ÌâÄ¿ÄÚÈÝ

(16·Ö)̼¡¢µª¡¢Áò¡¢ÂÈÊÇËÄÖÖÖØÒªµÄ·Ç½ðÊôÔªËØ¡£
£¨1£©CH4(g)ÔÚO2(g)ÖÐȼÉÕÉú³ÉCO(g)ºÍH2O(g)µÄ¡÷HÄÑÒÔÖ±½Ó²âÁ¿£¬Ô­ÒòÊÇ              ¡£
ÒÑÖª£ºa£®2CO(g)+O2(g)=2CO2(g)   ¡÷H =-566£®0 kJ¡¤mol-1
b£®CH4(g)+2O2(g)=CO2(g)+2H2O(g)  ¡÷H =-890£®0 kJ¡¤mol-1
ÔòCH4(g)ÔÚO2(g)ÖÐȼÉÕÉú³ÉCO(g)ºÍH2O(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ                                    ¡£
£¨2£©¹¤ÒµÉϺϳɰ±ÆøµÄ·´Ó¦Îª£ºN2(g)+3H2(g)2NH3(g) ¡÷H<0¡£ÏÖ½«10 mol N2ºÍ26 mol H2ÖÃÓÚÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖУ¬N2µÄƽºâת»¯ÂÊ()ÓëÌåϵ×Üѹǿ(P)¡¢Î¶È(T)µÄ¹ØϵÈçͼËùʾ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù·´Ó¦´ïµ½Æ½ºâ״̬Bʱ£¬ÈÝÆ÷µÄÈÝ»ý10 L£¬ÔòT1ʱ£¬ºÏ³É°±·´Ó¦µÄƽºâ³£ÊýK=       L2¡¤mol-1¡£
¢Úƽºâ״̬ÓÉA±äµ½Cʱ£¬¶ÔÓ¦µÄƽºâ³£ÊýK(A)   K(C)(Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±)¡£
£¨3£©ÔÚ25¡æʱ£¬HSCN¡¢HClO¡¢H2CO3µÄµçÀë³£ÊýÈçÏÂ±í£º

HClO
HSCN
H2CO3
K=3.210-8
K=0.13
Kl=4.210-7
K2=5.610-11
 
¢Ù1 mol¡¤L-1µÄKSCNÈÜÒºÖУ¬ËùÓÐÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ     >     >     >          ¡£
¢ÚÏòNa2CO3ÈÜÒºÖмÓÈë¹ýÁ¿HClOÈÜÒº£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ            ¡£
¢Û25¡æʱ£¬ÎªÖ¤Ã÷HClOΪÈõËᣬijѧϰС×éµÄͬѧû¼ÆÁËÒÔÏÂÈýÖÖʵÑé·½°¸¡£ÏÂÁÐÈýÖÖ·½°¸ÖУ¬ÄãÈÏΪÄܹ»´ïµ½ÊµÑéÄ¿µÄµÄÊÇ   (ÌîÏÂÁи÷ÏîÖÐÐòºÅ)¡£
a£®ÓÃpH¼Æ²âÁ¿0£®1 mol¡¤L-1NaClOÈÜÒºµÄpH£¬Èô²âµÃpH>7£¬¿ÉÖ¤Ã÷HClOΪÈõËá
b£®ÓÃpHÊÔÖ½²âÁ¿0£®01 mol¡¤L-1HClOÈÜÒºµÄpH£¬Èô²âµÃpH>2£¬¿ÉÖ¤Ã÷HClOΪÈõËá
c¡¢ÓÃÒÇÆ÷²âÁ¿Å¨¶È¾ùΪ0£®1 mol¡¤L-1µÄHClOÈÜÒººÍÑÎËáµÄµ¼µçÐÔ£¬Èô²âµÃHClOÈÜÒºµÄµ¼µçÐÔÈõÓÚÑÎËᣬ¿ÉÖ¤Ã÷HClOΪÈõËá

