12C±ÈÖµÊǵ±½ñ»î
33
×ŵÄÉúÎïµÄ0.109±¶£¬¹ÀËã¸Ã»¯Ê¯ÊǾà½ñ¶à¾Ã£¿Öܿڵ걱¾©Ô³È˾à½ñÔ¼50ÍòÄ꣬ÄÜ·ñÓÃ̼-14·¨²â¶¨ËüµÄÉú»îÄê´ú£¿ ½â£º
²»¿ÉÐУ¬ÒòΪC-14µÄ·ÅÉäÐÔÌ«Èõ£¬Ëü¿É½Ï׼ȷ¶Ï´úµÄÑùÆ·Ó¦¾à½ñ1000¡«50000Äê¼ä¡£
7-9. ij·´Ó¦ÔÚ273KºÍ313KϵÄËÙÂʳ£Êý·Ö±ðΪ1.06¡Á10-5 s-1ºÍ2.93¡Á10-3s-1£¬Çó¸Ã·´Ó¦ÔÚ298KϵÄËÙÂʳ£Êý¡£
k?0.6930.693??1.21?10?4a?1t1/25720lnct??kt ln0.109??1.21?10?4tc0 t?1.83?104aEa313?2732.93?10?34 -1 ½â£º ln?( ) 1 . 06 ? 10 ? 5 Ea = 9.98¡Á10J.mol8.314313?273
k9.98?104298?273?() k = 4.24¡Á10-4 s-1 ?5 ln 1.06?108.314298?273
7-10. ijһ¼¶·´Ó¦£¬ÔÚ300Kʱ·´Ó¦Íê³É50%Ðèʱ20min,ÔÚ350Kʱ·´Ó¦Íê³É50%Ðèʱ5.0min£¬¼ÆËã¸Ã·´Ó¦µÄ»î»¯ÄÜ¡£
Ea = 2.42¡Á104 J.mol-1
7-14. ÓÐÈËÌá³öÑõÆøÑõ»¯ä廯ÇâÆøÌåÉú³ÉË®ÕôÆøºÍäåÕôÆøµÄ·´Ó¦Àú³ÌÈçÏ£º HBr + O2 ¡ú HOOBr HOOBr + HBr ¡ú 2HOBr HOBr + HBr ¡ú H2O + Br2
¢ÙÔõÑùÓÉÕâÈý¸ö»ùÔª·´Ó¦¼ÓºÍÆðÀ´µÃ¸Ã·´Ó¦µÄ¼ÆÁ¿·½³Ìʽ£¿ ¢Úд³ö¸÷»ùÔª·´Ó¦µÄËÙÂÊ·½³Ì¡£ ¢ÛÖ¸³ö¸Ã·´Ó¦ÓÐÄÄЩÖмäÌ壿
¢ÜʵÑéÖ¸³ö£¬¸Ã·´Ó¦µÄ±íÏÖËÙÂÊ·½³Ì¶ÔÓÚHBrºÍO2¶¼ÊÇÒ»¼¶µÄ£¬ÔÚÉÏÊöÀú³ÌÖУ¬ÄÄÒ»²½»ùÔª·´Ó¦ÊÇËٿز½£¿
½â£º¢Ù A + B +2C µÃ£º4HBr + O2 ¡ú 2H2O + 2Br2 ¢Ú V1?k1CHBr?CO2 ¢ÜµÚÒ»²½ÊÇËٿز½
34
½â£º ?
v1t2?v2t1 ? ln20Ea350?300?()5.08.314350?300V3?k3CHBr?CHOBrV2?k2CHBr?CHOOBr ¢ÛÖмäÌåÓУº HOOBr ºÍ HOBr
µÚ8Õ ˮÈÜÒº
8-9 ½â´ð£º
?Tb?100.17?100?0.17K?Tb?Kbm?0.17?0.512??M?150gmol?1
8-12 ½â´ð
£¨1£©ÆÏÌÑÌÇ´óÓÚÕáÌÇ¡£ÒòΪÕáÌǵķÖ×ÓÁ¿´óÓÚÆÏÌÑÌÇ£¬¶þÕßµÄÖÊÁ¿·ÖÊýÏàͬ£¬ËùÒÔÆÏÌÑÌÇÈÜÒºµÄÖÊÁ¿Ä¦¶ûŨ¶È±ÈÕáÌÇÈÜÒº´ó¡£ £¨2£©ÆÏÌÑÌÇÈÜÒºµÄÉøÍ¸Ñ¹´ó¡£
£¨3£©NaClÈÜÒºµÄÉøÍ¸Ñ¹´ó¡£ÒòΪNaClÊÇÇ¿µç½âÖÊ£¬Êµ¼ÊÈÜÖʵÄŨ¶È½Ó½üÓÚNaClŨ¶ÈµÄ2±¶£¨¼´1mol.L-1£©¡£
£¨4£© CaCl2ÈÜÒºµÄÉøÍ¸Ñ¹´ó¡£
8-13 ½â´ð
º£Ë®µÄÖÊÁ¿·ÖÊýΪ3.5%£¬Òò¶ø¿ÉÒÔ¼ÆËã³öº£Ë®ÖÐÑεÄÖÊÁ¿Ä¦¶ûŨ¶È£¨ÒÔNaCl¼Æ£©Îª m¡£
0.4961000 ?M10m?58.5?100%?3.5%?m?0.62molKg?1
m?58.5?1000Ôòº£Ë®ÖÐʵ¼ÊÁ£×ÓµÄŨ¶ÈΪm* = 2¡Á0.62= 1.24mol.kg-1
?Tf?Kfm*?1.855?1.24?2.3?Tb?Kbm?0.512?1.24?0.64
*
ËùÒÔ£¬º£Ë®¿ªÊ¼½á±ùµÄζÈΪ-2.3¡æ£»¿ªÊ¼·ÐÌÚµÄζÈΪ106.4¡æ¡£
??cRT?1.2?8.314?298?30000kPa?3Mpa
35
µÚ9Õ Ëá¼îƽºâ
9-4. ±½¼×ËáµÄËá³£ÊýKa = 6.4¡Á10-5£¬ÊÔÇó£º
