ÌâÄ¿ÄÚÈÝ

ÇâÆø(H2)¡¢Ò»Ñõ»¯Ì¼(CO)¡¢ÐÁÍé(C8H18)¡¢¼×Íé(CH4)µÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º

H2(g)£«1/2O2(g)£½H2O(l)£»¦¤H£½£­285.8 kJ/mol£»CO(g)£«1/2O2(g) £½CO2(g)£»¦¤H£½£­283.0 kJ/mol¡¡C8H18(l)£«25/2O2(g)£½8CO2(g)£«9H2O(l)£»¦¤H£½£­5518 kJ/mol£»CH4(g)£«2O2(g)£½CO2(g)£«2H2O(l)£»¦¤H£½£­890.3 kJ/mol

ÏàͬÖÊÁ¿µÄH2¡¢CO¡¢C8H18¡¢CH4ÍêȫȼÉÕʱ£¬·Å³öÈÈÁ¿×îÉÙµÄÊÇ

[¡¡¡¡]

A£®

H2(g)

B£®

C8H18(l)

C£®

CO(g)

D£®

CH4(g)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
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
ÈôÓÐ17g °±Æø¾­´ß»¯Ñõ»¯ÍêÈ«Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøËù·Å³öµÄÈÈÁ¿Îª
226.3kJ
226.3kJ
£®
£¨2£©Ä³¿ÆÑÐС×éÑо¿£ºÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¸Ä±äÆðʼÎïÇâÆøµÄÎïÖʵÄÁ¿¶ÔN2£¨g£©+3H2£¨g£©?2NH3£¨g£©·´Ó¦µÄÓ°Ï죮
ʵÑé½á¹ûÈçͼËùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©
¢ÙͼÏóÖÐT2ºÍT1µÄ¹ØÏµÊÇ£ºT2
µÍÓÚ
µÍÓÚ
T1£¨Ìî¡°¸ßÓÚ¡±¡°µÍÓÚ¡±¡°µÈÓÚ¡±¡°ÎÞ·¨È·¶¨¡±£©
¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇ
c
c
£¨Ìî×Öĸ£©£®
£¨3£©N2O5ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖÆ±¸Êܵ½ÈËÃǵĹØ×¢£®Ò»¶¨Î¶ÈÏ£¬ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐN2O5¿É·¢ÉúÏÂÁз´Ó¦£º2N2O5£¨g£©?4NO2£¨g£©+O2£¨g£©¡÷H£¾0£®Ï±íΪ·´Ó¦ÔÚT1ζÈϵIJ¿·ÖʵÑéÊý¾Ý£º
t/s 0 500 1000
c£¨N2O5£©/mol?L-1 5.00 3.52 2.48
Ôò500sÄÚNO2µÄƽ¾ùÉú³ÉËÙÂÊΪ
0.00592mol?L-1?s-1
0.00592mol?L-1?s-1
£®
ÄÜÅжÏÉÏÊö·´Ó¦´ïµ½Æ½ºâµÄÒÀ¾ÝÊÇ
abd
abd
£¨ÌîдÐòºÅ£©
a£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä                  b£®µ¥Î»Ê±¼äÄÚÿÉú³É2mol NO2£¬Í¬Ê±Éú³É1mol N2O5
c£®ÈÝÆ÷ÄÚÆøÌåÃܶȱ£³Ö²»±ä              d£®NO2 ÓëN2O5ÎïÖʵÄÁ¿±ÈÀý±£³Ö²»±ä£®
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
N2£¨g£©+3H2£¨g£©¨T2NH3£¨g£©¡÷H=-92.4kJ/mol
2H2£¨g£©+O2£¨g£©¨T2H2O£¨g£©¡÷H=-483.6kJ/mol
ÈôÓÐ17g °±Æø¾­´ß»¯Ñõ»¯ÍêÈ«Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøËù·Å³öµÄÈÈÁ¿Îª
226.3kJ
226.3kJ