£¨1£©£¨5·Ö£©ÄÑÒÔ¿ØÖÆ·´Ó¦Ö»Éú³ÉCO(g)£¨2·Ö£¬ÌåÏÖ·´Ó¦ÄÑÒÔ¿ØÖƼ´¿É£©
2CH4(g)+3O2(g)=2CO(g)+4H2O(g) ?H=¡ª1214.0kJ?mol?1£¨3·Ö£©
£¨2£©£¨4·Ö£©¢Ù0.025£¨2·Ö£©¢Ú > £¨2·Ö£©
£¨3£©£¨7·Ö£©¢Ù c(K+)>C(SCN?)>c(OH?)>c(H+)£¨2·Ö£¬²»µ¥¶À¼Æ·Ö£©
¢ÚNa2CO3 + HClO = NaHCO3 +NaClO£¨3·Ö£©
¢Ûac£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©CH4ȼÉÕÈÝÒ×Éú³ÉCO2£¬ÄÑÒÔ¿ØÖÆ·´Ó¦Ö»Éú³ÉCO(g)£»Ê×ÏÈд³öCH4ÓëO2·´Ó¦Éú³ÉCOºÍH2OµÄ»¯Ñ§·½³Ìʽ£¬²¢×¢Ã÷״̬£¬È»ºó¸ù¾Ý¸Ç˹¶¨ÂÉÇó³öìʱ䣺?H=¡ª?H1+2?H2=¡ª1214.0kJ?mol?1£¬½ø¶ø¿Éд³öÈÈ»¯Ñ§·½³Ìʽ¡£
£¨2£©¢Ù·´Ó¦´ïµ½Æ½ºâ״̬Bʱ£¬N2µÄת»¯ÂÊΪ20%£¬¸ù¾ÝÈý¶Îʽ½øÐмÆËã
N2(g)+3H2(g)2NH3(g)
ÆðʼÎïÖʵÄÁ¿£¨mol£©  10     26          0
ת»¯ÎïÖʵÄÁ¿£¨mol£©   2     6           4
ƽºâÎïÖʵÄÁ¿£¨mol£©   8     20          4
Ôòƽºâ³£ÊýΪ£ºK=(0.4mol?L?1)2¡Â[0.8mol/L¡Á(2mol?L?1)3]=0.025 L2¡¤mol-1¡£
¢ÚBµãN2ת»¯ÂÊ´óÓÚCµã£¬ËµÃ÷Bµãƽºâ³£Êý´óÓÚCµãƽºâ³£Êý£¬AµãÓëBµãζÈÏàͬ£¬Ôòƽºâ³£ÊýÏàͬ£¬ËùÒÔK(A)  >  K(C)
£¨3£©¢ÙKSCNΪÈõËáÇ¿¼îÑΣ¬SCN?Ë®½âʹÈÜÒº³Ê¼îÐÔ£¬ËùÒÔÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºc(K+)>C(SCN?)>c(OH?)>c(H+)
¢ÚÒòΪK1(H2CO3)>K(HClO)>K2(H2CO3)£¬ËùÒÔHClOÓëNa2CO3·´Ó¦Éú³ÉNaClOºÍNaHCO3£¬»¯Ñ§·½³ÌʽΪ£ºNa2CO3 + HClO = NaHCO3 +NaClO
¢Ûa¡¢ÓÃpH¼Æ¿ÉÒÔ²âÁ¿0£®1 mol¡¤L-1NaClOÈÜÒºµÄpH£¬Èô²âµÃpH>7£¬ËµÃ÷NaClOË®½âÏÔ¼îÐÔ£¬¿ÉÖ¤Ã÷HClOΪÈõËᣬÕýÈ·£»b¡¢ÒòΪHClO¾ßÓÐƯ°×ÐÔ£¬ÎÞ·¨ÓÃpHÊÔÖ½²âÁ¿0£®01 mol¡¤L-1HClOÈÜÒºµÄpH£¬´íÎó£»c¡¢ÒòΪHClOºÍÑÎËáŨ¶ÈÏàͬ£¬ËùÒÔÈô²âµÃHClOÈÜÒºµÄµ¼µçÐÔÈõÓÚÑÎËᣬ¿ÉÖ¤Ã÷HClOΪÈõËᣬÕýÈ·¡£
¿¼µã£º±¾Ì⿼²éÈÈ»¯Ñ§·½³ÌʽµÄÊéд¡¢Æ½ºâ³£ÊýµÄÅжϺͼÆËã¡¢Èõµç½âÖʵĵçÀë¡¢ÑÎÀàµÄË®½â¡¢Àë×ÓŨ¶È±È½Ï¡¢ÊµÑé·½°¸µÄ·ÖÎö¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