¢ÙÖкÍ1.22g±½¼×ËáÐè0.4mol.L-1µÄNaOHÈÜÒº¶àÉÙºÁÉý. ¢ÚÇóÆä¹²éî¼îµÄ¼î³£ÊýKb¡£
¢Û±½¼×ËáÔÚË®ÖеÄÈܽâ¶ÈΪ2.06g.L-1£¬Ç󱥺ÍÈÜÒºpH¡£ ½â£º¢Ù 1.22/122 = 0.4¡Á10-3 x x = 25ml
¢Ú ¢Û
Kw1.0?10?14?10Kb???1.56?10Ka6.4?10?52.06?1.69?10?2mol/L122
?5?52?5?2?6.4?10?(6.4?10)?4?6.4?10?1.7?10? ? [H]?2C?pH = 3
9-5. ¼ÆËãÏÂÁÐÈÜÒºµÄpHÖµ£º ¢Ù10ml5.0¡Á10-3mol.L-1µÄNaOH¡£
¢Ú10ml0.4mol.L-1HClÓë10ml0.1mol.L-1NaOHµÄ»ìºÏÒº¡£ ¢Û10ml0.2mol.L-1NH3¡¤H2OÓë10ml0.1mol.L-1HCl»ìºÏÒº¡£ ¢Ü10ml0.2mol.L-1HAcÓë10ml0.2mol.L-1NH4ClµÄ»ìºÏÒº¡£ ½â£º¢Ù pOH= 2.3 pH = 11.7
¢Ú ·¢ÉúÖкͷ´Ó¦£¬pH È¡¾öÓÚÊ£ÓàµÄHCl¡£
9-8. 25¡æ±ê׼ѹÁ¦ÏµÄCO2ÆøÌåÔÚË®ÖеÄÈܽâ¶ÈΪ0.034mol.L-1,ÇóÈÜÒºµÄpHÖµºÍ[CO32-]¡£ ½â£º
[H?]?0.3?10?0.15mol/L pH?0.8220?¢Û¹ýÁ¿µÄNH3Óë²úÎïNH4Cl×é³É»º³åÌåϵ
C(NH4)pH?pKa?lg?9.24C(NH3?H2O)¢Ü pHÈ¡¾öÓÚHAcµÄµçÀë
[H?]?cKa?0.1?1.75?10?5?1.32?10?3mol/LpH= 2.88
[H?]?cK1?0.034?4.3?10?7?1.21?10?4mol.L?1pH = 3.92
[CO3]?Ka2?5.61?10?11mol.L?12?9-9. ½«15gP2O5ÈÜÓÚÈÈË®£¬Ï¡ÊÍÖÁ750ml, ÉèP2O5È«²¿×ª»¯ÎªH3PO4£¬¼ÆËãÈÜÒºÖеĸ÷×é·ÖŨ¶È¡£
K1=7.11¡Á10-3 K2=6.23¡Á10-8 K3=4.5¡Á10-13
36
½â£º
15?21000?1C(HPO)???0.28mol.L34 142750H3PO4 ==== H+ + H2PO4-
Cƽ 0.28 - x x x
x2K1??7.11?10?3 x?0.042mol.L?10.28?x¼´£º[H+] = [H2PO4-] = 0.042mol.L-1
[HPO42-] ¡Ö K2 = 6.23¡Á10-8
[H?][PO4]0.042?[PO4]K3???4.5?10?132??8[HPO4]6.23?10[PO43-] = 6.7¡Á10-19 mol.L-1
3?3?9-10. ijÈõËáHA£¬0.015mol.L-1µçÀë¶ÈΪ0.80%£¬Å¨¶ÈΪ0.1mol.L-1µçÀë¶È¶à´ó£¿ ½â£º
c?? 1?22c2?120.015?2? ? ?2?0.31%0.100.80"9-11. ¼ÆËã0.100mol.L-1Na2CO3ÈÜÒºµÄpHÖµºÍµçÀë¶È¡£ ½â£º CO32- + H2O ===== HCO3- + OH-
pOH = 2.37 pH = 11.36
9-12. ijδ֪Ũ¶ÈµÄÒ»ÔªÈõËáÓÃδ֪Ũ¶ÈµÄNaOHµÎ¶¨£¬µ±ÓÃÈ¥3.26mlNaOHʱ£¬»ìºÏÈÜÒºµÄpH=4.00£¬µ±ÓÃÈ¥18.30mlʱ£¬»ìºÏÈÜÒºµÄpH=5.00¡£Çó¸ÃÈõËáµÄµçÀë³£Êý¡£ ½â£º
9-13. »º³åÈÜÒºHAc-Ac-µÄ×ÜŨ¶ÈΪ1.0mol.L-1£¬µ±ÈÜÒºµÄpHΪ£º¢Ù4.00£»¢Ú5.00£¬HAcºÍAc-µÄŨ¶È·Ö±ðΪ¶à´ó£¿ ½â£º¢Ù
