£¨1£©¡¢¢ÙÓÃëÂ(N2H4)ΪȼÁÏ£¬ËÄÑõ»¯¶þµª×öÑõ»¯¼Á£¬Á½Õß·´Ó¦Éú³ÉµªÆøºÍÆø̬ˮ¡£
ÒÑÖª£ºN2(g)£«2O2(g)£½N2O4(g)  ¦¤H£½+10.7kJ¡¤mol-1
N2H4(g)£«O2(g)£½N2(g)£«2H2O(g)  ¦¤H£½-543kJ¡¤mol-1
д³öÆø̬ëºÍN2O4·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                                ¡£
¢ÚÒÑÖªËÄÑõ»¯¶þµªÔÚ´óÆøÖлòÔڽϸßζÈϺÜÄÑÎȶ¨´æÔÚ£¬ÆäºÜÈÝÒ×ת»¯Îª¶þÑõ»¯µª¡£ÊÔÍƶÏÓɶþÑõ»¯µªÖÆÈ¡ËÄÑõ»¯¶þµªµÄ·´Ó¦Ìõ¼þ(»ò´ëÊ©)£º                         ¡£
£¨2£©¿Æѧ¼ÒÖÆÔì³öÒ»ÖÖʹÓùÌÌåµç½âÖʵÄȼÁϵç³Ø£¬ÆäЧÂʸü¸ß£¬¿ÉÓÃÓÚº½Ì캽¿Õ¡£

ͼ¼×ËùʾװÖÃÖУ¬ÒÔÏ¡ÍÁ½ðÊô²ÄÁÏΪ¶èÐԵ缫£¬ÔÚÁ½¼«ÉÏ·Ö±ðͨÈëCH4ºÍ¿ÕÆø£¬ÆäÖйÌÌåµç½âÖÊÊDzôÔÓÁËY2O3µÄZrO2¹ÌÌ壬ËüÔÚ¸ßÎÂÏÂÄÜ´«µ¼Ñô¼«Éú³ÉµÄO2-(O2+4e ¡ú2O2-)
¢Ùcµç¼«Îª       £¬dµç¼«Éϵĵ缫·´Ó¦Ê½Îª                            ¡£
¢ÚͼÒÒÊǵç½â100mL 0.5mol¡¤L-1 CuSO4ÈÜÒº£¬aµç¼«Éϵĵ缫·´Ó¦Ê½Îª          ¡£Èôaµç¼«²úÉú56mL(±ê×¼×´¿ö)ÆøÌ壬ÔòËùµÃÈÜÒºµÄpH=     (²»¿¼ÂÇÈÜÒºÌå»ý±ä»¯)£¬ÈôҪʹµç½âÖÊÈÜÒº»Ö¸´µ½µç½âÇ°µÄ״̬£¬¿É¼ÓÈë       (Ñ¡Ìî×ÖĸÐòºÅ)
a.CuO    b.Cu(OH)2     c.CuCO3     d.Cu2(OH)2CO3

