£¨14·Ö£© ¼×Íé×÷ΪһÖÖÐÂÄÜÔ´ÔÚ»¯Ñ§ÁìÓòÓ¦Óù㷺£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺?
£¨1£©ÒÔÌìÈ»ÆøΪԭÁÏÖÆH2ÊǺϳɰ±µÄÒ»ÌõÖØÒªµÄ·Ïß¡£¼×ÍéµÄ²¿·ÖÑõ»¯¿ÉµÃµ½ºÏ³É°±µÄÔ­ÁÏÆøH2£¬Æ䷴ӦʽÈçÏ£º
¢ÙCH£¨g£©+1/2O£¨g£©£½CO£¨g£©+2H£¨g£© H1£½£­35.6kJ¡¤mol
ÊÔÅжϳ£ÎÂÏÂ,ÉÏÊö·´Ó¦ÄÜ·ñ×Ô·¢½øÐÐ:          (ÌÄÜ¡±»ò¡±·ñ¡±)¡£ÓÐÑо¿ÈÏΪ¼×Í鲿·ÖÑõ»¯µÄ»úÀíΪ£º
¢ÚCH£¨g£©+2O£¨g£©£½CO2£¨g£©+2H2O£¨g£© H2£½£­890.3kJ¡¤mol
¢ÛCH£¨g£©+CO£¨g£©£½2CO£¨g£©+2H£¨g£© H3£½247.3kJ¡¤mol
Çë½áºÏÒÔÉÏÌõ¼þд³öCH4ºÍH2O£¨g£©Éú³ÉCOºÍH2µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
                                                                          ¡£
¢ÆºãÎÂÏ£¬ÏòÒ»¸ö2LµÄÃܱÕÈÝÆ÷ÖгäÈë1 molN2ºÍ2.6 molH2£¬·´Ó¦¹ý³ÌÖжÔNH3µÄŨ¶È½øÐмì²â£¬µÃµ½µÄÊý¾ÝÈçϱíËùʾ£º
ʵÑéÊý¾Ý
ʱ¼ä/min
5
10
15
20
25
30
c(NH3)/( mol ¡¤/L-1)
0.08
0.14
0.18
0.20
0.20
0.20
´ËÌõ¼þÏÂ,¸Ã·´Ó¦´ïµ½»¯Ñ§Æ½ºâʱ,µªÆøµÄŨ¶ÈΪ              ¡£
£¨3£©ÈçÏÂͼËùʾ£¬×°ÖâñΪ¼×ÍéȼÁϵç³Ø£¨µç½âÖÊÈÜҺΪKOHÈÜÒº£©£¬Í¨¹ý×°ÖâòʵÏÖÌú°ôÉ϶ÆÍ­¡£?

¢Ùb´¦µç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½ÊÇ                            ¡£
¢Úµç¶Æ½áÊøºó£¬×°ÖâñÖÐÈÜÒºµÄpH          £¨Ìîд¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±£¬ÏÂͬ£©£¬×°ÖâòÖÐCuµÄÎïÖʵÄÁ¿Å¨¶È                  ¡£
¢ÛÈôÍêÈ«·´Ó¦ºó£¬×°ÖâòÖÐÒõ¼«ÖÊÁ¿Ôö¼Ó12.8g£¬Ôò×°ÖâñÖÐÀíÂÛÉÏÏûºÄ¼×Íé        L £¨±ê×¼×´¿öÏ£©¡£
£¨14·Ö£©£¨1£©ÄÜ£¨1·Ö£©
CH£¨g£©+H2O£¨g£©£½CO£¨g£©+3H2£¨g£©H£½250.3kJ¡¤mol£¨3·Ö£©ÆäÓàÿ¿Õ2·Ö
£¨2£©0.4 mol ¡¤L-1
£¨3£©¢Ù   O2£«2H2O£«4e£­=4OH-   ¢Ú±äС   ²»±ä    ¢Û  1.12
£¨1£©¸Ã·´Ó¦ÎªÒ»¸öìØÔöµÄ·ÅÈÈ·´Ó¦£¬¿ÉÒÔ×Ô·¢½øÐУ»Í¨¹ý¸Ç˹¶¨ÂÉ£¬Ê×ÏÈд³öÄ¿±ê·´Ó¦CH4(g)+H2O(g)=CO(g)+3H2(g)£¬¸Ã·´Ó¦µÈÓÚ(¢Ù¡Á4-¢Ú-¢Û)£¬¹Ê¦¤H=((-35.6)¡Á4-(-890.3)-247.3)=250.3KJ/mol£»
£¨2£©
 