Kw1.0?10?14?4Kb1???1.78?10Ka25.61?10?11[OH?]?c?Kb1?0.100?1.78?10?4?4.22?10?3mol.L?14.22?10?3???100%?4.22%0.1004.00?pKa?lgnËá?3.26x3.26x5.00?pKa?lgnËá?18.30x18.30xÁªÁ¢½âµÃ£ºnËá = 37.5x Ka = 9.52¡Á10-6
1?c¼î4.00?4.75?lgc¼î5.00?4.75?lg1?c¼îc¼îc¼î?0.15mol.L?1 cËá?0.85mol.L?137
¢Ú
c¼î?0.64mol.L?1 cËá?0.36mol.L?19-14. ÓûÅäpH=5.0µÄ»º³åÈÜÒº£¬Ðè³ÆÈ¡¶àÉÙ¿ËNaAc¡¤3H2O¹ÌÌåÈܽâÔÚ300ml0.5mol.L-1µÄHAcÈÜÒºÖУ¿ ½â£º
9-15. ijº¬ÔÓÖʵÄÒ»Ôª¼îÑùÆ·0.5000g(·Ö×ÓÁ¿Îª59.1)£¬ÓÃ0.1000mol.L-1HClµÎ¶¨£¬ÐèÓÃ75.00ml;Ôڵζ¨¹ý³ÌÖУ¬¼ÓÈë49.00mlËáʱ£¬ÈÜÒºµÄpH=10.65¡£Çó¸Ã¼îµÄµçÀë³£ÊýºÍÑùÆ·µÄ´¿¶È¡£ ½â£º
9-16. ½«Na2CO3ºÍNaHCO3»ìºÏÎï30gÅä³É1LÈÜÒº£¬²âµÃÈÜÒºµÄpH=10.62¡£¼ÆËãÈÜÒºº¬Na2CO3ºÍNaHCO3¸÷¶àÉÙ¿Ë£¿ ½â£ºÉèNaHCO3º¬x¿Ë
ÔòNa2CO3Ϊ: 30 ¨C 7.59 = 22.41g
9-19. ¼ÆËã10ml0.3mol.L-1µÄHAcºÍ20mlŨ¶ÈΪ0.15mol.L-1 HCN»ìºÏµÃµ½µÄÈÜÒºÖеÄ[H+]¡¢[Ac-]¡¢[CN-]¡£
½â£º KHAc = 1.76¡Á10-5 KHCN = 4.93¡Á10-10
¡ß KHAc >> KHCN ¡à [H+]È¡¾öÓÚHAcµÄµçÀë
9-20. ½ñÓÐClCH2COOH¡¢HCOOHºÍ(CH3)2AsO2H,ËüÃǵĵçÀë³£Êý·Ö±ðΪ£º1.40¡Á10-3¡¢1.77¡Á10-4¡¢6.40¡Á10-7¡¢ÊÔÎÊ£º¢ÙÅäÖÆpH=3.5µÄ»º³åÈÜҺѡÓÃÄÄÖÖËá×îºÃ£¿¢ÚÐèÒª¶àÉÙºÁÉýŨ¶ÈΪ4.0mol.L-1µÄËá
0.3?0.55.00?4.75?lg x / 136 x = 36.3g
10.65?pKa?lg0.1000?49.000.1000?(75.00?49.00)pKa?10.92 Ka?1.19?10?111.0?10?14Kb??8.4?10?4?111.19?100.5x?0.1?0.075 x?88.65Y.110.62?10.25?lgx/84 x?7.59g(30?x)/106[H?]?[Ac?]?10?0.3?1.76?10?5?1.33?10?3mol.L?130KHCN[H?][CN?]1.33?10?3[CN?]???4.93?10?1020[HCN]?0.1530[CN-] = 3.7¡Á10-8 mol.L-1
38
ºÍ¶àÉÙ¿ËNaOH²ÅÄÜÅä³É1L¹²éîËá¼î¶ÔµÄ×ÜŨ¶ÈΪ1.0mol.L-1µÄ»º³åÈÜÒº¡£ ½â£º¢ÙÑ¡ÓÃHCOOH×îºÃ¡£
9-22. ·Ö±ð¼ÆËãÏÂÁлìºÏÈÜÒºµÄpHÖµ£º
¢Ù50.0ml0.200mol.L-1NH4ClºÍ50.0ml0.200mol.L-1NaOH. ¢Ú50.0ml0.200mol.L-1NH4ClºÍ25.0ml0.200mol.L-1NaOH ¢Û 25.0ml0.200mol.L-1NH4ClºÍ50.0ml0.200mol.L-1NaOH ¢Ü 20.0ml1.00mol.L-1H2C2O4ºÍ30.0ml1. 00mol.L-1NaOH ½â£º¢Ù ÍêÈ«·´Ó¦£¬Éú³É°±Ë®¡£
V? ¢Ú Ëá1?1.0?0.25L4.01.0?x x?0.36mol.L?1xm¼î?0.36?40?14.4g3.50?p1.77?10?4?lg[OH?]?0.1?1.77?10?5?1.33?10?3mol.L?1 pOH = 2.88 pH = 11.12 ¢Ú Ê£ÓàµÄNH4ClºÍÉú³ÉµÄ°±Ë®×é³É»º³åÌåϵ
