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2-2. È·¶¨ÏÂÁи÷×éÎïÖÊìØÖµµÄ´óС˳Ðò¡£

(1)H2O(l)¡¢H2O(g)¡¢H2O(s)£» (3)CH4(g)¡¢C2H6 (g)£» (2)H2(g,310K )¡¢H2(g£¬298K) (4)Fe(s)¡¢Fe2O3(s)¡£

½â£º(1)S(H2O,g)>S(H2O,l)>S(H2O,s) (2)S(H2,g, 310K )>S(H2,g, 298K )

(3)S(C2H6,g) > S(CH4,g) (4)S(Fe2O3,s)>S(Fe,s)

2-3. ¼ÆËãÌåϵÈÈÁ¦Ñ§Äܵı仯 (1)Ìåϵ´Ó»·¾³ÎüÈÈ1000J£¬²¢¶Ô»·¾³×÷¹¦540J£»(2)ÌåϵÏò»·¾³·ÅÈÈ535J£¬»·¾³¶ÔÌåϵ×÷¹¦250J¡£

½â: (1)¦¤U=Q+W=(+1000)+(-540)=460(J) (2)¦¤U=Q+W= ( -535) +(+250) =-285(J)

2-4. ÇóÏÂÁз´Ó¦µÄ¦¤rHm¦È¡£[¦¤fHm¦È ( Fe2O3,s)=£­822.2kJ¡¤mol-1£¬¦¤fHm¦È ( Al2O3,s)=£­1670kJ¡¤mol-1£¬ÆäÓতfHm¦ÈÖµ²é¸½Â¼Ò» ]

(1)4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) (2)C2H4(g)+ H2O(g)= C2H5OH(l) (3)Fe2O3(s)+ 2Al(s) = 2Fe(s)+ Al2O3(s)

½â: (1) ¦¤rHm¦È =4¦¤fHm¦È(NO,g)+6¦¤fHm¦È(H2O,g)-[4¦¤fHm¦È( NH3,g)+5¦¤fHm¦È(O2,g)]

=4¡Á90.25+6¡Á(-241.8)-[4¡Á(-46.11)+5¡Á0] =-905.36(kJ¡¤mol-1)

(2) ¦¤rHm¦È=¦¤fHm¦È(C2H5OH,l)-[¦¤fHm¦È(C2H4,g)+¦¤fHm¦È(H2O,g)]

8

=-277.7-[(52.26+(-241.8)] =-88.16(kJ¡¤mol-1)

(3) ¦¤rHm¦È=2¦¤fHm¦È(Fe,s)+ ¦¤fHm¦È(Al2O3,s)- [¦¤fHm¦È(Fe2O3,s)+ 2¦¤fHm¦È(Al,s)]

=2¡Á0+(-1670)-[(-822.2)+2¡Á0] = -847.8 kJ¡¤mol-1

2-5. ÒÑÖª¦¤cHm¦È(C3H8,g)=-2220.9kJ¡¤mol-1£¬¦¤fHm¦È(H2O,l)=-285.8kJ¡¤mol-1£¬ ¦¤fHm¦È(CO2,g)=-393.5kJ¡¤mol-1£¬ÇóC3H8(g)µÄ¦¤fHm¦È¡£ ½â£ºC3H8(g)+5O2(g)=3CO2(g)+4H2O(g)

¦¤cHm¦È(C3H8,g)=3¦¤fHm¦È(CO2,g)+4¦¤fHm¦È(H2O,l)- ¦¤fHm¦È(C3H8£¬g) -2220.9 = 3¡Á(-393.5)+4¡Á(-285.8)-¦¤fHm¦È(C3H8£¬g) ¦¤fHm¦È(C3H8£¬g) =3¡Á(-393.5)+4¡Á(-285.8)+2220.9

=-102.8(kJ¡¤mol-1)

2-6. 1mol±û¶þËá[CH2(COOH)2]¾§ÌåÔÚµ¯Ê½Á¿ÈȼÆÖÐÍêȫȼÉÕ£¬298Kʱ·Å³öµÄÈÈÁ¿Îª866.5kJ£¬Çó1mol±û¶þËáÔÚ298KʱµÄµÈѹ·´Ó¦ÈÈ¡£

½â£º CH2(COOH)2(s) + 2O2(g) = 3CO2(g) + 2H2O(1) Òò QV £½£­866.5kJ¡¤mol-1£¬ Ôò Qp = QV + RT?¦¤ng

=-866.5 kJ¡¤mol-1+8.314¡Á10-3 kJ¡¤K-1¡¤mol-1¡Á298K¡Á(3-2) =-864.02kJ¡¤mol-1

2-7.ÒÑÖª£º(1) CH3COOH(l)+2O2(g)=2CO2(g) + 2H2O(l) ¦¤rHm¦È(1) = - 870.3kJ¡¤mol-1

(2) C(ʯī) + O2(g) = CO2(g) ¦¤rHm¦È(2)= - 393.5kJ¡¤mol-1 (3) H2(g) + 1/2 O2(g) = H2O(l) ¦¤rHm¦È(3)= - 285.5kJ¡¤mol-1

¼ÆËã 2C (ʯī) + 2H2(g) + O2(g) = CH3COOH(l)µÄ¦¤rHm¦È¡£ ½â£º¡ß¦¤rHm¦È=¦¤rHm¦È(2)¡Á 2 +¦¤rHm¦È(3)¡Á 2 -¦¤rHm¦È(1)