£¨2£©Ä³¿ÆÑÐС×éÑо¿£ºÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¸Ä±äÆðʼÎïÇâÆøµÄÎïÖʵÄÁ¿¶ÔN2£¨g£©+3H2£¨g£©¨T2NH3£¨g£©·´Ó¦µÄÓ°Ï죮
ʵÑé½á¹ûÈçͼËùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©
¢ÙͼÏóÖÐT2ºÍT1µÄ¹ØÏµÊÇ£ºT2
µÍÓÚ
µÍÓÚ
T1£¨Ìî¡°¸ßÓÚ¡±¡¢¡°µÍÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°ÎÞ·¨È·¶¨¡±£©£®
¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇ
c
c
£¨Ìî×Öĸ£©£®
¢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
2.08
£®
¢ÜÈô¸Ã·´Ó¦ÔÚ298K¡¢398KʱµÄƽºâ³£Êý·Ö±ðΪK1£¬K2£¬ÔòK1
£¾
£¾
K2£¨Ìî¡°£¾¡±¡°=¡±¡°£¼¡±£©
¢Ý¶ÔÓڸ÷´Ó¦ÓйرíÊöÕýÈ·µÄÊÇ
bd
bd
£®
a£®ÆäËûÌõ¼þ²»±ä£¬ËõСÈÝÆ÷Ìå»ýʱÕý·´Ó¦ËÙÂÊÔö´ó£¬Äæ·´Ó¦ËÙÂʼõС£¬¹Êƽºâ½«ÄæÏòÒÆ¶¯
b£®¾øÈÈÌõ¼þÏ£¬Èô²âµÃ¸ÃÌåϵζȲ»Ôٸı䣬Ôò·´Ó¦´¦ÓÚÆ½ºâ״̬
c£®ºãÈÝÌõ¼þÏ£¬ÈôÈÝÆ÷ÄÚÆøÌåµÄÃܶȱ£³Ö²»±ä£¬Ôò·´Ó¦´¦ÓÚÆ½ºâ״̬
d£®ÆäËûÌõ¼þ²»±ä£¬½«ÈÝÆ÷Ìå»ýÀ©´óΪԭÀ´µÄ2±¶£¬ÔòÖØÐÂÆ½ºâʱ£¬NH3µÄƽºâŨ¶È½«±ÈԭƽºâŨ¶ÈµÄÒ»°ë»¹ÒªÐ¡£®
¾«Ó¢¼Ò½ÌÍøÀûÓÃN2ºÍH2¿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
д³ö°±Æø¾­´ß»¯Ñõ»¯Éú³ÉÒ»Ñõ»¯µªÆøÌåºÍË®ÕôÆøµÄÈÈ»¯Ñ§·½³Ìʽ£º
 
£®
£¨2£©Ä³¿ÆÑÐС×éÑо¿£ºÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¸Ä±äÆðʼÎïÇâÆøµÄÎïÖʵÄÁ¿¶Ô·´Ó¦N2£¨g£©+3H2£¨g£©?2NH3£¨g£©µÄÓ°Ï죮ʵÑé½á¹ûÈçͼËùʾ£º£¨Í¼ÖÐT±íʾζȣ¬n±íʾÎïÖʵÄÁ¿£©
¢ÙͼÏóÖÐT2ºÍT1µÄ¹ØÏµÊÇ£ºT2
 
T2£¨Ìî¡°¸ßÓÚ¡±¡°µÍÓÚ¡±¡°µÈÓÚ¡±¡°ÎÞ·¨È·¶¨¡±£©
¢Ú±È½ÏÔÚa¡¢b¡¢cÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇ
 
£¨Ìî×Öĸ£©£®
£¨3£©ÔÚÒ»¶¨Î¶Ⱥʹ߻¯¼ÁÏ£¬½«3.2mol H2ºÍ1.2molN2»ìºÏÓÚÒ»¸öÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£¬ÔÚ2minĩʱ·´Ó¦Ç¡ºÃ´ïƽºâ£¬´ËʱÉú³ÉÁË0.8mol NH3£®¼ÆË㣺£¨Ð´³ö¼ÆËã¹ý³Ì£¬½á¹û±£ÁôСÊýµãºóһ룩
¢Ù2minÄÚÒÔH2±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊ£»
 