1.0?10?14?10 pH?pKa?9.25Ka??5.6?101.77?10?5
¢Û ¼î¶ÈÈ¡¾öÓÚ¹ýÊ£µÄ¼î
[OH?]?25.0?0.2?0.067mol.L?1 75.0 pOH = 1.17 pH = 12.83
¢Ü H2C2O4 + NaOH = NaHC2O4 + H2O Ê£Óà10mlNaOH NaHC2O4 + NaOH = Na2C2O4 + H2O
·´Ó¦²úÉúµÄNa2C2O4ÓëÊ£ÓàµÄNaHC2O4×é³É»º³åÌåϵ
pH?pK2?4.27
39
µÚ10Õ ³ÁµíÈÜ½âÆ½ºâ
10-1. ¢Ù25¡æÊ±PbI2ÔÚ´¿Ë®ÖÐÈܽâ¶ÈΪ1.29¡Á10-3 mol.L-1,ÇóPbI2µÄÈܶȻý¡£ ¢Ú25¡æÊ±BaCrO4ÔÚ´¿Ë®ÖÐÈܽâ¶ÈΪ2.91¡Á10-3 g.L-1,ÇóBaCrO4µÄÈܶȻý¡£ ½â£º¢Ù ¢Ú
Ksp(PbI2)?[Pb2?][I?]2 ?(1.29?10?3)(2?1.29?10?3)2?8.58?10?9Ksp(BaCrO4)?[Ba2?][CrO4]2.91?10?32 ?()?1.32?10?10253.322?10-3. AgIO3ºÍAg2CrO4µÄÈܶȻý·Ö±ðΪ9.2¡Á10-9 ºÍ1.12¡Á10-12 £¬Í¨¹ý¼ÆËã˵Ã÷£º ¢ÙÄÄÖÖÎïÖÊÔÚË®ÖеÄÈܽâ¶È´ó£¿
¢ÚÄÄÖÖÎïÖÊÔÚ0.01mol.L-1µÄAgNO3ÈÜÒºÖÐÈܽâ¶È´ó? ½â£º¢Ù AgIO3 ==== Ag+ + IO3-
S1?Ksp(AgIO3)?9.2?10?9?9.59?10?5mol.L?1Ag2CrO4 === 2Ag+ + CrO42-
Cƽ 2S2 S2
Ksp = [Ag+]2[CrO42-] = 4S23 S2 = 6.54¡Á10-5 mol.L-1
¢Ú AgIO3 ==== Ag+ + IO3- Cƽ 0.01+x x
Ksp =£¨0.01+x) x = 9.2¡Á10-9 x = 9.2¡Á10-7 mol.L-1
Ag2CrO4 === 2Ag+ + CrO42- Cƽ 0.01+2y y
Ksp =£¨0.01+2y)2 y = 1.12¡Á10-12 y = 1.12¡Á10-8 mol.L-1
10-4. ÏÖÓÐ100 ml Ca2+ºÍBa2+µÄ»ìºÏÒº£¬Á½ÖÖÀë×ÓµÄŨ¶È¶¼Îª0.01 mol.L-1¡£ ¢ÙÓÃNa2SO4×÷³Áµí¼ÁÄÜ·ñ½«Á½ÖÖÀë×Ó·ÖÀ룿 ¢Ú¼ÓÈë¶àÉÙNa2SO4²ÅÄÜʹBaSO4ÍêÈ«³Áµí£¿ ½â£º¢Ù Ksp(CaSO4) = 7.1¡Á10-5 Ksp(BaSO4) = 1.07¡Á10-10
Ca2+³Áµíʱ£º
´Ëʱ£º
7.1?10?5[SO4]??7.1?10?3mol.L?10.011.07?10?102?[Ba]??1.5?10?8mol.L?1?37.1?102? ¿É½«Á½ÖÖÀë×Ó·ÖÀë¡£
¢Ú W = 142.04 ¡Á 0.01¡Á0.1= 0.142g
10-8. 1L¶à´óŨ¶ÈµÄNH4ClÈÜÒº¿Éʹ1g Mg(OH)2³ÁµíÈܽâ?
40
½â£º
[Mg2?]?1?0.017mol.L?158.322NH4+ + Mg(OH)2 ==== Mg2+ + 2NH3¡¤H2O Cƽ x 0.017 2¡Á0.017
10-9. ÔÚ0.1 mol.L-1 HAcºÍ0.1 mol.L-1 CuSO4ÈÜÒºÖÐͨÈëH2S´ï±¥ºÍ£¬ÊÇ·ñÓÐCuS³ÁµíÉú³É£¿ ½â£º
10-10. ¼ÆËãÏÂÁз´Ó¦µÄƽºâ³£Êý£¬²¢ÌÖÂÛ·´Ó¦µÄ·½Ïò: ¢Ù PbS + 2HAc Pb2+ + H2S + 2Ac-
ƽºâ³£ÊýºÜС£¬ËùÒÔ·´Ó¦ÄæÏò½øÐÐ
10-12. ¶¨Á¿·ÖÎöÖÐÓÃAgNO3ÈÜÒºµÎ¶¨Cl-ÈÜÒº£¬ÓÃK2CrO4×÷ָʾ¼Á£¬Îʵζ¨ÖÕµãʱÈÜÒºÖеÄCrO42-Àë×ÓµÄŨ¶È¶à´óºÏÊÊ£¿ÉèÖÕµãʱÈÜÒºÌå»ýΪ50ml£¬Ôڵζ¨¿ªÊ¼Ê±Ó¦¼ÓÈë0.1mol.L-1µÄK2CrO4ÈÜÒº¶àÉÙºÁÉý£¿
