ÎïÀí»¯Ñ§ÉϲḴϰ ÏÂÔØ±¾ÎÄ

15. C 4ÖÖ»¯ºÏÎïºÍÁ½´¿×é·Ö¹²6ÖÖÎïÖÊ£¬×é³ÉÏàÁÚµÄÁ½ÎïÖÊÖ®¼äÄÜÐγÉÒ»µÍ¹²ÈÛÎï¹Ê¹²ÓÐ5ÖÖ¡£

16. D °´¡°Ò»´ÎÀäÈ´½á¾§£¬×î¶à¡±µÄÒªÇó£¬ÔÚ80oC100gÑÎÓ¦ÈÜÓÚË®Ðγɱ¥ºÍÈÜÒº£¬¼´¡ÁB=0.35£¬ÐγÉ100g/0.35=285.7gÈÜÒº¡£ÀäȴʱӦ¾¡¿ÉÄܽӽüµÍ¹²ÈÛµã-21oC£¬´Ó¶øÊÇÎïϵµã¡ÁB=0.35£¬ÒºÏà¡ÁB=0.19£¬¹ÌÏà¡ÁB=1 µÄ¶þÏàÆ½ºâ£¬°´¸Ü¸Ë¹æÔò

m(1-0.35)=( 285.7g-m)(0.35-0.19) m=56.4g

17. D ¶Ô²»ÏàºÏÈ۵㻯ºÏÎÔÚ²»ÏàºÏÈÛµãÉÏÊÇÈýÏàÆ½ºâ¡£ 18. B ´Ë¼´ËùνµÈÁ¿¹ØÏµ¡£

19. C TÈôСÓÚ»òµÈÓÚTe£¬¾Í»áÓйÌÌåÎö³ö¡£ÇÒÖ»ÓÐC=Ce,²ÅÄÜÔÚT±È TeÂÔ¸ßʱÎÞ¹ÌÌåÎö³ö¡£ 20. C 21. A 22. C Èý Ìî¿ÕÌâ

1. 2, 1, 1 2. 1, 2, 2 3. 273.15K 4. ´¿BºÍºã·ÐÎï 5. 2, 1 6. 2, 3, 4 7. ÕôÆøÑ¹ 8. ¸÷×é·Ö¼ä´ïµ½ÏàÆ½ºâ

µÚÁùÕÂ

Ò» ÅжÏÌâ

1. ´í ¦¤rGm ÊÇÓÉÖ¸¶¨Ì¬µÄ·´Ó¦ÎïÉú³ÉÖ¸¶¨Ì¬²úÎ·´Ó¦½ø¶ÈΪ1molʱÌåϵµÄ¼ª²¼Ë¹×ÔÓÉÄܵĸıäÁ¿¡£ÈçÌâËù˵µÄÔÚÒ»¶¨Á¿µÄ·´Ó¦ÎïÖнøÐУ¬Ôò²»¿ÉÄÜά³ÖɨӦÎïºÍ²úÎïÔÚÖ¸¶¨Ì¬¡£

O2. ´í ¦¤rGm ÊÇÓɱê׼̬µÄ·´Ó¦ÎïÉú³É±ê׼̬²úÎ·´Ó¦½ø¶ÈΪ1molʱÌåϵµÄ¼ª²¼Ë¹×ÔÓÉÄܵĸıäÁ¿¡£Æ½ºâʱ¦¤rGm = 0

3. ´í ¦¤rGm µÄ´óС¿É·´Ó³·´Ó¦ÔÚÖ¸¶¨Ìõ¼þQPϵķ´Ó¦Ç÷ÊÆ£¬µ«Ëæ×Å·´Ó¦½øÐУ¬QPÔڱ䣬¦¤rGmÒ²»á±ä¡£¹Ê²»ÄÜÓÉijһ¦¤rGmÖµ¾Í¶Ï¶¨·´Ó¦ÄܽøÐе½µ×¡£

4. ´í ¦¤rGm µÄ´óС¿É·´Ó³·´Ó¦Ç÷ÊÆµÄÇ¿Èõ£¬µ«²»ÄÜÈ·¶¨·´Ó¦µÄ¿ìÂý¡£ 5. ´í ¦¤rGm =£­RTlnKO£«RTlnQP ¿É¼û¦¤rGm £¾0£¬²»ÄܾͿ϶¨KO£¼1£¬»¹Òª¿´QPµÄ´óС¡£ 6. ¶Ô

7. ¶Ô µ±A£¬B£¬CµÄ·ÖѹÏàµÈʱ£¬QP£½1¡£¦¤rGm =¦¤rGmO£«RTlnQP =¦¤rGmO ¹Ê¿ÉÓæ¤rGmOÀ´ÅжϷ´Ó¦½øÐеķ½Ïò¡£ 8. ¶Ô. 9. ¶Ô

10. ¶Ô ÓÉÆ½ºâ×é³É¿ÉÇóµÃÉú³É·´Ó¦µÄƽºâ³£ÊýK£¬Óɦ¤rGmO£½¦¤fGmO£½-RTlnK¼´ÄÜÇóµÃ¡£ ¶þ Ñ¡ÔñÌâ