2013ÄêÎíö²ÌìÆø¶à´ÎËÁÅ°ÎÒ¹úÖж«²¿µØÇø¡£ÆäÖÐÆû³µÎ²ÆøºÍȼúβÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£
£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g)+2CO2CO2(g)+N2(g)
¢Ù¶ÔÓÚÆøÏà·´Ó¦£¬ÓÃij×é·Ö£¨B£©µÄƽºâѹǿ£¨PB£©´úÌæÎïÖʵÄÁ¿Å¨¶È£¨CB£©Ò²¿ÉÒÔ±íʾƽºâ³£Êý£¨¼Ç×÷KP£©£¬Ôò¸Ã·´Ó¦µÄKP=-                ¡£
¢Ú¸Ã·´Ó¦ÔÚµÍÎÂÏÂÄÜ×Ô·¢½øÐУ¬¸Ã·´Ó¦µÄ¦¤H            0¡££¨Ñ¡Ìî¡°>¡±¡¢¡°<¡±£©
¢ÛÔÚijһ¾øÈÈ¡¢ºãÈݵÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄNO¡¢CO·¢ÉúÉÏÊö·´Ó¦£¬²âµÃÕý·´Ó¦µÄËÙÂÊËæʱ¼ä±ä»¯µÄÇúÏßÈçͼËùʾ£¨ÒÑÖª£ºt2 --tl=t3£­t2£©¡£

ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ         ¡££¨Ìî±àºÅ£©
A£®·´Ó¦ÔÚcµãδ´ïµ½Æ½ºâ״̬
B£®·´Ó¦ËÙÂÊaµãСÓÚbµã
C£®·´Ó¦ÎïŨ¶Èaµã´óÓÚbµã
D£®NOµÄת»¯ÂÊ£ºtl¡«t2>t2¡«t3
£¨2£©ÃºµÄ×ÛºÏÀûÓá¢Ê¹ÓÃÇå½àÄÜÔ´µÈÓÐÀûÓÚ¼õÉÙ»·¾³ÎÛȾ¡£ºÏ³É°±¹¤ÒµÔ­ÁÏÆøµÄÀ´Ô´Ö®Ò»Ë®ÃºÆø·¨£¬ÔÚ´ß»¯¼Á´æÔÚÌõ¼þÏÂÓÐÏÂÁз´Ó¦£º
¢Ù
¢Ú
¢Û
¢Ù¡÷H3ºÍ¡÷H1¡¢¡÷H2µÄ¹ØϵΪ¡÷H3=            ¡£
¢ÚÔÚºãÎÂÌõ¼þÏ£¬½«l mol COºÍ1 mol H2O£¨g£©³äÈëij¹Ì¶¨ÈÝ»ýµÄ·´Ó¦ÈÝÆ÷£¬´ïµ½Æ½ºâʱÓÐ50%µÄCOת»¯ÎªCO2¡£ÔÚtlʱ±£³ÖζȲ»±ä£¬ÔÙ³äÈë1 mol H2O£¨g£©£¬ÇëÔÚͼÖл­³ötlʱ¿ÌºóH2µÄÌå»ý·ÖÊý±ä»¯Ç÷ÊÆÇúÏß¡£