½â£ºCl-¶¨Á¿³Áµíʱ£º
´ËʱAg2CrO4³ÁµíָʾÖÕµã [CrO4
2?[Mg2?][NH3?H2O]2[OH?]2Ksp5.61?10-12K???2??0.018?2?2-52 [OH](1.77?10)[NH4]Kb0.017?0.0342K??0.018 x?0.033mol.L?12xC0(NH4Cl) = 0.033 + 0.034 = 0.067 mol.L-1
[H?]?0.1?1.76?10?5?1.33?10?3mol.L?1[S2?]?K1K2[H2S][H?]25.7?10?8?1.2?10?15?0.1 ??3.9?10?18mol.L?1?32(1.33?10)J?[Cu2?][S2?]?0.1?3.9?10?18?3.9?10?19? J?Ksp?1.27?10?36 ? ÓгÁµíÉú³É¡£ [Pb2?][H2S][Ac?]2Ksp?Ka(HAc)?K?2K1?K2(H2S)[HAc]29.04?10?29?(1.76?10?5)2??4.09?10?16?8?155.7?10?1.2?101.77?10?10[Ag]???1.77?10?4mol.L?1??6[Cl]1.0?10?Ksp1.12?10?12?5?1]???3.6?10mol.L[Ag?]2(1.77?10?4)2Ksp 41
3.6¡Á10-5¡Á50 = 0.1x x = 0.02ml
10-14. ½ñÓÐÒ»ÈÜÒº£¬Ã¿ºÁÉýº¬Mg2+ºÍFe2+¸÷1mg£¬ÊÔ¼ÆË㣺Îö³öMg(OH)2ºÍFe(OH)2³ÁµíµÄ×îµÍpHÖµ¡£
½â£º [Mg2?]?1?0.0411mol.L?124.31
ҪʹMg(OH)2¡ý£º
ҪʹFe(OH)2¡ý£º
[OH]??[Fe2?]?1?0.0179mol.L?155.845Ksp[Mg2?]?5.61?10?12?1.17?10?5mol.L?10.0411 pOH = 4.93 pH = 9.07
4.87?10?17[OH]???5.21?10?8mol.L?12?[Fe]0.0179?KsppOH = 7.28 pH = 6.72
10-15. ÓÃNa2CO3ºÍNa2SÈÜÒº´¦ÀíAgI¹ÌÌ壬Äܲ»Äܽ«AgI¹ÌÌåת»¯ÎªAg2CO3ºÍAg2S£¿ ½â£º2AgI + CO32- === Ag2CO3 + 2I-
2?172[I?]2Ksp(AgI)(8.51?10)?22K???8.57?10?2?[CO3]Ksp(Ag2CO3)8.45?10?12 ²»ÄÜת»¯
2AgI + S2- === Ag2S + 2I-
(8.51?10?17)217?1.08?10 K ? ? ? ÄÜת»¯
Ksp(Ag2S)6.69?1050Ksp(AgI)2 42
µÚ11Õ Ñõ»¯»¹Ô·´Ó¦
11-9£®ÓÃÄÜË¹ÌØ·½³Ì¼ÆËãÀ´ËµÃ÷£¬Ê¹Fe + Cu2+ = Fe2+ + CuµÄ·´Ó¦Äæ×ªÊÇ·ñÓÐÏÖʵµÄ¿ÉÄÜÐÔ£¿
½â: Fe + Cu2+ = Fe2+ + Cu
0.05920.0592 E?{0.345?lg[Cu2?]}?{?0.44?lg[Fe2?]}22
2?0.0592[Cu] ?0.785?lg2[Fe2?]
[Fe2?]26?3.3?10ҪʹE£¼0 ±ØÐë
[Cu2?]
ʵ¼Ê¹¤×÷ÖкÜÄÑ´ïµ½£¬¹ÊҪʹ·´Ó¦Äæ×ªÊDz»¿ÉÄܵġ£
11-10. ÓÃÄÜË¹ÌØ·½³Ì¼ÆËãÓë¶þÑõ»¯ÃÌ·´Ó¦µÃµ½ÂÈÆøµÄÑÎËáÔÚÈÈÁ¦Ñ§ÀíÂÛÉϵÄ×îµÍŨ¶È ½â£º MnO2 + 4H+ +2e- = Mn2+ + 2H2O ??= 1.228V
Cl2 + 2e- = 2Cl- ?? = 1.358V ÉèҪʹ·´Ó¦½øÐеÄHClµÄ×îµÍŨ¶ÈΪx¡£
11-12. ÀûÓð뷴Ӧ2H+ + 2e- = H2µÄ±ê×¼µç¼«µçÊÆºÍ´×ËáµÄµçÀë³£Êý¼ÆËã°ë·´Ó¦µÄ±ê×¼µç¼«µçÊÆ¡£
2HAc + 2e- = H2 + 2Ac-
½â£º±ê̬ʱ£¬[HAc]=[Ac-]=1mol.L-1, p(H2) =p?
