ÌâÄ¿ÄÚÈÝ

ÔÚÒ»¹Ì¶¨ÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÄÚ¼ÓÈë0.2molµÄN2ºÍ0.6molµÄH2£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢ÉúÈçÏ·´Ó¦£ºN2(g)£«3H2(g)2NH3(g) ¡÷H£¼0 ¡£·´Ó¦ÖÐNH3µÄÎïÖʵÄÁ¿Å¨¶ÈµÄ±ä»¯Çé¿öÈçÏÂͼËùʾ£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¸ù¾Ýͼ£¬¼ÆËã´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâʱ£¬Æ½¾ù·´Ó¦ËÙÂÊv(NH3)=           ¡£
¢ÆÏÂÁÐÃèÊöÖÐÄÜ˵Ã÷ÉÏÊö·´Ó¦ÒÑ´ïƽºâµÄÊÇ     ¡£

A£®3vÕý£¨H2£©£½2vÄ棨NH3£©
B£®ÈÝÆ÷ÖÐÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÁ¿²»Ëæʱ¼ä¶ø±ä»¯
C£®ÈÝÆ÷ÖÐÆøÌåµÄÃܶȲ»Ëæʱ¼ä¶ø±ä»¯
D£®ÈÝÆ÷ÖÐÆøÌåµÄ·Ö×Ó×ÜÊý²»Ëæʱ¼ä¶ø±ä»¯
¢Ç µÚ5·ÖÖÓÄ©£¬±£³Öºãκãѹ£¬Èô¼ÌÐøͨÈë0.2molµÄN2ºÍ0.6molµÄH2£¬Æ½ºâ    Òƶ¯£¨Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡¢¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°²»¡±£©¡£

£¨1£©0.025mol/(L¡¤min)(2·Ö£¬Â©Ð´µ¥Î»¿Û1·Ö£©
£¨2£©BD£¨2·Ö£©
£¨3£©²»£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÓÉͼÅжϷ´Ó¦µÚ4·ÖÖÓʱ·´Ó¦´ïµ½Æ½ºâ£¬v(NH3)=0.1¡Â4=0.025mol/(L¡¤min)£»£¨2£©
A¡¢²»·ûºÏ»¯Ñ§¼ÆÁ¿Êý¹Øϵ£¬´íÎó£»B¡¢ÈÝÆ÷ÖÐÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÁ¿ÊÇÒ»¸ö²»¶Ï±ä»¯µÄÁ¿£¬µ±²»±äʱ¼´´ïµ½Æ½ºâ£¬ÕýÈ·£»C¡¢¸ÃÈÝÆ÷Ìå»ý²»±ä£¬ÃܶÈÊÇÒ»¸öºãÁ¿£¬´íÎó£»D¡¢·Ö×Ó¸öÊýÊÇÒ»¸ö±äÁ¿£¬µ±²»±äʱ¼´´ïµ½Æ½ºâ£¬ÕýÈ·¡££¨3£©¸Ä±äÌõ¼þºóµÄƽºâÓëԭƽºâºÃÊǵÈЧƽºâ£¬¹Êƽºâ²»Òƶ¯¡£
¿¼µã£º¿¼²é»¯Ñ§Æ½ºâµÄ½¨Á¢¡¢Òƶ¯µÈÓйØÎÊÌâ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÖйúÄòËغϳÉËþ(ÄòËþ)ʹÓÃÊÙÃü½öΪŷÃÀ¹ú¼ÒµÄ1/4¡£Îª´Ë±±¾©¸ÖÌúÑо¿ Ôº¶ÔËÄ´¨ãòÌ컯ÄòËþ¸¯Ê´¹ý³Ì½øÐÐÑо¿£¬µÃ³öÏÂÁи¯Ê´»úÀí£º