£¨14·Ö£©
£¨1£©¢Ù2 N2H4(g) +  N2O4(g)= 3N£²(g)£«4H20(g) ¡÷H£½£­1096.7KJ¡¤mol-1£¨2·Ö£©
¢Ú¼Óѹ¡¢½µÎ£¨¸÷1·Ö£©
£¨2£©¢ÙÕý¼«£¨2·Ö£©  CH4 - 8e- + 402-=CO2+2H2O £¨2·Ö£©
¢Ú4OH¡ª- 4e-=2H2O+O2 £¨2·Ö£©  1 £¨2·Ö£©  a¡¢c £¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢Ùa¡¢N2£¨g£©+2O2£¨g£©=N2O4£¨g£©¡÷H=10.7kJ¡¤mol£­1£»b¡¢N2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨g£©¡÷H=-543kJ¡¤mol£­1
ÒÀ¾Ý¸Ç˹¶¨ÂÉb¡Á2-aµÃµ½  2N2H4£¨g£©+N2O4£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1096.7KJ¡¤mol£­1£»
´ð°¸Îª£º2N2H4£¨g£©+N2O4£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1096.7KJ¡¤mol£­1£»
¢ÚËÄÑõ»¯¶þµªÔÚ´óÆøÖлòÔڽϸßζÈϺÜÄÑÎȶ¨´æÔÚ£¬ËüºÜÈÝÒ×ת»¯Îª¶þÑõ»¯µª£¬ÓɶþÑõ»¯µªÖÆÈ¡ËÄÑõ»¯¶þµª£¬2NO2N2O4£¬·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬·´Ó¦Ç°ºóÆøÌåÌå»ý¼õСËùÒÔ·´Ó¦Ìõ¼þΪ£ºÔö´óѹǿ¡¢½µÎ¶¼ÓÐÀûÓÚ·´Ó¦ÕýÏò½øÐУ»
¹Ê´ð°¸Îª£ºÔö´óѹǿ¡¢½µÎ£»
£¨2£©¢Ùͼ1ÊÇÔ­µç³Ø£¬ÒÀ¾ÝµçÁ÷Á÷ÏòÊÇ´ÓÕý¼«Á÷Ïò¸º¼«£¬cµç¼«ÎªÕý¼«£¬ÑõÆøµÃµ½µç×Ó·¢Éú»¹Ô­·´Ó¦£¬dµç¼«Îªµç³Ø¸º¼«¼×ÍéÊǵç×Ó·¢Éú»¹Ô­·´Ó¦£¬ÔÚÁ½¼«ÉÏ·Ö±ðͨÈëCH4ºÍ¿ÕÆø£¬ÆäÖйÌÌåµç½âÖÊÊDzôÔÓÁËY2O3µÄZrO2¹ÌÌ壬ËüÔÚ¸ßÎÂÏÂÄÜ´«µ¼Ñô¼«Éú³ÉµÄO2£­Àë×Ó£¬½áºÏµç×ÓÊغãд³öµç¼«·´Ó¦Îª£ºCH4- 8e-+4O2£­=CO2+2H2O£»
´ð°¸Îª£ºÕý¼«£» CH4- 8e-+4O2£­=CO2+2H2O£»
¢ÚÈçͼ2Ëùʾµç½â100mL0.5mol?L-1CuSO4ÈÜÒº£¬·¢ÉúµÄµç½â³Ø·´Ó¦Îª£º2CuSO4+2H2O2Cu+O2¡ü+2H2SO4£¬ÓëµçÔ´Õý¼«ÏàÁ¬µÄΪÑô¼«£¬ÈÜÒºÖÐ ÇâÑõ¸ùÀë×ÓÊǵç×Ó·¢ÉúÑõ»¯·´Ó¦£¬µç¼«·´Ó¦Îª£º4OH£­-4e£­=2H2O+O2¡ü£»Èôaµç¼«²úÉú56mL£¨±ê×¼×´¿ö£©ÆøÌåΪÑõÆø£¬ÎïÖʵÄÁ¿Îª0.0025mol£¬ÏûºÄÇâÑõ¸ùÀë×ÓÎïÖʵÄÁ¿Îª0.01mol£¬ÈÜÒºÖÐÉú³ÉÇâÀë×ÓÎïÖʵÄÁ¿Îª0.01mol£¬c£¨H£«£©==0.1mol¡¤L£­1£¬PH=-lg0.1=1£»ÔòËùµÃÈÜÒºµç½â¹ý³ÌÖÐCuSO4ÈÜҺÿËðʧ2¸öCuÔ­×Ó£¬¾ÍËðʧ2¸ö OÔ­×Ó£¬Ï൱ÓÚËðʧһ¸öCuO£¬ÎªÁËʹCuSO4ÈÜÒº£¬»Ö¸´Ô­Å¨¶È£¬Ó¦¼ÓÈëCuO£¬Ò²¿ÉÒÔ¼ÓÈëCuCO3£¬·ûºÏ»Ö¸´ÈÜҺŨ¶ÈµÄ¶¨Á¿¹Øϵ£¬µ«²»ÄܼÓÈëCu£¨OH£©2¡¢Cu2£¨OH£©2CO3£¬ÒòΪCuCO3+H2SO4CuSO4+CO2¡ü+H2O£¬Ï൱ÓÚ¼ÓCuO£¬¶øCu£¨OH£©2+H2SO4CuSO4+2H2O¡¢Cu2£¨OH£©2CO3+2H2SO4=2CuSO4 +CO2¡ü+3H2O£¬³ýÔö¼ÓÈÜÖÊÍ⻹Ôö¼ÓÁËË®£»Ñ¡ac£®
´ð°¸Îª£º4OH¡ª- 4e-=2H2O+O2 £»1£»ac£»
¿¼µã£ºÈÈ»¯Ñ§·½³Ìʽ£»»¯Ñ§µçÔ´ÐÂÐ͵ç³Ø