¢Û¼×´¼ÆûÓÍ¿É¡¯ÒÔ¼õÉÙÆû³µÎ²Æø¶Ô»·¾³µÄÎÛȾ¡£
ij»¯¹¤³§ÓÃˮúÆøΪԭÁϺϳɼ״¼£¬ºãÎÂÌõ¼þÏ£¬ÔÚÌå»ý¿É±äµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£ºCO(g)+2H2(g) CH3OH(g)µ½´ïƽºâʱ£¬²âµÃCO¡¢H2¡¢CH3OH·Ö±ðΪ1 mol¡¢1 mol¡¢1 mol£¬ÈÝÆ÷µÄÌå»ýΪ3L£¬ÏÖÍùÈÝÆ÷ÖмÌÐøͨÈË3 mol CO£¬´Ëʱv£¨Õý£©         v£¨Ä棩£¨Ñ¡Ìî¡®¡®>¡±¡¢¡°<¡¯¡¯»ò¡°=¡±£©£¬ÅжϵÄÀíÓÉ        ¡£

µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒª×÷Óá£
£¨1£©Ôڹ̶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºN2£¨g£©+3H2£¨g£© 2NH3£¨g£© ¡÷H=¡ª92£®4kJ/mol£¬
Æäƽºâ³£ÊýKÓëζÈTµÄ¹ØϵÈçÏÂ±í£º

T/K
298
398
498
ƽºâ³£ÊýK
4.1¡Á106
K1
K2
 
ÊÔÅжÏK1       K2£¨Ìîд¡°>¡± ¡° =¡±»ò¡°<¡±£©¡£
£¨2£©ÓÃ2mol N2ºÍ3mol H2ºÏ³É°±£¬ÈýÈÝÆ÷µÄ·´Ó¦Î¶ȷֱðΪT1¡¢T2¡¢T3ÇҺ㶨²»±ä£¬ÔÚÆäËüÌõ¼þÏàͬµÄÇé¿öÏ£¬ÊµÑé²âµÃ·´Ó¦¾ù½øÐе½t minʱN2µÄÖÊÁ¿·ÖÊýÈçͼËùʾ£¬´Ëʱ¼×¡¢ÒÒ¡¢±ûÈý¸öÈÝÆ÷ÖÐÒ»¶¨´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ        £¬¶¼´ïµ½Æ½ºâ״̬ʱ£¬N2ת»¯ÂÊ×îµÍµÄÊÇ   ¡£

£¨3£©NH3ÓëCO2ÔÚ120oC£¬´ß»¯¼Á×÷ÓÃÏ¿ÉÒԺϳɷ´Ó¦Éú³ÉÄòËØ£ºCO2 +2NH3£¨NH2£©2CO +H2O
ÔÚÃܱշ´Ó¦ÈÝÆ÷ÖУ¬»ìºÏÆøÌåÖÐNH3µÄº¬Á¿±ä»¯¹ØϵÈçͼËùʾ

£¨¸ÃÌõ¼þÏÂÄòËØΪ¹ÌÌ壩¡£ÔòAµãµÄÕý·´Ó¦/ËÙÂÊ£¨CO2£©      BµãµÄÄæ·´Ó¦ËÙÂÊ£¨CO2£©£¨Ìîд¡°>¡±¡°=¡±»ò¡°<¡±£©£¬NH3µÄƽºâת»¯ÂÊΪ____         £»
£¨4£©ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
2H2£¨g£©+O2£¨g£©=2H2O£¨1£©   ¡÷H = -571£®6kJ/mol
N2£¨g£©+O2£¨g£©2NO£¨g£©    ¡÷H =+180kJ/mol
Çëд³öÓÃNH3»¹Ô­NOµÄÈÈ»¯Ñ§·½³Ìʽ_                 £»
£¨5£©°±ÆøÔÚ´¿ÑõÖÐȼÉÕ£¬Éú³ÉÒ»ÖÖµ¥ÖʺÍË®£¬ÊÔд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ____     ¡£¿Æѧ¼ÒÀûÓôËÔ­Àí£¬Éè¼Æ³É°±Æø£­ÑõÆøȼÁϵç³Ø£¬ÔòͨÈË°±ÆøµÄµç¼«ÊÇ        £¨Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©£¬ÔÚ¼îÐÔÌõ¼þÏ£¬Í¨ÈË°±ÆøµÄµç¼«·¢ÉúµÄµç¼«·´Ó¦Ê½Îª                                  ¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø