£¨1£©ÒÑÖª£ºN2(g)+O2(g)=2NO(g)£»¡÷H= +180.5kJ/mol
4NH3(g)+5O2(g)=4NO(g)+6H2O(g)£»¡÷H=£­905kJ/mol
2H2(g)+O2(g)=2H2O(g)£»¡÷H=£­483.6kJ/mol
ÔòN2(g)+3H2(g)=2NH3(g)µÄ¡÷H=             ¡£
£¨2£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g)  2NH3(g)¡£ÔÚÒ»¶¨Î¶ÈÏ£¬½«Ò»¶¨Á¿µÄN2ºÍH2ͨÈë¹Ì¶¨Ìå»ýΪ1LµÄÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬¸Ä±äÏÂÁÐÌõ¼þ£¬ÄÜʹƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÊÇ           ¡£
¢ÙÔö´óѹǿ                     ¢ÚͨÈëHe
¢ÛʹÓô߻¯¼Á         ¢Ü½µµÍζÈ
£¨3£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g)  2NH3(g)¡£ÉèÔÚÈÝ»ýΪ2.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.60molN2(g)ºÍ1.60molH2(g)£¬·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâʱ£¬NH3µÄÎïÖʵÄÁ¿·ÖÊý£¨NH3µÄÎïÖʵÄÁ¿Óë·´Ó¦ÌåϵÖÐ×ܵÄÎïÖʵÄÁ¿Ö®±È£©Îª¡£¼ÆËã¸ÃÌõ¼þÏ´ﵽƽºâʱN2ת»¯ÂÊΪ      £»
£¨4£©ºÏ³É°±µÄÔ­ÁÏÇâÆøÊÇÒ»ÖÖÐÂÐ͵ÄÂÌÉ«ÄÜÔ´£¬¾ßÓйãÀ«µÄ·¢Õ¹Ç°¾°¡£ÏÖÓÃÇâÑõȼÁϵç³Ø½øÐÐÓÒͼËùʾʵÑ飺£¨ÆäÖÐc¡¢d¾ùΪ̼°ô£¬NaClÈÜÒºµÄÌå»ýΪ500ml£©

¢Ùb¼«Îª      ¼«£¬µç¼«·´Ó¦Ê½               £»
c¼«Îª      ¼«£¬µç¼«·´Ó¦Ê½                      
¢ÚÓÒͼװÖÃÖУ¬µ±b¼«ÉÏÏûºÄµÄO2ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ280mlʱ£¬ÔòÒÒ³ØÈÜÒºµÄPHΪ        £¨¼ÙÉ跴ӦǰºóÈÜÒºÌå»ý²»±ä£¬ÇÒNaClÈÜÒº×ãÁ¿£©

£¨1£©¡÷H=£­92.4kJ/mol   £¨2·Ö£©
£¨2£©¢Ù ¢Ü   £¨2·Ö£©
£¨3£©66.67% »ò2/3 £¨2·Ö£©
£¨4£©¢ÙÕý¼«£¬O2+4H++4e-=2H2O£»Ñô¼«£¬2Cl--2e-=Cl2¡ü£¨Ã¿¿Õ2·Ö£¬¹²8·Ö£©
¢Ú 13£¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÔÚÆû³µÉÏ°²×°ÈýЧ´ß»¯×ª»¯Æ÷£¬¿ÉʹÆû³µÎ²ÆøÖеÄÖ÷ÒªÎÛȾÎCO¡¢NOx¡¢Ì¼Ç⻯ºÏÎ½øÐÐÏ໥·´Ó¦£¬Éú³ÉÎÞ¶¾ÎïÖÊ£¬¼õÉÙÆû³µÎ²ÆøÎÛȾ¡£