N2+3H22NH3
ʼ̬mol/L
0.5
1.3
0
·´Ó¦mol/L
0.1
0.3
0.20
ÖÕ̬mol/L
0.4
1.0
0.20
 
£¨3£©Ìú°ô¶ÆÍ­£¬¿ÉÒÔÈ·¶¨CuΪÑô¼«£¬FeΪÒõ¼«£¬Õû¸öµç¶Æ¹ý³ÌÖеç½âÖÊŨ¶ÈûÓб仯£»
×ó±ßΪȼÁϵç³Ø£¬aΪ¸º¼«£¬¼×Íéʧµç×Ó£¬bΪÕý¼«£¬·¢Éú»¹Ô­·´Ó¦O2£«2H2O£«4e£­=4OH-£¬Õû¸ö¹ý³ÌÖÐKOHµÄÁ¿²»±ä£¬µ«ÓÐË®Éú³É£¬Å¨¶È¼õС£¬pH±äС¡£¸ù¾ÝµÃʧµç×ÓÊغã¿ÉÒÔÈ·¶¨£¬ÏûºÄ¼×ÍéΪ1.12L
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©£¨1£©»¯Ñ§·´Ó¦¿ÉÊÓΪ¾É¼ü¶ÏÁѺÍмüÐγɵĹý³Ì¡£»¯Ñ§¼üµÄ¼üÄÜÊÇÐγɣ¨»ò²ð¿ª£©1 mol»¯Ñ§¼üʱÊÍ·Å£¨»òÎüÊÕ£©µÄÄÜÁ¿¡£ÒÑÖª£ºN¡ÔN¼üµÄ¼üÄÜÊÇ948.9kJ¡¤mol£­1£¬H£­H¼üµÄ¼üÄÜÊÇ436.0 kJ¡¤mol£­1£»ÓÉN2ºÍH2ºÏ³É1molNH3ʱ¿É·Å³ö46.2kJµÄÈÈÁ¿¡£N£­H¼üµÄ¼üÄÜÊÇ      ¡¡       ¡£
£¨2£©¸Ç˹¶¨ÂÉÔÚÉú²úºÍ¿ÆѧÑо¿ÖÐÓкÜÖØÒªµÄÒâÒå¡£ÓÐЩ·´Ó¦µÄ·´Ó¦ÈÈËäÈ»ÎÞ·¨Ö±½Ó²âµÃ£¬µ«¿Éͨ¹ý¼ä½ÓµÄ·½·¨²â¶¨¡£ÏÖ¸ù¾ÝÏÂÁÐ3¸öÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
Fe2O3(s)+3CO(g)=2Fe(s)+3CO2(g)        ¡÷H£½£­24.8 kJ¡¤mol£­1
3Fe2O3(s)+ CO(g)==2Fe3O4(s)+ CO2(g)    ¡÷H£½£­47.2 kJ¡¤mol£­1
Fe3O4(s)+CO(g)==3FeO(s)+CO2(g)       ¡÷H£½£«640.5 kJ¡¤mol£­1
д³öCOÆøÌ廹ԭFeO¹ÌÌåµÃµ½Fe ¹ÌÌåºÍCO2ÆøÌåµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
_________________                                                           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨6·Ö£©×ÔÈ»½çÀﵪµÄ¹Ì¶¨Í¾¾¶Ö®Ò»ÊÇÔÚÉÁµçµÄ×÷ÓÃÏ£¬N2ÓëO2·´Ó¦Éú³ÉNO¡£
£¨1£©ÔÚ²»Í¬Î¶ÈÏ£¬·´Ó¦N2(g)£«O2(g)2NO(g) DH£½a kJ¡¤mol£­1µÄƽºâ³£ÊýKÈçÏÂ±í£º