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£© 8gҺ̬µÄCH3OHÔÚÑõÆøÖÐÍêȫȼÉÕ£¬Éú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱÊͷųöQ kJµÄÈÈÁ¿¡£ÊÔд³öҺ̬CH3OHȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ                             ¡£
£¨2£©ÔÚ»¯Ñ§·´Ó¦¹ý³ÌÖУ¬ÆÆ»µ¾É»¯Ñ§¼üÐèÒªÎüÊÕÄÜÁ¿£¬ÐγÉл¯Ñ§¼üÓÖ»áÊÍ·ÅÄÜÁ¿¡£

»¯Ñ§¼ü
H¡ªH
N¡ªH
N¡ÔN
¼üÄÜ/kJ·mol£­1
436
391
945
ÒÑÖª·´Ó¦N2£«3H2=2NH3¡¡¦¤H£½a KJ/mol¡£
ÊÔ¸ù¾Ý±íÖÐËùÁмüÄÜÊý¾Ý¼ÆËãaµÄÊýֵΪ£º               ¡£
£¨3£©ÒÑÖª£ºC(s£¬Ê¯Ä«)£«O2(g)=CO2(g)  ¦¤H1£½£­393.5 kJ/mol
2H2(g)£«O2(g)=2H2O(l)  ¦¤H2£½£­571.6 kJ/mol
2C2H2(g)£«5O2(g)=4CO2(g)£«2H2O(l)  ¦¤H3£½£­2599 kJ/mol
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬ÓÉC(s£¬Ê¯Ä«)ºÍH2(g)Éú³É1 mol C2H2(g)·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º
                                                          ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(1)CH4(g )+2O2(g )=CO2(g )+2H2O(g )    ¦¤H=-802.3kJ/mol
¸ÃÈÈ»¯Ñ§·´Ó¦·½³ÌʽµÄÒâÒåÊÇ_____________________________________¡£
(2)ÒÑÖª2gÒÒ´¼ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öQ kJµÄÈÈÁ¿£¬Ð´³ö±íʾÒÒ´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½
³Ìʽ£º____________________________________________________________.
(3)ÒÑÖª²ð¿ª1mol H-H¼ü£¬1mol N-H¼ü£¬1mol ¼ü·Ö±ðÐèÒªµÄÄÜÁ¿ÊÇ436kJ¡¢391KJ¡¢946kJ£¬ÔòN2ÓëH2·´Ó¦Éú³É1mol NH3(g)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ___________________.
(4)ÒÀ¾Ý¸Ç˹¶¨ÂÉ¿ÉÒÔ¶ÔijЩÄÑÒÔͨ¹ýʵÑéÖ±½Ó²â¶¨µÄ»¯Ñ§·´Ó¦µÄìʱä½øÐÐÍÆËã¡£
ÒÑÖª£ºC(ʯī£¬s)+O2(g)=CO2(g)       ¦¤H=-393.5kJ/mol   ¢Ù
2H2(g)+O2(g)=2H2O(l)               ¦¤H=-571.6kJ/mol   ¢Ú
2C2H2(g)+5O2(g)=4CO2(g)+2H2O(l)   ¦¤H=-2599kJ/mol   ¢Û
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¼ÆËã298KʱÓÉC£¨Ê¯Ä«£¬s£©ºÍH2(g)Éú³É1mol C2H2(g)·´Ó¦µÄìʱ䣺
____________________________.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)2014Äê10Ô³õ£¬Îíö²ÌìÆø¶à´ÎËÁÅ°ºÓ±±¡¢Ìì½ò¡¢±±¾©µÈµØÇø¡£ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£
£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g) + 2CO(g)2CO2(g)+ N2(g)¡£¡÷H£¼0
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽ                                                           ¡£
¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ                                   £¨Ìî´úºÅ£©¡£