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

2C(s)+ O2(g)=2CO(g) ¡÷H=-221.0 kJ/mol

C(s)+ O2(g)=CO2(g) ¡÷H=-393.5kJ/mol

    βÆøת»¯µÄ·´Ó¦Ö®Ò»£º2NO(g)+2CO(g)=N2(g)+2CO2(g) ¡÷H£½            ¡£

£¨2£©Ä³Ñо¿ÐÔѧϰС×éÔÚ¼¼ÊõÈËÔ±µÄÖ¸µ¼Ï£¬ÔÚijζÈʱ£¬°´ÏÂÁÐÁ÷³Ì̽¾¿Ä³ÖÖ´ß»¯¼Á×÷ÓÃϵķ´Ó¦ËÙÂÊ£¬ÓÃÆøÌå´«¸ÐÆ÷²âµÃ²»Í¬Ê±¼äµÄNOºÍCOŨ¶ÈÈç±í£º

Çë»Ø´ðÏÂÁÐÎÊÌâ(¾ù²»¿¼ÂÇζȱ仯¶Ô´ß»¯¼Á´ß»¯Ð§ÂʵÄÓ°Ïì)£º

¢ÙÇ°2sÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv (N2) = ___________________¡£

¢ÚÔÚ¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýK =           ¡££¨Ö»Ð´³ö¼ÆËã½á¹û£©

¢Û¶ÔÓڸÿÉÄæ·´Ó¦£¬Í¨¹ý×ۺϷÖÎöÒÔÉÏÐÅÏ¢£¬ÖÁÉÙ¿ÉÒÔ˵Ã÷      £¨Ìî×Öĸ£©¡£

A£®¸Ã·´Ó¦µÄ·´Ó¦Îï»ìºÏºóºÜ²»Îȶ¨

B£®¸Ã·´Ó¦Ò»µ©·¢Éú½«Ôں̵ܶÄʱ¼äÄÚÍê³É

C£®¸Ã·´Ó¦Ìåϵ´ïµ½Æ½ºâʱÖÁÉÙÓÐÒ»ÖÖ·´Ó¦ÎïµÄ°Ù·Öº¬Á¿½ÏС

D£®¸Ã·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏÂÄÜ×Ô·¢½øÐÐ

E£®¸Ã·´Ó¦Ê¹Óô߻¯¼ÁÒâÒå²»´ó

£¨3£©CO·ÖÎöÒÇÒÔȼÁϵç³ØΪ¹¤×÷Ô­Àí£¬Æä×°ÖÃÈçͼËùʾ£¬¸Ãµç³ØÖеç½âÖÊΪÑõ»¯îÆ£­Ñõ»¯ÄÆ£¬ÆäÖÐO2-¿ÉÒÔÔÚ¹ÌÌå½éÖÊNASICONÖÐ×ÔÓÉÒƶ¯¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ      ¡£

A£®¸º¼«µÄµç¼«·´Ó¦Ê½Îª£ºCO+O2¡ª¨D2e-£½CO2

B£®¹¤×÷ʱµç¼«b×÷Õý¼«£¬O2-Óɵ缫aÁ÷Ïòµç¼«b

C£®¹¤×÷ʱµç×ÓÓɵ缫aͨ¹ý´«¸ÐÆ÷Á÷Ïòµç¼«b

D£®´«¸ÐÆ÷ÖÐͨ¹ýµÄµçÁ÷Ô½´ó£¬Î²ÆøÖÐCOµÄº¬Á¿Ô½¸ß

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ºÏ³É°±¹¤ÒµµÄ²¿·Ö¹¤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

ÇëÄã»Ø´ðÏÂÁÐÎÊÌ⣺

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

          4NH3(g)+5O2(g)=4NO(g)+6H2O(g) £»¡÷H=¨C905kJ¡¤mol-1

          2H2(g)+O2(g)=2H2O(g)£»¡÷H=¨C483.6kJ¡¤mol-1

ÔòN2£¨g£©+3H2£¨g£©2NH3£¨g£©µÄ¡÷H=_________________¡£

£¨2£©¸Ä±äÏÂÁÐÌõ¼þ£¬ÄÜʹƽºâÏòÕý·´Ó¦·½Ïò½øÐÐÇÒƽºâ³£Êý²»±äµÄÊÇ_________¡£

A.Ôö´óѹǿ   B.½µµÍζȠ  C.ʹÓô߻¯¼Á   D. Ôö´ó·´Ó¦ÎïµÄŨ¶È 

£¨3£©ÔÚÒ»¶¨Ìõ¼þÏ£¬½«2molN2Óë5molH2»ìºÏÓÚÒ»¸ö10LµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦Çé¿öÈçÓÒͼËùʾ£º                                                                      