¡à¦¤rHm¦È=(-393.5)¡Á2 + (-285.5)¡Á2 -(-870.3)

=-488.3 kJ.mol-1

9

2-8. ÒÑÖª±½µÄÈÛ»¯ÈÈΪ10.67kJ¡¤mol-1 £¬±½µÄÈÛµãΪ5¡æ£¬Çó±½ÈÛ»¯¹ý³ÌµÄ¦¤Sm¦È¡£ ½â£º±½ÈÛ»¯¹ý³ÌÊÇÒ»µÈοÉÄæ¹ý³Ì£º

Qr10.67?103J?mol-1?S???38.36J?K-1?mol-1

T(273.15?5)K¦Èm2-9.ºË·´Ó¦¶ÑÖеĺËȼÁÏÊÇ235U£¬ËüÔÚÓË¿óÖеÄÖÊÁ¿·ÖÊý½öÕ¼0.7£¥£¬ÆäÓàΪ238U£¬ËüÃǺÜÄÑÓû¯Ñ§·½·¨·ÖÀë¡£·ÖÀëºÍ¸»¼¯235UÊÇͨ¹ýÏÂÁз´Ó¦Éú³ÉUF6£¬È»ºó¾­Æø»¯(b.p.56.54¡æ)½øÐÐÀ©É¢£¬¸ù¾ÝËüÃÇÀ©É¢ËÙÂʵIJ»Í¬¶ø´ï´ËÄ¿µÄ¡£ÊÔ¸ù¾ÝÏÂÁÐÒÑÖªÌõ¼þºÍ¸½Â¼µÄÊý¾Ý¼ÆËã·´Ó¦µÄ¦¤rHm¦È¡£

U(s)+ O2(g)+ 4HF(g)+ F2(g)£½UF6(g)+2 H2O(g) ÒÑÖª¸÷²½·´Ó¦ÈçÏÂ

U(s)+ O2(g)£½UO2(s)

¦¤rHm¦È=-1084.9 kJ¡¤mol-1 ¦¤rHm¦È=-2398.3 kJ¡¤mol-1 ¦¤rHm¦È=-233.2 kJ¡¤mol-1

UO2(s)+ 4 HF(g)£½UF4(s)+2 H2O(g) UF4(s)+ F2(g)£½UF6(g) ½â£º

U(s)+ O2(g)£½UO2(s) UO2(s)+ 4 HF(g)£½UF4(s)+2 H2O(g)

¦¤rHm¦È(1)=-1084.9 kJ¡¤mol-1 ¦¤rHm¦È(2)= -2398.3 kJ¡¤mol-1

+)UF4(s)+ F2(g)£½UF6(g) ¦¤rHm¦È(3)=-233.2 kJ¡¤mol-1 U(s)+ O2(g)+ 4HF(g)+ F2(g)£½UF6(g)+2 H2O(g)¦¤rHm¦È ¦¤rHm¦È=¦¤rHm¦È(1)+ ¦¤rHm¦È(2)+ ¦¤rHm¦È(3) =(-1084.9)+( -2398.3)+( -233.2) =-3716.4 (kJ¡¤mol-1)

2-10. ÓÉÌú¿óʯÉú²úÌúÓÐÁ½ÖÖ;¾¶£¬ÊÔ¼ÆËãËüÃǵÄתÏòζȡ£

(Sm¦È(Fe,s) = 27.28 J¡¤K-1¡¤mol-1£¬Sm¦È(Fe2O3,s) = 90.0 J¡¤K-1¡¤mol-1£¬¦¤fHm¦È(Fe2O3, s)=-822.2 kJ¡¤mol-1)

10

(1)Fe2O3(s) + 3C(ʯī) = 2Fe(s) + 3CO2(g)

22(2)Fe2O3(s) + 3H2(g) = 2Fe(s) + 3H2O(g) ½â£º

33¦È¦È¦È¦È?fHm(CO2,g)?2?fHm(Fe,s)??fHm(C,ʯī)??fHm(Fe2O3,s)2(1)T??rH?2

3¦È3¦È?rS¦È¦ÈSm(CO2,g)?2Sm(Fe,s)?Sm(C,ʯī)?Sm(Fe2O3,s)22¦Èm¦Èm33?(?393.5)?2?0??0?(?822.2)22=?838.9(K) 33(?213.7?2?27.28??5.74?90.0)?10?322¦È¦È¦È¦È¦È3?fHm(H2O,g)?2?fHm(Fe,s)?3?fHm(H2,g)??fHm(Fe2O3,s)?rHm (2) T? ?¦È¦È¦È¦È¦È?rSm3Sm(H2O,g)?2Sm(Fe,s)?3Sm(H2,g)?Sm(Fe2O3,s)?3?(?241.8)?2?0?3?0?(?822.2)?697(K) ?3(3?188.8?2?27.28?3?130.7?90.0)?102-11.ÀûÓñê׼Ħ¶ûȼÉÕÈȵÄÊý¾Ý¼ÆËãÏÂÁз´Ó¦µÄ¦¤rHm¦È CH3COOH (l) +C2H5OH (l) = CH3COO C2H5(l) + H2O(l)

½â£º CH3COOH (l) +C2H5OH (l) = CH3COO C2H5(l) + H2O(l) ¦¤cHm¦È -875 -1368 -2231 0

