Ò»ÖÖÐÂÐÍÈÛÈÚÑÎȼÁϵç³Ø¾ßÓи߷¢µçЧÂʶø±¶ÊÜÖØÊÓ¡£ÏÖÓÐLi2CO3ºÍNa2CO3µÄÈÛÈÚÑλìºÏÎï×÷µç½âÖÊ£¬Ò»¼«Í¨COÆøÌ壬ÁíÒ»¼«Í¨O2ºÍCO2µÄ»ìºÏÆøÌ壬ÖÆ×÷650¡æʱ¹¤×÷µÄȼÁϵç³Ø£¬Æäµç³Ø×Ü·´Ó¦ÊÇ2CO+O2=2CO2¡£ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ  £¨    £©

     A.ͨCOµÄÒ»¼«Êǵç³ØµÄÕý¼«                      

B.¸º¼«µç¼«·´Ó¦ÊÇ£ºO2+2CO2+4e¨D=2CO32¨D

     C.ÈÛÈÚÑÎÖÐCO32¨DµÄÎïÖʵÄÁ¿ÔÚ¹¤×÷ʱ±£³Ö²»±ä  

D.Õý¼«·¢ÉúÑõ»¯·´Ó¦

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¢ñ£®X¡¢Y¡¢ZÊǶÌÖÜÆÚÔªËصÄÈýÖÖ³£¼ûÑõ»¯ÎX¸úË®·´Ó¦¿ÉÉú³ÉÒ»ÖÖ¾ßÓл¹Ô­ÐԵIJ»Îȶ¨µÄ¶þÔªËᣬ¸ÃËáµÄ»¯Ñ§Ê½ÊÇ
H2SO3
H2SO3
£»YºÍX×é³ÉÔªËØÏàͬ£¬YÓëË®·´Ó¦Éú³ÉM£¬×ãÁ¿MµÄŨÈÜÒºÓë3.2g CuÔÚ¼ÓÈÈÌõ¼þϳä·Ö·´Ó¦£¬ÆäÖб»»¹Ô­µÄMµÄÎïÖʵÄÁ¿Îª
0.05mol
0.05mol
£®¹Ì̬Z¿ÉÓ¦ÓÃÓÚÈ˹¤½µÓ꣬0.5mol?L-1ZÓëNaOH·´Ó¦ËùµÃµÄÕýÑÎË®ÈÜÒºÖУ¬Àë×ÓŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòΪ
C£¨Na+£©£¾C£¨CO32- £©£¾C£¨OH-£©£¾C£¨HCO3-£©£¾C£¨H+£©
C£¨Na+£©£¾C£¨CO32- £©£¾C£¨OH-£©£¾C£¨HCO3-£©£¾C£¨H+£©
£®
¢ò£®N2O5ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢£®ÏÖÒÔH2¡¢O2¡¢ÈÛÈÚÑÎNa2CO3×é³ÉµÄȼÁϵç³Ø£¬ÆäÖÐYΪCO2£»²ÉÓõç½â·¨ÖƱ¸N2O5£¬×ܵķ´Ó¦·½³ÌʽΪ£ºN2O4+2HNO3=2N2O5+H2£¬×°ÖÃÈçͼËùʾ£®
д³öʯīIµç¼«ÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½
H2+CO32--2e-=CO2+H2O
H2+CO32--2e-=CO2+H2O
£®
ÔÚµç½â³ØÖÐÉú³ÉN2O5µÄµç¼«·´Ó¦Ê½Îª
N2O4+2HNO3-2e-=2N2O5+2H+
N2O4+2HNO3-2e-=2N2O5+2H+
£®
¢ó£®ÑÌÆøµÄÍÑÁò£¨³ýSO2£©¼¼ÊõºÍÍÑÏõ£¨³ýNOx£©¼¼Êõ¶¼ÊÇ»·¾³¿ÆѧÑо¿µÄÈȵ㣮
£¨1£©ÑÌÆøÍÑÁò¡¢ÍÑÏõ¶Ô»·¾³µÄÒâÒå
·ÀÖ¹ËáÓê¡¢¹â»¯Ñ§ÑÌÎíµÄ·¢Éú
·ÀÖ¹ËáÓê¡¢¹â»¯Ñ§ÑÌÎíµÄ·¢Éú
£®
£¨2£©Ä¿Ç°£¬¿Æѧ¼ÒÕýÔÚÑо¿Ò»ÖÖÒÔÒÒÏ©×÷Ϊ»¹Ô­¼ÁµÄÍÑÏõ£¨NO£©Ô­Àí£¬ÆäÍÑÏõ»úÀíʾÒâͼÈçͼ1£®