¢Ú¸ÃÌõ¼þÏÂµÄÆ½ºâ³£Êý£¿
 
£®
£¨1£©ÏÂÁÐËÄÖÖʵÑé×°Ö㨸ù¾ÝÐèÒª¿ÉÔÚÆäÖмÓÈëÒºÌå»ò¹ÌÌ壩
¾«Ó¢¼Ò½ÌÍø
¢ÙÓÃÓÚ¸ÉÔï¶þÑõ»¯ÁòÆøÌåµÄ×°ÖÃÓÐ
 
£¨Ìî×Öĸ£©£®
¢Ú×°ÖÃB³ý¿ÉÓÃÓÚÖÆÈ¡CO2¡¢H2£¬»¹¿ÉÖÆÈ¡
 
¡¢
 
£®£¨Ð´³öÁ½ÖÖÎïÖʵĻ¯Ñ§Ê½£©£®
¢Û¼ÈÄÜÓÃÓÚÊÕ¼¯ÂÈÆøÓÖÄÜÓÃÓÚÊÕ¼¯Ò»Ñõ»¯µªÆøÌåµÄ×°ÖÃÓÐ
 
£®£¨Ìî×Öĸ£©
¢Üµ±ÓÃA×°ÖòâÁ¿CO2ÆøÌåÌå»ýʱ£¬´Ó
 
¶Ë½øÆø£¬Æ¿ÖÐËùÊ¢µÄÒºÌåΪ
 
£¨ÌîÃû³Æ£©µ±ÓÃA×°ÖÃÊÕ¼¯ÇâÆøÊ±´Ó
 
¶Ë½øÆø£®
£¨2£©ÏÂÁÐʵÑé²Ù×÷»òʵÑéÊÂʵµÄÐðÊö²»ÕýÈ·µÄÊÇ
 
£¨ÌîÐòºÅ£©£®
A£®¹ýÂË¡¢Õô·¢²Ù×÷¹ý³ÌÖж¼ÐèÒªÓò£Á§°ô²»¶Ï½Á°èÒºÌ壮
B£®ÔÚÊÔ¹ÜÖзÅÈ뼸ƬËéÂËֽƬ£¬¼ÓÈ뼸µÎ90%µÄŨÁòËᣬµ·³Éºý×´£¬Î¢ÈȲ¢ÀäÈ´ºó£¬ÏȵÎÈ뼸µÎCuSO4ÈÜÒº£¬ÔÙ¼ÓÈë¹ýÁ¿µÄNaOHÈÜÒºÖкÍÖÁ³öÏÖCu£¨OH£©2³Áµí£®¼ÓÈÈÖÁ·ÐÌÚ£¬¿É¹Û²ìµ½ºìÉ«³ÁµíÉú³É£®
C£®ÖƱ¸ÁòËáÑÇÌú¾§Ìåʱ£¬Ïò·ÏÌúмÖмÓÈë¹ýÁ¿Ï¡ÁòËᣬ³ä·Ö·´Ó¦ºó£¬¹ýÂË£¬È»ºó¼ÓÈÈÕô¸ÉÂËÒº¼´¿ÉµÃÁòËáÑÇÌú¾§Ìå
D£®²â¶¨ÁòËáÍ­¾§Ìå½á¾§Ë®º¬Á¿µÄʵÑéÖУ¬³ÆÁ¿²Ù×÷ÖÁÉÙÒª½øÐÐ4´Î£®
E£®ÏòµÎÓзÓ̪µÄË®ÖмÓÈë×ãÁ¿µÄ¹ýÑõ»¯Äƿɹ۲쵽ÈÜÒº³ÊºìÉ«£®

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

¾«Ó¢¼Ò½ÌÍø