£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖصĻ·¾³ÎÊÌ⡣úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­NOx¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£
ÒÑÖª£ºCH4(g)+2NO2(g)£½N2(g)£«CO2(g)+2H2O(g) ¡÷H£½£­867 kJ/mol
2NO2(g)N2O4(g)  ¡÷H£½£­56.9 kJ/mol
H2O(g) £½ H2O(l)  ¦¤H £½ £­44.0 kJ£¯mol
д³öCH4´ß»¯»¹Ô­N2O4(g)Éú³ÉN2ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ£º                            ¡£
£¨3£©ÔÚÒ»¶¨Ìõ¼þÏ£¬Ò²¿ÉÒÔÓÃNH3´¦ÀíNOx¡£ÒÑÖªNOÓëNH3·¢Éú·´Ó¦Éú³ÉN2ºÍH2O£¬ÏÖÓÐNOºÍNH3µÄ»ìºÏÎï1mol£¬³ä·Ö·´Ó¦ºóµÃµ½µÄ»¹Ô­²úÎï±ÈÑõ»¯²úÎï¶à1.4 g£¬ÔòÔ­·´Ó¦»ìºÏÎïÖÐNOµÄÎïÖʵÄÁ¿¿ÉÄÜÊÇ_____________¡£
£¨4£©ÒÔ¼×ÍéΪԭÁÏÖÆÈ¡ÇâÆøÊǹ¤ÒµÉϳ£ÓõÄÖÆÇâ·½·¨¡£Ôò2 molCH4Óë×ãÁ¿H2O£¨g£©·´Ó¦×î¶à¿ÉÉú³É_______mol H2£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________________________________________¡£
(5)ÉÏÊö·½·¨ÖƵõÄH2¿ÉÒÔºÍCOÔÚÒ»¶¨Ìõ¼þϺϳɼ״¼ºÍ¶þ¼×ÃÑ£¨CH3OCH3£©¼°Ðí¶àÌþÀàÎïÖÊ¡£µ±Á½ÕßÒÔÎïÖʵÄÁ¿1:1´ß»¯·´Ó¦£¬ÆäÔ­×ÓÀûÓÃÂÊ´ï100%£¬ºÏ³ÉµÄÎïÖÊ¿ÉÄÜÊÇ                     ¡£
a.ÆûÓÍ        b.¼×´¼           c.¼×È©            d.ÒÒËá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¿Æѧ¼ÒÀûÓÃÌ«ÑôÄÜ·Ö½âË®Éú³ÉµÄÇâÆøÔÚ´ß»¯¼Á×÷ÓÃÏ   Óë¶þÑõ»¯Ì¼·´Ó¦Éú³É¼×´¼£¬²¢¿ª·¢³öÖ±½ÓÒÔ¼×´¼ÎªÈ¼ÁϵÄȼÁϵç³Ø¡£ÒÑÖª£ºH2(g)¡¢CO(g)ºÍCH3OH£¨1£©µÄȼÉÕÈÈ¡÷H·Ö±ðΪ-285.8 kJ£®¡¢Ò»283.0 kJºÍÒ»726.5£®kJ ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃÌ«ÑôÄÜ·Ö½â10mol H2O(1)ÏûºÄµÄÄÜÁ¿ÊÇ________kJ.
£¨2£©¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
__________________________________________________________________________.
£¨3£©ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬ÓÉCO2ºÍH2ºÏ³É¼×´¼£¬ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬
¿¼²éζȶԷ´Ó¦µÄÓ°Ï죬ʵÑé½á¹ûÈçÏÂͼËùʾ£¨×¢£º¡¢¾ù´óÓÚ300¡æ£©£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_______________£¨ÌîÐòºÅ£©
¢ÙζÈΪʱ£¬´Ó·´Ó¦¿ªÊ¼µ½·´Ó¦´ïµ½Æ½ºâ£¬Éú³É¼×´¼µÄƽ¾ùËÙÂÊΪ£º