ÍÑÏõÂÊÓëζȡ¢¸ºÔØÂÊ£¨·Ö×ÓɸÖд߻¯¼ÁµÄÖÊÁ¿·ÖÊý£©µÄ¹ØϵÈçͼ2Ëùʾ£®
¢Ùд³ö¸ÃÍÑÏõÔ­Àí×Ü·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
6NO+3O2+2C2H4
 ´ß»¯¼Á 
.
 
3N2+4CO2+4H2O
6NO+3O2+2C2H4
 ´ß»¯¼Á 
.
 
3N2+4CO2+4H2O
£®¢ÚΪ´ïµ½×î¼ÑÍÑÏõЧ¹û£¬Ó¦²ÉÈ¡µÄÌõ¼þÊÇ
350¡æ¡¢¸ºÔØÂÊ3%
350¡æ¡¢¸ºÔØÂÊ3%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2010?ʯ¾°É½Çøһģ£©ÀûÓÃN2ºÍH2¿ÉÒÔʵÏÖNH3µÄ¹¤ÒµºÏ³É£¬¶ø°±ÓÖ¿ÉÒÔ½øÒ»²½ÖƱ¸ÏõËᣬÔÚ¹¤ÒµÉÏÒ»°ã¿É½øÐÐÁ¬ÐøÉú²ú£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖª£ºN2£¨g£©+O2£¨g£©¨T2NO£¨g£©¡÷H=+180.5kJ?mol-1
N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-92.4kJ?mol-1
2H2£¨g£©+O2£¨g£©¨T2H2O£¨g£©¡÷H=-483.6kJ?mol-1
ÈôÓÐ17g°±Æø¾­´ß»¯Ñõ»¯ÍêÈ«Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøËù·Å³öµÄÈÈÁ¿Îª
226.3kJ
226.3kJ
£»
£¨2£©Ä³¿ÆÑÐС×éÑо¿ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¸Ä±äÆðʼÎïÇâÆøµÄÎïÖʵÄÁ¿¶ÔN2£¨g£©+3H2£¨g£©?2NH3£¨g£©·´Ó¦µÄÓ°Ï죬ʵÑé½á¹ûÈçͼ2Ëùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©
¢ÙͼÏóÖÐT1ºÍT2µÄ¹ØϵÊÇ£ºT1
£¼
£¼
T2£¨Ìî¡°£¾¡±¡¢¡°£¼¡±¡¢¡°=¡±»ò¡°ÎÞ·¨È·¶¨¡±£©
¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇ
c
c
£¨Ìî×Öĸ±êºÅ£©£¬
¢ÛÔÚÆðʼÌåϵÖмÓÈëN2µÄÎïÖʵÄÁ¿Îª
n
3
n
3
molʱ£¬·´Ó¦ºó°±µÄ°Ù·Öº¬Á¿×î´ó£®ÈôÈÝÆ÷ÈÝ»ýΪ1L£¬n=3mol·´Ó¦´ïµ½Æ½ºâʱH2µÄת»¯ÂÊΪ60%£¬Ôò´ËÌõ¼þÏ£¨T2£©£¬·´Ó¦µÄƽºâ³£ÊýK=
2.08£¨mol?L-1£©-2
2.08£¨mol?L-1£©-2
£»
£¨3£©N2O3ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢£®
¢ÙÒ»¶¨Î¶ÈÏ£¬ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐN2O3¿É·¢ÉúÏÂÁз´Ó¦£º2N2O3?4NO2£¨g£©+O2¡÷H£¾0ϱíΪ·´Ó¦ÔÚijζÈϵIJ¿·ÖʵÑéÊý¾ÝÔò50sÄÚNO2µÄƽ¾ùÉú³ÉËÙÂÊΪ
0.0592mol?L-1?s-1
0.0592mol?L-1?s-1