11-13. ÀûÓð뷴ӦCu2+ + 2e- = Cu ºÍCu(NH3)42+ +2e- = Cu + 4NH3µÄ±ê×¼µç¼«µçÊÆ(-0.065V)¼ÆËãÅäºÏ·´Ó¦ Cu2+ + 4NH3 = Cu(NH3)42+ µÄƽºâ³£Êý¡£ ½â£º¢Ù Cu2+ + 2e- = Cu
¢Ú Cu(NH3)42+ +2e- = Cu + 4NH3
¢Ù - ¢ÚµÃ£º Cu2+ + 4NH3 = Cu(NH3)42+
11-17. ÀûÓÃCr2O72- + H2O = 2CrO42- + 2H+ µÄK=1014ºÍAg2CrO4µÄÈܶȻýÒÔ¼°Ag+ + e- = AgµÄ±ê×¼µç¼«µçÊÆÇó2Ag2CrO4 + 2H+ + 4e- = 4Ag + Cr2O72- + H2OµÄ??¡£
43
1.228?0.05920.05921lgx4?1.358?lg222xx = 5.4 mol.L-1
[H?][Ac?]Ka??[H?]?1.75?10?5[HAc]??0?0.0592lg[H?]2?0.0592lg1.75?10?5??0.28V2nE?2?(0.345?0.065)lgK???13.850.05920.0592K = 7.1¡Á1013
½â£º¢Ù Cr2O72- + H2O = 2CrO42- + 2H+ K1=1014
¢Ú Ag2CrO4 = 2Ag+ + CrO42- K2=1.12¡Á10-12 2¡Á¢Ú - ¢Ù µÃ£º2Ag2CrO4 + 2H+ = Cr2O72- + 4Ag+ + H2O
[Cr2O7][Ag?]4K2(1.12?10?12)2???1.25?10?38 K??214[H]K110µ±[Cr2O72- ]=[H+ ]=1.0mol.L-1
2?2[Ag?]?41.25?10?38?3.35?10?10mol.L?1Ag+ + e- = Ag ?? = 0.7996V
??0.7996?0.0592lg3.35?10?10?0.239V
11-18. Óɱê×¼×ÔÓÉÄܼÆËãCl2(g) +2e- = 2Cl-(aq)µÄ±ê×¼µç¼«µçÊÆ¡£ ½â£º Cl2(g) +2e- = 2Cl-(aq) ¡÷fG?m/kJ.mol-1 0 -131.3
11-21. д³öÒÔK2CO3ÈÛÈÚÑÎΪµç½âÖʵÄÇâÑõȼÁÏµç³ØµÄµç¼«·´Ó¦ºÍµç³Ø·´Ó¦¡£ ½â£º ¸º¼«£ºH2 + CO32- = H2O + CO2 + 2e
Õý¼«£ºO2 + 2CO2 + 4e- = 2CO32- µç³Ø·´Ó¦£º2H2 + O2 = 2H2O
11-22. ¼îÐÔÒøÐ¿¿É³äµç¸Éµç³ØµÄÑõ»¯¼ÁΪAg2O£¬µç½âÖÊΪKOHË®ÈÜÒº£¬Ð´³öµç¼«·´Ó¦ºÍµç³Ø·´Ó¦¡£
??G?2?131.3?103????1.36VnF2?96485?½â£º ¸º¼«£ºZn + 2OH- = Zn(OH)2 +2e- Õý¼«£ºAg2O + H2O + 2e- = 2Ag + 2OH-
µç³Ø·´Ó¦£º Zn + Ag2O + H2O = 2Ag + Zn(OH)2
11-23. Ϊʲô¼ì²âǦÐîµç³Øµç½âÖÊÁòËáµÄŨ¶È¿ÉÒÔÈ·¶¨Ðîµç³Ø³äµçÊÇ·ñ³ä×㣿ǦÐîµç³Ø³äµçΪʲô»á·Å³öÆøÌ壿ÊÇÊ²Ã´ÆøÌ壿
½â£º2PbSO4 + 2H2O ==== Pb + PbO2 + 2H2SO4
ÒòΪ³äµç¹ý³Ì£¬ÓÐÁòËáÉú³É£¬Ê¹ÆäŨ¶ÈÔö´ó¡£ÓÉÓÚ³äµçʱˮ»á·¢Éúµç½â£¬¹Ê·Å³öµÄÆøÌåΪÇâÆøºÍÑõÆø¡£
11-25. µç½âº¬¸õôû0.23kg.dm3, Ìå»ý100dm3µÄË®ÈÜÒº¶Æ¸õ¡£Ê¹ÓÃ1500AµçÁ÷ͨµç10.0h£¬Òõ¼«µÄÖÊÁ¿Ôö¼ÓÁË0.679kg¡£Òõ¼«ºÍÑô¼«·Å³öµÄÆøÌåµÄÌå»ý±ÈΪ1.603¡£ÎÊ¢Ù³Á»ý³ö0.679kg¸õºÄÓõĵçÁ¿ÊÇ×ÜÓõçÁ¿µÄ°Ù·ÖÖ®¼¸£¿¢ÚÒõ¼«ºÍÑô¼«·Å³öµÄÆøÌåÔÚ±ê׼״̬ϵÄÌå»ý±È¡£¢ÛÒõ¼«ºÍÑô¼«·Å³öÆøÌåµÄ·´Ó¦µÄµçÁ÷ЧÂÊ·Ö±ð¶à´ó£¿ÊÔ½âÊÍÕâ¿ÉÄÜÊÇʲôԵ¹Ê£¿Ð´³öµç½âµÄ×Ü·´Ó¦¡£
44
