ÌâÄ¿ÄÚÈÝ

Ñо¿CO2µÄÀûÓöԴٽøµÍ̼Éç»áµÄ¹¹½¨¾ßÓÐÖØÒªÒâÒå¡£
£¨1£©½«CO2Ó뽹̿×÷ÓÃÉú³ÉCO£¬CO¿ÉÓÃÓÚÁ¶ÌúµÈ¡£
ÒÑÖª£ºFe2O3(s)+3C(ʯī)=2Fe(s)+3CO(g) ¡÷H 1=+489.0 kJ¡¤mol£­1
C(ʯī)+CO2(g)=2CO(g)  ¡÷H 2=+172.5 kJ¡¤mol£­1
ÔòCO»¹Ô­Fe2O3(s)µÄÈÈ»¯Ñ§·½³ÌʽΪ           ¡£
£¨2£©¶þÑõ»¯Ì¼ºÏ³É¼×´¼ÊÇ̼¼õÅŵÄз½Ïò£¬½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO2(g)+3H2(g)CH3OH(g)+H2O(g) ¡÷H
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK=          ¡£
¢ÚÈ¡Ò»¶¨Ìå»ýCO2ºÍH2µÄ»ìºÏÆøÌå(ÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã3)£¬¼ÓÈëºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦¡£·´Ó¦¹ý³ÌÖвâµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ(CH3OH)Ó뷴ӦζÈTµÄ¹ØϵÈçͼËùʾ£¬Ôò¸Ã·´Ó¦µÄ¦¤H      0(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±)¡£

¢ÛÔÚÁ½ÖÖ²»Í¬Ìõ¼þÏ·¢Éú·´Ó¦£¬²âµÃCH3OHµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼËùʾ£¬ÇúÏßI¡¢¢ò¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØϵΪK¢ñ     K¢ò(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±)¡£

£¨3£©ÒÔCO2ΪԭÁÏ»¹¿ÉÒԺϳɶàÖÖÎïÖÊ¡£
¢Ù¹¤ÒµÉÏÄòËØ[CO(NH2)2]ÓÉCO2ºÍNH3ÔÚÒ»¶¨Ìõ¼þϺϳɣ¬Æä·´Ó¦·½³ÌʽΪ    ¡£µ±°±Ì¼±È£½3£¬´ïƽºâʱCO2µÄת»¯ÂÊΪ60%£¬ÔòNH3µÄƽºâת»¯ÂÊΪ
        ¡£
¢ÚÓÃÁòËáÈÜÒº×÷µç½âÖʽøÐеç½â£¬CO2Ôڵ缫ÉÏ¿Éת»¯Îª¼×Í飬¸Ãµç¼«·´Ó¦µÄ·½³ÌʽΪ                        ¡£