V/s 0 50 100
c£¨N2O3£©/mol?L-1 5.00 3.52 2.48
¢ÚÏÖÒÔH2¡¢O2¡¢ÈÛÈÚÑÎNa2CO3×é³ÉµÄȼÁϵç³Ø£¬²ÉÓõç½â·¨ÖƱ¸N2O3£¬×°ÖÃÈçͼ1Ëùʾ£¬ÆäÖÐYΪCO2£®Ð´³öʯīIµç¼«ÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½
H2+CO32-¨TCO2+H2O+2e-
H2+CO32-¨TCO2+H2O+2e-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨16·Ö£©ÀûÓÃN2ºÍH2¿ÉÒÔʵÏÖNH3µÄ¹¤ÒµºÏ³É£¬¶ø°±ÓÖ¿ÉÒÔ½øÒ»²½ÖƱ¸ÏõËᣬÔÚ¹¤ÒµÉÏÒ»°ã¿É½øÐÐÁ¬ÐøÉú²ú¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

  £¨1£©ÒÑÖª£ºN2(g)+O2(g)=2NO(g)                  ¡÷H=+180.5kJ/mol

             N2(g)+3H2(g)  2NH3(g) ¡÷H=£­92.4kJ/mol

             2H2(g)+O2(g)=2H2O(g)             ¡÷H=£­483.6kJ/mo

ÈôÓÐ17g°±Æø¾­´ß»¯Ñõ»¯ÍêÈ«Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøËù·Å³öµÄÈÈÁ¿Îª                 ¡£

  £¨2£©Ä³¿ÆÑÐС×éÑо¿£ºÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¸Ä±äÆðʼÎïÇâÆøµÄÎïÖʵÄÁ¿¶ÔN2(g)+3H2(g)

2NH3(g)·´Ó¦µÄÓ°Ï졣ʵÑé½á¹ûÈçͼËùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©

         ¢ÙͼÏñÖÐT2ºÍT1µÄ¹ØϵÊÇ£ºT2        T2£¨Ìî¡°¸ßÓÚ¡±¡°µÍÓÚ¡±¡°µÈÓÚ¡±¡°ÎÞ·¨È·¶¨¡±£©

         ¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇ       £¨Ìî×Öĸ£©¡£

         ¢ÛÔÚÆðʼÌåϵÖмÓÈëN2µÄÎïÖʵÄÁ¿Îª          molʱ£¬·´Ó¦ºó°±µÄ°Ù·Öº¬Á¿×î´ó¡£ÈôÈÝÆ÷ÈÝ»ýΪH£¬n=3mol·´Ó¦´ïµ½Æ½ºâʱH2µÄת»¯ÂÊΪ60%£¬Ôò´ËÌõ¼þÏ£¨T2£©£¬·´Ó¦µÄƽºâ³£ÊýK=               ¡£

   £¨3£©N2O3ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢¡£

                                       ¢ÙÒ»¶¨Î¶ÈÏ£¬ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐN2O3¿É·¢ÉúÏÂÁз´Ó¦£º

                                       2N2O34NO2(g)+O2 ¡÷H>0ϱíΪ·´Ó¦ÔÚT1ζÈϵIJ¿·ÖʵÑéÊý¾Ý

Vs

0

500

1000

c(N2O3)/mol¡¤L£­1

5.00

3.52

2.48

Ôò500sÄÚNO2µÄƽ¾ùÉú³ÉËÙÂÊΪ                     .