¢Ú¸Ã·´Ó¦ÔÚʱµÄƽºâ³£Êý±ÈʱµÄС
¢Û¸Ã·´Ó¦Îª·ÅÈÈ·´Ó¦
¢Ü´¦ÓÚAµãµÄ·´Ó¦ÌåϵµÄζȴӱ䵽£¬´ïµ½Æ½ºâʱÔö´ó
£¨4£©ÔÚζÈʱ£¬½«1mol CO2ºÍ3mol H2³äÈëÒ»ÃܱպãÈÝÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦´ïµ½Æ½ºâºó£¬ÈôCO2µÄת»¯ÂÊΪa£¬Ôò´ËʱÈÝÆ÷ÄÚµÄѹǿÓëÆðʼѹǿ֮±ÈΪ___________¡£
£¨5£©ÔÚÖ±½ÓÒÔ¼×´¼ÎªÈ¼ÁϵÄȼÁϵç³ØÖУ¬µç½âÖÊÈÜҺΪËáÐÔ£¬¸º¼«µÄ·´Ó¦Ê½Îª___________________;Õý¼«µÄ·´Ó¦Ê½Îª_____________________________________.ÀíÏë״̬Ï£¬¸ÃȼÁϵç³ØÏûºÄlmol¼×´¼ËùÄܲúÉúµÄ×î´óµçÄÜΪ701.8kJ£¬Ôò¸ÃȼÁϵç³ØµÄÀíÂÛЧÂÊΪ_______________£¨È¼Áϵç³ØµÄÀíÂÛЧÂÊÊÇÖ¸µç³ØËù²úÉúµÄ×î´óµçÄÜÓëȼÁϵç³Ø·´Ó¦ËùÄÜÊͷŵÄÈ«²¿ÄÜÁ¿Ö®±È£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