(1)H2SÀ´×ԺϳÉÄòËصÄÌìÈ»Æø¡£ÔÚ380 K¡¢Ìå»ýΪ2 LµÄÃܱÕÈÝÆ÷ÖУ¬´æÔÚÈçÏ·´Ó¦£ºH2(g)£«S(s)H2S(g)  ¡÷H=£«21£®6kJ¡¤mol£­1¡£·´Ó¦´ïµ½Æ½ºâʱH2¡¢S¡¢H2SµÄÎïÖʵÄÁ¿¾ùΪ3 mol£¬Ôò380 Kʱ¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýΪ______£»ÏÂÁжԸ÷´Ó¦·ÖÎöÕýÈ·µÄÊÇ______(Ìî×ÖĸÐòºÅ)¡£

(2)ÔÚ·´Ó¦IÖз¢ÉúµÄ»¯Ñ§·´Ó¦Îª______¡£
(3)Ñо¿·¢ÏÖ·´Ó¦IIÊÇ·Ö±ðÒÔFe¡¢FeSΪµç¼«£¬ÒÔˮĤΪµç½âÖÊÈÜÒºµÄµç»¯Ñ§¸¯Ê´£¬Æä ¸º
¼«Îª______£»
ÒÑÖª£ºFe(s)£«S(s)=FeS(s) ¡÷H1=£­2£®5akJ¡¤mol£­1 
S(s)£«O2(g)=SO2(g)  ¡÷H2=£­5akJ¡¤mol£­1
4Fe(s)£«3O2(g)=2Fe2O3(s) ¡÷H3=£­6akJ¡¤mol£­1
Ôò·´Ó¦IIµÄÈÈ»¯Ñ§·½³ÌʽΪ_____                                         
(4)ÒÑÖªH2S2O3µÄK1=2£®2¡Á10-1¡¢K2=2£®8¡Á10-2¡£Na2S2O3Ë®ÈÜÒº³Ê______ÐÔ£¬¸ÃÈÜÒºÖеçºÉÊغãʽΪ_____                                                       £»·´Ó¦IYµÄ·´Ó¦ÀàÐÍΪ______         £»¸Ã·´Ó¦______(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£© ˵Ã÷FeSÈܽâÐÔÇ¿ÓÚFeS2O3
(5)ãòÌ컯ÄòËþµÄ×îÖÕ¸¯Ê´²úÎïΪ______£»ÎªÁËÓÐЧ·À¸¯£¬±±¸Ö½¨ÒéãòÌ컯ÔÚÉú²úÖÐÓà CuSO4ÈÜÒº¡°ÍÑÁò(H2S)¡±£¬ÆäÖÐÉæ¼°µÄÀë×Ó·½³ÌʽΪ
__________________                                                   

¼×´¼ÊÇÒ»ÖֺܺõÄȼÁÏ£¬¹¤ÒµÉÏÓÃCH4ºÍH2O ΪԭÁÏ£¬Í¨¹ý·´Ó¦IºÍ¢òÀ´ÖƱ¸¼×´¼¡£
¢Å½«1.0 mol CH4ºÍ2.0 mol H2O(g)ͨÈË·´Ó¦ÊÒ(ÈÝ»ýΪ100L)£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦:
CH4(g)£«H2O(g)CO(g)£«3H2(g)¡­¡­I¡£CH4µÄת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÓÒͼËùʾ¡£

¢ÙÒÑÖª100¡æʱ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬ÔòÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ                  ¡£
¢ÚͼÖеÄP1_      P2(Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±),100¡æʱƽºâ³£ÊýΪ          ¡£
¢Û¸Ã·´Ó¦µÄ¡÷H        0(Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±)¡£
(2)ÔÚÒ»¶¨Ìõ¼þÏ£¬½«a mol COÓë3a mol H2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃÏÂÄÜ×Ô·¢·´Ó¦Éú³É¼×´¼: CO(g)+2H2(g)CH3OH(g)  ¡÷H<0 ¡­¡­¢ò
¢ÜÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ
A£®Éý¸ßζȠ                    B£®½«CH3OH(g)´ÓÌåϵÖзÖÀë
C£®³äÈËHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó    D£®ÔÙ³äÈËlmol COºÍ3 mol H2
¢ÝΪÁËÑ°ÕҺϳɼ״¼µÄζȺÍѹǿµÄÊÊÒËÌõ¼þ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚÏÂÃæʵÑéÉè¼Æ±íÖС£
A£®Ï±íÖÐÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý:  a=_    £»b=_        ¡£