ζÈ/¡æ
1538
1760
2404
ƽºâ³£ÊýK
0.86¡Á10£­4
2.6¡Á10£­4
64¡Á10£­4
      ¸Ã·´Ó¦µÄ¡÷H   0£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©¡£
(2)2404¡æʱ£¬ÔÚÈÝ»ýΪ1.0LµÄÃܱÕÈÝÆ÷ÖÐͨÈë2.6mol N2ºÍ2.6mol O2£¬¼ÆËã·´Ó¦£º
N2(g)£«O2(g)2NO(g)´ïµ½Æ½ºâʱN2µÄŨ¶ÈΪ        ¡££¨´ËζÈϲ»¿¼ÂÇO2ÓëNOµÄ·´Ó¦¡£¼ÆËã½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£©
(3)¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦ÓÚÑо¿³£Î³£Ñ¹Ï¡°È˹¤¹Ìµª¡±µÄз½·¨¡£ÔøÓÐʵÑ鱨µÀ£ºÔÚ³£Î¡¢³£Ñ¹¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á(²ôÓÐÉÙÁ¿Fe2O3µÄTiO2)±íÃæÓëË®·¢Éú·´Ó¦£¬Éú³ÉµÄÖ÷Òª²úÎïΪNH3¡£ÏàÓ¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
2N2(g)£«6H2O(1)£½4NH3 (g)£«3O2(g)   DH£½£«1530kJ¡¤mol£­1
Ôò°±´ß»¯Ñõ»¯·´Ó¦4NH3(g)£«5O2(g)£½4NO(g)£«6H2O(1)µÄ·´Ó¦ÈÈDH£½        ¡£ £¨Óú¬aµÄ´úÊýʽ±íʾ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖª£ºCH4(g)+2O2(g)¡úCO2(g)+2H2O(g) ¦¤H£½£­Q1 kJ/mol£»
2H2(g)+O2(g) == 2H2O(g)        ¦¤H£½£­Q2 kJ/mol£»
H2O(g) == H2O(l)               ¦¤H£½£­Q3 kJ/mol  
 ³£ÎÂÏÂÈ¡Ìå»ý±ÈΪ4£º1µÄ¼×ÍéºÍH2µÄ»ìºÏÆøÌå112L£¨±ê×¼×´¿öÏ£©£¬¾­ÍêȫȼÉÕºó»Ö¸´µ½³£Î£¬Ôò·Å³öµÄÈÈÁ¿Îª£¨¡¡¡¡£©
A£®4Q1+0.5Q2B£®4Q1+Q2+10Q3C£®4Q1+2Q2D£®4Q1+0.5Q2+9Q3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