úµÄÆø»¯ÊǸßЧ¡¢Çå½àµØÀûÓÃú̿µÄÖØҪ;¾¶Ö®Ò»¡£
(1)ÔÚ250C 101kPaʱ£¬H2ÓëO2»¯ºÏÉú³É1mol H2O(g)·Å³ö241.8kJµÄÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ
___________
ÓÖÖª: ¢ÙC(s)£«O2(g)¨TCO2(g) ¡÷H£½£­393.5kJ/mol
¢ÚCO(g)£«O2(g)¨TCO2(g) ¡÷H£½£­283.0kJ/mol
½¹Ì¿ÓëË®ÕôÆø·´Ó¦Êǽ«¹ÌÌåú±äΪÆøÌåȼÁϵķ½·¨£¬C(s)£«H2O(g)¨TCO(g)£«H2(g) ¡÷H=____kJ/mol
(2) CO¿ÉÒÔÓëH2O(g)½øÒ»²½·¢Éú·´Ó¦: CO(g)£«H2O(g)CO2(g)£«H2(g) ¡÷H£¼0ÔÚºãÈÝÃܱÕÈÝÆ÷ÖУ¬Æðʼʱn(H2O)=0.20mol£¬n(CO)£½0.10 mol,ÔÚ8000Cʱ´ïµ½Æ½ºâ״̬£¬K£½1.0£¬Ôòƽºâʱ£¬ÈÝÆ÷ÖÐCOµÄת»¯ÂÊÊÇ_____________(¼ÆËã½á¹û±£ÁôһλСÊý)¡£
(3) ¹¤ÒµÉÏ´ÓúÆø»¯ºóµÄ»ìºÏÎïÖзÖÀë³öH2£¬½øÐа±µÄºÏ³É£¬ÒÑÖª·´Ó¦·´Ó¦N2(g)£«3H2(g2NH3(g)£¨¡÷H£¼0£©ÔÚµÈÈÝÌõ¼þϽøÐУ¬¸Ä±äÆäËû·´Ó¦Ìõ¼þ£¬ÔÚI¡¢II¡¢III½×¶ÎÌåϵÖи÷ÎïÖÊŨ¶ÈËæʱ¼ä±ä»¯µÄÇúÏßÈçÏÂͼËùʾ£º

¢ÙN2µÄƽ¾ù·´Ó¦ËÙÂÊv1(N2)¡¢vII(N2)¡¢vIII(N2)´Ó´óµ½Ð¡ÅÅÁдÎÐòΪ________£»
¢ÚÓɵÚÒ»´Îƽºâµ½µÚ¶þ´Îƽºâ£¬Æ½ºâÒƶ¯µÄ·½Ïò ÊÇ________£¬²ÉÈ¡µÄ´ëÊ©ÊÇ________¡£
¢Û±È½ÏµÚII½×¶Î·´Ó¦Î¶È(T2)ºÍµÚIII½×¶Î·´Ó¦Ëٶȣ¨T3)µÄ¸ßµÍ£ºT2________T3Ìî¡°¡µ¡¢=¡¢<¡±ÅжϵÄÀíÓÉÊÇ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

°±ÆøÊÇÉú²ú»¯·Ê¡¢ÏõËáµÈµÄÖØÒªÔ­ÁÏ£¬Î§Èƺϳɰ±ÈËÃǽøÐÐÁËһϵÁеÄÑо¿
£¨1£©ÇâÆø¼ÈÄÜÓ뵪ÆøÓÖÄÜÓëÑõÆø·¢Éú·´Ó¦£¬µ«ÊÇ·´Ó¦µÄÌõ¼þÈ´²»Ïàͬ¡£
ÒÑÖª£º2H2 (g) + O2 (g) = 2H2O (g)  ¦¤H =" -483.6" kJ/mol
3H2 (g) + N2 (g)  2NH3 (g) ¦¤H =" -92.4" kJ/mol

¼ÆËã¶ÏÁÑ1 mol N¡ÔN¼üÐèÒªÄÜÁ¿      kJ £¬ µªÆø·Ö×ÓÖл¯Ñ§¼ü±ÈÑõÆø·Ö×ÓÖеĻ¯Ñ§¼ü¼ü     £¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©£¬Òò´ËÇâÆøÓë¶þÕß·´Ó¦µÄÌõ¼þ²»Í¬¡£
£¨2£©¹ÌµªÊÇ¿Æѧ¼ÒÖÂÁ¦Ñо¿µÄÖØÒª¿ÎÌâ¡£×ÔÈ»½çÖдæÔÚÌìÈ»µÄ´óÆø¹Ìµª¹ý³Ì£ºN2 (g) + O2 (g) =" 2NO" (g) ¦¤H =" +180.8" kJ/mol £¬¹¤ÒµºÏ³É°±ÔòÊÇÈ˹¤¹Ìµª¡£
·ÖÎöÁ½Ö̵ֹª·´Ó¦µÄƽºâ³£Êý£¬ÏÂÁнáÂÛÕýÈ·µÄÊÇ       ¡£