         ¢ÚÏÖÒÔH2¡¢O2¡¢ÈÛÈÚÑÎNa2CO3×é³ÉµÄȼÁϵç³Ø£¬²ÉÓõç½â·¨ÖƱ¸N2O3£¬×°ÖÃÈçͼËùʾ£¬ÆäÖÐYΪCO2¡£

д³öʯīIµç¼«ÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½                             ¡£

ÔÚµç½â³ØÖÐÉú³ÉN2O3µÄµç¼«·´Ó¦Ê½Îª                                ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ìÕã½­Ê¡ÁùУ¸ßÈý2ÔÂÁª¿¼£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

¢ñ.¸Ç˹¶¨ÂÉÔÚÉú²úºÍ¿ÆѧÑо¿ÖÐÓкÜÖØÒªµÄÒâÒå¡£ÓÐЩ·´Ó¦µÄ·´Ó¦ÈÈËäÈ»ÎÞ·¨Ö±½Ó²âµÃ£¬µ«¿Éͨ¹ý¼ä½ÓµÄ·½·¨²â¶¨¡£ÏÖ¸ù¾ÝÏÂÁеÄ5¸ö·´Ó¦(ÓÉ°±Æø¡¢HClºÍË®ÖƱ¸NH4C1Ë®ÈÜÒº)¡£ÇëÅжϷ´Ó¦¢ÜµÄ·´Ó¦ÈÈ£º¦¤H£½                         ¡£
¢Ù NH3(g) + HCl(g) = NH4Cl(s) ¦¤H=£­176kJ¡¤mol¨C1
¢Ú NH3(g) + H2O(l) = NH3(aq) ¦¤H=£­35.1 kJ¡¤mol¨C1
¢Û HCl(g) + H2O(l) = HCl(aq) ¦¤H=£­72.3 kJ¡¤mol¨C1
¢Ü NH4C1(s) + H2O(1) = NH4C1(aq)
¢Ý NH3(aq) + HCl(aq) = NH4C1(aq) ¦¤H= £­52.3 kJ¡¤mol¨C1
¢ò.N2O5ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢¡£ÏÖÒÔH2¡¢O2¡¢ÈÛÈÚÑÎNa2CO3×é³ÉµÄȼÁϵç³Ø£¬²ÉÓõç½â·¨ÖƱ¸N2O5£¬×°ÖÃÈçͼËùʾ£¬ÆäÖÐYΪCO2¡£
д³öʯīIµç¼«ÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½                             ¡£
ÔÚµç½â³ØÖÐÉú³ÉN2O3µÄµç¼«·´Ó¦Ê½Îª                                ¡£
¢ó.ÑÌÆøµÄÍÑÁò£¨³ýSO2£©¼¼ÊõºÍÍÑÏõ£¨³ýNOx£©¼¼Êõ¶¼ÊÇ»·¾³¿ÆѧÑо¿µÄÈȵ㡣
¢ÅÑÌÆøÍÑÁò¡¢ÍÑÏõµÄ»·¾³ÒâÒåÊÇ         ¡£
(2)Ä¿Ç°£¬¿Æѧ¼ÒÕýÔÚÑо¿Ò»ÖÖÒÔÒÒÏ©×÷Ϊ»¹Ô­¼ÁµÄÍÑÏõ£¨NO£©Ô­Àí£¬ÆäÍÑÏõ»úÀíʾÒâͼÈçÏÂͼ1£¬ÍÑÏõÂÊÓëζȡ¢¸ºÔØÂÊ£¨·Ö×ÓɸÖд߻¯¼ÁµÄÖÊÁ¿·ÖÊý£©µÄ¹ØϵÈçͼ2Ëùʾ¡£