£¨1£©Fe2O3(s)+3CO(g)=2Fe(s)+3CO2(g) ¡÷H=¡ª28.5 kJ¡¤mol£­1£¨2·Ö£©
£¨2£©¢Ù £¨2·Ö£©   ¢Ú£¼£¨2·Ö£© ¢Û>£¨2·Ö£©
£¨3£©¢Ù2NH3+CO2CO(NH2)2+H2O£¨2·Ö£©    40%£¨2·Ö£©
¢ÚCO2+8e¡ª+8H+=CH4+2H2O£¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©·´Ó¦¢Ù£­ ¢Ú¡Á3¿ÉµÃFe2O3(s)+3CO(g)=2Fe(s)+3CO2(g) ¡÷H=¡ª28.5 kJ¡¤mol£­1£»£¨2£©¢Ú¸ÃͼÏñÊÇ·´Ó¦¹ý³Ìͼ£¬×î¸ßµãÇ°ÊÇδƽºâʱµÄ±ä»¯£¬×î¸ßµãºóµÄÊÇƽºâ±ä»¯£¬¼´Î¶ÈÉý¸ßƽºâÄæÏò½øÐУ¬·´Ó¦·ÅÈÈ£»¢ÛÓÉͼÅжÏÇúÏߢò¶ÔÓ¦µÄ·´Ó¦·´Ó¦ËÙÂʿ죬ζȸߣ¬¼×´¼º¬Á¿µÍ£¬KֵС£»£¨3£©¢Ù·´Ó¦Îª2NH3+CO2CO(NH2)2+H2O£¬Éè°±ÆøµÄÎïÖʵÄÁ¿Îª3mol£¬Ôò¶þÑõ»¯Ì¼µÄÁ¿Îª1mol£¬ÓÐ
2NH3+CO2CO(NH2)2+H2O
ʼÁ¿    3    1             0        0
ת»¯Á¿  1.2  0.6          0.6       0.6
ƽºâÁ¿  1.8  0.4           0.6       0.6
ÔòNH3µÄƽºâת»¯ÂÊΪ1.2¡Â3=0.4
¢Ú¸ù¾ÝÐÅÏ¢£¬ÔÚËáÐÔ½éÖÊÖз´Ó¦ÎªCO2+8e¡ª+8H+=CH4+2H2O¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¶þ¼×ÃÑ(CH3OCH3)±»³ÆΪ21ÊÀ¼ÍµÄÐÂÐÍȼÁÏ£¬ÔÚδÀ´¿ÉÄÜÌæ´úÆû³µÈ¼ÓÍ¡¢Ê¯ÓÍÒº»¯Æø¡¢³ÇÊÐúÆøµÈ£¬Êг¡Ç°¾°¼«Îª¹ãÀ«¡£ËüÇå½à¡¢¸ßЧ£¬¾ßÓÐÓÅÁ¼µÄ»·±£ÐÔÄÜ¡£
¹¤ÒµÉÏÖƶþ¼×ÃÑÊÇÔÚÒ»¶¨Î¶È(230¡«280 ¡æ)¡¢Ñ¹Ç¿(2.0¡«10.0 MPa)ºÍ´ß»¯¼Á×÷ÓÃϽøÐеģ¬·´Ó¦Æ÷Öз¢ÉúÁËÏÂÁз´Ó¦¡£
CO(g)£«2H2(g)CH3OH(g)    ¦¤H1£½£­90.7 kJ¡¤mol£­1¡¡¢Ù
2CH3OH(g)CH3OCH3(g)£«H2O(g)     ¦¤H2£½£­23.5 kJ¡¤mol£­1¡¡¢Ú
CO(g)£«H2O(g)CO2(g)£«H2(g)   ¦¤H3£½£­41.2 kJ¡¤mol£­1¡¡¢Û
(1)·´Ó¦Æ÷ÖеÄ×Ü·´Ó¦¿É±íʾΪ3CO(g)£«3H2(g)CH3OCH3(g)£«CO2(g)£¬Ôò¸Ã·´Ó¦µÄ¦¤H£½__________£¬Æ½ºâ³£Êý±í´ïʽΪ____________________£¬ÔÚºãΡ¢¿É±äÈÝ»ýµÄÃܱÕÈÝÆ÷ÖнøÐÐÉÏÊö·´Ó¦£¬Ôö´óѹǿ£¬¶þ¼×ÃѵIJúÂÊ»á________(ÌîÉý¸ß¡¢½µµÍ»ò²»±ä)¡£
(2)¶þÑõ»¯Ì¼ÊÇÒ»ÖÖÖØÒªµÄÎÂÊÒÆøÌ壬¼õÉÙ¶þÑõ»¯Ì¼µÄÅÅ·ÅÊǽâ¾öÎÂÊÒЧӦµÄÓÐЧ;¾¶¡£Ä¿Ç°£¬ÓɶþÑõ»¯Ì¼ºÏ³É¶þ¼×ÃѵÄÑо¿¹¤×÷ÒÑÈ¡µÃÁËÖØ´ó½øÕ¹£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ2CO2(g)£«6H2(g)??CH3OCH3(g)£«3H2O(g)¡¡¦¤H£¾0¡£
¸Ã·´Ó¦ÔÚºãΡ¢Ìå»ýºã¶¨µÄÃܱÕÈÝÆ÷ÖнøÐУ¬ÏÂÁв»ÄÜ×÷Ϊ¸Ã·´Ó¦ÒÑ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÅжÏÒÀ¾ÝµÄÊÇ________¡£
A£®ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄÃܶȲ»±ä
B£®ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄѹǿ±£³Ö²»±ä
C£®ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿±£³Ö²»±ä
D£®µ¥Î»Ê±¼äÄÚÏûºÄ2 mol CO2µÄͬʱÏûºÄ1 mol¶þ¼×ÃÑ
(3)¶þ¼×ÃÑÆøÌåµÄȼÉÕÈÈΪ1 455 kJ¡¤mol£­1£¬¹¤ÒµÉÏÓúϳÉÆø(CO¡¢H2)Ö±½Ó»ò¼ä½ÓºÏ³É¶þ¼×ÃÑ¡£ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ________¡£
A£®¶þ¼×ÃÑ·Ö×ÓÖк¬¹²¼Û¼ü
B£®¶þ¼×ÃÑ×÷ΪÆû³µÈ¼Áϲ»»á²úÉúÎÛȾÎï
C£®¶þ¼×ÃÑÓëÒÒ´¼»¥ÎªÍ¬ÏµÎï
D£®±íʾ¶þ¼×ÃÑȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪCH3OCH3(g)£«3O2(g)=2CO2(g)£«3H2O(g)¡¡¦¤H£½£­1 455 kJ¡¤mol£­1
(4)ÂÌÉ«µçÔ´¡°Ö±½Ó¶þ¼×ÃÑȼÁϵç³Ø¡±µÄ¹¤×÷Ô­ÀíʾÒâͼÈçͼËùʾ£ºÕý¼«Îª________(Ìî¡°Aµç¼«¡±»ò¡°Bµç¼«¡±)£¬Ð´³öAµç¼«µÄµç¼«·´Ó¦Ê½£º________________________________________¡£
µªÊÇ´óÆøÖк¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æ仯ºÏÎïÔÚÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒª×÷Ó㬼õÉÙµªÑõ»¯ÎïµÄÅÅ·ÅÊÇ»·¾³±£»¤µÄÖØÒªÄÚÈÝÖ®Ò»¡£Çë»Ø´ðÏÂÁеª¼°Æ仯ºÏÎïµÄÏà¹ØÎÊÌ⣺
(1)¾Ý±¨µÀ£¬Òâ´óÀû¿Æѧ¼Ò»ñµÃÁ˼«¾ßÑо¿¼ÛÖµµÄN4£¬Æä·Ö×ӽṹÓë°×Á×·Ö×ÓµÄÕýËÄÃæÌå½á¹¹ÏàËÆ¡£ÒÑÖª¶ÏÁÑ1 mol N£­N¼üÎüÊÕ167 kJÈÈÁ¿£¬Éú³É1 mol N¡ÔN¼ü·Å³ö942 kJÈÈÁ¿£¬Çëд³öN4ÆøÌåת±äΪN2·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                            ¡£
(2)¾Ý±¨µÀ£¬NH3¿ÉÖ±½ÓÓÃ×÷³µÓÃȼÁϵç³Ø£¬Ð´³ö¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½£º                          ¡£
(3)ÔÚT1¡æʱ£¬½«5 mol N2O5ÖÃÓÚ10L¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷Öз¢ÉúÏÂÁз´Ó¦£º2N2O5(g)4NO2(g)+O2(g)£»¡÷H£¾0¡£·´Ó¦ÖÁ5·ÖÖÓʱ¸÷ÎïÖʵÄŨ¶È²»ÔÙ·¢Éú±ä»¯£¬²âµÃNO2µÄÌå»ý·ÖÊýΪ50%¡£
¢ÙÇó¸Ã·´Ó¦µÄƽºâ³£ÊýK=        (Êý×Ö´úÈëʽ×Ó¼´¿É)£¬ÉÏÊöƽºâÌåϵÖÐO2µÄÌå»ý·ÖÊýΪ__________¡£
¢ÚÓÃO2±íʾ´Ó0¡«5 minÄڸ÷´Ó¦µÄƽ¾ùËÙÂʦÔ(O2)=        ¡£
¢Û½«ÉÏÊöƽºâÌåϵµÄζȽµÖÁT2¡æ£¬ÃܱÕÈÝÆ÷ÄÚ¼õСµÄÎïÀíÁ¿ÓР       ¡£
A£®Ñ¹Ç¿           B£®ÃܶȠ         C£®·´Ó¦ËÙÂÊ          D£®N2O5µÄŨ¶È
(4)ÔÚºãκãÈݵÄÃܱÕÈÝÆ÷ÖгäÈëNO2£¬½¨Á¢ÈçÏÂƽºâ£º2NO2(g)N2O4(g)£¬Æ½ºâʱN2O4ÓëNO2µÄÎïÖʵÄÁ¿Ö®±ÈΪa£¬ÆäËüÌõ¼þ²»±äµÄÇé¿öÏ£¬·Ö±ðÔÙ³äÈëNO2ºÍÔÙ³äÈëN2O4£¬Æ½ºâºóÒýÆðµÄ±ä»¯ÕýÈ·µÄÊÇ__________¡£
A£®¶¼ÒýÆða¼õС   B£®¶¼ÒýÆðaÔö´ó   C£®³äÈëNO2ÒýÆða¼õС£¬³äÈëN2O4ÒýÆðaÔö´ó
D£®³äÈëNO2ÒýÆðaÔö´ó£¬³äÈëN2O4ÒýÆða¼õС
ÄòËØ£¨H2NCONH2)ÊÇÓлú̬µª·Ê£¬ÔÚÅ©ÒµÉú²úÖÐÓÐ×ŷdz£ÖØÒªµÄ×÷Óá£
£¨1£©¹¤ÒµÉϺϳÉÄòËصķ´Ó¦·ÖÁ½²½½øÐУº
µÚÒ»²½£º2NH3(l)+CO2H2NCOONH4(°±»ù¼×Ëáï§)(l) ¡÷H1
µÚ¶þ²½£ºH2NCOONH4 (l) H2O+ H2NCONH2(l)¡÷H2
ij»¯Ñ§Ñ§Ï°Ð¡×éÄ£Ä⹤ҵÉϺϳÉÄòËصÄÌõ¼þ£¬ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖÐͶÈë4 mol NH3ºÍ1 mol CO2£¬ÊµÑé²âµÃ·´Ó¦Öи÷×é·ÖµÄÎïÖʵÄÁ¿Ëæʱ¼äµÄ±ä»¯ÈçÏÂͼIËùʾ¡£