 ·´Ó¦
´óÆø¹Ìµª
¹¤Òµ¹Ìµª
ζÈ/¡æ
27
2000
25
350
400
450
K
3.84¡Á10-31
0.1
5¡Á108
1.847
0.507
0.152
 
A£®³£ÎÂÏ£¬´óÆø¹Ìµª¼¸ºõ²»¿ÉÄܽøÐУ¬¶ø¹¤Òµ¹Ìµª·Ç³£ÈÝÒ×½øÐÐ
B£®ÈËÀà´ó¹æģģÄâ´óÆø¹ÌµªÊÇÎÞÒâÒåµÄ
C£®¹¤Òµ¹ÌµªÎ¶ÈÔ½µÍ£¬µªÆøÓëÇâÆø·´Ó¦Ô½ÍêÈ«
D£®KÔ½´ó˵Ã÷ºÏ³É°±·´Ó¦µÄËÙÂÊÔ½´ó
£¨3£©ÔÚºãκãÈÝÃܱÕÈÝÆ÷Öа´Õռס¢ÒÒ¡¢±ûÈýÖÖ·½Ê½·Ö±ðͶÁÏ, ·¢Éú·´Ó¦£º3H2 (g) + N2 (g)  2NH3 (g)²âµÃ¼×ÈÝÆ÷ÖÐH2µÄת»¯ÂÊΪ40%¡£
 
N2
H2
NH3
¼×
1
3
0
ÒÒ
0.5
1.5
1
±û
0
0
4
ÅжÏÒÒÈÝÆ÷Öз´Ó¦½øÐеķ½Ïò       ¡££¨Ìî¡°ÕýÏò¡±»ò¡°ÄæÏò¡±£©
´ïƽºâʱ£¬¼×¡¢ÒÒ¡¢±ûÈýÈÝÆ÷ÖÐNH3µÄÌå»ý·ÖÊý´óС˳ÐòΪ       ¡£
£¨4£©°±ÆøÊǺϳÉÏõËáµÄÔ­ÁÏ£¬Ð´³ö°±ÆøÓëÑõÆø·´Ó¦Éú³ÉÒ»Ñõ»¯µªºÍÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¿ª·¢¡¢Ê¹ÓÃÇå½àÄÜÔ´·¢Õ¹¡°µÍ̼¾­¼Ã¡±£¬Õý³ÉΪ¿Æѧ¼ÒÑо¿µÄÖ÷Òª¿ÎÌâ¡£ÇâÆø¡¢¼×´¼ÊÇÓÅÖʵÄÇå½àȼÁÏ£¬¿ÉÖÆ×÷ȼÁϵç³Ø¡£
£¨1£©ÒÑÖª£º¢Ù 2CH3OH(1) + 3O2(g) = 2CO2(g) + 4H2O(g)  ¦¤H1 =" ¨C" 1275.6 kJ/mol
¢Ú 2CO(g) + O2(g) = 2CO2(g)    ¦¤H2 =" ¨C" 566.0 kJ/mol
¢Û H2O(g) = H2O(1)  ¦¤H3 =" ¨C" 44.0 kJ/mol
д³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º___________¡£
£¨2£©Éú²ú¼×´¼µÄÔ­ÁÏCOºÍH2À´Ô´ÓÚ£ºCH4(g) + H2O(g) CO(g) + 3H2(g)  ¦¤H>0
¢ÙÒ»¶¨Ìõ¼þÏÂCH4µÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼa¡£ÔòTl ________T2(Ìî¡°<¡±¡¢¡°>¡±¡¢¡°=¡±£¬ÏÂͬ)£»A¡¢B¡¢CÈýµã´¦¶ÔӦƽºâ³£Êý£¨KA¡¢KB¡¢KC£©µÄ´óС¹ØϵΪ___________¡£
     