¦¤rHm¦È=[¦¤cHm¦È(CH3COOH,l)+¦¤cHm¦È(C2H5OH,l)]-[¦¤cHm¦È(H2O,l)- ¦¤cHm¦È(CH3COO C2H5, l)

=[ (-875)+(-1368)]-[0+(-2231)] =-12.0 (kJ¡¤mol-1)

2-12. ÊÔ¼ÆËãÏÂÁкϳɸʰ±ËáµÄ·´Ó¦ÔÚ298K¼°p¦È ϵĦ¤rGm¦È £¬²¢ÅжϴËÌõ¼þÏ·´

11

Ó¦µÄ×Ô·¢ÐÔ¡£NH3(g) + 2CH4(g) + 5/2 O2(g) = C2H5O2N(s) + 3H2O(l)

½â£º²é±íµÃ£º¸÷ÎïÖʵĦ¤fGm¦ÈΪ£º

NH3(g) + 2CH4(g) + 5/2 O2(g) = C2H5O2N(s) + 3H2O(l)

kJ¡¤mol-1 -16.45 -50.72 0 -377.3 -237.129 ¦¤rGm¦È=[¦¤fGm¦È(C2H5O2N,s)+3¦¤fGm¦È(H2O,l) ]- [¦¤fGm¦È(NH3,g) +2¦¤fGm¦È(CH4, g)+0] =-377.3+3(-237.129)-( -16.45)-2(-50.72) =-970.8 £¨kJ¡¤mol-1£©

2-13.ÌÇÔÚг´úл¹ý³ÌÖÐËù·¢ÉúµÄ×Ü·´Ó¦Îª C12H22O11(s) + 12O2(g) = 12CO2(g) + 11H2O(l) ÒÑÖª

¦¤fHm¦È(C12H22O11,s)=-2221kJ¡¤mol-1£¬Sm¦È(C12H22O11,s)=359.8 J¡¤K-1¡¤mol-1¡£Çó£º

(1)Èç¹ûʵ¼ÊÖ»ÓÐ30%µÄ×ÔÓÉÄÜת±äΪ·ÇÌå»ý¹¦£¬Ôò1gÌÇÔÚÌåÎÂ37¡æʱ½øÐÐг´úл£¬¿ÉÒԵõ½¶àÉÙkJµÄ·ÇÌå»ý¹¦£¿[Ìáʾ£º¸ù¾ÝÈÈÁ¦Ñ§ÍƵ¼£¬ÔÚµÈεÈѹÏÂ×ÔÓÉÄܵļõÉÙ(-¦¤G)µÈÓÚÌåϵ¶ÔÍâËù×÷µÄ×î´ó·ÇÌå»ý¹¦Wf]

(2) Ò»¸öÖÊÁ¿Îª75kgµÄÔ˶¯Ô±Ðè³Ô¶àÉÙ¿ËÌDzÅÄÜ»ñµÃµÇÉϸ߶ÈΪ2.0km¸ßɽµÄÄÜÁ¿¡£ ½â£º

(1)¦¤rHm¦È=12¦¤fHm¦È(CO2,g)+11¦¤fHm¦È(H2O,l)-[¦¤fHm¦È(C12H22O11,s) +12¦¤fHm¦È(O2,g)]

=12¡Á(-393.5)+11¡Á(-285.8)-[12¡Á0+(-2221)] =-5644.8(kJ¡¤mol-1) ¦¤rSm¦È=12Sm¦È(CO2,g)+11Sm¦È(H2O,l)-[Sm¦È(C12H22O11,s)+ 12Sm¦È(O2,g)]

=12¡Á213.7+11¡Á69.91-[359.8+12¡Á205.1] =512.4(J¡¤K-1¡¤mol-1)

¦¤rGm¦È=¦¤rHm¦È-¦¤rSm¦È =-5644.8-310¡Á(512.4¡Á10-3) =-5803.6(kJ¡¤mol-1) -¦¤rGm¦È=Wf=5803.6kJ¡¤mol-1 ¸ù¾ÝÌâÒâÿ¿ËÌÇÄÜÌṩ¸øÔ˶¯Ô±µÄÄÜÁ¿Îª ¦¤rGm¦È/M(C12H22O11)=5803.6¡Â342 =16.97(kJ¡¤g-1) ÄÜת»¯Îª·ÇÌå»ý¹¦µÄÄÜÁ¿Îª16.97¡Á30%=5.09(kJ¡¤g-1)

12

mgh75?9.81?2?103(2)Ô˶¯Ô±Ðè³ÔÌǵĿËÊýΪ G? ?=289£¨g£©35.095.09?102-14£®Ö²Îï½øÐйâºÏ×÷Óõķ´Ó¦Îª6CO2(g)+6H2O(l) C6H12O6(aq)+6O2(g)

Ò¶ÂÌËØ ¹â

¦¤rGm¦È£½2870kJ¡¤mol-1£¬ÉÖÐûÓйâÏßÓÉϸ¾úʹH2S×÷ΪÄÜÔ´ºÏ³ÉC6H12O6£¬·´Ó¦Ê½ÈçÏ 24H2S(g)+12 O2(g)£½24 H2O(l)+24S(s) ¦¤rGm¦È=-4885.7 kJ¡¤mol-1 Çó·´Ó¦