1. C ÓɼÆÁ¿·½³Ìʽ¿ÉÖªºÏ³É·´Ó¦µÄƽºâ³£ÊýÓ¦ÊǷֽⷴӦµÄƽºâ³£ÊýµÄµ¹ÊýµÄ1/2´Î·½¡£¼´Îª(1/0.25)1/2 = 2 .

2. B QP =( 1kPa)2/1kPa=1kPa>KP 3. C

4. D KP = K¡ÁP ¦²¦Ô§£

5. D ´ÓKPµÄµ¥Î»¿ÉÖª¦²¦Ô§££½1¡£ ¹ÊKC£½KP /RT = [831400/(8.314¡Á200)]mol?m-3=500 mol.m-3=0.5 mol?dm-3

¦ÔB ¦ÔB¦ÔB¦Ô§£

6. D K¡Á£½¦°¡ÁB£½¦°(nB/¦²nB) =¦°[(nB/V)V/¦²nB]= KC(V/¦²nB)7. B KO£½(20.9743¡Á106/101325)1/2=14.3875 ¦¤rGm0£½£­RTln KO = -8.314¡Á718Jln14.3875= -15.92kJ 8. D p(NH3)=p(HCl)=(1/2)¡Á100kPa=0.5 PO¹Ê KO£½[p(NH3)/ PO].[p(HCl)/ PO]=0.5¡Á0.5=0.25 9. B ×Üѹ100kPa.½üËÆ±ê×¼´óÆøÑ¹£¬¹ÊK¡Á£½KP¡£Q¡Á£½QP

¦¤rGm =¦¤rGmO£«RTlnQP

£½19.29¡Á103£«8.314¡Á1000ln(0.22/0.7) = 19.29¡Á103 ¨C 2.865¡Á8.314¡Á103 = -4.53¡Á103 < 0 10. D

11. D ¦Î£½¦¤nC/¦ÔC =0.9/3 =0.3 nA =1-2¡Á0.3= 0.4

nB= 2-0.3 = 1.7 nC =0.9 nD=1+2¡Á0.3= 1.6 ¦²n = 4.6 K¡Á = (0.9/4.6)3.(1.6/4.6)2/[(0.4/4.6)2.1.7/4.6] = 0.324 12. B ¦Î£½¦¤nB/¦ÔB =-0.2/-2 =0.1

13. A ÔÚT£¬Pºã¶¨Ï£¬»¯Ñ§·´Ó¦´ïµ½Æ½ºâ,±ØÓÐB£¬D¡£¶øCʽ£¬²»ÂÛ·´Ó¦ÊÇ·ñ´ïƽºâ×ÜÊdzÉÁ¢µÄ¡£Î¨ÓÐA£¬±ê×¼·´Ó¦ÈÈЧӦΪÁ㣬ֻÓÐÌØÊâµÄ·´Ó¦²Å³ÉÁ¢¡£

14. A Óɦ¤rGm0£½£¨980£­1.456T/K£©J?mol-1£¬T£½373K´úÈëµÃ¦¤rGm0>0 ¹Ê·´Ó¦¦Á£­HgS£½¦Â£­HgS µÄ·´Ó¦Îï¸üÎȶ¨¡£

15. A ¼´Ó¦ÔÚË®µÄÉú³É¼ª²¼Ë¹×ÔÓÉÄÜÉϼÓÉÏÓÉË®µÄ±¥ºÍÕôÆø±äΪ1´óÆøÑ¹µÄ¼ª²¼Ë¹×ÔÓÉÄܱä RTln(101325/3167.74)=8.59kJ ׼ȷֵӦÊÇ(-237.19£«8.59)kJ?mol-1=-228.6 kJ?mol-1 16. A ¸Ã·´Ó¦¦²¦ÔB£½-1 KP£½KC.RT-1 KP/KC.£½RT-1 =(8.314¡Á300)-1 = 4¡Á10-4 17. B

18. A ¦¤fGmO(CO)/kJ?mol-1£½-174.8+0.008314¡Á298ln(4/57)

¼ÆËãÖÐÂÔÈ¥ÁËÈÜҺѹÁ¦´Ó57mmHgÖÁ760mmHgµÄGµÄÔöÁ¿¡£ 19. B ¦¤fGmO(CO)/kJ?mol-1£½-90.52+0.008314¡Á298ln(1/0.1) 20. C ·ÅÈȶø·Ö×ÓÊý¼õÉٵķ´Ó¦¡£ 21. D lg KPO =-6690/(T/K)+17.3

ln KPO =2.303[-6690/(T/K)+17.3]=-15.41¡Á103/(T/K)+39.84

¶ÔÕÕ ln KPO =- (¦¤rHmO/RT) + (¦¤rSmO/R) ¦¤rHmO = 15.41¡Á103 K R= 128.1 ¡Á103J?mol-1 22. C lg KPO =-6690/(T/K)+17.3

ln KPO =2.303[-6690/(T/K)+17.3]=-15.41¡Á103/(T/K)+39.84

£­

¶ÔÕÕ ln KPO =- (¦¤rHmO/RT) + (¦¤rSmO/R) ¦¤rSmO = 39.84R =331.2 J?K1?mol-1 23. B ln KO=-10593.8K/T+6.470 ¶Ôln KO=-¦¤rHOm /RT+C µÃ¦¤rHOm =10593.8K¡Á8.314 J?.K-1?mol-1 =88.08 kJ?mol-1 24. B ¦¤rGm£½¦¤rHm -T¦¤rS m

=£­102kJ?mol-1 ¨CT(-330J?K-1?mol-1) =0 T=309.1K 25. C ln(S2/S1) = (¦¤H/R) ¡Á (1/T1-1/T2) ¦¤H = RT1T2 ln(S2/S1) / (T2-T1) = 8.314¡Á283¡Á303Jln(0.426/0.207)/20= 25726J