½â£º¢Ùµç½â·´Ó¦£º2CrO3 = 2Cr + 3O2
¢ÚÓÉÓÚÒõ¼«ÓÐÆøÌå²úÉú£¬ËµÃ÷µç½âʱ³ýÁ˸õ»¹ÔÍ⣬ˮ·¢ÉúÁ˵ç½â¡£Èô86%µÄµçÄÜÏûºÄÔÚË®µÄµç½âÉÏ£º
2CrO3 = 2Cr + 3O2 (Ñô¼«) 2H2O = 2H2 (Òõ¼«) + O2 (Ñô¼«)
11-26. ¾ÃÖÃ¿ÕÆøÖеÄÒøÆ÷»á±äºÚ£¬¾·ÖÎö֤ʵ£¬ºÚÉ«ÎïÖÊÊÇAg2S¡£Í¨¹ý¼ÆËã˵Ã÷£¬¿¼ÂÇÈÈÁ¦Ñ§Ç÷ÊÆ£¬ÒÔÏÂÄÄÒ»¸ö·´Ó¦µÄ¿ÉÄÜÐÔ¸ü´ó£¿
¢Ù 2Ag(s) + H2S(g) = Ag2S(s) + H2(g)
¢Ú 2Ag(s) + H2S(g) + 1/2 O2(g) = Ag2S(s) + H2O(l) ½â£º¢Ú - ¢ÙµÃ£º1/2 O2(g) + H2(g) = H2O(l)
¡÷rGm? = ¡÷fGm?(H2O,l) = - 231.129kJ.mol-1 ¡à ¢Ú·´Ó¦µÄ¿ÉÄÜÐÔ¸ü´ó¡£
11-28. ½âÊÍÈçÏÂÏÖÏó£º
¢Ù¶ÆÎýµÄÌú£¬ÌúÏȸ¯Ê´£¬¶ÆÐ¿µÄÌú£¬Ð¿Ïȸ¯Ê´¡£ ¢Ú﮵ĵçÀëÊÆºÍÉý»ªÈȶ¼±ÈÄÆ´ó£¬ÎªÊ²Ã´ï®µÄ??¸üС£¿
¢ÛͺÍпÔÚÔªËØÖÜÆÚϵÊÇÁÚ¾Ó£¬È»¶øËüÃÇÔÚ½ðÊô»î¶¯Ë³ÐòÖеÄλÖÃÈ´ÏàÈ¥ÉõÔ¶£¬ÊÔͨ¹ý²¨¶÷¡ª¹þ²®Ñ»··ÖÎöͺÍпµÄµç¼«µçÊÆÏà²îÕâô´óÖ÷ÒªÊÇÓÉʲôÄÜÁ¿Ïî¾ö¶¨£¿
¢ÜȼÁÏµç³ØÊÇ¡°Ò»ÖÖͨ¹ýȼÉÕ·´Ó¦Ê¹»¯Ñ§ÄÜÖ±½Óת»¯ÎªµçÄܵÄ×°Öá±µÄ˵·¨ÕýÈ·Âð£¿È¼ÁÏµç³ØµÄÀí
n(H2)?1500?10?3600?86%?240.66mol2?96485Q?1500?10?3600?5.4?107C0.679?103Q1??6?96485?7.56?106C52Q17.56?106??100%?14%7Q5.4?10n(O2)?6791?139.92moln(H2)?1.5n(Cr)?120.33?1.5?522V(H2)n(H2)240.66???1.72V(O2)n(O2)139.92¢Ûʵ¼ÊÉÏÒõ¼«ºÍÑô¼«·Å³öµÄÆøÌåÌå»ý±ÈΪ1.603£¬¹ÊÓÐÆäÓà¹ý³Ì·¢Éú¡£ÉèË®µç½âÏûºÄµÄµçÁ¿ÎªQ2
V(H2)n(H2)Q2/2F1.603???V(O2)n(O2)(Q1?Q2)/4F1½«Q1 = 7.56¡Á106 C´úÈëµÃ£º Q2 = 3.05¡Á107C
3.05?107?B??100%?56.5u.4?10»¹ÓÐ29.5%µÄµçÁ÷ÏûºÄÔÚÆäËü·´Ó¦ÉÏ£¬¿ÉÄÜÊÇCr(¢ó)ÓëCr(¢ö)Ö®¼äµÄÑ»·×ª»¯ÉÏ¡£
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ÂÛЧÂÊÓпÉÄܳ¬¹ý100%Âð£¿Æä¹¤×÷ζÈÓëÀíÂÛЧÂʳÊʲô¹ØÏµ£¿ ¢ÝÌúÄÜÖû»Í¶øÈýÂÈ»¯ÌúÄÜÈܽâÍ¡£
¢ÞZnO22-µÄ¼îÐÔÈÜÒºÄܰÑÍת»¯ÎªCu(OH)42-,ʹÆäÈܽ⠢߽«MnSO4ÈÜÒºµÎÈëKMnO4ËáÐÔÈÜÒºµÃµ½MnO2³Áµí¡£ ¢àCu+(aq)ÔÚË®ÈÜÒºÖлáÆç»¯ÎªÍºÍCu2+(aq) ¢áCr2+ÔÚË®ÈÜÒºÖв»Îȶ¨£¬»áÓëË®·´Ó¦¡£
¢â½«Cl2Ë®ÈÜÒºµÎÈëI-¡¢Br-µÄ»ìºÏÒºÖУ¬Ïà¼ÌµÃµ½µÄ²úÎïÊÇI2¡¢HIO3ºÍBr2£¬¶ø²»ÊÇI2£¬Br2ºÍHIO3¡£ ½â£º¢Ù ¡ß ??(Zn2+/Zn) £¼ ??(Fe2+/Fe) £¼ ??(Sn2+/Sn) »îÆÃÐÔ´óµÄ½ðÊôÏȱ»¸¯Ê´ ¢Ú ÕâÊÇÓÉÓÚLi+°ë¾¶Ð¡£¬Ë®ºÏÈÈÌØ±ð´óµÄÔµ¹Ê¡£ ¢ÛÓÉÓÚZnµÄÉý»ªÈÈÌØ±ðС£¬Ê¹Æä×ÜÄÜÁ¿¼õС¡£
¢Ü²»ÕýÈ·.µç³ØÖз¢ÉúµÄÑõ»¯»¹Ô·´Ó¦ÓëȼÉÕ·´Ó¦Ïàͬ£¬È´²»ÊÇȼÉÕ·´Ó¦£¨È¼ÉÕ·´Ó¦µÄ±êÖ¾ÊÇ·¢¹â·¢ÈÈ£©¡£