¢Ú100¡æʱ£¬½«1 mol CH4ºÍ2 mol H2OͨÈëÈÝ»ýΪ1 LµÄ¶¨ÈÝÃÜ·âÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£¬ÄÜ˵Ã÷¸Ã·´Ó¦ÒѾ­´ïµ½Æ½ºâ״̬µÄÊÇ__________
a£®ÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨  
b£®µ¥Î»Ê±¼äÄÚÏûºÄ0.1 mol CH4ͬʱÉú³É0.3 mol H2
c£®ÈÝÆ÷µÄѹǿºã¶¨      
d£®3vÕý(CH4) = vÄæ(H2)
Èç¹û´ïµ½Æ½ºâʱCH4µÄת»¯ÂÊΪ0.5£¬Ôò100¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK =___________
£¨3£©Ä³ÊµÑéС×éÀûÓÃCO(g) ¡¢ O2(g) ¡¢KOH£¨aq£©Éè¼Æ³ÉÈçͼbËùʾµÄµç³Ø×°Öã¬Ôò¸Ãµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½Îª___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

£¨14·Ö£©Ñо¿Òø¼°Æ仯ºÏÎï¾ßÓÐÖØÒªÒâÒå¡£
£¨l£©ÒÑÖª£º
Ag2O£¨s£©+2HC1£¨g£©   2AgC1£¨s£©+H2O£¨1£© ¡÷H1=£­324£®4 kJ¡¤mo1£­1
2Ag£¨s£©+1/2O2£¨g£©   Ag2O£¨s£©            ¡÷H2=£­30£®6 kJ¡¤mo1£­1
H2£¨g£©+C12£¨g£©    2HC1£¨g£©           ¡÷H3=£­184£®4 kJ£®¡¤mo1£­1
2H2£¨9£©+O2£¨g£©   2H2O£¨1£©           ¡÷H4=£­571£®2 l¡¤mo1£­1
д³öÂÈÆøÓëÒøÉú³É¹ÌÌåÂÈ»¯ÒøµÄÈÈ»¯Ñ§·½³Ìʽ________¡£
£¨2£©ÃÀÀöµÄÒøÊγ£ÓÃFe(NO3)3ÈÜҺʴ¿Ì£¬Ð´³öFe3+ÓëAg·´Ó¦µÄÀë×Ó·½³Ìʽ___        _£»ÒªÅж¨Fe(NO3)3ÈÜÒºÖÐNO3¡ªÊÇ·ñÔÚÒøÊÎÊ´¿ÌÖз¢Éú·´Ó¦£¬¿ÉÈ¡                   µÄÏõËáÈÜÒº£¬È»ºó¸ù¾ÝÆäÊÇ·ñÓëAg·¢Éú·´Ó¦À´Åж¨¡£
£¨3£©Òøп¼îÐÔµç³ØµÄµç½âÖÊÈÜҺΪKOHÈÜÒº£¬·Åµçʱ£¬Õý¼«Ag2O2ת»¯ÎªAg£¬¸º¼«Znת»¯ÎªZn(OH)2£¬ÔòÕý¼«·´Ó¦Ê½Îª           £¬¸º¼«¸½½üÈÜÒºµÄpH ___  £¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©¡£
£¨4£©µç½â·¨¾«Á¶Òøʱ£¬´ÖÒøÓ¦ÓëÖ±Á÷µçÔ´µÄ       ¼«ÏàÁ¬£¬µ±ÓÃAgNO3ºÍHNO3»ìºÏÈÜÒº×öµç½âÖÊÈÜҺʱ£¬·¢ÏÖÒõ¼«ÓÐÉÙÁ¿ºì×ØÉ«ÆøÌ壬Ôò²úÉú¸ÃÏÖÏóµÄµç¼«·´Ó¦Ê½Îª____¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