ʵÑé±àºÅ
T£¨¡æ£©
n(CO)£¯n(H2)
P(Mpa)
1
150
1£¯3
0.1
2
a
1£¯3
5
3
350
b
5
B£®¸ù¾Ý·´Ó¦¢òµÄÌص㣬ÓÒÏÂͼÊÇÔÚѹǿ·Ö±ðΪ0.1 MPaºÍ5MPaÏÂCOµÄת»¯ÂÊËæζȱ仯µÄÇúÏßͼ£¬ÇëÖ¸Ã÷ͼÖеÄѹǿPx =           Mpa¡£

¡°½à¾»Ãº¼¼Êõ¡±Ñо¿ÔÚÊÀ½çÉÏÏ൱Æձ飬¿ÆÑÐÈËԱͨ¹ýÏòµØÏÂú²ãÆø»¯Â¯Öн»Ìæ¹ÄÈë¿ÕÆøºÍË®ÕôÆøµÄ·½·¨£¬Á¬Ðø²ú³öÁ˸ßÈÈÖµµÄú̿Æø£¬ÆäÖ÷Òª³É·ÖÊÇCOºÍH2¡£COºÍH2¿É×÷ΪÄÜÔ´ºÍ»¯¹¤Ô­ÁÏ£¬Ó¦ÓÃÊ®·Ö¹ã·º¡£Éú²úú̿ÆøµÄ·´Ó¦Ö®Ò»ÊÇ£ºC(s)+H2O(g)CO(g)+H2(g)?131.4 kJ¡£
£¨1£©ÔÚÈÝ»ýΪ3LµÄÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬5minºóÈÝÆ÷ÄÚÆøÌåµÄÃܶÈÔö´óÁË0.12g£¯L£¬ÓÃH2O±íʾ0¡«5miinµÄƽ¾ù·´Ó¦ËÙÂÊΪ_________________________¡£
£¨2£©ÄÜ˵Ã÷¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ________£¨Ñ¡Ìî±àºÅ£©¡£
a£®vÕý (C)= vÄæ(H2O)   b£®ÈÝÆ÷ÖÐCOµÄÌå»ý·ÖÊý±£³Ö²»±ä
c£®c(H2)=c(CO)        d£®Ì¿µÄÖÊÁ¿±£³Ö²»±ä
£¨3£©ÈôÉÏÊö·´Ó¦ÔÚt0ʱ¿Ì´ïµ½Æ½ºâ(Èçͼ)£¬ÔÚt1ʱ¿Ì¸Ä±äijһÌõ¼þ£¬ÇëÔÚÓÒͼÖмÌÐø»­³öt1ʱ¿ÌÖ®ºóÕý·´Ó¦ËÙÂÊËæʱ¼äµÄ±ä»¯£º

¢ÙËõСÈÝÆ÷Ìå»ý£¬t2ʱµ½´ïƽºâ(ÓÃʵÏß±íʾ)£»
¢Út3ʱƽºâ³£ÊýKÖµ±ä´ó£¬t4µ½´ïƽºâ(ÓÃÐéÏß±íʾ)¡£
£¨4£©ÔÚÒ»¶¨Ìõ¼þÏÂÓÃCOºÍH2¾­ÈçÏÂÁ½²½·´Ó¦ÖƵü×Ëá¼×õ¥£º
¢ÙCO(g) + 2H2(g)CH3OH(g)   ¢ÚCO(g) + CH3OH(g)HCOOCH3(g)
¢Ù·´Ó¦¢ÙÖÐCOµÄƽºâת»¯ÂÊ(¦Á)Óëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ¡£ÔÚ²»¸Ä±ä·´Ó¦ÎïÓÃÁ¿µÄÇé¿öÏ£¬ÎªÌá¸ßCOµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ                                ¡£