¢ÙÒÑÖª×Ü·´Ó¦µÄ¿ìÂýÊÇÓɽÏÂýµÄÒ»²½·´Ó¦¾ö¶¨µÄ¡£ÔòºÏ³ÉÄòËØ×Ü·´Ó¦µÄ¿ìÂýÓɵÚ______²½·´Ó¦¾ö¶¨£¬×Ü·´Ó¦½øÐе½______minʱµ½´ïƽºâ¡£
¢ÚµÚ¶þ²½·´Ó¦µÄƽºâ³£ÊýKËæζȵı仯ÈçÉÏÓÒͼIIËùʾ£¬Ôò¦¤H2______0(Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±¡££©
£¨2£©¸ÃС×齫һ¶¨Á¿´¿¾»µÄ°±»ù¼×Ëáï§ÖÃÓÚÌØÖƵÄÃܱÕÕæ¿ÕÈÝÆ÷ÖÐ(¼ÙÉèÈÝÆ÷Ìå»ý²»±ä£¬¹ÌÌåÊÔÑùÌå»ýºöÂÔ²»¼Æ£©£¬Ôں㶨ζÈÏÂʹÆä´ïµ½·Ö½âƽºâ£º¡£ÊµÑé²âµÃ²»Í¬Î¶ÈϵÄƽºâÊý¾ÝÁÐÓÚÏÂ±í£º
ζÈ/¡æ
15.0
20.0
25.0
30.0
35.0
ƽºâ×Üѹǿ/Kpa
5.7
8.3
12.0
17.1
24.0
ƽºâÆøÌå×ÜŨ¶È/10-3mol/L
2.4
3.4
4.8
6.8
9.4
 