24H2S(g)+6CO2(g)£½C6H12O6(aq)+18 H2O(l)+24 S(s)µÄ¦¤rGm¦È ½â£º6CO2(g)+6H2O(l) C6H12O6(aq)+6O2(g)

Ò¶ÂÌËØ ¹â

¦¤rGm¦È(1)£½2870kJ¡¤mol-1 ¦¤rGm¦È(2)=-4885.7 kJ¡¤mol-1 ¦¤rGm¦È

+)24H2S(g)+12 O2(g)£½24 H2O(l)+24S(s) 24H2S(g)+6CO2(g)£½C6H12O6(aq)+18 H2O(l)+24 S(s)

¦¤rGm¦È=¦¤rGm¦È(1)+ ¦¤rGm¦È(2)=2870-4885.7= -2016(kJ¡¤mol-1)

2-15.¼×´¼ÊÇÖØÒªµÄÄÜÔ´ºÍ»¯¹¤Ô­ÁÏ£¬Óø½Â¼µÄÊý¾Ý¼ÆËãËüµÄÈ˹¤ºÏ³É·´Ó¦µÄ¦¤rHm¦È¡¢¦¤rSm¦ÈºÍ¦¤rGm¦È£¬ÅжÏÔÚ±ê׼״̬Ï·´Ó¦×Ô·¢½øÐеķ½Ïò²¢¹ÀËãתÏòζȡ£

CO(g) + 2H2(g) = CH3OH(l)

½â£º CO(g) + 2H2(g) = CH3OH(l) Sm¦È/J¡¤K-1¡¤mol-1 198.0 130.7 126.8 ¦¤cHm¦È/ kJ¡¤mol-1 -283.0 -285.8 -726.6 ¦¤rHm¦È=(-283.0)+2¡Á (-285.8)-(-726.6)=-128.0 kJ¡¤mol-1 ¦¤rSm¦È=126.8-(2¡Á130.7+198.0)=-332.6 J¡¤K-1¡¤mol-1

¦¤rGm¦È=-128.0 kJ¡¤mol-1-(298.15K¡Á-332.6 J¡¤K-1¡¤mol-1¡Á10-3)=-28.83 kJ¡¤mol-1 ¦¤rGm¦È<0£¬±ê׼״̬ÏÂÕý·´Ó¦·½Ïò×Ô·¢¡£

תÏòÎÂ¶È T=¦¤rHm¦È/¦¤rSm¦È=-128.0¡Á103/(-332.6)=384.8K

2-16.ÒÑÖª2MnO42-+10Cl-+16H+=2Mn2++5Cl2+8H2O ¦¤rGm¦È(1)=-142.0 kJ¡¤mol-1

Cl2+2Fe2+=2Fe3++2Cl- ¦¤rGm¦È(2)=-113.6 kJ¡¤mol-1

ÇóÏÂÁз´Ó¦µÄ¦¤rGm¦È

MnO42-+5Fe2++8H+= Mn2++5Fe3+ +4H2O

13

½â£º¦¤rGm¦È=[¦¤rGm¦È(1)+5¡Á¦¤rGm¦È(2)]/2

=[-142.0 kJ¡¤mol-1 +5¡Á(-113.6 kJ¡¤mol-1)]/2 =-710.0 kJ¡¤mol-1/2 =-355.0 kJ¡¤mol-1

14

µÚ3Õ »¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâ

3-1. ºÏ³É°±Ô­ÁÏÆøÖÐÇâÆøºÍµªÆøµÄÌå»ý±ÈÊÇ3¡Ã1£¬³ýÕâÁ½ÖÖÆøÌåÍ⣬ԭÁÏÆøÖл¹º¬ÓÐÆäËüÔÓÖÊÆøÌå4£¥(Ìå»ý·ÖÊý)¡£Ô­ÁÏÆø×ÜѹÁ¦Îª15195kPa£¬¼ÆË㵪¡¢ÇâµÄ·Öѹ¡£

½â£ºx(H2) = (1-0.04)¡Á=0.72 x(N2) = (1-0.04)¡Á=0.24

p(H2) = 15195 kPa¡Á0.72= 10940kPa p(N2) = 15195 kPa¡Á0.24= 3647kPa 3-2. 660KʱµÄ·´Ó¦£º2NO+ O2 = 2NO2£¬NOºÍO2µÄ³õʼŨ¶Èc(NO)ºÍc(O2)¼°·´Ó¦³õʼËÙÂÊvµÄʵÑéÊý¾ÝÈçÏ£º

c(NO)/ mol¡¤L-1 0.10 0.10 0.20 c(O2)/ mol¡¤L-1 0.10 0.20 0.20 L-1¡¤s-1 v/ mol¡¤0.030 0.060 0.240 3414(1) д³öÉÏÊö·´Ó¦µÄËÙÂÊ·½³Ì£¬Ö¸³ö·´Ó¦µÄ¼¶Êý£» (2) ¼ÆËã¸Ã·´Ó¦µÄËÙÂʳ£ÊýkÖµ£»