¢ÙÇó5minÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv(NH3)=_______________

¢Ú´ïµ½Æ½ºâʱNH3µÄÌå»ý·ÖÊýΪ___________%

(4)½üÄêÀ´¿Æѧ¼ÒÌá³öÁ˵ç½âºÏ³É°±µÄ·½·¨£º²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É£¨ÄÜ´«µÝH+£©Îª½éÖÊ£¬ÓÃÎü¸½ÔÚËüÄÚÍâ±íÃæÉϵĽðÊôîٶྦྷ±¡Ä¤×öµç¼«£¬ÊµÏÖÁ˸ßת»¯Âʵĵç½â·¨ºÏ³É°±£¨×°ÖÃÈçͼ£©

Çë»Ø´ð£ºîٵ缫AÊǵç½â³ØµÄ_______¼«£¨Ìî¡°Ñô¡±»ò¡°Òõ¡±£©£¬¸Ã¼«Éϵĵ缫·´Ó¦Ê½ÊÇ_________________________

 

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010½ì½­ËÕÆô¶«ÖÐѧ¸ßÈý¿¼Ç°¸¨µ¼ÑµÁ·»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

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/mol
ÈôÓÐ17 g°±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°Ïì¡£
ʵÑé½á¹ûÈçͼËùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©

¢ÙͼÏñÖÐT2ºÍT1µÄ¹ØϵÊÇ£ºT2 ______T1£¨Ìî¡°¸ßÓÚ¡±¡¢¡°µÍÓÚ¡±¡¢
¡°µÈÓÚ¡±»ò¡°ÎÞ·¨È·¶¨¡±£©¡£
¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2 µÄת»¯ÂÊ×î¡¡¡¡¸ßµÄÊÇ______£¨Ìî×Öĸ£©¡£
¢ÛÔÚÆðʼÌåϵÖмÓÈëN2µÄÎïÖʵÄÁ¿Îª________molʱ£¬·´Ó¦ºó°±µÄ°Ù·Öº¬Á¿×î´ó£»ÈôÈÝÆ÷ÈÝ»ýΪ1L£¬n=3mol·´Ó¦´ïµ½Æ½ºâʱH2µÄת»¯ÂÊΪ60%£¬Ôò´ËÌõ¼þÏ£¨T2£©£¬·´Ó¦µÄƽºâ³£ÊýK=________________________¡£
£¨3£©N2O5ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢¡£
¢ÙÒ»¶¨Î¶ÈÏ£¬ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐN2O5¿É·¢ÉúÏÂÁз´Ó¦£º
2N2O5(g)4NO2(g)£«O2(g) ¦¤H£¾0ϱíΪ·´Ó¦ÔÚT1ζÈϵIJ¿·ÖʵÑéÊý¾Ý

Ôò500sÄÚNO2µÄƽ¾ùÉú³ÉËÙÂÊΪ                                    ¡£
¢ÚÏÖÒÔH2¡¢O2¡¢ÈÛÈÚÑÎW#W$W%.K**S*&5^UNa2CO3×é³ÉµÄȼÁϵç³Ø£¬²ÉÓõç½â·¨ÖƱ¸N2O5£¬
×°ÖÃÈçͼËùʾ£¬ÆäÖÐYΪCO2¡£