¢ÚÒÑÖª·´Ó¦¢ÙÖÐCOµÄת»¯ÂÊΪ80%£¬·´Ó¦¢ÚÖÐÁ½ÖÖ·´Ó¦ÎïµÄת»¯ÂʾùΪ85%£¬Ôò5.04kgCO×î¶à¿ÉÖƵü×Ëá¼×õ¥        kg¡£

£¨14·Ö£©I£®ÒÑÖª£ºC(s)£«H2O(g)CO(g)£«H2(g)  ¦¤H
Ò»¶¨Î¶ÈÏ£¬ÔÚ1.0 LÃܱÕÈÝÆ÷ÖзÅÈë1 mol C£¨s£©¡¢1 mol H2O(g)½øÐз´Ó¦,·´Ó¦Ê±¼ä(t)ÓëÈÝÆ÷ÄÚÆøÌå×Üѹǿ(p)µÄÊý¾Ý¼ûÏÂ±í£º

ʱ¼ät/h
0
1
2
4
8
16
20
25
30
×Üѹǿp/100 kPa
4.56
5.14
5.87
6.30
7.24
8.16
8.18
8.20
8.20
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÏÂÁÐÄÄЩѡÏî¿ÉÒÔ˵Ã÷¸Ã¿ÉÄæ·´Ó¦ÒÑ´ïƽºâ״̬      ¡£
A£®»ìºÏÆøÌåµÄÃܶȲ»ÔÙ·¢Éú¸Ä±ä      B£®ÏûºÄ1 mol H2O£¨g£©µÄͬʱÉú³É1 mol H2
C£®¦¤H²»±ä                         D£®vÕý(CO) = vÄæ(H2)
£¨2£©ÓÉ×ÜѹǿPºÍÆðʼѹǿP0±íʾ·´Ó¦ÌåϵµÄ×ÜÎïÖʵÄÁ¿n×Ü£¬n×Ü£½____ mol£»ÓɱíÖÐÊý¾Ý¼ÆËã·´Ó¦´ïƽºâʱ£¬·´Ó¦ÎïH2O(g)µÄת»¯ÂʦÁ =_____£¨¾«È·µ½Ð¡ÊýµãºóµÚ¶þ룩¡£
¢ò£®Áòµ¥Öʼ°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£
£¨1£©ÒÑÖª25¡æʱ£ºxSO2 (g)£«2xCO(g)£½2xCO2 (g)£«Sx (s)     ¦¤H£½ax kJ/mol     ¢Ù
2xCOS(g)£«xSO2 (g)£½2xCO2 (g)£«3Sx (s)   ¦¤H£½bx kJ/mol¡£    ¢Ú
Ôò·´Ó¦COS(g)Éú³ÉCO(g)¡¢Sx (s)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                            ¡£
£¨2£©ÏòµÈÎïÖʵÄÁ¿Å¨¶ÈNa2S¡¢NaOH»ìºÏÈÜÒºÖеμÓÏ¡ÑÎËáÖÁ¹ýÁ¿¡£ÆäÖÐH2S¡¢HS?¡¢S2?µÄ·Ö²¼·ÖÊý£¨Æ½ºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý£©ÓëµÎ¼ÓÑÎËáÌå»ýµÄ¹ØϵÈçͼËùʾ£¨ºöÂԵμӹý³ÌH2SÆøÌåµÄÒݳö£©¡£ÊÔ·ÖÎö£º

¢ÙBÇúÏß´ú±í      ·ÖÊý±ä»¯(ÓÃ΢Á£·ûºÅ±íʾ)£»µÎ¼Ó¹ý³ÌÖУ¬ÈÜÒºÖÐÒ»¶¨³ÉÁ¢£º
c(Na+)=                               ¡£
¢ÚMµã£¬ÈÜÒºÖÐÖ÷ÒªÉæ¼°µÄÀë×Ó·½³Ìʽ                             ¡£

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

¾«Ó¢¼Ò½ÌÍø