µ±µç³Ø·´Ó¦µÄ¡÷S£¾0ʱ£¬¡÷GµÄ¾ø¶ÔÖµ´óÓÚ¡÷H£¬ËùÒÔȼÁÏµç³ØµÄÀíÂÛЧÂÊÓпÉÄܳ¬¹ý100%¡£ ¡÷G =¡÷H-T¡÷S ¡÷S£¾0 T¡ü Ôò?¡ü
?G??¡÷S£¼0 T¡ü Ôò?¡ý
?H ¢Ý Fe + Cu2+ = Fe2+ + Cu E? = ??(Cu2+/Cu) ¨C ??(Fe2+/Fe) = 0.345 - (-0.44)£¾0 2Fe3+ + Cu2+ = 2Fe2+ +Cu
E? = ??(Fe3+/Fe2+) ¨C ??(Cu2+/Cu) = 0.771 ¨C0.345 > 0
¢Þ Zn(OH)42- + Cu = Cu(OH)42- + Zn ¢ß 3Mn2+ + 2MnO4- + 2H2O = 5MnO2¡ý+ 4H+
E? = ??(MnO4-/MnO2)¨C??(MnO2/Mn2+) =1.679¨C1.228£¾0 ¢à 2Cu+ = Cu2+ + Cu
E? = ??(Cu+/Cu) ¨C ??(Cu2+/Cu+) = 0.522 ¨C 0.152£¾0 ¢á 2Cr2+ + 2H+ = Cr3+ + H2
E? = ??(H+/H2) ¨C ??(Cr3+/Cr2+) = 0 ¨C£¨-0.41£©£¾0 ¢â ÒòΪCl2 ÓëBr-·´Ó¦´æÔÚ¶¯Á¦Ñ§Õϰ£¬·´Ó¦ËÙÂÊÂý¡£
??(Br2/Br-) = 1.085V ??(IO3-/I2) =1.195V
11-29. ÒÔM´ú±í´¢Çâ²ÄÁÏ£¬MHΪ¸º¼«²ÄÁÏ£¬KOHΪµç½âÖÊ£¬Ð´³öÄøÇâµç³ØµÄµç¼«·´Ó¦ºÍµç³Ø·´Ó¦¡£
½â£º¸º¼«£ºMH + OH- = H2O + M + e-
Õý¼«£ºNiO(OH) + H2O + e- = Ni(OH)2 + OH- µç³Ø·´Ó¦£ºMH + NiO(OH) = Ni(OH)2 + M
11-30.ÈçÈô¸Ê¹¯µç¼«µÄµç¼«µçÊÆÎªÁ㣬ÑõµÄµç¼«µçÊÆ¶à´ó. ½â£º??(O2/H2O) = 1.229 - ??(Hg2Cl2/Hg) = 1.229 - 0.2801 = 0.949V
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²¹³äÁ·Ï°Ìâ
1. ½âÊÍÏÂÁÐÏÖÏó:
¢ÙAg»î¶¯Ë³ÐòλÓÚHÖ®ºó,µ«Ëü¿É´ÓHIÖÐÖû»³öH2.
¢Ú·Ö±ðÓÃÏõËáÄÆºÍÏ¡ÁòËá¾ù²»ÄÜÑõ»¯Fe2+,µ«¶þÕߵĻìºÏÈÜҺȴ¿ÉÒÔ. ¢Û¾ÃÖÃÓÚ¿ÕÆøÖеÄÇâÁòËáÈÜÒº»á±ä»ì×Ç. ¢ÜµÃ²»µ½FeI3ÕâÖÖ»¯ºÏÎï.
½â£º¢Ù ¡ß AgÔÚHIÖÐÉú³ÉAgI³Áµí ¦Õ ?(AgI/Ag)=-0.152V
¡à 2Ag + 2HI = 2AgI¡ý+ H2¡ü
¢Ú ¡ß ¦Õ ?(HNO3/NO)=0.96V £¾ ¦Õ ?(Fe3+/Fe2+)=0.77V ¡à 4H+ + NO3- + 3Fe2+ = 3Fe3+ + NO¡ü+ 2H2O ¢Û ¡ß ¦Õ ?(O2/H2O)=1.229V £¾ ¦Õ ?(S/H2S)=0.15V ¡à O2 + 2H2S = 2S¡ý+ 2H2O
¢Ü ¡ß ¦Õ ?(Fe3+/Fe2+)=0.77V £¾ ¦Õ ?(I2/I-)=0.54V ¡à 2 Fe3+ + 2I- = 2Fe2+ + I2
2. ¼ÆËã·´Ó¦Fe + Cu2+ = Fe2+ + CuµÄƽºâ³£Êý,Èô·´Ó¦½áÊøºóÈÜÒºÖÐ[Fe2+]=0.1mol/L,ÊÔÎÊ´ËʱÈÜÒºÖеÄCu2+Ũ¶ÈΪ¶àÉÙ?
½â: Fe + Cu2+ = Fe2+ + Cu Cƽ X 0.1
lgK?2(0.337?0.44)?26.25 K?1.8?10260.0592K?0.1 X?5.6?10?28mol.L?1X17. ÒÑÖªE?(F2/F-)=2.87V,Ka(HF)=6.6¡Á10-4,Çó¦Õ ?(F2/HF) ½â: ¢Ù F2+ 2e- = 2F- ¦Õ? = 2.87V ¢Ú F2+ 2e- +2H+ = 2HF ¦Õ? =? µ±c(HF) = c(H+) =1.0 mol?L-1
[H?][F?]?[F?]?6.6?10?4mol/l Ka?[HF]
??(F2/HF)?2.87?0.05921lg?3.06V2(6.6?10?4)2 47