£®ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
(1)CH3COOH(l)+2O2(g) ="==" 2CO2(g)+2H2O(l)    ¡÷H1= £­870.3 kJ¡¤mol-1
(2)C(s)+O2(g) ="==" CO2(g)                     ¡÷H2= £­393.5 kJ¡¤mol-1
(3)H2(g)+O2(g) ===  H2O(l)                   ¡÷H3= £­285.8 kJ¡¤mol-1
Ôò·´Ó¦2C(s)+2H2(g)+O2(g)="==CH3COOH(l)" µÄìʱä¡÷HΪ£¨    £©kJ¡¤mol-1
A£®488.3B£®£­244.15C£®244.15D£®£­488.3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö£©»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦A(g)+B(g)C(g)+D(g)¹ý³ÌÖеÄÄÜÁ¿±ä»¯ÈçÓÒͼËùʾ£¬Åжϸ÷´Ó¦¡÷H   0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±¡¢
»ò¡°ÎÞ·¨È·¶¨¡±£©¡£

£¨2£©ÔÚAl2O3¡¢Ni´ß»¯ÏÂÆø̬¼×Ëá·¢ÉúÏÂÁз´Ó¦£º
¼×Ëá(g)=" CO" (g)+ H2O (g)¡¡ ¡÷H1=" +34.0" kJ/mol
¼×Ëá(g)= CO2 (g)+ H2(g)    ¡÷H2=" ¡ª7.0" kJ/mol
Ôò¼×ËáµÄ·Ö×ÓʽΪ           £¬ÔÚ¸ÃÌõ¼þÏ£¬Æø̬CO2ºÍÆø̬H2·´Ó¦Éú³ÉÆø̬COºÍÆø̬H2OµÄÈÈ»¯Ñ§·½³ÌʽΪ                             ¡£
£¨3£©ÈçͼËùʾ£¬Ë®²ÛÖÐÊÔ¹ÜÄÚÓÐһöÌú¶¤£¬·ÅÖÃÊýÌì¹Û²ì£º

IÈôÊÔ¹ÜÄÚÒºÃæÉÏÉý£¬ÔòÕý¼«·´Ó¦£º                         ¡£
IIÈôÊÔ¹ÜÄÚÒºÃæϽµ£¬Ôò·¢Éú           ¸¯Ê´¡£
IIIÈôÈÜÒº¼×Ϊˮ£¬ÈÜÒºÒÒΪº£Ë®£¬ÔòÌú¶¤ÔÚ   £¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©ÈÜÒºÖи¯Ê´µÄËٶȿ졣

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©£®2009Äê10ÔÂ16ÈÕµÚʮһ½ìÈ«ÔË»áÔÚ¼ÃÄϾÙÐУ¬È«ÔË»áʹÓõĻð¾æµÄȼÁÏÊDZûÍé¡£
£¨1£©ÒÑÖª11g±ûÍéÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·Å³öµÄÈÈÁ¿Îª555kJ£¬Çëд³ö±ûÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ
                                                 
(2)ijζÈÏ£¬´¿Ë®ÖеÄc£¨H+£©=2.0¡Á10-7mol/L£¬ÈôζȲ»±ä£¬µÎÈëÏ¡ÁòËáʹc£¨H+£©=5.0¡Á10-6mol/L£¬Ôò¸ÃÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©Îª
                             

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£º£¬Ôò¶ÔÓÚÈÈ»¯Ñ§·½³Ìʽ£º£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A£®ÈÈ»¯Ñ§·´Ó¦·½³ÌʽÖл¯Ñ§¼ÆÁ¿Êý±íʾ·Ö×Ó¸öÊý
B£®¸Ã·´Ó¦µÄ
C£®
D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(11·Ö)
£¨1£©ÒÑÖªH2(g)+Br2(1)2HBr(g) ¡÷H=£­72kJ/mol£¬Õô·¢1mol Br2(1)ÐèÒªÎüÊÕµÄÄÜÁ¿Îª30kJ£¬ÆäËüÏà¹ØÊý¾ÝÈçÏÂ±í£º
 
H2(g)
Br2(g)
HBr(g)
¼üÄÜ/kJ¡¤mol£­1
436
a
369
¢ÙBr2(1)£½Br2(g)  ¡÷H1£¬Ôò¡÷H1=_____________£»
¢Úд³öH2(g)ÓëBr2(g)·´Ó¦Éú³ÉHBr(g)µÄÈÈ»¯Ñ§·½³Ìʽ£º_____________________________
¢ÛÊÔ¼ÆËãa=_________________¡£
£¨2£©Á×ÔÚÑõÆøÖÐȼÉÕ£¬¿ÉÄÜÉú³ÉÁ½ÖÖ¹Ì̬Ñõ»¯Îï¡£3.1gµÄµ¥ÖÊÁ×£¨P£©ÔÚÒ»¶¨Á¿ÑõÆøÖÐȼÉÕÉú³ÉP2O3¡¢P2O5¸÷0.025mol£¬·´Ó¦ÎïÍêÈ«ºÄ¾¡£¬²¢·Å³öX kJÈÈÁ¿¡£
¢ÙÒÑÖª1molÁ×ÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿ÎªY kJ£¬Ôò1molPÓëO2·´Ó¦Éú³É¹Ì̬P2O3µÄ·´Ó¦ÈÈ¡÷H=________________________________¡£
¢Úд³ö1mol PÓëO2·´Ó¦Éú³É¹Ì̬P2O3µÄÈÈ»¯Ñ§·½³Ìʽ£º
___________________________________________________________________

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