¢Ùд³ö¸ÃÍÑÏõÔ­Àí×Ü·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º     ¡£¢ÚΪ´ïµ½×î¼ÑÍÑÏõЧ¹û£¬Ó¦²ÉÈ¡µÄÌõ¼þÊÇ    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º±±¾©ÊÐʯ¾°É½Çø2010Äê¸ßÈýһģ£¨Àí¿Æ×ۺϣ©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

£¨16·Ö£©ÀûÓÃN2ºÍH2¿ÉÒÔʵÏÖNH3µÄ¹¤ÒµºÏ³É£¬¶ø°±ÓÖ¿ÉÒÔ½øÒ»²½ÖƱ¸ÏõËᣬÔÚ¹¤ÒµÉÏÒ»°ã¿É½øÐÐÁ¬ÐøÉú²ú¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

  £¨1£©ÒÑÖª£ºN2(g)+O2(g)=2NO(g)                  ¡÷H=+180.5kJ/mol

              N2(g)+3H2(g)  2NH3(g) ¡÷H=£­92.4kJ/mol

              2H2(g)+O2(g)=2H2O(g)             ¡÷H=£­483.6kJ/mo

ÈôÓÐ17g°±Æø¾­´ß»¯Ñõ»¯ÍêÈ«Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøËù·Å³öµÄÈÈÁ¿Îª                  ¡£

  £¨2£©Ä³¿ÆÑÐС×éÑо¿£ºÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¸Ä±äÆðʼÎïÇâÆøµÄÎïÖʵÄÁ¿¶ÔN2(g)+3H2(g)

2NH3(g)·´Ó¦µÄÓ°Ï졣ʵÑé½á¹ûÈçͼËùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©

         ¢ÙͼÏñÖÐT2ºÍT1µÄ¹ØϵÊÇ£ºT2        T2£¨Ìî¡°¸ßÓÚ¡±¡°µÍÓÚ¡±¡°µÈÓÚ¡±¡°ÎÞ·¨È·¶¨¡±£©

         ¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇ        £¨Ìî×Öĸ£©¡£

         ¢ÛÔÚÆðʼÌåϵÖмÓÈëN2µÄÎïÖʵÄÁ¿Îª          molʱ£¬·´Ó¦ºó°±µÄ°Ù·Öº¬Á¿×î´ó¡£ÈôÈÝÆ÷ÈÝ»ýΪH£¬n=3mol·´Ó¦´ïµ½Æ½ºâʱH2µÄת»¯ÂÊΪ60%£¬Ôò´ËÌõ¼þÏ£¨T2£©£¬·´Ó¦µÄƽºâ³£ÊýK=                ¡£

   £¨3£©N2O3ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢¡£

                                       ¢ÙÒ»¶¨Î¶ÈÏ£¬ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐN2O3¿É·¢ÉúÏÂÁз´Ó¦£º

                                       2N2O34NO2(g)+O2 ¡÷H>0ϱíΪ·´Ó¦ÔÚT1ζÈϵIJ¿·ÖʵÑéÊý¾Ý

Vs

0

500

1000

c(N2O3)/mol¡¤L£­1

5.00

3.52

2.48

Ôò500sÄÚNO2µÄƽ¾ùÉú³ÉËÙÂÊΪ                      .

         ¢ÚÏÖÒÔH2¡¢O2¡¢ÈÛÈÚÑÎNa2CO3×é³ÉµÄȼÁϵç³Ø£¬²ÉÓõç½â·¨ÖƱ¸N2O3£¬×°ÖÃÈçͼËùʾ£¬ÆäÖÐYΪCO2¡£

д³öʯīIµç¼«ÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½                              ¡£

ÔÚµç½â³ØÖÐÉú³ÉN2O3µÄµç¼«·´Ó¦Ê½Îª                                 ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