(3) ¼ÆËãc(NO) = c(O2) = 0.15 mol¡¤L-1ʱµÄ·´Ó¦ËÙÂÊ¡£

½â£º(1)v?kcx(NO)?cy(O2) ½«µÚÒ»¡¢¶þ×éÊý¾Ý´úÈëºóÏà³ý£¬µÃµ½

v20.060mol?L-1?s?1k(0.10mol?L-1)x(0.20mol?L-1)y ??-1?1-1x-1yv10.030mol?L?sk(0.10mol?L)(0.10mol?L)2y?2 y = 1

v30.240mol?L-1?s?1k(0.20mol?L-1)x(0.20mol?L-1)yͬÀí£¬ ??-1?1-1x-1yv20.060mol?L?sk(0.10mol?L)(0.20mol?L)2x?4 x = 2

v?kc2(NO)?c(O2) Èý¼¶·´Ó¦

(2) ½«µÚÒ»×éÊý´úÈëËÙÂÊ·½³Ì£º0.030 mol¡¤L-1¡¤s-1 = k(0.10 mol¡¤L-1)2(0.10 mol¡¤L-1)

15

k =30 L2?mol-2?s-1

(3) v? 30 L2?mol-2?s-1¡Á(0.15mol?L-1)3 = 0.101 mol¡¤L-1¡¤s-1 3-3. ÒÑÖªÒÒÈ©µÄ´ß»¯·Ö½âÓë·Ç´ß»¯·Ö½â·´Ó¦·Ö±ðΪ ¢Ù CH3CHO = CH4 + CO Ea´ß = 136 kJ¡¤mol-1 ¢Ú CH3CHO = CH4 + CO Ea = 190 kJ¡¤mol-1

Èô·´Ó¦¢ÙÓë¢ÚµÄָǰÒò×Ó½üËÆÏàµÈ£¬ÊÔ¼ÆËã300Kʱ£¬·´Ó¦¢Ù µÄ·´Ó¦ËÙÂÊÊÇ·´Ó¦¢ÚµÄ¶àÉÙ±¶¡£

Ae?136RT54?103J?mol-1½â£º ?e??vkAe?190RTkΪ4.74 L ?mol-1?s-1£¬¼ÆËã¸Ã·´Ó¦µÄ»î»¯ÄÜ¡£

v´ßk´ß8.314J?K-1?mol?1?300K= 2.5¡Á109

3-4. ÒÑÖª·´Ó¦ 2NO2 = 2NO+O2 ÔÚ592KʱËÙÂʳ£ÊýkΪ0.498 L?mol-1? s-1£¬ÔÚ656ʱ

?1?1Ea4.74L?mol?s656K?592K ½â£º ln?()?1?1?1?10.498L?mol?s8.314J?K?mol656K?592K Ea = 113672J?mol-1 = 113.7 kJ?mol-1

3-5. ÔÚ100kPaºÍ298Kʱ£¬HCl(g)µÄÉú³ÉÈÈΪ-92.3 kJ?mol-1£¬Éú³É·´Ó¦µÄ»î»¯ÄÜΪ113kJ?mol-1£¬¼ÆËãÆäÄæ·´Ó¦µÄ»î»¯ÄÜ¡£

½â£º 1/2H2(g) + 1/2Cl2(g) = HCl(g)

¡÷fHm? = ¡÷rHm? = - 92.3 kJ?mol-1£¬ EaÕý = 113 kJ?mol-1 - 92.3 kJ?mol-1 = 113 kJ?mol-1 ¨C EaÄæ EaÄæ = 205.3 kJ?mol-1

3-6. ÒÑÖª·´Ó¦C2H5Br = C2H4 + HBr£¬ÔÚ650Kʱ£¬ËÙÂʳ£ÊýkΪ2.0¡Á10-3s-1£¬¸Ã·´Ó¦µÄ»î»¯ÄÜEaΪ225kJ?mol-1£¬¼ÆËã700KʱµÄkÖµ¡£

k700225?103J?mol?1700K?650K½â£º ln?()

2.0?10?3s?18.314J?K?1?mol?1700K?650K

k700?19.57

2.0?10?3s?1 k700 = 3.9¡Á10-2 s-1

3-7. ÏÂÁÐ˵·¨ÊÇ·ñÕýÈ·£¬¼òҪ˵Ã÷ÀíÓÉ¡£

(1) ËÙÂÊ·½³ÌʽÖи÷ÎïÖÊŨ¶ÈµÄÖ¸ÊýµÈÓÚ·´Ó¦·½³ÌʽÖи÷ÎïÖʵļÆÁ¿Êýʱ£¬¸Ã·´Ó¦¼´

16

Ϊ»ùÔª·´Ó¦¡£

(2) »¯Ñ§·´Ó¦µÄ»î»¯ÄÜÔ½´ó£¬ÔÚÒ»¶¨µÄÌõ¼þÏÂÆä·´Ó¦ËÙÂÊÔ½¿ì¡£

(3) ij¿ÉÄæ·´Ó¦£¬ÈôÕý·´Ó¦µÄ»î»¯ÄÜ´óÓÚÄæ·´Ó¦µÄ»î»¯ÄÜ£¬ÔòÕý·´Ó¦ÎªÎüÈÈ·´Ó¦¡£ (4) ÒÑ֪ij·´Ó¦µÄËÙÂʳ£Êýµ¥Î»ÎªL?mol -1?s-1£¬¸Ã·´Ó¦ÎªÒ»¼¶·´Ó¦¡£

(5) ijһ·´Ó¦ÔÚÒ»¶¨Ìõ¼þϵÄƽºâת»¯ÂÊΪ25.3%£¬µ±¼ÓÈë´ß»¯¼Áʱ£¬Æäת»¯ÂÊ´óÓÚ25.3%¡£

(6) Éý¸ßζȣ¬Ê¹ÎüÈÈ·´Ó¦ËÙÂÊÔö´ó£¬·ÅÈÈ·´Ó¦ËÙÂʽµµÍ¡£ ½â£º(1) ´í£¬Àý£º¸´ÔÓ·´Ó¦ H2(g) + I2(g) = 2HI(g) v?kc(H2)c(I2)

(2) ´í£¬»î»¯ÄÜÊǾö¶¨·´Ó¦ËÙÂʵÄÖØÒªÒòËØ£¬Ea´ó£¬»î»¯·Ö×Ó·ÖÊýС£¬·´Ó¦ËÙÂÊvÂý¡£ (3) ¶Ô£¬ÒòΪ ¦¤H =EaÕý £­ EaÄæ

(4) ´í£¬Ò»¼¶·´Ó¦µÄËÙÂʳ£ÊýÁ¿¸ÙΪ£ºs-1»òmin-1£¬h-1¡£ (5) ´í£¬´ß»¯¼Á²»¸Ä±äƽºâ̬£¬Òò¶ø²»¸Ä±äת»¯ÂÊ¡£ (6) ´í£¬ÉýΣ¬ÎüÈÈ·´Ó¦ºÍ·ÅÈÈ·´Ó¦µÄ·´Ó¦ËÙÂʶ¼¼Ó¿ì¡£

3-8. ÒÑ֪ij·´Ó¦µÄ»î»¯ÄÜΪ40kJ?mol-1£¬·´Ó¦ÓÉ300KÉýÖÁ¶à¸ßʱ£¬·´Ó¦ËÙÂÊ¿ÉÔö¼Ó99±¶£¿ ½â£ºÉèζÈÉýÖÁx K

40?103J?mol?1xK?300Kln99?()

8.314J?K?1?mol?1300xK2x = 420.5K

3-9. ·´Ó¦2NO(g) + Br2(g) = 2NOBr(g) £¬Æä»úÀíÈçÏ£º (1) NO(g) + Br2(g) ¡ú NOBr2(g) (Âý) (2) NOBr2(g) + NO(g)¡ú 2NOBr(g) (¿ì)

½«ÈÝÆ÷Ìå»ýËõСΪԭÀ´µÄ1/2ʱ£¬¼ÆËã·´Ó¦ËÙÂÊÔö¼Ó¶àÉÙ±¶£¿ ½â£º?1?kc(NO)c(Br2)

Ìå»ýËõСºóŨ¶ÈÔö´ó£¬?2?k2c(NO)2c(Br2)=4kc(NO)c(Br2)

17

v2£½ 4 (±¶) v13-10. ÈËÌåÖÐijÖÖøµÄ´ß»¯·´Ó¦µÄ»î»¯ÄÜΪ50kJ?mol-1£¬Õý³£È˵ÄÌåÎÂΪ37¡æ£¬ÎÊ·¢ÉÕµ½40¡æµÄ²¡ÈË£¬¸Ã·´Ó¦µÄËÙÂÊÔö¼ÓÁË°Ù·ÖÖ®¼¸£¿

½â£ºÉè40¡æʱËÙÂÊÊÇ37¡æʱËÙÂʵÄx±¶

50?103J?mol?1313K?310K()?0.186 lnx??1?18.314J?K?mol313K?310Kx = 1.20 ´Ó37¡æµ½40¡æ£¬¸Ã·´Ó¦µÄËÙÂÊÔö´ó20% 3-11. ÅжÏÌâ(¶ÔµÄ¼Ç¨D¡Ì¡¬£¬´íµÄ¼Ç¨D¡Á¡¬)£º

(1)ÖÊÁ¿×÷Óö¨ÂÉÖ»ÊÊÓÃÓÚ»ùÔª·´Ó¦£¬¹Ê¶Ô¼òµ¥·´Ó¦²»ÊÊÓᣠ( ) (2) Á㼶·´Ó¦µÄËÙÂÊΪÁã¡£ ( ) (3) »î»¯ÄÜËæζȵÄÉý¸ß¶ø¼õС ( ) (4)·´Ó¦¼¶ÊýÓú¸ß£¬Ôò·´Ó¦ËÙÂÊÊÜ·´Ó¦ÎïŨ¶ÈµÄÓ°ÏìÓú´ó¡£ ( ) (5)ËÙÂʳ£ÊýÈ¡¾öÓÚ·´Ó¦ÎïµÄ±¾ÐÔ£¬Ò²ÓëζȺʹ߻¯¼ÁÓйء£ ( ) (6) ½µµÍζȿɽµµÍ·´Ó¦µÄ»î»¯ÄÜ¡£ ( ) ½â£º(1)´í (2)´í (3)´í (4)¶Ô (5)¶Ô (6)´í 3-12. д³öÏÂÁз´Ó¦µÄ±ê׼ƽºâ³£ÊýK¦È±íʾʽ (1) BaSO4(s)

Ba2+(aq) + SO42-(aq) (2) Zn(s) + CO2(g)

HAc(aq) + OH-(aq) (4) Mg(s)+2H+(aq)

¦ÈZnO(s) + CO(g) Mg2+(aq) + H2(g)

(3) Ac-(aq) + H2O(l)

2-½â£º(1) K¦È?[Ba2+]r[SO4]r (2) K?pr(CO)

pr(CO2)

[HAc]r[OH-]rpr(H2)?[Mg2+]r¦È (3) K? (4) K?-[Ac]r[H+]2r¦È3-13. ÒÑÖªÏÂÁз´Ó¦£º C(s) + H2O(g) CO(g) + H2(g) ¡÷rHm?> 0

ÔÚ·´Ó¦´ïƽºâºóÔÙ½øÐÐÈçϲÙ×÷ʱ£¬Æ½ºâÔõÑùÒƶ¯£¿

18

(1) Éý¸ß·´Ó¦ÎÂ¶È (2) Ôö´ó×ÜѹÁ¦ (3) Ôö´ó·´Ó¦Æ÷ÈÝ»ý (4) ¼ÓÈë´ß»¯¼Á

½â£º(1) ÕýÏòÒƶ¯ (2) ÄæÏòÒƶ¯ (3) ÕýÏòÒƶ¯ (4) ²»Òƶ¯ 3-14. ÒÑÖªÏÂÁз´Ó¦µÄ±ê׼ƽºâ³£Êý£º (1) HAc

H+ + Ac- K1? = 1.76¡Á10-5

NH4+ +OH- K2? = 1.77¡Á10-5

(2) NH3 + H2O(3) H2O

H+ + OH- K3? = 1.0¡Á10-14

NH4+ + Ac-

Çó·´Ó¦(4)µÄ±ê׼ƽºâ³£ÊýK4?£¬(4) NH3 + HAc

½â£º·´Ó¦(4) =·´Ó¦(1)+·´Ó¦(2) -·´Ó¦(3)

¦ÈK1¦ÈK21.76?10?5?1.77?10?54

3.1¡Á10 K???¦È?14K31.0?10¦È43-15. ÒÑ֪ij·´Ó¦¡÷rHm?(298K)= 20kJ?mol-1£¬ÔÚ300KµÄ±ê׼ƽºâ³£ÊýK?Ϊ103£¬Çó·´Ó¦µÄ±ê×¼ìرäÁ¿¡÷rSm?(298K) ¡£

½â£º¡÷rGm?(300K) = -RTlnK? = -8.314¡Á10-3 kJ?mol-1¡Á300K¡Áln103 = -17.23 kJ?mol-1 ¡÷rGm?) (300K) ¡Ö¡÷rHm? - T¡÷rSm? -17.23 kJ?mol-1 = 20kJ?mol-1 - 298K¡Á¡÷rSm? ¡÷rSm? = 0.125 kJ?K-1?mol-1 = 125 J?K-1?mol-1 3-16. HIµÄ·Ö½â·´Ó¦Îª2HI(g)

H2(g) + I2(g) ¿ªÊ¼Ê±HI(g)µÄŨ¶ÈΪ1.00mol?L-1£¬

µ±´ïµ½Æ½ºâʱÓÐ24.4%HI·¢ÉúÁ˷ֽ⣬Èô½«·Ö½âÂʽµµÍµ½10%£¬ÆäËüÌõ¼þ²»±äʱ£¬µâµÄŨ¶ÈÓ¦Ôö¼Ó¶àÉÙ£¿

½â£ºÉè·Ö½âÂʽµµÍµ½10%£¬µâµÄŨ¶ÈÓ¦Ôö¼Óx mol?L-1 2HI(g)

H2(g) + I2(g)

ƽºâʱ/mol?L-1 0.756 0.122 0.122 Öؽ¨Æ½ºâ/mol?L-1 0.900 0.050 0.050+x

19

(0.122mol?L-1)2 Kc??0.0260

(0.756mol?L-1)2(0.050mol?L-1)(0.050mol?L-1?xmol?L-1) Kc??0.0260 -12(0.900mol?L) x = 0.371 mol?L-1

3-17. ÔÚ25¡æʱÕáÌÇË®½âÉú³ÉÆÏÌÑÌǺ͹ûÌÇ£º C12H22O11+H2O

C6H12O6 (ÆÏÌÑÌÇ) + C6H12O6(¹ûÌÇ)£¬ÌåϵÖÐË®µÄŨ¶ÈÊÓΪ³£Êý¡£

(1) ÕáÌǵÄÆðʼŨ¶ÈΪ2a mol?L-1£¬´ïƽºâʱÕáÌÇË®½âÁË50%£¬KcÊǶàÉÙ£¿

(2) ÕáÌǵÄÆðʼŨ¶ÈΪa mol?L-1,£¬ÔòÔÚͬһζÈÏÂƽºâʱ£¬Ë®½â²úÎïµÄŨ¶ÈÊǶàÉÙ£¿ ½â£ºÉèÕáÌǵÄÆðʼŨ¶ÈΪa mol?L-1£¬Í¬Ò»Î¶ÈÏÂƽºâʱ£¬Ë®½â²úÎïµÄŨ¶ÈÊÇx mol?L-1. C12H22O11+H2O

C6H12O6 (ÆÏÌÑÌÇ) + C6H12O6(¹ûÌÇ)

(1) ƽºâʱ/mol?L-1 a a a (2) Öؽ¨Æ½ºâ/mol?L-1 a£­ x x x

(amol?L-1)2 Kc??amol?L-1 -1amol?L(xmol?L-1)2-1 Kc? ?amol?L-1-1amol?L?xmol?L x = 0.618a mol?L-1

3-18. Ë®µÄ·Ö½â·´Ó¦Îª 2H2O(g)

2H2(g) + O2(g)£¬ÔÚ1227¡æºÍ727¡æÏ·´Ó¦µÄ±ê×¼

ƽºâ³£ÊýK?·Ö±ðΪ1.9¡Á10-11ºÍ3.9¡Á10-19£¬¼ÆËã¸Ã·´Ó¦µÄ·´Ó¦ÈÈ¡÷rHm?¡£(441.6kJ?mol-1)

¦È¦ÈK2¦¤rHmT?T½â£º lg¦È?(21)

K12.303RTT12¦È¦¤rHm1.9?10?111500K?1000K lg?()

3.9?10?192.303?8.314?10?3kJ?K-1?mol-11500K?1000K ¡÷rHm? = 441.6kJ?mol-1

3-19. ÒÑÖªÏÂÁз´Ó¦¸÷ÎïÖʵġ÷fHm?ºÍSm?£¬¼ÆËã·´Ó¦ÔÚ298KºÍ373KʱµÄK? NH3(aq) + H2O(l)

NH4+(aq) + OH-(aq)

20

¡÷fHm? /kJ?mol -80.3 -285.8 -132.5 -230.0 Sm? /J?K-1?mol-1 111.3 69.9 113.4 -10.75

½â£º¡÷rHm? = (-132.5kJ?mol-230.0kJ?mol) - (-80.3kJ?mol-285.8kJ?mol)

= 3.6kJ?mol

¡÷rSm? = 113.4 J?K-1?mol-1£­10.75 J?K-1?mol-1£­111.3 J?K-1?mol-1£­69.9 J?K-1?mol-1

= £­78.6 J?K-1?mol-1

¡÷rGm? = ¡÷rHm?£­T¡÷rSm? = 3.6kJ?mol£­298K¡Á(£­78.6)¡Á10-3 kJ?K-1?mol-1

= 27.0kJ?mol-1

27.0kJ?mol-1 = £­8.314¡Á10-3 kJ?K-1?mol-1¡Á298¡ÁlnK?(298K) K?(298K) = 1.85¡Á10-5

?1?1?1?1?1?1?1K¦È(373K)3.6?103J?mol?1373K-298Klg?() 1.85?10?52.303?8.314J?K?1?mol?1298K?373KK¦È(373K)K¦È(373K)lg?0.127 ?1.34 1.85?10?51.85?10?5K? (373) = 2.48¡Á10-5

3-20. ÒÑ֪ˮÔÚ373KʱÆø»¯ìÊΪ40kJ?mol-1£¬ÈôѹÁ¦¹øÄÚѹÁ¦×î¸ß¿É´ï150kPa£¬Çó´Ëʱ¹øÄÚµÄζȡ£

½â£º ·´Ó¦ H2O(l)

H2O(g) ¦¤rHm?= 40kJ?mol-1

K¦È(373K)?p(H2O)/p¦È?100kPa/100kPa?1K¦È(xK)?p(H2O)/p¦È?150kPa/100kPa?1.5

1.540.60kJ?mol?1xK?373Kln?() 18.314?10?3kJ?K?1mol?1373xK2 x = 385K

21

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27

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30

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32

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9-1. ³Æȡij¿ÉÈÜÐÔÑÎ0.1616g£¬ÓÃBaSO4ÖØÁ¿·¨²â¶¨ÆäÁòº¬Á¿£¬³ÆµÃBaSO4³ÁµíΪ0.1491g£¬¼ÆËãÊÔÑùÖÐSO3µÄÖÊÁ¿·ÖÊý¡£

½â£ºw(SO3)=m(SO2) m1M(SO3).m(BaSO4)

M(BaSO4)M(SO3)m(BaSO4) M(BaSO4).m m(SO3)= w(SO3)=

=80.06¡ä0.1491=0.3665

233.39¡ä0.16169-2. ³ÆÈ¡Á׿óʯÊÔÑù0.4530g£¬ÈܽâºóÒÔMgNH4PO4ÐÎʽ³Áµí£¬×ÆÉÕºóµÃMg2P2O7 0.2825g£¬¼ÆËãÊÔÑùÖÐP¼°P2O5µÄÖÊÁ¿·ÖÊý¡£

½â£º

m(P)m2M(P).m(Mg2P2O7)m(P)=M(Mg2P2O7)w(P)=w(P)=2¡äM(P).m(Mg2P2O7)2´´30.970.2825==0.1735M(Mg2P2O7).m222.60¡ä0.4530

w(P2O5)=M(P2O5).m(Mg2P2O7)141.95¡ä0.2825==0.3977

M(Mg2P2O7).m222.60¡ä0.45309-3. ³ÆÈ¡´¿NaCl 0.1169g£¬¼ÓË®Èܽâºó£¬ÒÔK2CrO4Ϊָʾ¼Á£¬ÓÃAgNO3±ê×¼ÈÜÒºµÎ¶¨Ê±¹²ÓÃÈ¥20.00mL£¬Çó¸ÃAgNO3ÈÜÒºµÄŨ¶È¡£

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Ag++Cl-=AgCl¡¥n(AgCl)=n(NaCl)c(AgCl)?V(AgCl)m(NaCl)M(NaCl)

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