¢Ù¿ÉÒÔÅжϸ÷ֽⷴӦÒѾ­´ïµ½»¯Ñ§Æ½ºâ״̬µÄ±êÖ¾ÊÇ____________¡£
A£®2V(NH3)=V(CO2)                   B£®ÃܱÕÈÝÆ÷ÖÐ×Üѹǿ²»±ä
C£®ÃܱÕÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȲ»±ä      D£®ÃܱÕÈÝÆ÷Öа±ÆøµÄÌå»ý·ÖÊý²»±ä
¢Ú¸ù¾Ý±íÖÐÊý¾Ý£¬¼ÆËã25.0¡ãCʱ¸Ã·Ö½â·´Ó¦µÄƽºâ³£ÊýΪ______(±£ÁôСÊýµãºóһλ)¡£
£¨3£©ÒÑÖª£º
N2(g)+O2(g)=2NO(g)   ¡÷H1=+180.6KJ/mol
N2(g)+3H2(g)=2NH3(g)   ¡÷H2=-92.4KJ/mol
2H2(g)+O2(g)=2H2O(g)  ¡÷H3=-483.6KJ/mol
Ôò4NO(g)+4NH3(g) +O2(g)= 4N2(g)+6 H2O(g)µÄ¡÷H=___kJ ? mol-1¡£
£¨4£©ÄòËØȼÁϵç³ØµÄ½á¹¹ÈçͼËùʾ¡£Æ乤×÷ʱ¸º¼«µç¼«·´Ó¦Ê½¿É±íʾΪ______¡£

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

¾«Ó¢¼Ò½ÌÍø