д³öʯīIµç¼«ÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½                              ¡£
ÔÚµç½â³ØÖÐÉú³ÉN2O5µÄµç¼«·´Ó¦Ê½Îª                                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÕã½­Ê¡¸ßÈýµÚһѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨11·Ö£©ÔÚÆû³µÉÏ°²×°ÈýЧ´ß»¯×ª»¯Æ÷£¬¿ÉʹÆû³µÎ²ÆøÖеÄÖ÷ÒªÎÛȾÎCO¡¢NOx¡¢Ì¼Ç⻯ºÏÎ½øÐÐÏ໥·´Ó¦£¬Éú³ÉÎÞ¶¾ÎïÖÊ£¬¼õÉÙÆû³µÎ²ÆøÎÛȾ¡£

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

2C(s)+ O2(g)=2CO(g)    ¡÷H=£­221.0 kJ¡¤mol£­1

C(s)+ O2(g)=CO2(g)     ¡÷H=£­393.5 kJ¡¤mol£­1

¢ÙβÆøת»¯µÄ·´Ó¦Ö®Ò»£º2NO(g)+2CO(g)=N2(g)+2CO2(g)  ¡÷H£½             ¡£

¢ÚÒÑÖª£ºN2¡¢O2·Ö×ÓÖл¯Ñ§¼üµÄ¼üÄÜ·Ö±ðÊÇ946 kJ¡¤mol¡ª1¡¢497 kJ¡¤mol¡ª1£¬ÔòNO·Ö×ÓÖл¯Ñ§¼üµÄ¼üÄÜΪ   ¡¡ ¡¡¡¡¡¡¡¡    kJ¡¤mol¡ª1¡£

£¨2£©Ä³Ñо¿ÐÔѧϰС×éÔÚ¼¼ÊõÈËÔ±µÄÖ¸µ¼Ï£¬ÔÚijζÈʱ£¬°´ÏÂÁÐÁ÷³Ì̽¾¿Ä³ÖÖ´ß»¯¼Á×÷ÓÃϵķ´Ó¦ËÙÂÊ£¬ÓÃÆøÌå´«¸ÐÆ÷²âµÃ²»Í¬Ê±¼äµÄNOºÍCOŨ¶ÈÈç±í£º

Çë»Ø´ðÏÂÁÐÎÊÌâ(¾ù²»¿¼ÂÇζȱ仯¶Ô´ß»¯¼Á´ß»¯Ð§ÂʵÄÓ°Ïì)£º

¢ÙÇ°3sÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv (N2) = _________________________¡£

¢ÚÔÚ¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýK =       ¡¡¡¡¡¡     ¡££¨Ö»Ð´³ö¼ÆËã½á¹û£©

¢Û¸Ã¿ÉÄæ·´Ó¦¡÷S   0£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±£©£¬ÔÚ_______________£¨Ìî¡°¸ßΡ±¡¢¡°µÍΡ±»ò¡°ÈκÎζȡ±£©ÏÂÄÜ×Ô·¢½øÐС£

£¨3£©CO·ÖÎöÒÇÒÔȼÁϵç³ØΪ¹¤×÷Ô­Àí£¬Æä×°ÖÃÈçͼËùʾ£¬¸Ãµç³ØÖеç½âÖÊΪÑõ»¯îÆ£­Ñõ»¯ÄÆ£¬ÆäÖÐO2-¿ÉÒÔÔÚ¹ÌÌå½éÖÊNASICONÖÐ×ÔÓÉÒƶ¯¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ     

A£®¸º¼«µÄµç¼«·´Ó¦Ê½Îª£º

CO+O2¡ª¨D2e-£½CO2

B£®¹¤×÷ʱµç¼«b×÷Õý¼«£¬O2-Óɵ缫aÁ÷Ïòµç¼«b

C£®¹¤×÷ʱµç×ÓÓɵ缫aͨ¹ý´«¸ÐÆ÷Á÷Ïòµç¼«b

D£®´«¸ÐÆ÷ÖÐͨ¹ýµÄµçÁ÷Ô½´ó£¬Î²ÆøÖÐCOµÄº¬Á¿Ô½¸ß

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